This preview shows page 1. Sign up to view the full content.
Unformatted text preview: cen58933_ch07.qxd 9/4/2002 12:12 PM Page 367 CHAPTER EXTERNAL FORCED
CONVECTION
n Chapter 6 we considered the general and theoretical aspects of forced
convection, with emphasis on differential formulation and analytical solutions. In this chapter we consider the practical aspects of forced convection
to or from flat or curved surfaces subjected to external flow, characterized by
the freely growing boundary layers surrounded by a free flow region that
involves no velocity and temperature gradients.
We start this chapter with an overview of external flow, with emphasis on
friction and pressure drag, flow separation, and the evaluation of average drag
and convection coefficients. We continue with parallel flow over flat plates.
In Chapter 6, we solved the boundary layer equations for steady, laminar, parallel flow over a flat plate, and obtained relations for the local friction coefficient and the Nusselt number. Using these relations as the starting point, we
determine the average friction coefficient and Nusselt number. We then extend
the analysis to turbulent flow over flat plates with and without an unheated
starting length.
Next we consider cross flow over cylinders and spheres, and present graphs
and empirical correlations for the drag coefficients and the Nusselt numbers,
and discuss their significance. Finally, we consider cross flow over tube banks
in aligned and staggered configurations, and present correlations for the pressure drop and the average Nusselt number for both configurations. I 7
CONTENTS
7–1
Drag and Heat Transfer in
External Flow
368
7–2
Parallel Flow over Flat
Plates
371
7–3
Flow across Cylinders and
Spheres
380
7–4
Flow across Tube
Banks
389
Topic of Special Interest:
Reducing Heat Transfer
through Surfaces:
Thermal Insulation
395 367 cen58933_ch07.qxd 9/4/2002 12:12 PM Page 368 368
HEAT TRANSFER 7–1 I DRAG AND HEAT TRANSFER
IN EXTERNAL FLOW Fluid flow over solid bodies frequently occurs in practice, and it is responsible for numerous physical phenomena such as the drag force acting on the automobiles, power lines, trees, and underwater pipelines; the lift developed by
airplane wings; upward draft of rain, snow, hail, and dust particles in high
winds; and the cooling of metal or plastic sheets, steam and hot water pipes,
and extruded wires. Therefore, developing a good understanding of external
flow and external forced convection is important in the mechanical and thermal design of many engineering systems such as aircraft, automobiles, buildings, electronic components, and turbine blades.
The flow fields and geometries for most external flow problems are too complicated to be solved analytically, and thus we have to rely on correlations
based on experimental data. The availability of highspeed computers has
made it possible to conduct series of “numerical experimentations” quickly by
solving the governing equations numerically, and to resort to the expensive and
timeconsuming testing and experimentation only in the final stages of design.
In this chapter we will mostly rely on relations developed experimentally.
The velocity of the fluid relative to an immersed solid body sufficiently far
from the body (outside the boundary layer) is called the freestream velocity,
and is denoted by u . It is usually taken to be equal to the upstream velocity
also called the approach velocity, which is the velocity of the approaching
fluid far ahead of the body. This idealization is nearly exact for very thin bodies, such as a flat plate parallel to flow, but approximate for blunt bodies such
as a large cylinder. The fluid velocity ranges from zero at the surface (the noslip condition) to the freestream value away from the surface, and the subscript “infinity” serves as a reminder that this is the value at a distance where
the presence of the body is not felt. The upstream velocity, in general, may
vary with location and time (e.g., the wind blowing past a building). But in the
design and analysis, the upstream velocity is usually assumed to be uniform
and steady for convenience, and this is what we will do in this chapter. Friction and Pressure Drag Wind tunnel
60 mph FD FIGURE 7–1
Schematic for measuring the drag
force acting on a car in a wind tunnel. You may have seen high winds knocking down trees, power lines, and even
trailers, and have felt the strong “push” the wind exerts on your body. You experience the same feeling when you extend your arm out of the window of a
moving car. The force a flowing fluid exerts on a body in the flow direction is
called drag (Fig. 7–1)
A stationary fluid exerts only normal pressure forces on the surface of a
body immersed in it. A moving fluid, however, also exerts tangential shear
forces on the surface because of the noslip condition caused by viscous
effects. Both of these forces, in general, have components in the direction
of flow, and thus the drag force is due to the combined effects of pressure and
wall shear forces in the flow direction. The components of the pressure
and wall shear forces in the normal direction to flow tend to move the body in
that direction, and their sum is called lift.
In general, both the skin friction (wall shear) and pressure contribute to
the drag and the lift. In the special case of a thin flat plate aligned parallel
to the flow direction, the drag force depends on the wall shear only and is cen58933_ch07.qxd 9/4/2002 12:12 PM Page 369 369
CHAPTER 7 independent of pressure. When the flat plate is placed normal to the flow direction, however, the drag force depends on the pressure only and is independent of the wall shear since the shear stress in this case acts in the direction
normal to flow (Fig. 7–2). For slender bodies such as wings, the shear force
acts nearly parallel to the flow direction. The drag force for such slender bodies is mostly due to shear forces (the skin friction).
The drag force FD depends on the density of the fluid, the upstream velocity , and the size, shape, and orientation of the body, among other things.
The drag characteristics of a body is represented by the dimensionless drag
coefficient CD defined as
CD Drag coefficient: FD
1
2 (71) 2 A where A is the frontal area (the area projected on a plane normal to the direction of flow) for blunt bodies—bodies that tends to block the flow. The frontal
area of a cylinder of diameter D and length L, for example, is A LD. For
parallel flow over flat plates or thin airfoils, A is the surface area. The drag coefficient is primarily a function of the shape of the body, but it may also depend on the Reynolds number and the surface roughness.
The drag force is the net force exerted by a fluid on a body in the direction
of flow due to the combined effects of wall shear and pressure forces. The part
of drag that is due directly to wall shear stress w is called the skin friction
drag (or just friction drag) since it is caused by frictional effects, and the part
that is due directly to pressure P is called the pressure drag (also called the
form drag because of its strong dependence on the form or shape of the body).
When the friction and pressure drag coefficients are available, the total drag
coefficient is determined by simply adding them,
CD CD, friction CD, pressure (72) The friction drag is the component of the wall shear force in the direction of
flow, and thus it depends on the orientation of the body as well as the magnitude of the wall shear stress w. The friction drag is zero for a surface normal
to flow, and maximum for a surface parallel to flow since the friction drag in
this case equals the total shear force on the surface. Therefore, for parallel
flow over a flat plate, the drag coefficient is equal to the friction drag coefficient, or simply the friction coefficient (Fig. 7–3). That is,
Flat plate: CD CD, friction Cf (73) Once the average friction coefficient Cf is available, the drag (or friction)
force over the surface can be determined from Eq. 71. In this case A is the
surface area of the plate exposed to fluid flow. When both sides of a thin
plate are subjected to flow, A becomes the total area of the top and bottom
surfaces. Note that the friction coefficient, in general, will vary with location
along the surface.
Friction drag is a strong function of viscosity, and an “idealized” fluid with
zero viscosity would produce zero friction drag since the wall shear stress
would be zero (Fig. 7–4). The pressure drag would also be zero in this case
during steady flow regardless of the shape of the body since there will be no
pressure losses. For flow in the horizontal direction, for example, the pressure
along a horizontal line will be constant (just like stationary fluids) since the High pressure
Low pressure
+
+
+
+
+
+
+
+ –
–
–
–
–
–
–
–
Wall shear FIGURE 7–2
Drag force acting on a flat
plate normal to flow depends on
the pressure only and is
independent of the wall shear,
which acts normal to flow. CD, pressure = 0
CD = CD, friction = Cf
FD, pressure = 0
FD = FD, friction = Ff = Cf A ρ 2 2 FIGURE 7–3
For parallel flow over a flat plate,
the pressure drag is zero, and thus the
drag coefficient is equal to the friction
coefficient and the drag force is
equal to the friction force. FD = 0 if µ = 0 FIGURE 7–4
For the flow of an “idealized”
fluid with zero viscosity past a body,
both the friction drag and pressure
drag are zero regardless of the
shape of the body. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 370 370
HEAT TRANSFER Separation
point Reattachment
point upstream velocity is constant, and thus there will be no net pressure force
acting on the body in the horizontal direction. Therefore, the total drag is zero
for the case of ideal inviscid fluid flow.
At low Reynolds numbers, most drag is due to friction drag. This is especially the case for highly streamlined bodies such as airfoils. The friction drag
is also proportional to the surface area. Therefore, bodies with a larger surface
area will experience a larger friction drag. Large commercial airplanes, for example, reduce their total surface area and thus drag by retracting their wing
extensions when they reach the cruising altitudes to save fuel. The friction
drag coefficient is independent of surface roughness in laminar flow, but is a
strong function of surface roughness in turbulent flow due to surface roughness elements protruding further into the highly viscous laminar sublayer.
The pressure drag is proportional to the difference between the pressures
acting on the front and back of the immersed body, and the frontal area.
Therefore, the pressure drag is usually dominant for blunt bodies, negligible
for streamlined bodies such as airfoils, and zero for thin flat plates parallel to
the flow.
When a fluid is forced to flow over a curved surface at sufficiently high velocities, it will detach itself from the surface of the body. The lowpressure region behind the body where recirculating and back flows occur is called the
separation region. The larger the separation area is, the larger the pressure
drag will be. The effects of flow separation are felt far downstream in the form
of reduced velocity (relative to the upstream velocity). The region of flow
trailing the body where the effect of the body on velocity is felt is called the
wake (Fig. 7–5). The separated region comes to an end when the two flow
streams reattach, but the wake keeps growing behind the body until the fluid
in the wake region regains its velocity. The viscous effects are the most significant in the boundary layer, the separated region, and the wake. The flow
outside these regions can be considered to be inviscid. Heat Transfer
Separation
region
Wake region
(within dashed line) FIGURE 7–5
Separation and reattachment
during flow over a cylinder,
and the wake region. The phenomena that affect drag force also affect heat transfer, and this effect
appears in the Nusselt number. By nondimensionalizing the boundary layer
equations, it was shown in Chapter 6 that the local and average Nusselt numbers have the functional form
Nux f1(x*, Rex, Pr) and Nu f2(ReL, Pr) (74a, b) The experimental data for heat transfer is often represented conveniently
with reasonable accuracy by a simple powerlaw relation of the form
Nu m
C Re L Pr n (75) where m and n are constant exponents, and the value of the constant C depends on geometry and flow.
The fluid temperature in the thermal boundary layer varies from Ts at the
surface to about T at the outer edge of the boundary. The fluid properties also
vary with temperature, and thus with position across the boundary layer. In
order to account for the variation of the properties with temperature, the fluid
properties are usually evaluated at the socalled film temperature, defined as cen58933_ch07.qxd 9/4/2002 12:12 PM Page 371 371
CHAPTER 7 T Ts Tf (76) 2 which is the arithmetic average of the surface and the freestream temperatures. The fluid properties are then assumed to remain constant at those values
during the entire flow. An alternative way of accounting for the variation of
properties with temperature is to evaluate all properties at the free stream temperature and to multiply the Nusselt number relation in Eq. 75 by (Pr /Prs)r
or ( / s)r.
The local drag and convection coefficients vary along the surface as a result
of the changes in the velocity boundary layers in the flow direction. We are
usually interested in the drag force and the heat transfer rate for the entire surface, which can be determined using the average friction and convection coefficient. Therefore, we present correlations for both local (identified with the
subscript x) and average friction and convection coefficients. When relations
for local friction and convection coefficients are available, the average friction and convection coefficients for the entire surface can be determined by
integration from
L CD 1
L 0 h 1
L 0 CD, x dx (77) and
L h x dx (78) When the average drag and convection coefficients are available, the drag
force can be determined from Eq. 71 and the rate of heat transfer to or from
an isothermal surface can be determined from
˙
Q h As(Ts T) (79) where As is the surface area. 7–2 I PARALLEL FLOW OVER FLAT PLATES Consider the parallel flow of a fluid over a flat plate of length L in the flow direction, as shown in Figure 7–6. The xcoordinate is measured along the plate
surface from the leading edge in the direction of the flow. The fluid approaches the plate in the xdirection with uniform upstream velocity and
temperature T . The flow in the velocity boundary layer starts out as laminar,
but if the plate is sufficiently long, the flow will become turbulent at a distance xcr from the leading edge where the Reynolds number reaches its critical
value for transition.
The transition from laminar to turbulent flow depends on the surface geometry, surface roughness, upstream velocity, surface temperature, and the type
of fluid, among other things, and is best characterized by the Reynolds number. The Reynolds number at a distance x from the leading edge of a flat plate
is expressed as T Turbulent
Laminar y
x xcr Ts
L FIGURE 7–6
Laminar and turbulent regions
of the boundary layer during
flow over a flat plate. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 372 372
HEAT TRANSFER x Rex x
v (710) Note that the value of the Reynolds number varies for a flat plate along the
L /v at the end of the plate.
flow, reaching ReL
For flow over a flat plate, transition from laminar to turbulent is usually
taken to occur at the critical Reynolds number of
xcr Recr 105 5 (711) The value of the critical Reynolds number for a flat plate may vary from 105
to 3 106, depending on the surface roughness and the turbulence level of the
free stream. Friction Coefficient
Based on analysis, the boundary layer thickness and the local friction coefficient at location x for laminar flow over a flat plate were determined in Chapter 6 to be
1L
Cf = –– Cf, x dx
L0
1 L 0.664
= –– ––––– dx
L 0 Re1/2
x
0.664
= –––––
L ()
()
()
L
0 x
––––
ν 0.664
= ––––– ––––
ν
L
–––––––
= 2 × 0.664
L Laminar: 5x
Re1/2
x and Cf, x 0.664
,
Re1/2
x Rex 5 10 5 (712a, b) The corresponding relations for turbulent flow are –1/2 dx Turbulent: –1/2 L
ν v, x L
x1/2 –1
–––
––
2 0 –1/2 1.328
= –––––
1/2
Re L FIGURE 7–7
The average friction coefficient over
a surface is determined by integrating
the local friction coefficient over the
entire surface. v, x 0.382x
and
Re1/5
x 0.0592
,
Re1/5
x Cf, x 10 5 5 Rex 107 (713a, b) where x is the distance from the leading edge of the plate and Rex
x/v is the
Reynolds number at location x. Note that Cf, x. is proportional to Rex 1/2 and thus
to x 1/2 for laminar flow. Therefore, Cf, x is supposedly infinite at the leading
edge (x 0) and decreases by a factor of x 1/2 in the flow direction. The local
friction coefficients are higher in turbulent flow than they are in laminar flow
because of the intense mixing that occurs in the turbulent boundary layer.
Note that Cf, x reaches its highest values when the flow becomes fully turbulent, and then decreases by a factor of x 1/5 in the flow direction.
The average friction coefficient over the entire plate is detennined by substituting the relations above into Eq. 77 and performing the integrations
(Fig.7–7). We get
Laminar: Turbulent: Cf Cf 1.328
Re 1/2
L 0.074
Re 1/5
L ReL 5 10 5 5 10 5 ReL 10 7 (714) (715) The first relation gives the average friction coefficient for the entire plate when
the flow is laminar over the entire plate. The second relation gives the average
friction coefficient for the entire plate only when the flow is turbulent over the
entire plate, or when the laminar flow region of the plate is too small relative to
the turbulent flow region (that is, xcr L where the length of the plate xcr over
xcr/v).
which the flow is laminar can be determined from Recr 5 105 cen58933_ch07.qxd 9/4/2002 12:12 PM Page 373 373
CHAPTER 7 In some cases, a flat plate is sufficiently long for the flow to become turbulent, but not long enough to disregard the laminar flow region. In such cases,
the average friction coefficient over the entire plate is determined by performing the integration in Eq. 77 over two parts: the laminar region 0 x xcr
and the turbulent region xcr x L as
1
L Cf xcr
0 L Cf, x laminar dx Cf, x, turbulent dx (716) xcr Note that we included the transition region with the turbulent region. Again
taking the critical Reynolds number to be Recr 5 105 and performing the
integrations of Eq. 716 after substituting the indicated expressions, the average friction coefficient over the entire plate is determined to be
0.074
Re 1/5
L Cf 1742
ReL 5 10 5 107 ReL (717) The constants in this relation will be different for different critical Reynolds
numbers. Also, the surfaces are assumed to be smooth, and the free stream to
be turbulent free. For laminar flow, the friction coefficient depends on only
the Reynolds number, and the surface roughness has no effect. For turbulent
flow, however, surface roughness causes the friction coefficient to increase
severalfold, to the point that in fully turbulent regime the friction coefficient
is a function of surface roughness alone, and independent of the Reynolds
number (Fig. 7–8).
A curve fit of experimental data for the average friction coefficient in this
regime is given by Schlichting as
Cf Rough surface, turbulent: 1.89 1.62 log 2.5 (718) L were is the surface roughness, and L is the length of the plate in the flow direction. In the absence of a better relation, the relation above can be used for
turbulent flow on rough surfaces for Re l06, especially when L 10 4. Relative
roughness,
L Friction
coefficient
Cf 0.0*
1 10
1 10
1 10 0.0029
0.0032
0.0049
0.0084 hx x
k 0.332 Re0.5 Pr1/3
x Pr 0.60 3 FIGURE 7–8
For turbulent flow, surface roughness
may cause the friction coefficient
to increase severalfold.
h
Cf The local Nusselt number at a location x for laminar flow over a flat plate was
determined in Chapter 6 by solving the differential energy equation to be
Nu x 4 *Smooth surface for Re = 107. Others
calculated from Eq. 718. Heat Transfer Coefficient Laminar: 5 hx or Cf, x (719)
δx The corresponding relation for turbulent flow is
Turbulent: Nu x hx x
k 0.0296 Re 0.8
x 1/3 Pr 0.6 Pr
5 105 60
Rex T 10 7 (720) Note that hx is proportional to Re 0.5 and thus to x 0.5 for laminar flow. Therex
fore, hx is infinite at the leading edge (x 0) and decreases by a factor of x 0.5
in the flow direction. The variation of the boundary layer thickness and the
friction and heat transfer coefficients along an isothermal flat plate are shown
in Figure 7–9. The local friction and heat transfer coefficients are higher in Laminar Transition Turbulent x FIGURE 7–9
The variation of the local friction
and heat transfer coefficients for
flow over a flat plate. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 374 374
HEAT TRANSFER turbulent flow than they are in laminar flow. Also, hx reaches its highest values when the flow becomes fully turbulent, and then decreases by a factor of
x 0.2 in the flow direction, as shown in the figure.
The average Nusselt number over the entire plate is determined by substituting the relations above into Eq. 78 and performing the integrations. We get
hL
k Nu Laminar:
Nu Turbulent: hL
k 0.664 Re 0.5 Pr 1/3
L 0.037 Re 0.8 Pr1/3
L ReL
0.6 Pr
5 10 5 5
60
ReL 10 5 (721) (722) 10 7 The first relation gives the average heat transfer coefficient for the entire plate
when the flow is laminar over the entire plate. The second relation gives the
average heat transfer coefficient for the entire plate only when the flow is turbulent over the entire plate, or when the laminar flow region of the plate is too
small relative to the turbulent flow region.
In some cases, a flat plate is sufficiently long for the flow to become turbulent, but not long enough to disregard the laminar flow region. In such cases,
the average heat transfer coefficient over the entire plate is determined by performing the integration in Eq. 78 over two parts as
h 1
L xcr
0 L hx, laminar dx hx, trubulent dx (723) xcr Again taking the critical Reynolds number to be Recr 5 105 and performing the integrations in Eq. 723 after substituting the indicated expressions, the
average Nusselt number over the entire plate is determined to be (Fig. 7–10)
h hx,turbulent Nu
h average hx, laminar Laminar
0 xcr Turbulent
Lx FIGURE 7–10
Graphical representation of the
average heat transfer coefficient for a
flat plate with combined laminar and
turbulent flow. hL
k (0.037 Re0.8
L 871)Pr1/3 0.6 Pr
5 10 5 60
ReL 10 7 (724) The constants in this relation will be different for different critical Reynolds
numbers.
Liquid metals such as mercury have high thermal conductivities, and are
commonly used in applications that require high heat transfer rates. However,
they have very small Prandtl numbers, and thus the thermal boundary layer
develops much faster than the velocity boundary layer. Then we can assume
the velocity in the thermal boundary layer to be constant at the free stream
value and solve the energy equation. It gives
Nu x 0.565(Re x P r) 1/2 Pr 0.05 (725) It is desirable to have a single correlation that applies to all fluids, including
liquid metals. By curvefitting existing data, Churchill and Ozoe (Ref. 3) proposed the following relation which is applicable for all Prandtl numbers and
is claimed to be accurate to 1%,
Nu x hx x
k 0.3387 Pr 1/3 Re1/2
x
[1 (0.0468/Pr)2/3]1/4 (726) These relations have been obtained for the case of isothermal surfaces
but could also be used approximately for the case of nonisothermal surfaces
by assuming the surface temperature to be constant at some average value. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 375 375
CHAPTER 7 Also, the surfaces are assumed to be smooth, and the free stream to be turbulent free. The effect of variable properties can be accounted for by evaluating
all properties at the film temperature. Flat Plate with Unheated Starting Length
So far we have limited our consideration to situations for which the entire
plate is heated from the leading edge. But many practical applications involve
surfaces with an unheated starting section of length , shown in Figure 7–11,
and thus there is no heat transfer for 0 x
. In such cases, the velocity
boundary layer starts to develop at the leading edge (x 0), but the thermal
boundary layer starts to develop where heating starts (x = ).
Consider a flat plate whose heated section is maintained at a constant tem). Using integral solution methods (see
perature (T
Ts constant for x
Kays and Crawford, 1994), the local Nusselt numbers for both laminar and
turbulent flows are determined to be
Laminar:
Turbulent: Nux (for 0)
Nux
[1 ( /x)3/4]1/3
Nux (for 0 )
Nux
[1 ( / x) 9/10 ]1/9 0.332 Re0.5 Pr1/3
x
[1 ( /x)3/4]1/3 (728) Thermal boundary layer
Velocity boundary layer Ts (727) 0.0296 Re0.8 Pr1/3
x
[1 ( / x) 9/10]1/9 T ξ
x FIGURE 7–11 for x
. Note that for
0, these Nux relations reduce to Nux (for 0), which
is the Nusselt number relation for a flat plate without an unheated starting
length. Therefore, the terms in brackets in the denominator serve as correction
factors for plates with unheated starting lengths.
The determination of the average Nusselt number for the heated section of
a plate requires the integration of the local Nusselt number relations above,
which cannot be done analytically. Therefore, integrations must be done numerically. The results of numerical integrations have been correlated for the
average convection coefficients [Thomas, (1977) Ref. 11] as
Laminar: h Turbulent: h 2 [1 ( / x) 3/4 ]
hx
1
/L 5[1 ( / x )9/10]
hx
4(1
/ L) L (729) L (730) The first relation gives the average convection coefficient for the entire
heated section of the plate when the flow is laminar over the entire plate. Note
that for
0 it reduces to hL 2hx L , as expected. The second relation gives
the average convection coefficient for the case of turbulent flow over the entire plate or when the laminar flow region is small relative to the turbulent
region. Uniform Heat Flux
When a flat plate is subjected to uniform heat flux instead of uniform temperature, the local Nusselt number is given by
Laminar: Nu x Turbulent: Nu x 0.453 Re 0.5 Pr 1/3
x (731) 0.8
0.0308 Re x Pr1/3 (732) Flow over a flat plate with an unheated
starting length. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 376 376
HEAT TRANSFER These relations give values that are 36 percent higher for laminar flow and
4 percent higher for turbulent flow relative to the isothermal plate case. When
the plate involves an unheated starting length, the relations developed for the
uniform surface temperature case can still be used provided that Eqs. 731 and
732 are used for Nux(for 0) in Eqs. 727 and 728, respectively.
·
When heat flux qs is prescribed, the rate of heat transfer to or from the plate
and the surface temperature at a distance x are determined from
˙
Q ˙
qs A s (733) and
˙
qs h x [Ts( x) → T] Ts(x) ˙
qs
hx T (734) where As is the heat transfer surface area.
EXAMPLE 7–1 Flow of Hot Oil over a Flat Plate Engine oil at 60°C flows over the upper surface of a 5mlong flat plate whose
temperature is 20°C with a velocity of 2 m/s (Fig. 7–12). Determine the total
drag force and the rate of heat transfer per unit width of the entire plate. T = 60°C
= 2 m/s . Oil Q
A Ts = 20°C L=5m FIGURE 7–12
Schematic for Example 71. SOLUTION Engine oil flows over a flat plate. The total drag force and the rate
of heat transfer per unit width of the plate are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The critical Reynolds
number is Recr 5 105.
Properties The properties of engine oil at the film temperature of Tf (Ts T )/
2 (20 60)/2 40°C are (Table A–14).
876 kg/m3
0.144 W/m °C k Analysis Noting that L ReL Pr 2870
242 10 6 m2/s 5 m, the Reynolds number at the end of the plate is L (2 m/s)(5 m)
0.242 10 5 m 2/s 4.13 104 which is less than the critical Reynolds number. Thus we have laminar flow over
the entire plate, and the average friction coefficient is Cf 1.328 Re L 0.5 1.328 (4.13 Noting that the pressure drag is zero and thus CD
force acting on the plate per unit width becomes
2 FD Cf As 2 0.0207 (5 1 m2) 103) 0.5 0.0207 Cf for a flat plate, the drag (876 kg/m3)(2 m/s) 2
1N
2
1 kg · m/s2 181 N
The total drag force acting on the entire plate can be determined by multiplying
the value obtained above by the width of the plate.
This force per unit width corresponds to the weight of a mass of about 18 kg.
Therefore, a person who applies an equal and opposite force to the plate to keep cen58933_ch07.qxd 9/4/2002 12:12 PM Page 377 377
CHAPTER 7 it from moving will feel like he or she is using as much force as is necessary to
hold a 18kg mass from dropping.
Similarly, the Nusselt number is determined using the laminar flow relations
for a flat plate, Nu hL
k 0.5
0.664 ReL Pr1/3 0.664 (4.13 104)0.5 28701/3 1918 Then, h k
Nu
L 0.144 W/m · °C
(1918)
5m 55.2 W/m2 · °C and
·
Q hAs(T Ts) (55.2 W/m 2 · °C)(5 1 m2)(60 20)°C 11,040 W Discussion Note that heat transfer is always from the highertemperature
medium to the lowertemperature one. In this case, it is from the oil to the
plate. The heat transfer rate is per m width of the plate. The heat transfer for
the entire plate can be obtained by multiplying the value obtained by the actual
width of the plate. EXAMPLE 7–2 Cooling of a Hot Block by Forced Air at High
Elevation
The local atmospheric pressure in Denver, Colorado (elevation 1610 m), is
83.4 kPa. Air at this pressure and 20°C flows with a velocity of 8 m/s over a
1.5 m 6 m flat plate whose temperature is 140°C (Fig. 713). Determine the
rate of heat transfer from the plate if the air flows parallel to the (a) 6mlong
side and (b) the 1.5m side. k
@ 1 atm 0.02953 W/m · °C
2.097 10 5 m2/s Pr 0.7154 The atmospheric pressure in Denver is P
(83.4 kPa)/(101.325 kPa/atm)
0.823 atm. Then the kinematic viscosity of air in Denver becomes
@ 1 atm /P (2.097 10 5 m2/s)/0.823 2.548 10 5 Analysis (a ) When air flow is parallel to the long side, we have L
the Reynolds number at the end of the plate becomes ReL L (8 m/s)(6 m)
2.548 10 5 m2/s 1.884 106 m2/s
6 m, and Ts = 140°C T = 20°C
= 8 m/s
·
Q m Air 1.5 SOLUTION The top surface of a hot block is to be cooled by forced air. The
rate of heat transfer is to be determined for two cases.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr 5 105. 3 Radiation effects are negligible. 4 Air is an ideal gas.
Properties The properties k, , Cp, and Pr of ideal gases are independent of
pressure, while the properties and are inversely proportional to density and
thus pressure. The properties of air at the film temperature of Tf (Ts T )/2
(140 20)/2 80°C and 1 atm pressure are (Table A–15) Patm = 83.4 kPa 6m FIGURE 7–13
Schematic for Example 72. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 378 378
HEAT TRANSFER which is greater than the critical Reynolds number. Thus, we have combined
laminar and turbulent flow, and the average Nusselt number for the entire plate
is determined to be Nu hL
0.8
(0.037 ReL
871)Pr1/3
k
[0.037(1.884 106)0.8 871]0.71541/3
2687 Then 0.02953 W/m · °C
k
Nu
(2687)
L
6m
wL (1.5 m)(6 m) 9 m2 h
As 13.2 W/m2 · °C and ·
Q
20°C
8 m/s
140°C Air hAs(Ts T) (13.2 W/m2 · °C)(9 m2)(140 20)°C 104 W Note that if we disregarded the laminar region and assumed turbulent flow over
the entire plate, we would get Nu 3466 from Eq. 7–22, which is 29 percent
higher than the value calculated above.
(b) When air flow is along the short side, we have L
number at the end of the plate becomes 1.5 m ·
Qconv = 14,300 W 1.43 L ReL 6m (8 m/s)(1.5 m)
2.548 10 5 m2/s 1.5 m, and the Reynolds 10 5 4.71 which is less than the critical Reynolds number. Thus we have laminar flow over
the entire plate, and the average Nusselt number is (a) Flow along the long side
Air
20°C
8 m/s Nu 140°C 6m
(b) Flow along the short side FIGURE 7–14
The direction of fluid flow can
have a significant effect on
convection heat transfer. 0.5
0.664 ReL Pr1/3 0.664 (4.71 105)0.5 0.71541/3 408 Then
k
Nu
L ·
Q hAs(Ts 0.02953 W/m · °C
(408)
1.5 m 8.03 W/m2 · °C m h
1.5 ·
Qconv = 8670 W hL
k and T) (8.03 W/m2 · °C)(9 m2)(140 20)°C 8670 W which is considerably less than the heat transfer rate determined in case (a).
Discussion Note that the direction of fluid flow can have a significant effect on
convection heat transfer to or from a surface (Fig. 714). In this case, we can
increase the heat transfer rate by 65 percent by simply blowing the air along the
long side of the rectangular plate instead of the short side. EXAMPLE 7–3 Cooling of Plastic Sheets by Forced Air The forming section of a plastics plant puts out a continuous sheet of plastic
that is 4 ft wide and 0.04 in. thick at a velocity of 30 ft/min. The temperature
of the plastic sheet is 200°F when it is exposed to the surrounding air, and a
2ftlong section of the plastic sheet is subjected to air flow at 80°F at a velocity of 10 ft/s on both sides along its surfaces normal to the direction of motion cen58933_ch07.qxd 9/4/2002 12:12 PM Page 379 379
CHAPTER 7 of the sheet, as shown in Figure 7–15. Determine (a) the rate of heat transfer
from the plastic sheet to air by forced convection and radiation and (b) the temperature of the plastic sheet at the end of the cooling section. Take the density,
specific heat, and emissivity of the plastic sheet to be
75 lbm/ft3, Cp 0.4
Btu/lbm · °F, and
0.9. SOLUTION Plastic sheets are cooled as they leave the forming section of
a plastics plant. The rate of heat loss from the plastic sheet by convection and
radiation and the exit temperature of the plastic sheet are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The critical Reynolds number is Recr 5 105. 3 Air is an ideal gas. 4 The local atmospheric pressure is
1 atm. 5 The surrounding surfaces are at the temperature of the room air.
Properties The properties of the plastic sheet are given in the problem statement. The properties of air at the film temperature of Tf (Ts T )/2 (200
80)/2 140°F and 1 atm pressure are (Table A–15E)
k 0.01623 Btu/h · ft · °F
Pr 0.7202
0.7344 ft2/h 0.204 10 3 ft2/s Analysis (a) We expect the temperature of the plastic sheet to drop somewhat
as it flows through the 2ftlong cooling section, but at this point we do not
know the magnitude of that drop. Therefore, we assume the plastic sheet to be
isothermal at 200°F to get started. We will repeat the calculations if necessary
to account for the temperature drop of the plastic sheet.
Noting that L
4 ft, the Reynolds number at the end of the air flow across
the plastic sheet is ReL L (10 ft /s)(4 ft)
0.204 10 3 f t 2/s 1.961 10 5 which is less than the critical Reynolds number. Thus, we have laminar flow
over the entire sheet, and the Nusselt number is determined from the laminar
flow relations for a flat plate to be hL
k Nu 0.5
0.664 ReL Pr1/3 0.664 (1.961 105)0.5 (0.7202)1/3 263.6 Then, h
As 0.01623 Btu/ h ft
k
Nu
L
4 ft
(2 ft)(4 ft)(2 sides) 16 ft2 °F (263.6) 1.07 Btu/h · ft2 · °F and ·
Q conv ·
Q rad hAs(Ts T )
(1.07 Btu/h · ft2 · °F)(16 ft2)(200
2054 Btu/h 4
As(Ts4 Tsurr)
(0.9)(0.1714 10
2584 Btu/h 8 80)°F Btu/h · ft2 · R4)(16 ft2)[(660 R)4 (540 R)4] 2 ft
200°F
Plastic sheet Air
80°F, 10 ft/s 4 ft
0.04 in
30 ft/min FIGURE 7–15
Schematic for Example 7–3. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 380 380
HEAT TRANSFER Therefore, the rate of cooling of the plastic sheet by combined convection and
radiation is
·
Qtotal ·
Q conv ·
Q rad 2054 2584 4638 Btu/h (b) To find the temperature of the plastic sheet at the end of the cooling section, we need to know the mass of the plastic rolling out per unit time (or the
mass flow rate), which is determined from ·
m Ac plastic (75 lbm/ft3) 4 0.04 3 30
ft
ft /s
12
60 0.5 lbm/s Then, an energy balance on the cooled section of the plastic sheet yields
·
Q ·
m Cp(T2 T1) → T2 T1 ·
Q
·
m Cp ·
Noting that Q is a negative quantity (heat loss) for the plastic sheet and substituting, the temperature of the plastic sheet as it leaves the cooling section is
determined to be T2 200°F 4638 Btu/h
1h
(0.5 lbm/s)(0.4 Btu/lbm · °F) 3600 s 193.6°F Discussion The average temperature of the plastic sheet drops by about 6.4°F
as it passes through the cooling section. The calculations now can be repeated
by taking the average temperature of the plastic sheet to be 196.8°F instead of
200°F for better accuracy, but the change in the results will be insignificant because of the small change in temperature. 7–3 Midplane θ
D Wake Separation
Stagnation
point
point
Boundary layer FIGURE 7–16
Typical flow patterns in
cross flow over a cylinder. I FLOW ACROSS CYLINDERS AND SPHERES Flow across cylinders and spheres is frequently encountered in practice. For
example, the tubes in a shellandtube heat exchanger involve both internal
flow through the tubes and external flow over the tubes, and both flows must
be considered in the analysis of the heat exchanger. Also, many sports such as
soccer, tennis, and golf involve flow over spherical balls.
The characteristic length for a circular cylinder or sphere is taken to be the
external diameter D. Thus, the Reynolds number is defined as Re
D/
where is the uniform velocity of the fluid as it approaches the cylinder or
sphere. The critical Reynolds number for flow across a circular cylinder or
sphere is about Recr 2 105. That is, the boundary layer remains laminar
for about Re 2 105 and becomes turbulent for Re 2 105.
Cross flow over a cylinder exhibits complex flow patterns, as shown in Figure 7–16. The fluid approaching the cylinder branches out and encircles the
cylinder, forming a boundary layer that wraps around the cylinder. The fluid
particles on the midplane strike the cylinder at the stagnation point, bringing
the fluid to a complete stop and thus raising the pressure at that point. The
pressure decreases in the flow direction while the fluid velocity increases.
At very low upstream velocities (Re 1), the fluid completely wraps around
the cylinder and the two arms of the fluid meet on the rear side of the cylinder cen58933_ch07.qxd 9/4/2002 12:12 PM Page 381 381
CHAPTER 7
400
200
100
60
40 CD 20
10
6
4
2 Smooth cylinder 1
0.6
0.4
Sphere 0.2
0.1
0.06
10–1 100 101 102 103 104 105 106 Re FIGURE 7–17
Average drag coefficient for cross flow over a smooth circular
cylinder and a smooth sphere (from Schlichting, Ref. 10). in an orderly manner. Thus, the fluid follows the curvature of the cylinder. At
higher velocities, the fluid still hugs the cylinder on the frontal side, but it is too
fast to remain attached to the surface as it approaches the top of the cylinder.
As a result, the boundary layer detaches from the surface, forming a separation
region behind the cylinder. Flow in the wake region is characterized by random
vortex formation and pressures much lower than the stagnation point pressure.
The nature of the flow across a cylinder or sphere strongly affects the total
drag coefficient CD. Both the friction drag and the pressure drag can be significant. The high pressure in the vicinity of the stagnation point and the low
pressure on the opposite side in the wake produce a net force on the body in
the direction of flow. The drag force is primarily due to friction drag at low
Reynolds numbers (Re 10) and to pressure drag at high Reynolds numbers
(Re 5000). Both effects are significant at intermediate Reynolds numbers.
The average drag coefficients CD for cross flow over a smooth single circular cylinder and a sphere are given in Figure 7–17. The curves exhibit different behaviors in different ranges of Reynolds numbers:
• For Re 1, we have creeping flow, and the drag coefficient decreases
with increasing Reynolds number. For a sphere, it is CD 24/Re. There is
no flow separation in this regime.
• At about Re 10, separation starts occurring on the rear of the body with
vortex shedding starting at about Re 90. The region of separation
increases with increasing Reynolds number up to about Re 103. At this
point, the drag is mostly (about 95 percent) due to pressure drag. The drag
coefficient continues to decrease with increasing Reynolds number in this
range of 10 Re 103. (A decrease in the drag coefficient does not
necessarily indicate a decrease in drag. The drag force is proportional to
the square of the velocity, and the increase in velocity at higher Reynolds
numbers usually more than offsets the decrease in the drag coefficient.) cen58933_ch07.qxd 9/4/2002 12:12 PM Page 382 382
HEAT TRANSFER • In the moderate range of 103 Re 105, the drag coefficient remains
relatively constant. This behavior is characteristic of blunt bodies. The
flow in the boundary layer is laminar in this range, but the flow in the
separated region past the cylinder or sphere is highly turbulent with a
wide turbulent wake.
• There is a sudden drop in the drag coefficient somewhere in the range of
105 Re 106 (usually, at about 2 105). This large reduction in CD is
due to the flow in the boundary layer becoming turbulent, which moves
the separation point further on the rear of the body, reducing the size of
the wake and thus the magnitude of the pressure drag. This is in contrast
to streamlined bodies, which experience an increase in the drag
coefficient (mostly due to friction drag) when the boundary layer
becomes turbulent.
Flow separation occurs at about
80 (measured from the stagnation
point) when the boundary layer is laminar and at about
140 when it is
turbulent (Fig. 7–18). The delay of separation in turbulent flow is caused by
the rapid fluctuations of the fluid in the transverse direction, which enables the
turbulent boundary layer to travel further along the surface before separation
occurs, resulting in a narrower wake and a smaller pressure drag. In the range
of Reynolds numbers where the flow changes from laminar to turbulent, even
the drag force FD decreases as the velocity (and thus Reynolds number) increases. This results in a sudden decrease in drag of a flying body and instabilities in flight. Laminar
boundary
layer Separation
(a) Laminar flow (Re < 2 × 10 5) Laminar
boundary
layer Transition Effect of Surface Roughness
Turbulent
boundary
layer Separation
(b) Turbulence occurs (Re > 2 × 10 5) FIGURE 7–18
Turbulence delays flow separation. We mentioned earlier that surface roughness, in general, increases the drag
coefficient in turbulent flow. This is especially the case for streamlined bodies.
For blunt bodies such as a circular cylinder or sphere, however, an increase in
the surface roughness may actually decrease the drag coefficient, as shown in
Figure 7–19 for a sphere. This is done by tripping the flow into turbulence at
a lower Reynolds number, and thus causing the fluid to close in behind the
body, narrowing the wake and reducing pressure drag considerably. This
results in a much smaller drag coefficient and thus drag force for a roughsurfaced cylinder or sphere in a certain range of Reynolds number compared
to a smooth one of identical size at the same velocity. At Re 105, for example, CD 0.1 for a rough sphere with /D 0.0015, whereas CD 0.5 for
a smooth one. Therefore, the drag coefficient in this case is reduced by a factor of 5 by simply roughening the surface. Note, however, that at Re 106,
CD 0.4 for the rough sphere while CD 0.1 for the smooth one. Obviously,
roughening the sphere in this case will increase the drag by a factor of 4
(Fig. 7–20).
The discussion above shows that roughening the surface can be used to
great advantage in reducing drag, but it can also backfire on us if we are not
careful—specifically, if we do not operate in the right range of Reynolds number. With this consideration, golf balls are intentionally roughened to induce
turbulence at a lower Reynolds number to take advantage of the sharp drop in
the drag coefficient at the onset of turbulence in the boundary layer (the typical velocity range of golf balls is 15 to 150 m/s, and the Reynolds number
is less than 4 105). The critical Reynolds number of dimpled golf balls is cen58933_ch07.qxd 9/4/2002 12:12 PM Page 383 383
CHAPTER 7
0.6
ε = relative roughness
D (( 0.4 2 π D2
4 CD = 1ρ
2 FD 0.5 Golf
ball
0.3 0.2 0.1 0 ε = 1.25×10–2
D
ε = 5×10–3
D
ε = 1.5×10–3
D
4×104 ε = 0 (smooth)
D 4×105 105 4×106 106 D
Re = v FIGURE 7–19
The effect of surface roughness on the drag coefficient of a sphere (from Blevins, Ref. 1). about 4
104. The occurrence of turbulent flow at this Reynolds number
reduces the drag coefficient of a golf ball by half, as shown in Figure 7–19.
For a given hit, this means a longer distance for the ball. Experienced golfers
also give the ball a spin during the hit, which helps the rough ball develop a
lift and thus travel higher and further. A similar argument can be given for a
tennis ball. For a table tennis ball, however, the distances are very short, and
the balls never reach the speeds in the turbulent range. Therefore, the surfaces
of table tennis balls are made smooth.
Once the drag coefficient is available, the drag force acting on a body
in cross flow can be determined from Eq. 71 where A is the frontal area
(A LD for a cylinder of length L and A
D2/4 for a sphere). It should be
kept in mind that the freestream turbulence and disturbances by other bodies
in flow (such as flow over tube bundles) may affect the drag coefficients
significantly. EXAMPLE 7–4 Drag Force Acting on a Pipe in a River A 2.2cmouterdiameter pipe is to cross a river at a 30mwide section while
being completely immersed in water (Fig. 7–21). The average flow velocity of
water is 4 m/s and the water temperature is 15 C. Determine the drag force exerted on the pipe by the river. SOLUTION A pipe is crossing a river. The drag force that acts on the pipe is to
be determined.
Assumptions 1 The outer surface of the pipe is smooth so that Figure 7–17 can
be used to determine the drag coefficient. 2 Water flow in the river is steady.
3 The direction of water flow is normal to the pipe. 4 Turbulence in river flow is
not considered. CD
Re Smooth
surface Rough surface,
/D 0.0015 105
106 0.5
0.1 0.1
0.4 FIGURE 7–20
Surface roughness may increase
or decrease the drag coefficient
of a spherical object, depending on
the value of the Reynolds number. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 384 384
HEAT TRANSFER
River
Pipe FIGURE 7–21
Schematic for Example 7–4. 30 m Properties The density and dynamic viscosity of water at 15 C are
999.1 kg/m3 and
1.138 10 3 kg/m · s (Table A9).
Analysis Noting that D
0.022 m, the Reynolds number for flow over the
pipe is D D Re (999.1 kg/m3)(4 m/s)(0.022 m)
1.138 10 3 kg/m · s 7.73 104 The drag coefficient corresponding to this value is, from Figure 717, CD 1.0.
Also, the frontal area for flow past a cylinder is A LD. Then the drag force acting on the pipe becomes
2 800 FD
θ 700 CD A 2 1.0(30 0.022 m2) (999.1 kg/m3)(4 m/s)2
1N
2
1 kg · m/s2 5275 N
D Discussion Note that this force is equivalent to the weight of a mass over 500
kg. Therefore, the drag force the river exerts on the pipe is equivalent to hanging
a total of over 500 kg in mass on the pipe supported at its ends 30 m apart. The
necessary precautions should be taken if the pipe cannot support this force. 600 Nuθ Re = 219
,0
186,00 00
0
500
170,000
140,0
00
400
101,300
300 Heat Transfer Coefficient 70,800 Flows across cylinders and spheres, in general, involve flow separation,
which is difficult to handle analytically. Therefore, such flows must be studied
experimentally or numerically. Indeed, flow across cylinders and spheres has
been studied experimentally by numerous investigators, and several empirical
correlations have been developed for the heat transfer coefficient.
The complicated flow pattern across a cylinder greatly influences heat
transfer. The variation of the local Nusselt number Nu around the periphery
of a cylinder subjected to cross flow of air is given in Figure 7–22. Note that,
for all cases, the value of Nu starts out relatively high at the stagnation point
(
0°) but decreases with increasing as a result of the thickening of the
laminar boundary layer. On the two curves at the bottom corresponding to
80°, which is the
Re 70,800 and 101,300, Nu reaches a minimum at
separation point in laminar flow. Then Nu increases with increasing as a result of the intense mixing in the separated flow region (the wake). The curves 200 100
0
0° 40°
80°
120°
θ from stagnation point 160° FIGURE 7–22
Variation of the local
heat transfer coefficient along the
circumference of a circular cylinder in
cross flow of air (from Giedt, Ref. 5). cen58933_ch07.qxd 9/4/2002 12:12 PM Page 385 385
CHAPTER 7 at the top corresponding to Re 140,000 to 219,000 differ from the first two
curves in that they have two minima for Nu . The sharp increase in Nu at
about
90° is due to the transition from laminar to turbulent flow. The later
decrease in Nu is again due to the thickening of the boundary layer. Nu
reaches its second minimum at about
140°, which is the flow separation
point in turbulent flow, and increases with as a result of the intense mixing
in the turbulent wake region.
The discussions above on the local heat transfer coefficients are insightful;
however, they are of little value in heat transfer calculations since the
calculation of heat transfer requires the average heat transfer coefficient over
the entire surface. Of the several such relations available in the literature for
the average Nusselt number for cross flow over a cylinder, we present the one
proposed by Churchill and Bernstein:
Nucyl hD
k 0.62 Re1/2 Pr1/3
1
[1 (0.4/Pr)2/3]1/4 0.3 Re
282,000 5/8 4/5 (735) This relation is quite comprehensive in that it correlates available data well for
Re Pr
0.2. The fluid properties are evaluated at the film tempera1 (T
Ts), which is the average of the freestream and surface
ture Tf
2
temperatures.
For flow over a sphere, Whitaker recommends the following comprehensive correlation:
Nusph hD
k 1/4 2 [0.4 Re1/2 0.06 Re2/3] Pr0.4 s (736) which is valid for 3.5 Re 80,000 and 0.7 Pr 380. The fluid properties in this case are evaluated at the freestream temperature T , except for s,
which is evaluated at the surface temperature Ts. Although the two relations
above are considered to be quite accurate, the results obtained from them can
be off by as much as 30 percent.
The average Nusselt number for flow across cylinders can be expressed
compactly as
Nucyl hD
k C Rem Prn (737) where n 1 and the experimentally determined constants C and m are given
3
in Table 7–1 for circular as well as various noncircular cylinders. The characteristic length D for use in the calculation of the Reynolds and the Nusselt
numbers for different geometries is as indicated on the figure. All fluid properties are evaluated at the film temperature.
The relations for cylinders above are for single cylinders or cylinders oriented such that the flow over them is not affected by the presence of others.
Also, they are applicable to smooth surfaces. Surface roughness and the freestream turbulence may affect the drag and heat transfer coefficients significantly. Eq. 737 provides a simpler alternative to Eq. 735 for flow over
cylinders. However, Eq. 735 is more accurate, and thus should be preferred
in calculations whenever possible. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 386 386
HEAT TRANSFER TABLE 71
Empirical correlations for the average Nusselt number for forced convection
over circular and noncircular cylinders in cross flow (from Zukauskas, Ref. 14,
and Jakob, Ref. 6)
Crosssection
of the cylinder Fluid Range of Re Nusselt number Gas or
liquid 0.4–4
4–40
40–4000
4000–40,000
40,000–400,000 Nu
Nu
Nu
Nu
Nu 0.989Re0.330
0.911Re0.385
0.683Re0.466
0.193Re0.618
0.027Re0.805 Gas 5000–100,000 Nu 0.102Re0.675 Pr1/3 Gas 5000–100,000 Nu 0.246Re0.588 Pr1/3 Gas 5000–100,000 Nu 0.153Re0.638 Pr1/3 Gas 5000–19,500
19,500–100,000 Nu
Nu 0.160Re0.638 Pr1/3
0.0385Re0.782 Pr1/3 Gas 4000–15,000 Nu 0.228Re0.731 Pr1/3 Gas 2500–15,000 Nu 0.248Re0.612 Pr1/3 Circle
D Square Pr1/3
Pr1/3
Pr1/3
Pr1/3
Pr1/3 D Square
(tilted
45°) D Hexagon
D Hexagon
(tilted
45°) Vertical
plate D D Ellipse
Ts = 110°C D Wind
= 8 m/s
T = 10°C .1 D =0 m EXAMPLE 7–5 FIGURE 7–23
Schematic for Example 7–5. Heat Loss from a Steam Pipe in Windy Air A long 10cmdiameter steam pipe whose external surface temperature is
110°C passes through some open area that is not protected against the winds
(Fig. 7–23). Determine the rate of heat loss from the pipe per unit of its length cen58933_ch07.qxd 9/4/2002 12:12 PM Page 387 387
CHAPTER 7 when the air is at 1 atm pressure and 10°C and the wind is blowing across the
pipe at a velocity of 8 m/s. SOLUTION A steam pipe is exposed to windy air. The rate of heat loss from the
steam is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas.
Properties The properties of air at the average film temperature of Tf
(Ts T )/2 (110 10)/2 60°C and 1 atm pressure are (Table A15)
k 0.02808 W/m · °C
1.896 10 5 m2/s Pr 0.7202 Analysis The Reynolds number is Re D (8 m/s)(0.1 m)
1.896 10 5 m2/s 104 4.219 The Nusselt number can be determined from 0.62 Re1/2 Pr1/3
1
[1 (0.4/Pr)2/3]1/4 Re
282,000 hD
k 0.3 0.3 Nu 0.62(4.219
10 4)1/2 (0.7202)1/3
1
[1 (0.4/0.7202)2/3]1/4 5/8 4/5 4.219 10 4
282,000 5 /8 4 /5 124
and h k
Nu
D 0.02808 W/m · °C
(124)
0.1 m 34.8 W/m2 · °C Then the rate of heat transfer from the pipe per unit of its length becomes As
·
Q pL
DL
hAs(Ts T ) (0.1 m)(1 m) 0.314 m2
(34.8 W/m2 · C)(0.314 m2)(110 10)°C 1093 W The rate of heat loss from the entire pipe can be obtained by multiplying the
value above by the length of the pipe in m.
Discussion The simpler Nusselt number relation in Table 7–1 in this case
would give Nu 128, which is 3 percent higher than the value obtained above
using Eq. 735. EXAMPLE 7–6 Cooling of a Steel Ball by Forced Air Air 3 A 25cmdiameter stainless steel ball (
8055 kg/m , Cp 480 J/kg · °C) is
removed from the oven at a uniform temperature of 300°C (Fig. 7–24). The ball
is then subjected to the flow of air at 1 atm pressure and 25°C with a velocity
of 3 m/s. The surface temperature of the ball eventually drops to 200°C. Determine the average convection heat transfer coefficient during this cooling
process and estimate how long the process will take. T = 25°C
= 3 m/s Steel
ball
300°C FIGURE 7–24
Schematic for Example 7–6. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 388 388
HEAT TRANSFER SOLUTION A hot stainless steel ball is cooled by forced air. The average convection heat transfer coefficient and the cooling time are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas. 4 The outer surface temperature of the ball is uniform at all times. 5 The surface temperature of the ball during cooling is
changing. Therefore, the convection heat transfer coefficient between the ball
and the air will also change. To avoid this complexity, we take the surface temperature of the ball to be constant at the average temperature of (300 200)/2
250°C in the evaluation of the heat transfer coefficient and use the value obtained for the entire cooling process.
Properties The dynamic viscosity of air at the average surface temperature is
2.76 10 5 kg/m · s. The properties of air at the freestream
temperature of 25°C and 1 atm are (Table A15)
s @ 250°C k 0.02551 W/m · °C
1.849 10 5 kg/m · s Pr 5 1.562 10
0.7296 m2/s Analysis The Reynolds number is determined from D Re (3 m/s)(0.25 m)
1.562 10 5 m2/s 104 4.802 The Nusselt number is
1/4 hD
k Nu 2 2 [0.4 Re1/2 [0.4(4.802
1.849
2.76 10 5
10 5 0.06 Re2/3] Pr0.4
4 1/2 10 ) 0.06(4.802 s 104)2/3](0.7296)0.4 1/4 135
Then the average convection heat transfer coefficient becomes h k
Nu
D 0.02551 W/m · °C
(135)
0.25 m 13.8 W/m2 °C In order to estimate the time of cooling of the ball from 300°C to 200°C, we determine the average rate of heat transfer from Newton’s law of cooling by using
the average surface temperature. That is, D2
(0.25 m)2 0.1963 m2
hAs(Ts, ave T ) (13.8 W/m2 · °C)(0.1963 m2)(250 As
·
Q ave 25)°C 610 W Next we determine the total heat transferred from the ball, which is simply the
change in the energy of the ball as it cools from 300°C to 200°C: m
Qtotal 1 D3
V
6
mCp(T2 T1) (8055 kg/m3) 1 (0.25 m)3 65.9 kg
6
(65.9 kg)(480 J/kg · °C)(300 200)°C 3,163,000 J In this calculation, we assumed that the entire ball is at 200°C, which is not
necessarily true. The inner region of the ball will probably be at a higher temperature than its surface. With this assumption, the time of cooling is determined to be cen58933_ch07.qxd 9/4/2002 12:12 PM Page 389 389
CHAPTER 7 t Q
·
Q ave 3,163,000 J
610 J/s 5185 s 1 h 26 min Discussion The time of cooling could also be determined more accurately using the transient temperature charts or relations introduced in Chapter 4. But
the simplifying assumptions we made above can be justified if all we need is a
ballpark value. It will be naive to expect the time of cooling to be exactly 1 h 26
min, but, using our engineering judgment, it is realistic to expect the time of
cooling to be somewhere between one and two hours.
Flow
direction
↑ 7–4 I FLOW ACROSS TUBE BANKS Crossflow over tube banks is commonly encountered in practice in heat
transfer equipment such as the condensers and evaporators of power plants,
refrigerators, and air conditioners. In such equipment, one fluid moves
through the tubes while the other moves over the tubes in a perpendicular
direction.
In a heat exchanger that involves a tube bank, the tubes are usually placed
in a shell (and thus the name shellandtube heat exchanger), especially when
the fluid is a liquid, and the fluid flows through the space between the tubes
and the shell. There are numerous types of shellandtube heat exchangers,
some of which are considered in Chap. 13. In this section we will consider the
general aspects of flow over a tube bank, and try to develop a better and more
intuitive understanding of the performance of heat exchangers involving a
tube bank.
Flow through the tubes can be analyzed by considering flow through a single tube, and multiplying the results by the number of tubes. This is not the
case for flow over the tubes, however, since the tubes affect the flow pattern
and turbulence level downstream, and thus heat transfer to or from them, as
shown in Figure 7–25. Therefore, when analyzing heat transfer from a tube
bank in cross flow, we must consider all the tubes in the bundle at once.
The tubes in a tube bank are usually arranged either inline or staggered in
the direction of flow, as shown in Figure 7–26. The outer tube diameter D is
taken as the characteristic length. The arrangement of the tubes in the tube
bank is characterized by the transverse pitch ST, longitudinal pitch SL , and the
diagonal pitch SD between tube centers. The diagonal pitch is determined from
SD S2
L (S T /2) 2 (738) As the fluid enters the tube bank, the flow area decreases from A1 STL to
AT (ST D)L between the tubes, and thus flow velocity increases. In staggered
arrangement, the velocity may increase further in the diagonal region if the
tube rows are very close to each other. In tube banks, the flow characteristics
are dominated by the maximum velocitiy max that occurs within the tube
bank rather than the approach velocity . Therefore, the Reynolds number is
defined on the basis of maximum velocity as
ReD max D max D (739) FIGURE 725
Flow patterns for staggered and
inline tube banks (photos by
R. D. Willis, Ref 12). cen58933_ch07.qxd 9/4/2002 12:12 PM Page 390 390
HEAT TRANSFER The maximum velocity is determined from the conservation of mass requirement for steady incompressible flow. For inline arrangement, the maximum velocity occurs at the minimum flow area between the tubes, and the
conservation of mass can be expressed as (see Fig. 726a)
A1
maxAT
or ST
(ST D). Then the maximum velocity becomes
max SL , T1
ST D
A1 ST AT
max 1st row 2nd row 3rd row (a) Inline SL , T1 SD ST D
AD A1 AT
AD A1 = ST L
AT = (ST D)L
AD = (SD D)L (b) Staggered FIGURE 7–26
Arrangement of the tubes in inline
and staggered tube banks (A1, AT, and
AD are flow areas at indicated
locations, and L is the length of the
tubes). ST D (740) In staggered arrangement, the fluid approaching through area A1 in Figure 7–26b passes through area AT and then through area 2AD as it wraps
around the pipe in the next row. If 2AD AT, maximum velocity will still occur at AT between the tubes, and thus the max relation Eq. 740 can also be
[or, if 2(SD D) (ST D)],
used for staggered tube banks. But if 2AD
maximum velocity will occur at the diagonal cross sections, and the maximum
velocity in this case becomes
Staggered and SD (ST D)/2: max ST
2(SD D) (741) since
A1
ST 2 max(SD D).
max(2AD) or
The nature of flow around a tube in the first row resembles flow over a single tube discussed in section 7–3, especially when the tubes are not too close
to each other. Therefore, each tube in a tube bank that consists of a single
transverse row can be treated as a single tube in crossflow. The nature of flow
around a tube in the second and subsequent rows is very different, however,
because of wakes formed and the turbulence caused by the tubes upstream.
The level of turbulence, and thus the heat transfer coefficient, increases with
row number because of the combined effects of upstream rows. But there is no
significant change in turbulence level after the first few rows, and thus the
heat transfer coefficient remains constant.
Flow through tube banks is studied experimentally since it is too complex
to be treated analytically. We are primarily interested in the average heat transfer coefficient for the entire tube bank, which depends on the number of tube
rows along the flow as well as the arrangement and the size of the tubes.
Several correlations, all based on experimental data, have been proposed for
the average Nusselt number for cross flow over tube banks. More recently,
Zukauskas has proposed correlations whose general form is
Nu D hD
k C Re m Pr n(Pr/Prs) 0.25
D (742) where the values of the constants C, m, and n depend on value Reynolds number. Such correlations are given in Table 7–2 explicitly for 0.7 Pr 500 and
0 ReD 2 106. The uncertainty in the values of Nusselt number obtained
from these relations is 15 percent. Note that all properties except Prs are to
be evaluated at the arithmetic mean temperature of the fluid determined from
Tm Ti Te
2 (743) where Ti and Te are the fluid temperatures at the inlet and the exit of the tube
bank, respectively. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 391 391
CHAPTER 7 TABLE 7–2
Nusselt number correlations for cross flow over tube banks for N
0.7 Pr 500 (from Zukauskas, Ref. 15, 1987)*
Arrangement Range of ReD Correlation 0–100
1000–2
105–2 2 NuD 0.63
0.27 Re D Pr 0.36(Pr/Prs )0.25 106 NuD 0.8
0.033 Re D Pr 0.4(Pr/Prs )0.25 NuD 0.4
1.04 Re D Pr 0.36(Pr/Prs )0.25 NuD 0.5
0.71 Re D Pr 0.36(Pr/Prs )0.25 10 500–1000
1000–2
2 105–2 0.5
0.52 Re D Pr 0.36(Pr/Prs )0.25 NuD
5 0–500
Staggered 0.4
0.9 Re D Pr 0.36(Pr/Prs )0.25 NuD 100–1000 Inline 16 and 5 10 0.6
0.35(ST /SL)0.2 Re D Pr 0.36(Pr/Prs )0.25 NuD 106 NuD 0.8
0.031(ST /SL)0.2 Re D Pr 0.36(Pr/Prs )0.25 *All properties except Prs are to be evaluated at the arithmetic mean of the inlet and outlet temperatures
of the fluid (Prs is to be evaluated at Ts ). The average Nusselt number relations in Table 7–2 are for tube banks with
16 or more rows. Those relations can also be used for tube banks with NL provided that they are modified as
F NuD NuD, NL (744) where F is a correction factor F whose values are given in Table 7–3. For
ReD 1000, the correction factor is independent of Reynolds number.
Once the Nusselt number and thus the average heat transfer coefficient for
the entire tube bank is known, the heat transfer rate can be determined from
Newton’s law of cooling using a suitable temperature difference T. The first
thought that comes to mind is to use T Ts Tm Ts (Ti Te)/2. But
this will, in general, over predict the heat transfer rate. We will show in the
next chapter that the proper temperature difference for internal flow (flow
over tube banks is still internal flow through the shell) is the 1ogarithmic
mean temperature difference Tln defined as
Tln (Ts Te) (Ts
ln[(Ts Te)/(Ts Ti)
Ti)] Ti
Te
ln( Te / Ti ) (745) We will also show that the exit temperature of the fluid Te can be determined
from TABLE 7–3
Correction factor F to be used in NuD, NL, = F NuD for NL
(from Zukauskas, Ref 15, 1987).
NL 1 2 3 4 5 16 and ReD
7 10 1000
13 Inline 0.70 0.80 0.86 0.90 0.93 0.96 0.98 0.99 Staggered 0.64 0.76 0.84 0.89 0.93 0.96 0.98 0.99 cen58933_ch07.qxd 9/4/2002 12:12 PM Page 392 392
HEAT TRANSFER Te Ts (Ts Ti) exp As h
·
m Cp (746) ·
where As N DL is the heat transfer surface area and m
(N T S T L) is the
mass flow rate of the fluid. Here N is the total number of tubes in the bank, NT
is the number of tubes in a transverse plane, L is the length of the tubes, and
is the velocity of the fluid just before entering the tube bank. Then the heat
transfer rate can be determined from
˙
Q h As Tln ˙
mCp(Te Ti) (747) The second relation is usually more convenient to use since it does not require
the calculation of Tln. Pressure Drop
Another quantity of interest associated with tube banks is the pressure drop
P, which is the difference between the pressures at the inlet and the exit of
the tube bank. It is a measure of the resistance the tubes offer to flow over
them, and is expressed as
P NL f 2
max 2 (748) where f is the friction factor and is the correction factor, both plotted in Figures 7–27a and 7–27b against the Reynolds number based on the maximum
velocity max. The friction factor in Figure 7–27a is for a square inline tube
bank (ST SL), and the correction factor given in the insert is used to account
for the effects of deviation of rectangular inline arrangements from square
arrangement. Similarly, the friction factor in Figure 7–27b is for an equilateral
staggered tube bank (ST SD), and the correction factor is to account for the
effects of deviation from equilateral arrangement. Note that
1 for both
square and equilateral triangle arrangements. Also, pressure drop occurs in the
flow direction, and thus we used NL (the number of rows) in the P relation.
The power required to move a fluid through a tube bank is proportional to
the pressure drop, and when the pressure drop is available, the pumping power
required can be determined from
˙
Wpump ˙
VP ˙
mP (749) ˙
˙
˙
where V
(NT S T L) is the volume flow rate and m
V
(NT S T L) is
the mass flow rate of the fluid through the tube bank. Note that the power required to keep a fluid flowing through the tube bank (and thus the operating
cost) is proportional to the pressure drop. Therefore, the benefits of enhancing
heat transfer in a tube bank via rearrangement should be weighed against the
cost of additional power requirements.
In this section we limited our consideration to tube banks with base surfaces
(no fins). Tube banks with finned surfaces are also commonly used in practice, especially when the fluid is a gas, and heat transfer and pressure drop correlations can be found in the literature for tube banks with pin fins, plate fins,
strip fins, etc. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 393 393
CHAPTER 7
60
40
PL
PT 20
10 1.5 8
6 SL SL/D
ST /D ST PT 10
6
2
1
0.6 PL 4 PL 103
104 1.25 106 0.2 Friction factor, ƒ 2 0.1 0.2 2.0 1
0.8
0.6
0.4 105 ReD,max (PT 0.6 1 2 1)/(PL 6 10 1) 0.2 2.5 3.0 0.1
8
6
4 6 8 101 2 4 6 8 102 2 4 6 8 103 2 4 6 8104 2 4 6 8 105 2 4 6 8 106 ReD,max
(a) Inline arrangement
80
60
40 1.6 SD ReD,max 20
1.5 10 SD 8
6 ST 1.2 Friction factor, ƒ 103
104 104
103 4
1.0 2 PT 2.0
1
0.8
0.6
0.4 105 102 1.25
0.4 0.6 0.8 1 2 4 PT /PL 2.5
3.5 0.2
0.1 102 105 1.4 2 4 6 8101 2 4 6 8 102 2 4 6 8103 2 4 6 8 104 2 4 6 8105 2 4 6 8 106 ReD,max
(b) Staggered arrangement EXAMPLE 7–7 Preheating Air by Geothermal Water in a Tube
Bank In an industrial facility, air is to be preheated before entering a furnace by geothermal water at 120ºC flowing through the tubes of a tube bank located in a
duct. Air enters the duct at 20ºC and 1 atm with a mean velocity of 4.5 m/s,
and flows over the tubes in normal direction. The outer diameter of the tubes is
1.5 cm, and the tubes are arranged inline with longitudinal and transverse
pitches of SL ST 5 cm. There are 6 rows in the flow direction with 10 tubes
in each row, as shown in Figure 7–28. Determine the rate of heat transfer per
unit length of the tubes, and the pressure drop across the tube bank. SOLUTION Air is heated by geothermal water in a tube bank. The rate of heat
transfer to air and the pressure drop of air are to be determined. 2 FIGURE 7–27
Friction factor f and correction
factor for tube banks (from
Zukauskas, Ref. 16, 1985). cen58933_ch07.qxd 9/4/2002 12:12 PM Page 394 394
HEAT TRANSFER
AIR
= 4.5 m/s
T1 = 20°C Ts = 120°C Assumptions 1 Steady operating conditions exist. 2 The surface temperature of
the tubes is equal to the temperature of geothermal water.
Properties The exit temperature of air, and thus the mean temperature, is not
known. We evaluate the air properties at the assumed mean temperature of
60ºC (will be checked later) and 1 atm are Table A–15): k 1.06 kg/m3 0.02808 W/m K, Cp 1.007 kJ/kg K,
2.008 10 5 kg/m s Pr 0.7202 Prs Pr@Ts 0.7073 Also, the density of air at the inlet temperature of 20ºC (for use in the mass flow
rate calculation at the inlet) is 1 1.204 kg/m3
Analysis It is given that D 0.015 m, SL ST 0.05 m, and
4.5 m/s.
Then the maximum velocity and the Reynolds number based on the maximum
velocity become ST
SL = ST = 5 cm D = 1.5 cm ST
max FIGURE 7–28
Schematic for Example 7–7. ST ReD D
max D 0.05
(4.5 m /s) 6.43 m /s
0.05 0.015
(1.06 kg /m 3)(6.43 m /s)(0.015 m)
5091
2.008 10 5 kg /m s The average Nusselt number is determined using the proper relation from Table
7–2 to be 0.27 Re 0.63 Pr 0.36(Pr/Prs) 0.25
D
0.27(5091)0.63(0.7202)0.36(0.7202/0.7073)0.25 NuD 52.2 This Nusselt number is applicable to tube banks with NL 16. In our case, the
number of rows is NL
6, and the corresponding correction factor from Table
7–3 is F 0.945. Then the average Nusselt number and heat transfer coefficient for all the tubes in the tube bank become Nu D, NL F Nu D (0.945)(52.2) 49.3
NuD, NLk 49.3(0.02808 W/m ºC)
D
0.015 m h 92.2 W/m2 ºC NT
6
10
60. For a unit tube
The total number of tubes is N
NL
length (L 1 m), the heat transfer surface area and the mass flow rate of air
(evaluated at the inlet) are 60 (0.015 m)(1 m) 2.827 m2 As N DL ˙
m ˙
m1
1 (NT S T L)
(1.204 kg/m3)(4.5 m/s)(10)(0.05 m)(1 m) 2.709 kg/s Then the fluid exit temperature, the log mean temperature difference, and the
rate of heat transfer become Te Ts
120 (Ts Ash
˙
mCp
(2.827 m 2)(92.2 W/m 2 ºC)
20) exp
(2.709 kg/s)(1007 J/kg ºC) Ti) exp (120 29.11º C cen58933_ch07.qxd 9/4/2002 12:12 PM Page 395 395
CHAPTER 7 Tln
˙
Q (Ts Te) (Ts Ti)
(120 29.11) (120 20)
95.4º C
ln[(Ts Te)/(Ts Ti)] ln[(120 29.11)/(120 20)]
hAs Tln (92.2 W/m2 ºC)(2.827 m2)(95.4ºC) 2.49 104 W The rate of heat transfer can also be determined in a simpler way from ·
Q ·
hAs Tin m Cp(Te Ti)
(2.709 kg/s)(1007 J/kg · °C)(29. 11 20)°C 2.49 104 W For this square inline tube bank, the friction coefficient corresponding to
ReD
5088 and SL/D
5/1.5
3.33 is, from Fig. 7–27a, f
0.16. Also,
1 for the square arrangements. Then the pressure drop across the tube
bank becomes P V2
max
2
(1.06 kg/m3)(6.43 m/s)3
1N
6(0.16)(1)
2
1 kg m /s 2 NL f 21 Pa Discussion The arithmetic mean fluid temperature is (Ti + Te)/2
(20
110.9)/2 65.4ºC, which is fairly close to the assumed value of 60°C. Therefore, there is no need to repeat calculations by reevaluating the properties at
65.4°C (it can be shown that doing so would change the results by less
than 1 percent, which is much less than the uncertainty in the equations and
the charts used). TOPIC OF SPECIAL INTEREST Reducing Heat Transfer through Surfaces: Thermal Insulation
Thermal insulations are materials or combinations of materials that are
used primarily to provide resistance to heat flow (Fig. 7–29). You are probably familiar with several kinds of insulation available in the market. Most
insulations are heterogeneous materials made of low thermal conductivity
materials, and they involve air pockets. This is not surprising since air has
one of the lowest thermal conductivities and is readily available. The Styrofoam commonly used as a packaging material for TVs, VCRs, computers, and just about anything because of its light weight is also an excellent
insulator.
Temperature difference is the driving force for heat flow, and the greater
the temperature difference, the larger the rate of heat transfer. We can slow
down the heat flow between two mediums at different temperatures by
putting “barriers” on the path of heat flow. Thermal insulations serve as
such barriers, and they play a major role in the design and manufacture of
all energyefficient devices or systems, and they are usually the cornerstone
of energy conservation projects. A 1991 Drexel University study of the
energyintensive U.S. industries revealed that insulation saves the U.S. *This section can be skipped without a loss in continuity. Insulation
Heat
loss
Heat FIGURE 7–29
Thermal insulation retards heat
transfer by acting as a barrier in the
path of heat flow. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 396 396
HEAT TRANSFER Combustion
gases FIGURE 7–30
Insulation also helps the environment
by reducing the amount of fuel burned
and the air pollutants released. FIGURE 7–31
In cold weather, we minimize heat loss
from our bodies by putting on thick
layers of insulation (coats or furs). industry nearly 2 billion barrels of oil per year, valued at $60 billion a year
in energy costs, and more can be saved by practicing better insulation techniques and retrofitting the older industrial facilities.
Heat is generated in furnaces or heaters by burning a fuel such as coal,
oil, or natural gas or by passing electric current through a resistance heater.
Electricity is rarely used for heating purposes since its unit cost is much
higher. The heat generated is absorbed by the medium in the furnace and its
surfaces, causing a temperature rise above the ambient temperature. This
temperature difference drives heat transfer from the hot medium to the ambient, and insulation reduces the amount of heat loss and thus saves fuel
and money. Therefore, insulation pays for itself from the energy it saves.
Insulating properly requires a onetime capital investment, but its effects
are dramatic and long term. The payback period of insulation is often less
than one year. That is, the money insulation saves during the first year is
usually greater than its initial material and installation costs. On a broader
perspective, insulation also helps the environment and fights air pollution
and the greenhouse effect by reducing the amount of fuel burned and thus
the amount of CO2 and other gases released into the atmosphere (Fig.
7–30).
Saving energy with insulation is not limited to hot surfaces. We can also
save energy and money by insulating cold surfaces (surfaces whose temperature is below the ambient temperature) such as chilled water lines,
cryogenic storage tanks, refrigerated trucks, and airconditioning ducts.
The source of “coldness” is refrigeration, which requires energy input, usually electricity. In this case, heat is transferred from the surroundings to the
cold surfaces, and the refrigeration unit must now work harder and longer
to make up for this heat gain and thus it must consume more electrical energy. A cold canned drink can be kept cold much longer by wrapping it in
a blanket. A refrigerator with wellinsulated walls will consume much less
electricity than a similar refrigerator with little or no insulation. Insulating
a house will result in reduced cooling load, and thus reduced electricity
consumption for airconditioning.
Whether we realize it or not, we have an intuitive understanding and
appreciation of thermal insulation. As babies we feel much better in our
blankies, and as children we know we should wear a sweater or coat when
going outside in cold weather (Fig. 7–31). When getting out of a pool after
swimming on a windy day, we quickly wrap in a towel to stop shivering.
Similarly, early man used animal furs to keep warm and built shelters using
mud bricks and wood. Cork was used as a roof covering for centuries. The
need for effective thermal insulation became evident with the development
of mechanical refrigeration later in the nineteenth century, and a great deal
of work was done at universities and government and private laboratories
in the 1910s and 1920s to identify and characterize thermal insulation.
Thermal insulation in the form of mud, clay, straw, rags, and wood strips
was first used in the eighteenth century on steam engines to keep workmen
from being burned by hot surfaces. As a result, boiler room temperatures
dropped and it was noticed that fuel consumption was also reduced. The realization of improved engine efficiency and energy savings prompted the
search for materials with improved thermal efficiency. One of the first such
materials was mineral wool insulation, which, like many materials, was cen58933_ch07.qxd 9/4/2002 12:12 PM Page 397 397
CHAPTER 7 discovered by accident. About 1840, an iron producer in Wales aimed a
stream of highpressure steam at the slag flowing from a blast furnace, and
manufactured mineral wool was born. In the early 1860s, this slag wool
was a byproduct of manufacturing cannons for the Civil War and quickly
found its way into many industrial uses. By 1880, builders began installing
mineral wool in houses, with one of the most notable applications being
General Grant’s house. The insulation of this house was described in an article: “it keeps the house cool in summer and warm in winter; it prevents
the spread of fire; and it deadens the sound between floors” [Edmunds
(1989), Ref. 4]. An article published in 1887 in Scientific American detailing the benefits of insulating the entire house gave a major boost to the use
of insulation in residential buildings.
The energy crisis of the 1970s had a tremendous impact on the public
awareness of energy and limited energy reserves and brought an emphasis
on energy conservation. We have also seen the development of new and
more effective insulation materials since then, and a considerable increase
in the use of insulation. Thermal insulation is used in more places than you
may be aware of. The walls of your house are probably filled with some
kind of insulation, and the roof is likely to have a thick layer of insulation.
The “thickness” of the walls of your refrigerator is due to the insulation
layer sandwiched between two layers of sheet metal (Fig. 7–32). The walls
of your range are also insulated to conserve energy, and your hot water tank
contains less water than you think because of the 2 to 4cmthick insulation in the walls of the tank. Also, your hot water pipe may look much
thicker than the cold water pipe because of insulation. Insulation Heat
transfer Reasons for Insulating
If you examine the engine compartment of your car, you will notice that the
firewall between the engine and the passenger compartment as well as the
inner surface of the hood are insulated. The reason for insulating the hood
is not to conserve the waste heat from the engine but to protect people from
burning themselves by touching the hood surface, which will be too hot if
not insulated. As this example shows, the use of insulation is not limited to
energy conservation. Various reasons for using insulation can be summarized as follows:
• Energy Conservation Conserving energy by reducing the rate of heat
flow is the primary reason for insulating surfaces. Insulation materials
that will perform satisfactorily in the temperature range of 268°C to
1000°C ( 450°F to 1800°F) are widely available.
• Personnel Protection and Comfort A surface that is too hot poses a
danger to people who are working in that area of accidentally touching
the hot surface and burning themselves (Fig. 7–33). To prevent this danger and to comply with the OSHA (Occupational Safety and Health
Administration) standards, the temperatures of hot surfaces should be
reduced to below 60°C (140°F) by insulating them. Also, the excessive
heat coming off the hot surfaces creates an unpleasant environment in
which to work, which adversely affects the performance or productivity
of the workers, especially in summer months. FIGURE 7–32
The insulation layers in the walls of a
refrigerator reduce the amount of heat
flow into the refrigerator and thus the
running time of the refrigerator, saving
electricity. Insulation FIGURE 7–33
The hood of the engine compartment
of a car is insulated to reduce its
temperature and to protect people
from burning themselves. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 398 398
HEAT TRANSFER Insulation z z z FIGURE 7–34
Insulation materials absorb vibration
and sound waves, and are used to
minimize sound transmission. • Maintaining Process Temperature Some processes in the chemical
industry are temperaturesensitive, and it may become necessary to insulate the process tanks and flow sections heavily to maintain the same
temperature throughout.
• Reducing Temperature Variation and Fluctuations The temperature in an enclosure may vary greatly between the midsection and the
edges if the enclosure is not insulated. For example, the temperature
near the walls of a poorly insulated house is much lower than the temperature at the midsections. Also, the temperature in an uninsulated enclosure will follow the temperature changes in the environment closely
and fluctuate. Insulation minimizes temperature nonuniformity in an enclosure and slows down fluctuations.
• Condensation and Corrosion Prevention Water vapor in the air condenses on surfaces whose temperature is below the dew point, and the
outer surfaces of the tanks or pipes that contain a cold fluid frequently
fall below the dewpoint temperature unless they have adequate insulation. The liquid water on exposed surfaces of the metal tanks or pipes
may promote corrosion as well as algae growth.
• Fire Protection Damage during a fire may be minimized by keeping
valuable combustibles in a safety box that is well insulated. Insulation
may lower the rate of heat flow to such levels that the temperature in the
box never rises to unsafe levels during fire.
• Freezing Protection Prolonged exposure to subfreezing temperatures
may cause water in pipes or storage vessels to freeze and burst as a result of heat transfer from the water to the cold ambient. The bursting of
pipes as a result of freezing can cause considerable damage. Adequate
insulation will slow down the heat loss from the water and prevent
freezing during limited exposure to subfreezing temperatures. For example, covering vegetables during a cold night will protect them from
freezing, and burying water pipes in the ground at a sufficient depth will
keep them from freezing during the entire winter. Wearing thick gloves
will protect the fingers from possible frostbite. Also, a molten metal or
plastic in a container will solidify on the inner surface if the container is
not properly insulated.
• Reducing Noise and Vibration An added benefit of thermal insulation is its ability to dampen noise and vibrations (Fig. 7–34). The
insulation materials differ in their ability to reduce noise and vibration,
and the proper kind can be selected if noise reduction is an important
consideration.
There are a wide variety of insulation materials available in the market,
but most are primarily made of fiberglass, mineral wool, polyethylene,
foam, or calcium silicate. They come in various trade names such as
Ethafoam Polyethylene Foam Sheeting, Solimide Polimide Foam Sheets,
FPC Fiberglass Reinforced Silicone Foam Sheeting, Silicone Sponge Rubber Sheets, fiberglass/mineral wool insulation blankets, wirereinforced cen58933_ch07.qxd 9/4/2002 12:12 PM Page 399 399
CHAPTER 7 mineral wool insulation, ReflectAll Insulation, granulated bulk mineral
wool insulation, cork insulation sheets, foilfaced fiberglass insulation,
blended sponge rubber sheeting, and numerous others.
Today various forms of fiberglass insulation are widely used in process
industries and heating and airconditioning applications because of their
low cost, light weight, resiliency, and versatility. But they are not suitable
for some applications because of their low resistance to moisture and fire
and their limited maximum service temperature. Fiberglass insulations
come in various forms such as unfaced fiberglass insulation, vinylfaced
fiberglass insulation, foilfaced fiberglass insulation, and fiberglass insulation sheets. The reflective foilfaced fiberglass insulation resists vapor penetration and retards radiation because of the aluminum foil on it and is
suitable for use on pipes, ducts, and other surfaces.
Mineral wool is resilient, lightweight, fibrous, woollike, thermally
efficient, fire resistant up to 1100°C (2000°F), and forms a sound barrier.
Mineral wool insulation comes in the form of blankets, rolls, or blocks.
Calcium silicate is a solid material that is suitable for use at high temperatures, but it is more expensive. Also, it needs to be cut with a saw during installation, and thus it takes longer to install and there is more waste. Superinsulators
You may be tempted to think that the most effective way to reduce heat
transfer is to use insulating materials that are known to have very low thermal conductivities such as urethane or rigid foam (k 0.026 W/m · °C) or
fiberglass (k 0.035 W/m · °C). After all, they are widely available, inexpensive, and easy to install. Looking at the thermal conductivities of materials, you may also notice that the thermal conductivity of air at room
temperature is 0.026 W/m · °C, which is lower than the conductivities of
practically all of the ordinary insulating materials. Thus you may think that
a layer of enclosed air space is as effective as any of the common insulating materials of the same thickness. Of course, heat transfer through the air
will probably be higher than what a pure conduction analysis alone would
indicate because of the natural convection currents that are likely to occur
in the air layer. Besides, air is transparent to radiation, and thus heat will
also be transferred by radiation. The thermal conductivity of air is practically independent of pressure unless the pressure is extremely high or extremely low. Therefore, we can reduce the thermal conductivity of air and
thus the conduction heat transfer through the air by evacuating the air
space. In the limiting case of absolute vacuum, the thermal conductivity
will be zero since there will be no particles in this case to “conduct” heat
from one surface to the other, and thus the conduction heat transfer will be
zero. Noting that the thermal conductivity cannot be negative, an absolute
vacuum must be the ultimate insulator, right? Well, not quite.
The purpose of insulation is to reduce “total” heat transfer from a surface, not just conduction. A vacuum totally eliminates conduction but offers zero resistance to radiation, whose magnitude can be comparable to
conduction or natural convection in gases (Fig. 7–35). Thus, a vacuum is T1 T2
Vacuum Radiation FIGURE 7–35
Evacuating the space between two
surfaces completely eliminates heat
transfer by conduction or convection
but leaves the door wide open for
radiation. cen58933_ch07.qxd 9/4/2002 12:12 PM Page 400 400
HEAT TRANSFER Thin metal
sheets
T1 T2 Vacuum FIGURE 7–36
Superinsulators are built by closely
packing layers of highly reflective thin
metal sheets and evacuating the space
between them. no more effective in reducing heat transfer than sealing off one of the lanes
of a twolane road is in reducing the flow of traffic on a oneway road.
Insulation against radiation heat transfer between two surfaces is
achieved by placing “barriers” between the two surfaces, which are highly
reflective thin metal sheets. Radiation heat transfer between two surfaces is
inversely proportional to the number of such sheets placed between the surfaces. Very effective insulations are obtained by using closely packed layers of highly reflective thin metal sheets such as aluminum foil (usually 25
sheets per cm) separated by fibers made of insulating material such as glass
fiber (Fig. 7–36). Further, the space between the layers is evacuated to form
a vacuum under 0.000001 atm pressure to minimize conduction or convection heat transfer through the air space between the layers. The result is an
insulating material whose apparent thermal conductivity is below 2 10 5
W/m · °C, which is one thousand times less than the conductivity of air or
any common insulating material. These specially built insulators are called
superinsulators, and they are commonly used in space applications and
cryogenics, which is the branch of heat transfer dealing with temperatures
below 100 K ( 173°C) such as those encountered in the liquefaction, storage, and transportation of gases, with helium, hydrogen, nitrogen, and oxygen being the most common ones. The Rvalue of Insulation
The effectiveness of insulation materials is given by some manufacturers in
terms of their Rvalue, which is the thermal resistance of the material per
unit surface area. For flat insulation the Rvalue is obtained by simply dividing the thickness of the insulation by its thermal conductivity. That is,
Rvalue k,
Btu/h·ft·°F
Insulation L, ft L
Rvalue = —
k FIGURE 7–37
The Rvalue of an insulating material
is simply the ratio of the thickness of
the material to its thermal conductivity
in proper units. L
k (flat insulation) (750) where L is the thickness and k is the thermal conductivity of the material.
Note that doubling the thickness L doubles the Rvalue of flat insulation.
For pipe insulation, the Rvalue is determined using the thermal resistance
relation from
r2 r2
Rvalue
(pipe insulation)
(751)
ln
k r1
where r1 is the inside radius of insulation and r2 is the outside radius of insulation. Once the Rvalue is available, the rate of heat transfer through the
insulation can be determined from
·
Q T
Rvalue Area (752) where T is the temperature difference across the insulation and Area is the
outer surface area for a cylinder.
In the United States, the Rvalues of insulation are expressed without any
units, such as R19 and R30. These Rvalues are obtained by dividing
the thickness of the material in feet by its thermal conductivity in the unit
Btu/h · ft · °F so that the Rvalues actually have the unit h · ft2 · °F/Btu. For
example, the Rvalue of 6in.thick glass fiber insulation whose thermal
conductivity is 0.025 Btu/h · ft · °F is (Fig. 7–37) cen58933_ch07.qxd 9/4/2002 12:13 PM Page 401 401
CHAPTER 7 Rvalue L
k 0.5 ft
0.025 Btu/h · ft · °F 20 h · ft2 · °F/Btu Thus, this 6in.thick glass fiber insulation would be referred to as R20 insulation by the builders. The unit of Rvalue is m2 · °C/W in SI units, with
the conversion relation 1 m2 · °C/W 5.678 h · ft2 · °F/Btu. Therefore, a
small Rvalue in SI corresponds to a large Rvalue in English units. Optimum Thickness of Insulation
It should be realized that insulation does not eliminate heat transfer; it
merely reduces it. The thicker the insulation, the lower the rate of heat
transfer but also the higher the cost of insulation. Therefore, there should
be an optimum thickness of insulation that corresponds to a minimum combined cost of insulation and heat lost. The determination of the optimum
thickness of insulation is illustrated in Figure 7–38. Notice that the cost of
insulation increases roughly linearly with thickness while the cost of heat
loss decreases exponentially. The total cost, which is the sum of the insulation cost and the lost heat cost, decreases first, reaches a minimum, and
then increases. The thickness corresponding to the minimum total cost is
the optimum thickness of insulation, and this is the recommended thickness
of insulation to be installed.
If you are mathematically inclined, you can determine the optimum thickness by obtaining an expression for the total cost, which is the sum of the
expressions for the lost heat cost and insulation cost as a function of thickness; differentiating the total cost expression with respect to the thickness;
and setting it equal to zero. The thickness value satisfying the resulting
equation is the optimum thickness. The cost values can be determined from
an annualized lifetime analysis or simply from the requirement that the insulation pay for itself within two or three years. Note that the optimum
thickness of insulation depends on the fuel cost, and the higher the fuel
cost, the larger the optimum thickness of insulation. Considering that insulation will be in service for many years and the fuel prices are likely to escalate, a reasonable increase in fuel prices must be assumed in calculations.
Otherwise, what is optimum insulation today will be inadequate insulation
in the years to come, and we may have to face the possibility of costly
retrofitting projects. This is what happened in the 1970s and 1980s to insulations installed in the 1960s.
The discussion above on optimum thickness is valid when the type and
manufacturer of insulation are already selected, and the only thing to be
determined is the most economical thickness. But often there are several
suitable insulations for a job, and the selection process can be rather confusing since each insulation can have a different thermal conductivity, different installation cost, and different service life. In such cases, a selection
can be made by preparing an annualized cost versus thickness chart like
Figure 7–39 for each insulation, and determining the one having the lowest minimum cost. The insulation with the lowest annual cost is obviously
the most economical insulation, and the insulation thickness corresponding to the minimum total cost is the optimum thickness. When the optimum
thickness falls between two commercially available thicknesses, it is a
good practice to be conservative and choose the thicker insulation. The Cost
per
year
Total
cost
Insu st n co latio Lost
heat
cost Minimum
total cost
Optimum
thickness 0 Insulation thickness FIGURE 7–38
Determination of the optimum
thickness of insulation on the basis of
minimum total cost. Cost
per
year Total cost curves
for different insulation A
B
C Optimum
thickness for D
0 D Minimum
cost Insulation thickness FIGURE 7–39
Determination of the most economical
type of insulation and its
optimum thickness. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 402 402
HEAT TRANSFER TABLE 7–4
Recommended insulation
thicknesses for flat hot surfaces as a
function of surface temperature
(from TIMA Energy Savings Guide)
Surface
temperature Insulation
thickness 150°F
250°F
350°F
550°F
750°F
950°F 2
3
4
6
9
10 (66°C)
(121°C)
(177°C)
(288°C)
(400°C)
(510°C) (5.1 cm)
(7.6 cm)
(10.2 cm)
(15.2 cm)
(22.9 cm)
(25.44 cm) extra thickness will provide a little safety cushion for any possible decline
in performance over time and will help the environment by reducing the
production of greenhouse gases such as CO2.
The determination of the optimum thickness of insulation requires a heat
transfer and economic analysis, which can be tedious and timeconsuming.
But a selection can be made in a few minutes using the tables and charts prepared by TIMA (Thermal Insulation Manufacturers Association) and member companies. The primary inputs required for using these tables or charts
are the operating and ambient temperatures, pipe diameter (in the case of
pipe insulation), and the unit fuel cost. Recommended insulation thicknesses
for hot surfaces at specified temperatures are given in Table 7–4. Recommended thicknesses of pipe insulations as a function of service temperatures
are 0.5 to 1 in. for 150°F, 1 to 2 in. for 250°F, 1.5 to 3 in. for 350°F, 2 to 4.5
in. for 450°F, 2.5 to 5.5 in. for 550°F, and 3 to 6 in. for 650°F for nominal
pipe diameters of 0.5 to 36 in. The lower recommended insulation thicknesses are for pipes with small diameters, and the larger ones are for pipes
with large diameters. EXAMPLE 7–8 Effect of Insulation on Surface Temperature Hot water at Ti
120°C flows in a stainless steel pipe (k
15 W/m · °C)
whose inner diameter is 1.6 cm and thickness is 0.2 cm. The pipe is to be
covered with adequate insulation so that the temperature of the outer surface
of the insulation does not exceed 40°C when the ambient temperature is
To 25°C. Taking the heat transfer coefficients inside and outside the pipe to
be hi
70 W/m2 · °C and ho
20 W/m2 · °C, respectively, determine the
thickness of fiberglass insulation (k
0.038 W/m · °C) that needs to be installed on the pipe. SOLUTION A steam pipe is to be covered with enough insulation to reduce the
exposed surface temperature. The thickness of insulation that needs to be installed is to be determined.
Assumptions 1 Heat transfer is steady since there is no indication of any
change with time. 2 Heat transfer is onedimensional since there is thermal
symmetry about the centerline and no variation in the axial direction. 3 Thermal
conductivities are constant. 4 The thermal contact resistance at the interface is
negligible. To
ho r1
r2
Ti
hi
Steam Properties The thermal conductivities are given to be k 15 W/m · °C for the
steel pipe and k 0.038 W/m · °C for fiberglass insulation.
40°C r3
Insulation Ti T1
Ri T2
R1 T3
R2 FIGURE 7–40
Schematic for Example 7–8. To
Ro Analysis The thermal resistance network for this problem involves four resistances in series and is given in Figure 7–40. The inner radius of the pipe is
r1 0.8 cm and the outer radius of the pipe and thus the inner radius of the insulation is r2 1.0 cm. Letting r3 represent the outer radius of the insulation,
the areas of the surfaces exposed to convection for an L 1mlong section of
the pipe become A1 2 r1L 2 (0.008 m)(1 m) A3 2 r3L 2 r3 (1 m) 0.0503 m2 6.28r3 m2 cen58933_ch07.qxd 9/4/2002 12:13 PM Page 403 403
CHAPTER 7 Then the individual thermal resistances are determined to be 1
hi A1 1
(70 W/m2 · °C)(0.0503 m2) ln(r2 /r1)
2 k1 L ln(0.01/0.008)
2 (15 W/m · °C)(1 m) Ri Rconv, 1 R1 Rpipe R2 Rinsulation ln(r3 /r2)
2 k2 L 0.0024°C/ W ln(r3 /0.01)
2 (0.038 W/m · °C)(1 m) 4.188 ln(r3 /0.01)°C/ W
1
1
Rconv, 2
ho A3 (20 W/m2 · °C)(6.28r3 m2) Ro 0.284°C/ W 1
°C/ W
125.6r3 Noting that all resistances are in series, the total resistance is determined to be Rtotal Ri R1 R2 R0
[0.284 0.0024 4.188 ln(r3/0.01) 1/125.6r3]°C/ W Then the steady rate of heat loss from the steam becomes ·
Q Ti To
Rtotal [0.284 0.0024 (120 125)°C
4.188 ln(r3 /0.01) 1/125.6r3 ]°C/ W Noting that the outer surface temperature of insulation is specified to be 40°C,
the rate of heat loss can also be expressed as T3 ·
Q To
Ro (40 25)°C
(1/125.6r3)°C/ W 1884r3 Setting the two relations above equal to each other and solving for r3 gives
r3 0.0170 m. Then the minimum thickness of fiberglass insulation required is t r3 r2 0.0170 0.0100 0.0070 m 0.70 cm Discussion Insulating the pipe with at least 0.70cmthick fiberglass insulation will ensure that the outer surface temperature of the pipe will be at 40°C
or below. EXAMPLE 7–9 Optimum Thickness of Insulation During a plant visit, you notice that the outer surface of a cylindrical curing
oven is very hot, and your measurements indicate that the average temperature
of the exposed surface of the oven is 180°F when the surrounding air temperature is 75°F. You suggest to the plant manager that the oven should be insulated, but the manager does not think it is worth the expense. Then you propose
to the manager to pay for the insulation yourself if he lets you keep the savings
from the fuel bill for one year. That is, if the fuel bill is $5000/yr before insulation and drops to $2000/yr after insulation, you will get paid $3000. The manager agrees since he has nothing to lose, and a lot to gain. Is this a smart bet
on your part?
The oven is 12 ft long and 8 ft in diameter, as shown in Figure 7–41. The
plant operates 16 h a day 365 days a year, and thus 5840 h/yr. The insulation 180°F T = 75°F Curing
oven
8 ft 12 ft FIGURE 7–41
Schematic for Example 7–9. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 404 404
HEAT TRANSFER to be used is fiberglass (kins 0.024 Btu/h · ft · °F), whose cost is $0.70/ft2 per
inch of thickness for materials, plus $2.00/ft2 for labor regardless of thickness.
The combined heat transfer coefficient on the outer surface is estimated to be
ho
3.5 Btu/h · ft2 · °F. The oven uses natural gas, whose unit cost is
$0.75/therm input (1 therm 100,000 Btu), and the efficiency of the oven is
80 percent. Disregarding any inflation or interest, determine how much money
you will make out of this venture, if any, and the thickness of insulation (in
whole inches) that will maximize your earnings. SOLUTION A cylindrical oven is to be insulated to reduce heat losses. The optimum thickness of insulation and the potential earnings are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat transfer through the insulation is onedimensional. 3 Thermal conductivities are constant. 4 The thermal
contact resistance at the interface is negligible. 5 The surfaces of the cylindrical
oven can be treated as plain surfaces since its diameter is greater than 3 ft.
Properties The thermal conductivity of insulation is given to be k 0.024 Btu/
h · ft · °F.
Analysis The exposed surface area of the oven is
As 2Abase 2 r2 Aside 2 (4 ft)2 2 rL 2 (4 ft)(12 ft) 402 ft2 The rate of heat loss from the oven before the insulation is installed is determined from ·
Q ho As(Ts (3.5 Btu/h · ft2 · °F)(402 ft2)(180 T) 75)°F 147,700 Btu/h Noting that the plant operates 5840 h/yr, the total amount of heat loss from the
oven per year is Q ·
Qt (147,700 Btu/h)(5840 h/yr) 109 Btu/yr 0.863 The efficiency of the oven is given to be 80 percent. Therefore, to generate
this much heat, the oven must consume energy (in the form of natural gas) at
a rate of Qin Q/ oven (0.863
10,790 therms 109 Btu/yr)/0.80 1.079 109 Btu/yr since 1 therm 100,000 Btu. Then the annual fuel cost of this oven before insulation becomes Annual cost Qin Unit cost
(10,790 therm/yr)($0.75/therm) $8093/yr That is, the heat losses from the exposed surfaces of the oven are currently
costing the plant over $8000/yr.
When insulation is installed, the rate of heat transfer from the oven can be
determined from ·
Q ins Ts T
Rtotal Ts
Rins T
Rconv Ts
As t
ins
kins T
1
ho We expect the surface temperature of the oven to increase and the heat transfer coefficient to decrease somewhat when insulation is installed. We assume cen58933_ch07.qxd 9/4/2002 12:13 PM Page 405 405
CHAPTER 7 these two effects to counteract each other. Then the relation above for 1in.thick insulation gives the rate of heat loss to be As(Ts T )
(402 ft2)(180 75)°F
tins
1/12 ft
1
1
0.024 Btu/h · ft · °F 3.5 Btu/h · ft2 · °F
kins ho
11,230 Btu/h ·
Q ins Also, the total amount of heat loss from the oven per year and the amount and
cost of energy consumption of the oven become Qins
Qin, ins
Annual cost ·
Q ins t (11,230 Btu/h)(5840 h/yr) Qins / oven (0.6558
820 therms 108 Btu/yr)/0.80 Qin, ins Unit cost
(820 therm/yr)($0.75/therm) 108 Btu/yr 0.6558 108 Btu/yr 0.820 $615/yr Therefore, insulating the oven by 1in.thick fiberglass insulation will reduce the
fuel bill by $8093
$615
$7362 per year. The unit cost of insulation is
given to be $2.70/ft2. Then the installation cost of insulation becomes Insulation cost (Unit cost)(Surface area) ($2.70/ft2)(402 ft2) $1085 The sum of the insulation and heat loss costs is Total cost Insulation cost Heat loss cost $1085 $615 $1700 Then the net earnings will be Earnings Income Expenses $8093 $1700 $6393 To determine the thickness of insulation that maximizes your earnings, we
repeat the calculations above for 2, 3, 4, and 5in.thick insulations, and list
the results in Table 7–5. Note that the total cost of insulation decreases first
with increasing insulation thickness, reaches a minimum, and then starts to
increase. TABLE 7–5
The variation of total insulation cost with insulation thickness
Insulation
thickness
1
2
3
4
5 in.
in.
in.
in.
in. Heat loss,
Btu/h
11,230
5838
3944
2978
2392 Lost fuel,
therms/yr
820
426
288
217
175 Lost fuel
cost, $/yr
615
320
216
163
131 Insulation
cost, $
1085
1367
1648
1930
2211 Total cost,
$
1700
1687
1864
2093
2342 We observe that the total insulation cost is a minimum at $1687 for the case of
2in.thick insulation. The earnings in this case are Maximum earnings Income Minimum expenses
$8093 $1687 $6406 cen58933_ch07.qxd 9/4/2002 12:13 PM Page 406 406
HEAT TRANSFER which is not bad for a day’s worth of work. The plant manager is also a big winner in this venture since the heat losses will cost him only $320/yr during the
second and consequent years instead of $8093/yr. A thicker insulation could
probably be justified in this case if the cost of insulation is annualized over the
lifetime of insulation, say 20 years. Several energy conservation measures are
being marketed as explained above by several power companies and private
firms. SUMMARY
The force a flowing fluid exerts on a body in the flow direction
is called drag. The part of drag that is due directly to wall shear
stress w is called the skin friction drag since it is caused by
frictional effects, and the part that is due directly to pressure is
called the pressure drag or form drag because of its strong dependence on the form or shape of the body.
The drag coefficient CD is a dimensionless number that represents the drag characteristics of a body, and is defined as
CD Cf Combined: Cf x C f, x
Nu x Turbulent: Cf, x Nu x hx x
k 5 10 5 Re L 1742
ReL
Cf 5
1.89 10 7 10 5 ReL 1.62 log 10 7
2.5 L The average Nusselt number relations for flow over a flat plate
are:
Laminar: hL
k Nu x 0.664 Re 0.5 P r 1/3
L ReL 5 10 5 Turbulent: xcr 5 10 5 For parallel flow over a flat plate, the local friction and convection coefficients are
Laminar: 10 5 A Transition from laminar to turbulent occurs at the critical
Reynolds number of
Re x , cr 0.074
Re 1/5
L
0.074
Re1/5
L 5 Re L 2 where A is the frontal area for blunt bodies, and surface area
for parallel flow over flat plates or thin airfoils. For flow over
a flat plate, the Reynolds number is
Rex Turbulent: 1.328
Re 1/2
L Cf Rough surface, turbulent: FD
1
2 Laminar: 0.664
Re x 5 10 5
Re 1/2
x
hx x
Pr 0.6
0.332 Re0.5 Pr1/3
x
k
0.0592
,
5 10 5 Re x 10 7
Re 1/5
x
0.6 Pr 60
0.0296 Re 0.8 Pr1/3
x
5 10 5 Re x 10 7 The average friction coefficient relations for flow over a flat
plate are: Nu hL
k 0.037 Re0.8 Pr1/3
L Combined:
hL
Nu
(0.037 Re0.8
L
k 871) Pr 1/3, 0.6 Pr
5 10 5 60
ReL 10 7 0.6 Pr
5 10 5 60
ReL 10 7 For isothermal surfaces with an unheated starting section of
length , the local Nusselt number and the average convection
coefficient relations are
Laminar: Nux Turbulent: Nu x Laminar: h Turbulent: h Nux (for 0)
0.332 Re0.5 Pr 1/3
x
3/4 1/3
[1 ( / x) ]
[1 ( / x) 3/4 ]1/3
Nu x (for 0)
0.0296 Re0.8 Pr1/3
x
9/10 1/ 9
[1 ( /x) ]
[1 ( /x) 9/10 ]1/9
2[1 ( / x ) 3/4]
hx L
1
/L
9/10
5[1 ( / x )
hx L
(1
/ L) cen58933_ch07.qxd 9/4/2002 12:13 PM Page 407 407
CHAPTER 7 These relations are for the case of isothermal surfaces. When a
flat plate is subjected to uniform heat flux, the local Nusselt
number is given by
Laminar:
Turbulent: 0.453 Re 0.5 P r 1/3
x
0.0308 Re 0.8 P r 1/3
x Nux
Nux The average Nusselt numbers for cross flow over a cylinder
and sphere are
Nucyl hD
k 0.62 Re1/2 Pr1/3
1
[1 (0.4/ Pr)2/3]1/4 0.3 which is valid for Re Pr
Nu sph hD
k 2 0.06 Re 2/3]Pr 0.4 m
C R e D P r n(Pr/Prs) 0.25 where the values of the constants C, m, and n depend on value
Reynolds number. Such correlations are given in Table 7–2.
All properties except Prs are to be evaluated at the arithmetic
mean of the inlet and outlet temperatures of the fluid defined as
Tm (Ti Te)/2.
The average Nusselt number for tube banks with less than
16 rows is expressed as
Nu D, NL 1/4 Inline and Staggered with SD (ST D)/2:
ST
max
ST D
(ST D)/2:
ST
max
2(SD D) where ST the transverse pitch and SD is the diagonal pitch. The
average Nusselt number for cross flow over tube banks is expressed as F NuD where F is the correction factor whose values are given in
Table 73. The heat transfer rate to or from a tube bank is determined from
˙
Q s which is valid for 3.5
Re
80,000 and 0.7
Pr
380.
The fluid properties are evaluated at the film temperature
(T
Ts)/2 in the case of a cylinder, and at the freeTf
stream temperature T (except for s , which is evaluated at the
surface temperature Ts) in the case of a sphere.
In tube banks, the Reynolds number is based on the maximum velocity max that is related to the approach velocity as Staggered with SD hD
k 5/8 4 /5 Re
282,000 0.2, and [0.4 Re1/2 Nu D h A s Tln ˙
mCp (Te Ti) where Tln is the logarithmic mean temperature difference defined as
Tln (Ts Te ) (Ts
ln[(Ts Te )/(Ts Ti )
Ti )] Ti
Te
ln( Te / Ti) and the exit temperature of the fluid Te is
Te Ts (Ts Ti ) exp Ash
˙
mCp ·
where As N DL is the heat transfer surface area and m
(NT ST L) is the mass flow rate of the fluid. The pressure
drop P for a tube bank is expressed as
P NL f where f is the friction factor and
given in Figs. 7–27. 2
max 2
is the correction factor, both REFERENCES AND SUGGESTED READING
1. R. D. Blevin. Applied Fluid Dynamics Handbook.
New York: Van Nostrand Reinhold, 1984.
2. S. W. Churchill and M. Bernstein. “A Correlating
Equation for Forced Convection from Gases and Liquids
to a Circular Cylinder in Cross Flow.” Journal of Heat
Transfer 99 (1977), pp. 300–306. 5. W. H. Giedt. “Investigation of Variation of Point UnitHeat Transfer Coefficient around a Cylinder Normal to an
Air Stream.” Transactions of the ASME 71 (1949),
pp. 375–381.
6. M. Jakob. Heat Transfer. Vol. l. New York: John Wiley &
Sons, 1949. 3. S. W. Churchill and H. Ozoe. “Correlations for Laminar
Forced Convection in Flow over an Isothermal Flat Plate
and in Developing and Fully Developed Flow in an
Isothermal Tube.” Journal of Heat Transfer 95 (Feb.
1973), pp. 78–84. 7. W. M. Kays and M. E. Crawford. Convective Heat and
Mass Transfer. 3rd ed. New York: McGrawHill, 1993. 4. W. M. Edmunds. “Residential Insulation.” ASTM
Standardization News (Jan. 1989), pp. 36–39. 9. H. Schlichting. Boundary Layer Theory, 7th ed. New
York, McGrawHill, 1979. 8. F. Kreith and M. S. Bohn. Principles of Heat Transfer, 6th
ed. Pacific Grove, CA: Brooks/Cole, 2001. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 408 408
HEAT TRANSFER 10. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul,
MN: West, 1995.
11. W. C. Thomas. “Note on the Heat Transfer Equation for
Forced Convection Flow over a Flat Plate with and
Unheated Starting Length.” Mechanical Engineering
News, 9, no.1 (1977), p. 361.
12. R. D. Willis. “Photographic Study of Fluid Flow Between
Banks of Tubes.” Engineering (1934), pp. 423–425.
13. A. Zukauskas, “Convection Heat Transfer in Cross Flow.”
In Advances in Heat Transfer, J. P. Hartnett and T. F.
Irvine, Jr., Eds. New York: Academic Press, 1972, Vol. 8,
pp. 93–106. 14. A. Zukauskas. “Heat Transfer from Tubes in Cross Flow.”
In Advances in Heat Transfer, ed. J. P. Hartnett and T. F.
Irvine, Jr. Vol. 8. New York: Academic Press, 1972.
15. A. Zukauskas. “Heat Transfer from Tubes in Cross Flow.”
In Handbook of Single Phase Convective Heat Transfer,
Eds. S. Kakac, R. K. Shah, and Win Aung. New York:
Wiley Interscience, 1987.
16. A. Zukauskas and R. Ulinskas, “Efficiency Parameters for
Heat Transfer in Tube Banks.” Heat Transfer Engineering
no. 2 (1985), pp. 19–25. PROBLEMS*
Drag Force and Heat Transfer in External Flow
7–1C What is the difference between the upstream velocity
and the freestream velocity? For what types of flow are these
two velocities equal to each other?
7–2C What is the difference between streamlined and blunt
bodies? Is a tennis ball a streamlined or blunt body?
7–3C What is drag? What causes it? Why do we usually try
to minimize it?
7–4C What is lift? What causes it? Does wall shear contribute to the lift?
7–5C During flow over a given body, the drag force, the upstream velocity, and the fluid density are measured. Explain
how you would detennine the drag coefficient. What area
would you use in calculations?
7–6C Define frontal area of a body subjected to external
flow. When is it appropriate to use the frontal area in drag and
lift calculations?
7–7C What is the difference between skin friction drag and
pressure drag? Which is usually more significant for slender
bodies such as airfoils?
7–8C What is the effect of surface roughness on the friction
drag coefficient in laminar and turbulent flows?
7–9C What is the effect of streamlining on (a) friction drag
and (b) pressure drag? Does the total drag acting on a body
necessarily decrease as a result of streamlining? Explain.
*Problems designated by a “C” are concept questions, and
students are encouraged to answer them all. Problems designated
by an “E” are in English units, and the SI users can ignore them.
Problems with an EESCD icon
are solved using EES, and
complete solutions together with parametric studies are included
on the enclosed CD. Problems with a computerEES icon
are
comprehensive in nature, and are intended to be solved with a
computer, preferably using the EES software that accompanies
this text. 7–10C What is flow separation? What causes it? What is the
effect of flow separation on the drag coefficient? Flow Over Flat Plates
7–11C What does the friction coefficient represent in flow
over a flat plate? How is it related to the drag force acting on
the plate?
7–12C Consider laminar flow over a flat plate. Will the friction coefficient change with distance from the leading edge?
How about the heat transfer coefficient?
7–13C How are the average friction and heat transfer coefficients determined in flow over a flat plate?
7–14 Engine oil at 80°C flows over a 6mlong flat plate
whose temperature is 30°C with a velocity of 3 m/s. Determine
the total drag force and the rate of heat transfer over the entire
plate per unit width.
7–15 The local atmospheric pressure in Denver, Colorado
(elevation 1610 m), is 83.4 kPa. Air at this pressure and
at 30°C flows with a velocity of 6 m/s over a 2.5m 8m flat
plate whose temperature is 120°C. Determine the rate of
heat transfer from the plate if the air flows parallel to the
(a) 8mlong side and (b) the 2.5m side.
7–16 During a cold winter day, wind at 55 km/h is blowing
parallel to a 4mhigh and 10mlong wall of a house. If the air
outside is at 5°C and the surface temperature of the wall is Attic
space Air
5°C
55 km/h FIGURE P7–16 4m
10 m 12°C cen58933_ch07.qxd 9/4/2002 12:13 PM Page 409 409
CHAPTER 7 12°C, determine the rate of heat loss from that wall by convection. What would your answer be if the wind velocity was douAnswers: 9081 W, 16,200 W
bled?
7–17
Reconsider Problem 7–16. Using EES (or other)
software, investigate the effects of wind velocity
and outside air temperature on the rate of heat loss from the
wall by convection. Let the wind velocity vary from 10 km/h to
80 km/h and the outside air temperature from 0ºC to 10ºC. Plot
the rate of heat loss as a function of the wind velocity and of
the outside temperature, and discuss the results.
7–18E Air at 60°F flows over a 10ftlong flat plate at 7 ft/s.
Determine the local friction and heat transfer coefficients at intervals of 1 ft, and plot the results against the distance from the
leading edge.
7–19E
Reconsider Problem 7–18. Using EES (or
other) software, evaluate the local friction and
heat transfer coefficients along the plate at intervals of 0.1 ft,
and plot them against the distance from the leading edge.
7–20 Consider a hot automotive engine, which can be approximated as a 0.5mhigh, 0.40mwide, and 0.8mlong rectangular block. The bottom surface of the block is at a
temperature of 80°C and has an emissivity of 0.95. The ambient air is at 20°C, and the road surface is at 25°C. Determine
the rate of heat transfer from the bottom surface of the engine
block by convection and radiation as the car travels at a velocity of 80 km/h. Assume the flow to be turbulent over the entire
surface because of the constant agitation of the engine block.
7–21 The forming section of a plastics plant puts out a continuous sheet of plastic that is 1.2 m wide and 2 mm thick at a
rate of 15 m/min. The temperature of the plastic sheet is 90°C
when it is exposed to the surrounding air, and the sheet is subjected to air flow at 30°C at a velocity of 3 m/s on both sides
along its surfaces normal to the direction of motion of the
sheet. The width of the air cooling section is such that a fixed
point on the plastic sheet passes through that section in 2 s. Determine the rate of heat transfer from the plastic sheet to the air. Plastic
sheet 15 m/min FIGURE P7–21 Air
30°C 200 W/ m2
70 km/h FIGURE P7–22
7–23 Reconsider Problem 7–22. Using EES (or other) software, investigate the effects of the train velocity and the rate
of absorption of solar radiation on the equilibrium temperature of the top surface of the car. Let the train velocity vary
from 10 km/h to 120 km/h and the rate of solar absorption from
100 W/m2 to 500 W/m2. Plot the equilibrium temperature as
functions of train velocity and solar radiation absorption rate,
and discuss the results.
7–24 A 15cm
15cm circuit board dissipating 15 W of
power uniformly is cooled by air, which approaches the circuit
board at 20°C with a velocity of 5 m/s. Disregarding any heat
transfer from the back surface of the board, determine the surface temperature of the electronic components (a) at the leading edge and (b) at the end of the board. Assume the flow to be
turbulent since the electronic components are expected to act
as turbulators.
7–25 Consider laminar flow of a fluid over a flat plate maintained at a constant temperature. Now the freestream velocity
of the fluid is doubled. Determine the change in the drag force
on the plate and rate of heat transfer between the fluid and the
plate. Assume the flow to remain laminar. Air
30°C, 3 m/s 90°C 7–22 The top surface of the passenger car of a train moving
at a velocity of 70 km/h is 2.8 m wide and 8 m long. The top
surface is absorbing solar radiation at a rate of 200 W/m2, and
the temperature of the ambient air is 30°C. Assuming the roof
of the car to be perfectly insulated and the radiation heat exchange with the surroundings to be small relative to convection, determine the equilibrium temperature of the top surface
Answer: 35.1°C
of the car. 7–26E Consider a refrigeration truck traveling at 55 mph at
a location where the air temperature is 80°F. The refrigerated
compartment of the truck can be considered to be a 9ftwide,
8fthigh, and 20ftlong rectangular box. The refrigeration
system of the truck can provide 3 tons of refrigeration (i.e., it
can remove heat at a rate of 600 Btu/min). The outer surface
of the truck is coated with a lowemissivity material, and thus
radiation heat transfer is very small. Determine the average
temperature of the outer surface of the refrigeration compartment of the truck if the refrigeration system is observed to be cen58933_ch07.qxd 9/4/2002 12:13 PM Page 410 410
HEAT TRANSFER passages between the fins. The heat sink is to dissipate 20 W of
heat and the base temperature of the heat sink is not to exceed
60°C. Assuming the fins and the base plate to be nearly isothermal and the radiation heat transfer to be negligible, determine
the minimum freestream velocity the fan needs to supply to
avoid overheating. Air, 80°F
= 55 mph
20 ft 8 ft Refrigeration
truck Air
25°C FIGURE P7–26E
60°C operating at half the capacity. Assume the air flow over the entire outer surface to be turbulent and the heat transfer coefficient at the front and rear surfaces to be equal to that on side
surfaces.
7–27 Solar radiation is incident on the glass cover of a solar
collector at a rate of 700 W/m2. The glass transmits 88 percent
of the incident radiation and has an emissivity of 0.90. The entire hot water needs of a family in summer can be met by two
collectors 1.2 m high and 1 m wide. The two collectors are attached to each other on one side so that they appear like a single collector 1.2 m 2 m in size. The temperature of the glass
cover is measured to be 35°C on a day when the surrounding
air temperature is 25°C and the wind is blowing at 30 km/h.
The effective sky temperature for radiation exchange between
the glass cover and the open sky is 40°C. Water enters the
tubes attached to the absorber plate at a rate of 1 kg/min. Assuming the back surface of the absorber plate to be heavily insulated and the only heat loss to occur through the glass cover,
determine (a) the total rate of heat loss from the collector,
(b) the collector efficiency, which is the ratio of the amount of
heat transferred to the water to the solar energy incident on the
collector, and (c) the temperature rise of water as it flows
through the collector.
Tsky = – 40°C Air
25°C Solar
collector
35°C Fins 0.5 cm
10 cm 7–28 A transformer that is 10 cm long, 6.2 cm wide, and
5 cm high is to be cooled by attaching a 10 cm 6.2 cm wide
polished aluminum heat sink (emissivity 0.03) to its top surface. The heat sink has seven fins, which are 5 mm high, 2 mm
thick, and 10 cm long. A fan blows air at 25°C parallel to the 5 cm 6.2 cm FIGURE P7–28
7–29 Repeat Problem 7–28 assuming the heat sink to be
blackanodized and thus to have an effective emissivity of
0.90. Note that in radiation calculations the base area (10 cm
6.2 cm) is to be used, not the total surface area.
7–30 An array of power transistors, dissipating 6 W of power
each, are to be cooled by mounting them on a 25cm 25cm
square aluminum plate and blowing air at 35°C over the plate
with a fan at a velocity of 4 m/s. The average temperature of
the plate is not to exceed 65°C. Assuming the heat transfer
from the back side of the plate to be negligible and disregarding radiation, determine the number of transistors that can be
placed on this plate. Solar
radiation FIGURE P7–27 Transformer
20 W Aluminum
plate Power
transistor, 6 W 35°C
Air
4 m/s 25 cm
65°C
25 cm FIGURE P7–30 cen58933_ch07.qxd 9/4/2002 12:13 PM Page 411 411
CHAPTER 7 7–31 Repeat Problem 7–30 for a location at an elevation of
1610 m where the atmospheric pressure is 83.4 kPa.
Answer: 4 7–32 Air at 25°C and 1 atm is flowing over a long flat plate
with a velocity of 8 m/s. Determine the distance from the leading edge of the plate where the flow becomes turbulent, and the
thickness of the boundary layer at that location.
7–33 Repeat Problem 7–32 for water.
7–34 The weight of a thin flat plate 50 cm 50 cm in size is
balanced by a counterweight that has a mass of 2 kg, as shown
in the figure. Now a fan is turned on, and air at 1 atm and 25°C
flows downward over both surfaces of the plate with a freestream velocity of 10 m/s. Determine the mass of the counterweight that needs to be added in order to balance the plate in
this case. of air at 1 atm pressure and 30°C with a velocity of 6 m/s. The
surface temperature of the ball eventually drops to 250°C. Determine the average convection heat transfer coefficient during this
cooling process and estimate how long this process has taken.
7–41 Reconsider Problem 7–40. Using EES (or other)
software, investigate the effect of air velocity on
the average convection heat transfer coefficient and the cooling
time. Let the air velocity vary from 1 m/s to 10 m/s. Plot the
heat transfer coefficient and the cooling time as a function of
air velocity, and discuss the results. 7–42E A person extends his uncovered arms into the windy
air outside at 54°F and 20 mph in order to feel nature closely.
Initially, the skin temperature of the arm is 86°F. Treating the
arm as a 2ftlong and 3in.diameter cylinder, determine the
rate of heat loss from the arm. Air
25°C, 10 m /s Air
54°F, 20 mph 86°F
Plate 50 cm 50 cm FIGURE P7–34
Flow across Cylinders and Spheres
7–35C Consider laminar flow of air across a hot circular
cylinder. At what point on the cylinder will the heat transfer be
highest? What would your answer be if the flow were turbulent?
7–36C In flow over cylinders, why does the drag coefficient
suddenly drop when the flow becomes turbulent? Isn’t turbulence supposed to increase the drag coefficient instead of
decreasing it?
7–37C In flow over blunt bodies such as a cylinder, how
does the pressure drag differ from the friction drag?
7–38C Why is flow separation in flow over cylinders delayed in turbulent flow?
7–39 A long 8cmdiameter steam pipe whose external surface temperature is 90°C passes through some open area that is
not protected against the winds. Determine the rate of heat loss
from the pipe per unit of its length when the air is at 1 atm pressure and 7°C and the wind is blowing across the pipe at a velocity of 50 km/h.
7–40 A stainless steel ball (
8055 kg/m3, Cp 480 J/kg ·
°C) of diameter D 15 cm is removed from the oven at a uniform temperature of 350°C. The ball is then subjected to the flow FIGURE P7–42E
7–43E Reconsider Problem 7–42E. Using EES (or
other) software, investigate the effects of air
temperature and wind velocity on the rate of heat loss from the
arm. Let the air temperature vary from 20°F to 80°F and the
wind velocity from 10 mph to 40 mph. Plot the rate of heat loss
as a function of air temperature and of wind velocity, and discuss the results.
7–44 An average person generates heat at a rate of 84 W
while resting. Assuming onequarter of this heat is lost from
the head and disregarding radiation, determine the average surface temperature of the head when it is not covered and is subjected to winds at 10°C and 35 km/h. The head can be
approximated as a 30cmdiameter sphere. Answer: 12.7°C
7–45 Consider the flow of a fluid across a cylinder maintained at a constant temperature. Now the freestream velocity
of the fluid is doubled. Determine the change in the drag force
on the cylinder and the rate of heat transfer between the fluid
and the cylinder. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 412 412
HEAT TRANSFER 7–46 A 6mmdiameter electrical transmission line carries an
electric current of 50 A and has a resistance of 0.002 ohm per
meter length. Determine the surface temperature of the wire
during a windy day when the air temperature is 10°C and the
wind is blowing across the transmission line at 40 km/h. the person. The average human body can be treated as a 1ftdiameter cylinder with an exposed surface area of 18 ft2. Disregard any heat transfer by radiation. What would your answer
Answers: 95.1°F, 91.6°F
be if the air velocity were doubled? Wind, 40 km/h
10°C 85°F
6 ft /s
300 Btu/h Transmission
lines FIGURE P7–46
7–47 Reconsider Problem 7–46. Using EES (or other)
software, investigate the effect of the wind velocity on the surface temperature of the wire. Let the wind velocity vary from 10 km/h to 80 km/h. Plot the surface temperature
as a function of wind velocity, and discuss the results.
7–48 A heating system is to be designed to keep the wings of
an aircraft cruising at a velocity of 900 km/h above freezing
temperatures during flight at 12,200m altitude where the standard atmospheric conditions are 55.4°C and 18.8 kPa. Approximating the wing as a cylinder of elliptical cross section
whose minor axis is 30 cm and disregarding radiation, determine
the average convection heat transfer coefficient on the wing surface and the average rate of heat transfer per unit surface area.
7–49 A long aluminum wire of diameter 3 mm is
extruded at a temperature of 370°C. The wire
is subjected to cross air flow at 30°C at a velocity of 6 m/s.
Determine the rate of heat transfer from the wire to the air per
meter length when it is first exposed to the air. 370°C 30°C
6 m/s FIGURE P7–50E
751 An incandescent lightbulb is an inexpensive but highly
inefficient device that converts electrical energy into light. It
converts about 10 percent of the electrical energy it consumes
into light while converting the remaining 90 percent into heat.
(A fluorescent lightbulb will give the same amount of light
while consuming only onefourth of the electrical energy, and
it will last 10 times longer than an incandescent lightbulb.) The
glass bulb of the lamp heats up very quickly as a result of absorbing all that heat and dissipating it to the surroundings by
convection and radiation.
Consider a 10cmdiameter 100W lightbulb cooled by a fan
that blows air at 25°C to the bulb at a velocity of 2 m/s. The
surrounding surfaces are also at 25°C, and the emissivity of the
glass is 0.9. Assuming 10 percent of the energy passes through
the glass bulb as light with negligible absorption and the rest of
the energy is absorbed and dissipated by the bulb itself, determine the equilibrium temperature of the glass bulb. 3 mm Aluminum
wire FIGURE P7–49
7–50E Consider a person who is trying to keep cool on a hot
summer day by turning a fan on and exposing his entire body
to air flow. The air temperature is 85°F and the fan is blowing
air at a velocity of 6 ft/s. If the person is doing light work and
generating sensible heat at a rate of 300 Btu/h, determine the
average temperature of the outer surface (skin or clothing) of Air
25°C
2 m/s 100 W
ε = 0.9
Light, 10 W FIGURE P7–51
7–52 During a plant visit, it was noticed that a 12mlong
section of a 10cmdiameter steam pipe is completely exposed
to the ambient air. The temperature measurements indicate that cen58933_ch07.qxd 9/4/2002 12:13 PM Page 413 413
CHAPTER 7 the average temperature of the outer surface of the steam pipe
is 75°C when the ambient temperature is 5°C. There are also
light winds in the area at 10 km/h. The emissivity of the outer
surface of the pipe is 0.8, and the average temperature of the
surfaces surrounding the pipe, including the sky, is estimated to
be 0°C. Determine the amount of heat lost from the steam during a 10hlong work day.
Steam is supplied by a gasfired steam generator that has an
efficiency of 80 percent, and the plant pays $0.54/therm of natural gas (1 therm 105,500 kJ). If the pipe is insulated and 90
percent of the heat loss is saved, determine the amount of
money this facility will save a year as a result of insulating the
steam pipes. Assume the plant operates every day of the year
for 10 h. State your assumptions. at 30°C flowing over the duct with a velocity of 200 m/min.
If the surface temperature of the duct is not to exceed 65°C, determine the total power rating of the electronic devices that can
Answer: 640 W
be mounted into the duct.
Electronic
components
inside
30°C
200 m/min
Air 1.5 m Tsurr = 0°C
ε = 0.8
75°C
10 cm 65°C 20 cm FIGURE P7–55 Steam pipe 7–56 Repeat Problem 7–55 for a location at 4000m altitude
where the atmospheric pressure is 61.66 kPa.
5°C
10 km/h FIGURE P7–52
7–53 Reconsider Problem 7–52. There seems to be some uncertainty about the average temperature of the surfaces surrounding the pipe used in radiation calculations, and you are
asked to determine if it makes any significant difference in
overall heat transfer. Repeat the calculations for average surrounding and surface temperatures of 20°C and 25°C, respectively, and determine the change in the values obtained.
7–54E A 12ftlong, 1.5kW electrical resistance wire is
made of 0.1in.diameter stainless steel (k 8.7 Btu/h · ft · °F).
The resistance wire operates in an environment at 85°F. Determine the surface temperature of the wire if it is cooled by a fan
blowing air at a velocity of 20 ft/s.
85°F
20 ft/s 1.5 kW
resistance
heater FIGURE P7–54E
7–55 The components of an electronic system are located in
a 1.5mlong horizontal duct whose cross section is 20 cm
20 cm. The components in the duct are not allowed to come
into direct contact with cooling air, and thus are cooled by air 7–57 A 0.4W cylindrical electronic component with diameter 0.3 cm and length 1.8 cm and mounted on a circuit board is
cooled by air flowing across it at a velocity of 150 m/min. If
the air temperature is 40°C, determine the surface temperature
of the component.
7–58 Consider a 50cmdiameter and 95cmlong hot water
tank. The tank is placed on the roof of a house. The water inside the tank is heated to 80ºC by a flatplate solar collector
during the day. The tank is then exposed to windy air at 18ºC
with an average velocity of 40 km/h during the night. Estimate
the temperature of the tank after a 45mm period. Assume the
tank surface to be at the same temperature as the water inside,
and the heat transfer coefficient on the top and bottom surfaces
to be the same as that on the side surface.
7–59 Reconsider Problem 7–58. Using EES (or other)
software, plot the temperature of the tank as a
function of the cooling time as the time varies from 30 mm to
5 h, and discuss the results. 7–60 A 1.8mdiameter spherical tank of negligible thickness
contains iced water at 0ºC. Air at 25ºC flows over the tank with
a velocity of 7 m/s. Determine the rate of heat transfer to the
tank and the rate at which ice melts. The heat of fusion of water at 0ºC is 333.7 kJ/kg.
761 A 10cmdiameter, 30cmhigh cylindrical bottle contains cold water at 3ºC. The bottle is placed in windy air at
27ºC. The water temperature is measured to be 11ºC after 45
minutes of cooling. Disregarding radiation effects and heat
transfer from the top and bottom surfaces, estimate the average
wind velocity. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 414 414
HEAT TRANSFER Flow across Tube Banks Water
15°C
0.8 m/s 7–62C In flow across tube banks, why is the Reynolds number based on the maximum velocity instead of the uniform approach velocity? SL = 4 cm D = 1 cm
Ts = 90°C
ST = 3 cm 7–63C In flow across tube banks, how does the heat transfer
coefficient vary with the row number in the flow direction?
How does it vary with in the transverse direction for a given
row number?
7–64 Combustion air in a manufacturing facility is to be preheated before entering a furnace by hot water at 90ºC flowing
through the tubes of a tube bank located in a duct. Air enters
the duct at 15ºC and 1 atm with a mean velocity of 3.8 m/s, and
flows over the tubes in normal direction. The outer diameter
of the tubes is 2.1 cm, and the tubes are arranged inline with
longitudinal and transverse pitches of SL ST 5 cm. There
are eight rows in the flow direction with eight tubes in each
row. Determine the rate of heat transfer per unit length of the
tubes, and the pressure drop across the tube bank.
7–65 Repeat Problem 7–64 for staggered arrangement with
SL ST 5 cm.
7–66 Air is to be heated by passing it over a bank of
3mlong tubes inside which steam is condensing at 100ºC. Air
approaches the tube bank in the normal direction at 20ºC and
1 atm with a mean velocity of 5.2 m/s. The outer diameter of
the tubes is 1.6 cm, and the tubes are arranged staggered with
longitudinal and transverse pitches of SL ST 4 cm. There
are 20 rows in the flow direction with 10 tubes in each row.
Determine (a) the rate of heat transfer, (b) and pressure drop
across the tube bank, and (c) the rate of condensation of steam
inside the tubes. FIGURE P7–69
7–70 Air is to be cooled in the evaporator section of a refrigerator by passing it over a bank of 0.8cmouterdiameter
and 0.4mlong tubes inside which the refrigerant is evaporating at 20ºC. Air approaches the tube bank in the normal direction at 0ºC and 1 atm with a mean velocity of 4 m/s. The
tubes are arranged inline with longitudinal and transverse
pitches of SL ST 1.5 cm. There are 30 rows in the flow direction with 15 tubes in each row. Determine (a) the refrigeration capacity of this system and (b) and pressure drop across
the tube bank.
0°C
1 atm
4 m/s
Air 0.4 m ST = 1.5 cm
0.8 cm SL = 1.5 cm 7–67 Repeat Problem 7–66 for inline arrangement with
SL ST 5 cm.
7–68 Exhaust gases at 1 atm and 300ºC are used to preheat
water in an industrial facility by passing them over a bank of
tubes through which water is flowing at a rate of 6 kg/s. The
mean tube wall temperature is 80ºC. Exhaust gases approach
the tube bank in normal direction at 4.5 m/s. The outer diameter of the tubes is 2.1 cm, and the tubes are arranged inline
with longitudinal and transverse pitches of SL ST 8 cm.
There are 16 rows in the flow direction with eight tubes in each
row. Using the properties of air for exhaust gases, determine
(a) the rate of heat transfer per unit length of tubes, (b) and
pressure drop across the tube bank, and (c) the temperature rise
of water flowing through the tubes per unit length of tubes.
7–69 Water at 15ºC is to be heated to 65ºC by passing it over
a bundle of 4mlong 1cmdiameter resistance heater rods
maintained at 90ºC. Water approaches the heater rod bundle in
normal direction at a mean velocity of 0.8 m/s. The rods arc
arranged inline with longitudinal and transverse pitches of
SL 4 cm and ST 3 cm. Determine the number of tube rows
NL in the flow direction needed to achieve the indicated temperature rise. Refrigerant, 20°C FIGURE P7–70
7–71 Repeat Problem 7–70 by solving it for staggered
arrangement with SL ST 1.5 cm, and compare the performance of the evaporator for the inline and staggered arrangements.
7–72 A tube bank consists of 300 tubes at a distance of 6 cm
between the centerlines of any two adjacent tubes. Air approaches the tube bank in the normal direction at 40ºC and
1 atm with a mean velocity of 7 m/s. There are 20 rows in the
flow direction with 15 tubes in each row with an average surface temperature of 140ºC. For an outer tube diameter of 2 cm,
determine the average heat transfer coefficient. Special Topic: Thermal Insulation
7–73C What is thermal insulation? How does a thermal insulator differ in purpose from an electrical insulator and from a
sound insulator?
7–74C Does insulating cold surfaces save energy? Explain. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 415 415
CHAPTER 7 7–75C What is the Rvalue of insulation? How is it determined? Will doubling the thickness of flat insulation double its
Rvalue?
7–76C How does the Rvalue of an insulation differ from its
thermal resistance?
7–77C Why is the thermal conductivity of superinsulation
orders of magnitude lower than the thermal conductivities of
ordinary insulations?
7–78C Someone suggests that one function of hair is to insulate the head. Do you agree with this suggestion?
7–79C Name five different reasons for using insulation in industrial facilities.
7–80C What is optimum thickness of insulation? How is it
determined?
7–81 What is the thickness of flat R8 (in SI units) insulation
whose thermal conductivity is 0.04 W/m · °C?
7–82E What is the thickness of flat R20 (in English units)
insulation whose thermal conductivity is 0.02 Btu/h · ft · °F?
7–83 Hot water at 110°C flows in a cast iron pipe (k 52
W/m · °C) whose inner radius is 2.0 cm and thickness is 0.3
cm. The pipe is to be covered with adequate insulation so that
the temperature of the outer surface of the insulation does not
exceed 30°C when the ambient temperature is 22°C. Taking
the heat transfer coefficients inside and outside the pipe to be
hi 80 W/m2 · °C and ho 22 W/m2 · °C, respectively, determine the thickness of fiber glass insulation (k 0.038 W/m ·
°C) that needs to be installed on the pipe.
Answer: 1.32 cm 7–84 Reconsider Problem 7–83. Using EES (or other)
software, plot the thickness of the insulation as a
function of the maximum temperature of the outer surface of
insulation in the range of 24ºC to 48ºC. Discuss the results.
7–85 Consider a furnace whose average outer surface
temperature is measured to be 90°C when the average surrounding air temperature is 27°C. The furnace is 6 m
long and 3 m in diameter. The plant operates 80 h per week for
52 weeks per year. You are to insulate the furnace using fiberglass insulation (kins 0.038 W/m · °C) whose cost is $10/m2
per cm of thickness for materials, plus $30/m2 for labor regardless of thickness. The combined heat transfer coefficient on the
outer surface is estimated to be ho 30 W/m2 · °C. The furnace
uses natural gas whose unit cost is $0.50/therm input (1 therm
105,500 kJ), and the efficiency of the furnace is 78 percent. The
management is willing to authorize the installation of the thickest insulation (in whole cm) that will pay for itself (materials and
labor) in one year. That is, the total cost of insulation should be
roughly equal to the drop in the fuel cost of the furnace for one
year. Determine the thickness of insulation to be used and the
money saved per year. Assume the surface temperature of the
furnace and the heat transfer coefficient are to remain constant.
Answer: 14 cm 7–85 Repeat Problem 7–85 for an outer surface temperature
of 75°C for the furnace.
7–87E Steam at 400°F is flowing through a steel pipe (k 8.7
Btu/h · ft · °F) whose inner and outer diameters are 3.5 in. and
4.0 in., respectively, in an environment at 60°F. The pipe is insulated with 1in.thick fiberglass insulation (k 0.020 Btu/h · ft ·
°F), and the heat transfer coefficients on the inside and the outside of the pipe are 30 Btu/h · ft2 · °F and 5 Btu/h · ft2 · °F, respectively. It is proposed to add another 1in.thick layer of
fiberglass insulation on top of the existing one to reduce the heat
losses further and to save energy and money. The total cost of
new insulation is $7 per ft length of the pipe, and the net fuel cost
of energy in the steam is $0.01 per 1000 Btu (therefore, each
1000 Btu reduction in the heat loss will save the plant $0.01).
The policy of the plant is to implement energy conservation
measures that pay for themselves within two years. Assuming
continuous operation (8760 h/year), determine if the proposed
additional insulation is justified.
7–88 The plumbing system of a plant involves a section of a
plastic pipe (k 0.16 W/m · °C) of inner diameter 6 cm and
outer diameter 6.6 cm exposed to the ambient air. You are to
insulate the pipe with adequate weatherjacketed fiberglass
insulation (k 0.035 W/m · °C) to prevent freezing of water in
the pipe. The plant is closed for the weekends for a period of
60 h, and the water in the pipe remains still during that period.
The ambient temperature in the area gets as low as 10°C in
winter, and the high winds can cause heat transfer coefficients
as high as 30 W/m2 · °C. Also, the water temperature in the
pipe can be as cold as 15°C, and water starts freezing when its
temperature drops to 0°C. Disregarding the convection resistance inside the pipe, determine the thickness of insulation that
will protect the water from freezing under worst conditions.
7–89 Repeat Problem 7–88 assuming 20 percent of the water
in the pipe is allowed to freeze without jeopardizing safety.
Answer: 27.9 cm Review Problems
7–90 Consider a house that is maintained at 22°C at all times.
The walls of the house have R3.38 insulation in SI units (i.e.,
an L/k value or a thermal resistance of 3.38 m2 · °C/W). During
a cold winter night, the outside air temperature is 4°C and wind
at 50 km/h is blowing parallel to a 3mhigh and 8mlong wall
of the house. If the heat transfer coefficient on the interior
surface of the wall is 8 W/m2 · °C, determine the rate of heat
loss from that wall of the house. Draw the thermal resistance
network and disregard radiation heat transfer.
Answer: 122 W 7–91 An automotive engine can be approximated as a 0.4mhigh, 0.60mwide, and 0.7mlong rectangular block. The
bottom surface of the block is at a temperature of 75°C and has
an emissivity of 0.92. The ambient air is at 5°C, and the road
surface is at 10°C. Determine the rate of heat transfer from the
bottom surface of the engine block by convection and radiation cen58933_ch07.qxd 9/4/2002 12:13 PM Page 416 416
HEAT TRANSFER as the car travels at a velocity of 60 km/h. Assume the flow to
be turbulent over the entire surface because of the constant
agitation of the engine block. How will the heat transfer be
affected when a 2mmthick gunk (k
3 W/m · °C) has
formed at the bottom surface as a result of the dirt and oil
collected at that surface over time? Assume the metal
temperature under the gunk still to be 75°C.
Engine
block
Air
60 km/h
5°C 75°C
Gunk
ε = 0.92 2 mm Road
10°C FIGURE P7–91
7–92E The passenger compartment of a minivan traveling at
60 mph can be modeled as a 3.2fthigh, 6ftwide, and 11ftlong rectangular box whose walls have an insulating value
of R3 (i.e., a wall thickness–to–thermal conductivity ratio of
3 h · ft2 · °F/Btu). The interior of a minivan is maintained at an
average temperature of 70°F during a trip at night while the
outside air temperature is 90°F. The average heat transfer
coefficient on the interior surfaces of the van is 1.2 Btu/h · ft2 ·
°F. The air flow over the exterior surfaces can be assumed to be
turbulent because of the intense vibrations involved, and the
heat transfer coefficient on the front and back surfaces can be
taken to be equal to that on the top surface. Disregarding any
heat gain or loss by radiation, determine the rate of heat
transfer from the ambient air to the van. 7–94 Consider a person who is trying to keep cool on a hot
summer day by turning a fan on and exposing his body to air
flow. The air temperature is 32°C, and the fan is blowing air at
a velocity of 5 m/s. The surrounding surfaces are at 40°C, and
the emissivity of the person can be taken to be 0.9. If the
person is doing light work and generating sensible heat at a rate
of 90 W, determine the average temperature of the outer
surface (skin or clothing) of the person. The average human
body can be treated as a 30cmdiameter cylinder with an
Answer: 36.2°C
exposed surface area of 1.7 m2.
7–95 Four power transistors, each dissipating 12 W, are
mounted on a thin vertical aluminum plate (k 237 W/m · °C)
22 cm 22 cm in size. The heat generated by the transistors is
to be dissipated by both surfaces of the plate to the surrounding
air at 20°C, which is blown over the plate by a fan at a velocity
of 250 m/min. The entire plate can be assumed to be nearly
isothermal, and the exposed surface area of the transistor can
be taken to be equal to its base area. Determine the temperature
of the aluminum plate.
7–96 A 3minternaldiameter spherical tank made of 1cmthick stainless steel (k
15 W/m · °C) is used to store iced
water at 0°C. The tank is located outdoors at 30°C and is
subjected to winds at 25 km/h. Assuming the entire steel tank
to be at 0°C and thus its thermal resistance to be negligible,
determine (a) the rate of heat transfer to the iced water in the
tank and (b) the amount of ice at 0°C that melts during a 24h
period. The heat of fusion of water at atmospheric pressure is
hif 333.7 kJ/kg. Disregard any heat transfer by radiation.
Troom = 30°C
25 km/h
Iced
Water
Di = 3 m Air
60 mph
90°F 1 cm Tin = 0°C FIGURE P7–96
FIGURE P7–92E
7–93 Consider a house that is maintained at a constant
temperature of 22°C. One of the walls of the house has three
singlepane glass windows that are 1.5 m high and 1.2 m long.
The glass (k 0.78 W/m · °C) is 0.5 cm thick, and the heat
transfer coefficient on the inner surface of the glass is
8 W/m2 · C. Now winds at 60 km/h start to blow parallel to the
surface of this wall. If the air temperature outside is 2°C,
determine the rate of heat loss through the windows of this
wall. Assume radiation heat transfer to be negligible. 7–97 Repeat Problem 7–96, assuming the inner surface of
the tank to be at 0°C but by taking the thermal resistance of the
tank and heat transfer by radiation into consideration. Assume
the average surrounding surface temperature for radiation
exchange to be 15°C and the outer surface of the tank to have
an emissivity of 0.9.
Answers: (a) 9630 W, (b) 2493 kg
7–98E A transistor with a height of 0.25 in. and a diameter of
0.22 in. is mounted on a circuit board. The transistor is cooled
by air flowing over it at a velocity of 500 ft/min. If the air
temperature is 120°F and the transistor case temperature is not
to exceed 180°F, determine the amount of power this transistor
can dissipate safely. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 417 417
CHAPTER 7
Air, 500 ft/min
120°F Power
transistor
Ts ≤ 180°F 0.22 in. 0.25 in. FIGURE P7–98E
7–99 The roof of a house consists of a 15cmthick concrete
slab (k
2 W/m · °C) 15 m wide and 20 m long. The
convection heat transfer coefficient on the inner surface of the
roof is 5 W/m2 · °C. On a clear winter night, the ambient air is
reported to be at 10°C, while the night sky temperature is 100
K. The house and the interior surfaces of the wall are
maintained at a constant temperature of 20°C. The emissivity
of both surfaces of the concrete roof is 0.9. Considering both
radiation and convection heat transfer, determine the rate of
heat transfer through the roof when wind at 60 km/h is blowing
over the roof.
If the house is heated by a furnace burning natural gas with
an efficiency of 85 percent, and the price of natural gas is
$0.60/therm (1 therm
105,500 kJ of energy content),
determine the money lost through the roof that night during a
Answers: 28 kW, $9.44
14h period. 7–101 The boiling temperature of nitrogen at atmospheric
pressure at sea level (1 atm pressure) is 196°C. Therefore,
nitrogen is commonly used in lowtemperature scientific
studies, since the temperature of liquid nitrogen in a tank open
to the atmosphere will remain constant at 196°C until it is
depleted. Any heat transfer to the tank will result in the
evaporation of some liquid nitrogen, which has a heat of
vaporization of 198 kJ/kg and a density of 810 kg/m3 at 1 atm.
Consider a 4mdiameter spherical tank that is initially filled
with liquid nitrogen at 1 atm and 196°C. The tank is exposed
to 20°C ambient air and 40 km/h winds. The temperature of the
thinshelled spherical tank is observed to be almost the same as
the temperature of the nitrogen inside. Disregarding any radiation heat exchange, determine the rate of evaporation of the
liquid nitrogen in the tank as a result of heat transfer from the
ambient air if the tank is (a) not insulated, (b) insulated with
5cmthick fiberglass insulation (k
0.035 W/m · °C), and
(c) insulated with 2cmthick superinsulation that has an effective thermal conductivity of 0.00005 W/m · °C. N2 vapor
Tair = 20°C
40 km/h
1 atm
Liquid N2
–196°C Tsky = 100 K
Tair = 10°C
Concrete
roof
60 km/h
20 m
ε = 0.9 15 cm 15 m ·
Q Insulation FIGURE P7–101 Tin = 20°C FIGURE P7–99
7–100 Steam at 250°C flows in a stainless steel pipe (k 15
W/m · °C) whose inner and outer diameters are 4 cm and 4.6
cm, respectively. The pipe is covered with 3.5cmthick glass
wool insulation (k 0.038 W/m · °C) whose outer surface has
an emissivity of 0.3. Heat is lost to the surrounding air and
surfaces at 3°C by convection and radiation. Taking the heat
transfer coefficient inside the pipe to be 80 W/m2 · °C,
determine the rate of heat loss from the steam per unit length of
the pipe when air is flowing across the pipe at 4 m/s. 7–102 Repeat Problem 7–101 for liquid oxygen, which has a
boiling temperature of 183°C, a heat of vaporization of 213
kJ/kg, and a density of 1140 kg/m3 at 1 atm pressure.
7–103 A 0.3cmthick, 12cmhigh, and 18cmlong circuit
board houses 80 closely spaced logic chips on one side, each
dissipating 0.06 W. The board is impregnated with copper
fillings and has an effective thermal conductivity of 16
W/m · °C. All the heat generated in the chips is conducted
across the circuit board and is dissipated from the back side of
the board to the ambient air at 30°C, which is forced to flow
over the surface by a fan at a freestream velocity of 400
m/min. Determine the temperatures on the two sides of the
circuit board. cen58933_ch07.qxd 9/4/2002 12:13 PM Page 418 418
HEAT TRANSFER 7–104E It is well known that cold air feels much
colder in windy weather than what the thermometer reading indicates because of the “chilling effect” of
the wind. This effect is due to the increase in the convection
heat transfer coefficient with increasing air velocities. The
equivalent windchill temperature in °F is given by (1993
ASHRAE Handbook of Fundamentals, Atlanta, GA, p. 8.15) Design and Essay Problems Tequiv ) 7–106 On average, superinsulated homes use just 15 percent
of the fuel required to heat the same size conventional home
built before the energy crisis in the 1970s. Write an essay on
superinsulated homes, and identify the features that make them
so energy efficient as well as the problems associated with
them. Do you think superinsulated homes will be economically
attractive in your area? where is the wind velocity in mph and Tambient is the ambient
air temperature in °F in calm air, which is taken to be air with
light winds at speeds up to 4 mph. The constant 91.4°F in the
above equation is the mean skin temperature of a resting person
in a comfortable environment. Windy air at a temperature Tambient
and velocity
will feel as cold as calm air at a temperature
Tequiv. The equation above is valid for winds up to 43 mph.
Winds at higher velocities produce little additional chilling
effect. Determine the equivalent wind chill temperature of an
environment at 10°F at wind speeds of 10, 20, 30, and 40 mph.
Exposed flesh can freeze within one minute at a temperature
below 25°F in calm weather. Does a person need to be
concerned about this possibility in any of the cases above? 7–107 Conduct this experiment to determine the heat loss coefficient of your house or apartment in W/ºC or But/h ºF. First
make sure that the conditions in the house are steady and the
house is at the set temperature of the thermostat. Use an outdoor thermometer to monitor outdoor temperature. One
evening, using a watch or timer, determine how long the heater
was on during a 3h period and the average outdoor temperature during that period. Then using the heat output rating of
your heater, determine the amount of heat supplied. Also, estimate the amount of heat generation in the house during that period by noting the number of people, the total wattage of lights
that were on, and the heat generated by the appliances and
equipment. Using that information, calculate the average rate
of heat loss from the house and the heat loss coefficient. 91.4 (91.4 Tambient)(0.475 Winds
40°F
35 mph 0.0203 0.304 It feels
like 11°F FIGURE P7–104E 7–l05E Reconsider Problem 7–104E. Using EES (or
other) software, plot the equivalent wind chill
temperatures in ºF as a function of wind velocity in the range
of 4 mph to 100 mph for ambient temperatures of 20ºF, 40ºF
and 60ºF. Discuss the results. 7–108 The decision of whether to invest in an energysaving
measure is made on the basis of the length of time for it to pay
for itself in projected energy (and thus cost) savings. The easiest way to reach a decision is to calculate the simple payback
period by simply dividing the installed cost of the measure by
the annual cost savings and comparing it to the lifetime of the
installation. This approach is adequate for short payback periods (less than 5 years) in stable economies with low interest
rates (under 10 percent) since the error involved is no larger
than the uncertainties. However, if the payback period is long,
it may be necessary to consider the interest rate if the money is
to be borrowed, or the rate of return if the money is invested
elsewhere instead of the energy conservation measure. For example, a simple payback period of five years corresponds to
5.0, 6.12, 6.64, 7.27, 8.09, 9.919, 10.84, and 13.91 for an interest rate (or return on investment) of 0, 6, 8, 10, 12, 14, 16, and
18 percent, respectively. Finding out the proper relations from
engineering economics books, determine the payback periods
for the interest rates given above corresponding to simple payback periods of 1 through 10 years.
7–109 Obtain information on frostbite and the conditions
under which it occurs. Using the relation in Problem 7–104E,
prepare a table that shows how long people can stay in cold and
windy weather for specified temperatures and wind speeds
before the exposed flesh is in danger of experiencing frostbite.
7–110 Write an article on forced convection cooling with air,
helium, water, and a dielectric liquid. Discuss the advantages
and disadvantages of each fluid in heat transfer. Explain the
circumstances under which a certain fluid will be most suitable
for the cooling job. ...
View
Full
Document
 Spring '10
 Ghaz

Click to edit the document details