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Unformatted text preview: cen58933_ch08.qxd 9/4/2002 11:29 AM Page 419 CHAPTER INTERNAL FORCED
CONVECTION
iquid or gas flow through pipes or ducts is commonly used in heating and
cooling applications. The fluid in such applications is forced to flow by a
fan or pump through a tube that is sufficiently long to accomplish the
desired heat transfer. In this chapter we will pay particular attention to the determination of the friction factor and convection coefficient since they are directly related to the pressure drop and heat transfer rate, respectively. These
quantities are then used to determine the pumping power requirement and the
required tube length.
There is a fundamental difference between external and internal flows. In
external flow, considered in Chapter 7, the fluid has a free surface, and thus
the boundary layer over the surface is free to grow indefinitely. In internal
flow, however, the fluid is completely confined by the inner surfaces of the
tube, and thus there is a limit on how much the boundary layer can grow.
We start this chapter with a general physical description of internal flow,
and the mean velocity and mean temperature. We continue with the discussion
of the hydrodynamic and thermal entry lengths, developing flow, and fully
developed flow. We then obtain the velocity and temperature profiles for
fully developed laminar flow, and develop relations for the friction factor and
Nusselt number. Finally we present empirical relations for developing
and fully developed flows, and demonstrate their use. L 8
CONTENTS
8–1
Introduction 420
8–2
Mean Velocity and
Mean Temperature 420
8–3
The Entrance Region 423
8–4
General Thermal Analysis 426
8–5
Laminar Flow in Tubes 431
8–6
Turbulent Flow in Tubes 441 419 cen58933_ch08.qxd 9/4/2002 11:29 AM Page 420 420
HEAT TRANSFER 8–1
Circular pipe Water
50 atm Rectangular
duct Air
1.2 atm FIGURE 8–1
Circular pipes can withstand large
pressure differences between the
inside and the outside without
undergoing any distortion, but
the noncircular pipes cannot. I INTRODUCTION You have probably noticed that most fluids, especially liquids, are transported
in circular pipes. This is because pipes with a circular cross section can withstand large pressure differences between the inside and the outside without
undergoing any distortion. Noncircular pipes are usually used in applications
such as the heating and cooling systems of buildings where the pressure difference is relatively small and the manufacturing and installation costs are
lower (Fig. 8–1). For a fixed surface area, the circular tube gives the most heat
transfer for the least pressure drop, which explains the overwhelming popularity of circular tubes in heat transfer equipment.
The terms pipe, duct, tube, and conduit are usually used interchangeably for
flow sections. In general, flow sections of circular cross section are referred to
as pipes (especially when the fluid is a liquid), and the flow sections of noncircular cross section as ducts (especially when the fluid is a gas). Small diameter pipes are usually referred to as tubes. Given this uncertainty, we will
use more descriptive phrases (such as a circular pipe or a rectangular duct)
whenever necessary to avoid any misunderstandings.
Although the theory of fluid flow is reasonably well understood, theoretical
solutions are obtained only for a few simple cases such as the fully developed
laminar flow in a circular pipe. Therefore, we must rely on the experimental
results and the empirical relations obtained for most fluid flow problems
rather than closed form analytical solutions. Noting that the experimental results are obtained under carefully controlled laboratory conditions, and that no
two systems are exactly alike, we must not be so naive as to view the results
obtained as “exact.” An error of 10 percent (or more) in friction or convection
coefficient calculated using the relations in this chapter is the “norm” rather
than the “exception.”
Perhaps we should mention that the friction between the fluid layers in a
tube may cause a slight rise in fluid temperature as a result of mechanical energy being converted to thermal energy. But this frictional heating is too small
to warrant any consideration in calculations, and thus is disregarded. For example, in the absence of any heat transfer, no noticeable difference will be detected between the inlet and exit temperatures of a fluid flowing in a tube. The
primary consequence of friction in fluid flow is pressure drop. Thus, it is reasonable to assume that any temperature change in the fluid is due to heat
transfer. But frictional heating must be considered for flows that involve
highly viscous fluids with large velocity gradients.
In most practical applications, the flow of a fluid through a pipe or duct can
be approximated to be onedimensional, and thus the properties can be assumed to vary in one direction only (the direction of flow). As a result, all
properties are uniform at any cross section normal to the flow direction, and
the properties are assumed to have bulk average values over the cross section.
But the values of the properties at a cross section may change with time unless
the flow is steady. 8–2 I MEAN VELOCITY AND MEAN TEMPERATURE In external flow, the freestream velocity served as a convenient reference
velocity for use in the evaluation of the Reynolds number and the friction cen58933_ch08.qxd 9/4/2002 11:29 AM Page 421 421
CHAPTER 8 coefficient. In internal flow, there is no free stream and thus we need an alternative. The fluid velocity in a tube changes from zero at the surface because
of the noslip condition, to a maximum at the tube center. Therefore, it is convenient to work with an average or mean velocity m, which remains constant for incompressible flow when the cross sectional area of the tube is
constant.
The mean velocity in actual heating and cooling applications may change
somewhat because of the changes in density with temperature. But, in practice, we evaluate the fluid properties at some average temperature and treat
them as constants. The convenience in working with constant properties usually more than justifies the slight loss in accuracy.
The value of the mean velocity m in a tube is determined from the requirement that the conservation of mass principle be satisfied (Fig. 8–2). That is,
·
m m Ac (r, x)dAc (81) =0 max
max (a) Actual m (b) Idealized FIGURE 8–2
Actual and idealized velocity profiles
for flow in a tube (the mass flow rate
of the fluid is the same for both cases). Ac ·
where m is the mass flow rate, is the density, Ac is the cross sectional area,
and (r, x) is the velocity profile. Then the mean velocity for incompressible
flow in a circular tube of radius R can be expressed as
Ac R (r, x)dAc m (r, x)2 rdr
0 Ac 2 R 2
R2 R (r, x)rdr (82) 0 Therefore, when we know the mass flow rate or the velocity profile, the mean
velocity can be determined easily.
When a fluid is heated or cooled as it flows through a tube, the temperature
of the fluid at any cross section changes from Ts at the surface of the wall to
some maximum (or minimum in the case of heating) at the tube center. In
fluid flow it is convenient to work with an average or mean temperature Tm
that remains uniform at a cross section. Unlike the mean velocity, the mean
temperature Tm will change in the flow direction whenever the fluid is heated
or cooled.
The value of the mean temperature Tm is determined from the requirement
that the conservation of energy principle be satisfied. That is, the energy transported by the fluid through a cross section in actual flow must be equal to the
energy that would be transported through the same cross section if the fluid
were at a constant temperature Tm. This can be expressed mathematically as
(Fig. 8–3)
·
E fluid ·
m CpTm ·
m ·
Cp T m Ac CpT dAc Tm ·
m ·
m Cp R
0 CpT(
m( 2 rdr)
2 R )Cp 2
2
mR R T(r, x)
0 Tmin (a) Actual (83) ·
where Cp is the specific heat of the fluid. Note that the product mCpTm at any
cross section along the tube represents the energy flow with the fluid at that
cross section. Then the mean temperature of a fluid with constant density and
specific heat flowing in a circular pipe of radius R can be expressed as
·
CpT m Ts (r, x) rdr (84) Tm
(b) Idealized FIGURE 8–3
Actual and idealized temperature
profiles for flow in a tube (the rate at
which energy is transported with the
fluid is the same for both cases). cen58933_ch08.qxd 9/4/2002 11:29 AM Page 422 422
HEAT TRANSFER Note that the mean temperature Tm of a fluid changes during heating or cooling. Also, the fluid properties in internal flow are usually evaluated at the bulk
mean fluid temperature, which is the arithmetic average of the mean temperatures at the inlet and the exit. That is, Tb (Tm, i Tm, e)/2. Laminar and Turbulent Flow In Tubes
Flow in a tube can be laminar or turbulent, depending on the flow conditions.
Fluid flow is streamlined and thus laminar at low velocities, but turns turbulent as the velocity is increased beyond a critical value. Transition from laminar to turbulent flow does not occur suddenly; rather, it occurs over some
range of velocity where the flow fluctuates between laminar and turbulent
flows before it becomes fully turbulent. Most pipe flows encountered in practice are turbulent. Laminar flow is encountered when highly viscous fluids
such as oils flow in small diameter tubes or narrow passages.
For flow in a circular tube, the Reynolds number is defined as
mD Re
Circular tube:
Dh = 4(πD2/4)
=D
πD a Rectangular duct:
Dh = b 4ab
2ab
=
2(a + b)
a+b Circular tubes: FIGURE 8–4
The hydraulic diameter Dh 4Ac /p
is defined such that it reduces to
ordinary diameter for circular tubes.
Turbulent Dh Pipe wall FIGURE 8–5
In the transitional flow region of
2300 Re 4000, the flow
switches between laminar
and turbulent randomly. 4Ac
p 4 D2/4
D D It certainly is desirable to have precise values of Reynolds numbers for
laminar, transitional, and turbulent flows, but this is not the case in practice.
This is because the transition from laminar to turbulent flow also depends on
the degree of disturbance of the flow by surface roughness, pipe vibrations,
and the fluctuations in the flow. Under most practical conditions, the flow in a
tube is laminar for Re 2300, turbulent for Re 10,000, and transitional in
between. That is,
2300 Die trace (86) where Ac is the cross sectional area of the tube and p is its perimeter. The
hydraulic diameter is defined such that it reduces to ordinary diameter D for
circular tubes since a Laminar 4Ac
p Dh a 4a2
Dh =
=a
4a (85) where m is the mean fluid velocity, D is the diameter of the tube, and
/ is the kinematic viscosity of the fluid.
For flow through noncircular tubes, the Reynolds number as well as the
Nusselt number and the friction factor are based on the hydraulic diameter
Dh defined as (Fig. 8–4) D Square duct: mD Re
Re
Re 2300
10,000
10,000 laminar flow
transitional flow
turbulent flow In transitional flow, the flow switches between laminar and turbulent
randomly (Fig. 8–5). It should be kept in mind that laminar flow can be
maintained at much higher Reynolds numbers in very smooth pipes by
avoiding flow disturbances and tube vibrations. In such carefully controlled cen58933_ch08.qxd 9/4/2002 11:29 AM Page 423 423
CHAPTER 8 experiments, laminar flow has been maintained at Reynolds numbers of up to
100,000. 8–3 I THE ENTRANCE REGION Consider a fluid entering a circular tube at a uniform velocity. As in external
flow, the fluid particles in the layer in contact with the surface of the tube will
come to a complete stop. This layer will also cause the fluid particles in the
adjacent layers to slow down gradually as a result of friction. To make up for
this velocity reduction, the velocity of the fluid at the midsection of the tube
will have to increase to keep the mass flow rate through the tube constant. As
a result, a velocity boundary layer develops along the tube. The thickness of
this boundary layer increases in the flow direction until the boundary layer
reaches the tube center and thus fills the entire tube, as shown in Figure 8–6.
The region from the tube inlet to the point at which the boundary layer
merges at the centerline is called the hydrodynamic entrance region, and the
length of this region is called the hydrodynamic entry length Lh. Flow in the
entrance region is called hydrodynamically developing flow since this is the
region where the velocity profile develops. The region beyond the entrance region in which the velocity profile is fully developed and remains unchanged
is called the hydrodynamically fully developed region. The velocity profile
in the fully developed region is parabolic in laminar flow and somewhat flatter in turbulent flow due to eddy motion in radial direction.
Now consider a fluid at a uniform temperature entering a circular tube
whose surface is maintained at a different temperature. This time, the fluid
particles in the layer in contact with the surface of the tube will assume the
surface temperature. This will initiate convection heat transfer in the tube and
the development of a thermal boundary layer along the tube. The thickness of
this boundary layer also increases in the flow direction until the boundary
layer reaches the tube center and thus fills the entire tube, as shown in
Figure 8–7.
The region of flow over which the thermal boundary layer develops and
reaches the tube center is called the thermal entrance region, and the length
of this region is called the thermal entry length Lt. Flow in the thermal entrance region is called thermally developing flow since this is the region where
the temperature profile develops. The region beyond the thermal entrance region in which the dimensionless temperature profile expressed as (Ts T)/
(Ts Tm) remains unchanged is called the thermally fully developed region.
The region in which the flow is both hydrodynamically and thermally developed and thus both the velocity and dimensionless temperature profiles remain unchanged is called fully developed flow. That is,
Velocity boundary layer Velocity profile r x
Hydrodynamic entrance region Hydrodynamically
fully developed region FIGURE 8–6
The development of the
velocity boundary layer in a
tube. (The developed mean velocity
profile will be parabolic in laminar
flow, as shown, but somewhat
blunt in turbulent flow.) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 424 424
HEAT TRANSFER
Thermal
boundary layer
Ti FIGURE 8–7
The development of the
thermal boundary layer in a tube.
(The fluid in the tube is being cooled.) Temperature profile Ts x
Thermal
entrance region Thermally
fully developed region (r, x)
0 →
x
Ts(x) T(r, x)
0
x Ts(x) Tm(x) Hydrodynamically fully developed:
Thermally fully developed: (r) (87)
(88) The friction factor is related to the shear stress at the surface, which is related to the slope of the velocity profile at the surface. Noting that the velocity profile remains unchanged in the hydrodynamically fully developed
region, the friction factor also remains constant in that region. A similar argument can be given for the heat transfer coefficient in the thermally fully developed region.
In a thermally fully developed region, the derivative of (Ts T)/(Ts Tm)
with respect to x is zero by definition, and thus (Ts T)/(Ts Tm) is independent of x. Then the derivative of (Ts T)/(Ts Tm) with respect r must also
be independent of x. That is,
Ts
r Ts T
Tm ( T/ r) r
Ts Tm rR R f(x) (89) Surface heat flux can be expressed as
q·s hx(Ts Tm) k T
r → hx
rR k( T/ r) r
Ts Tm R (810) which, from Eq. 8–9, is independent of x. Thus we conclude that in the thermally fully developed region of a tube, the local convection coefficient is constant (does not vary with x). Therefore, both the friction and convection
coefficients remain constant in the fully developed region of a tube.
Note that the temperature profile in the thermally fully developed region
may vary with x in the flow direction. That is, unlike the velocity profile, the
temperature profile can be different at different cross sections of the tube in
the developed region, and it usually is. However, the dimensionless temperature profile defined above remains unchanged in the thermally developed region when the temperature or heat flux at the tube surface remains constant.
During laminar flow in a tube, the magnitude of the dimensionless Prandtl
number Pr is a measure of the relative growth of the velocity and thermal
boundary layers. For fluids with Pr 1, such as gases, the two boundary layers essentially coincide with each other. For fluids with Pr 1, such as oils,
the velocity boundary layer outgrows the thermal boundary layer. As a result, cen58933_ch08.qxd 9/4/2002 11:29 AM Page 425 425
CHAPTER 8 the hydrodynamic entry length is smaller than the thermal entry length. The
opposite is true for fluids with Pr 1 such as liquid metals.
Consider a fluid that is being heated (or cooled) in a tube as it flows through
it. The friction factor and the heat transfer coefficient are highest at the tube
inlet where the thickness of the boundary layers is zero, and decrease gradually to the fully developed values, as shown in Figure 8–8. Therefore, the
pressure drop and heat flux are higher in the entrance regions of a tube, and
the effect of the entrance region is always to enhance the average friction and
heat transfer coefficients for the entire tube. This enhancement can be significant for short tubes but negligible for long ones. h
or
f
hx
fx Entrance
Fully
region developed
region
x Entry Lengths Lh The hydrodynamic entry length is usually taken to be the distance from the
tube entrance where the friction coefficient reaches within about 2 percent of
the fully developed value. In laminar flow, the hydrodynamic and thermal
entry lengths are given approximately as [see Kays and Crawford (1993),
Ref. 13, and Shah and Bhatti (1987), Ref. 25] Lt
Fully developed
flow
Thermal boundary layer
Velocity boundary layer Lh, laminar
Lt, laminar 0.05 Re D
0.05 Re Pr D (811) Pr Lh, laminar (812) For Re 20, the hydrodynamic entry length is about the size of the diameter,
but increases linearly with the velocity. In the limiting case of Re 2300, the
hydrodynamic entry length is 115D.
In turbulent flow, the intense mixing during random fluctuations usually
overshadows the effects of momentum and heat diffusion, and therefore the
hydrodynamic and thermal entry lengths are of about the same size and independent of the Prandtl number. Also, the friction factor and the heat transfer
coefficient remain constant in fully developed laminar or turbulent flow since
the velocity and normalized temperature profiles do not vary in the flow direction. The hydrodynamic entry length for turbulent flow can be determined
from [see Bhatti and Shah (1987), Ref. 1, and Zhiqing (1982), Ref. 31]
Lh, turbulent 1.359 Re1/4 (813) The hydrodynamic entry length is much shorter in turbulent flow, as expected,
and its dependence on the Reynolds number is weaker. It is 11D at Re
10,000, and increases to 43D at Re 105. In practice, it is generally agreed
that the entrance effects are confined within a tube length of 10 diameters, and
the hydrodynamic and thermal entry lengths are approximately taken to be
Lh, turbulent Lt, turbulent 10D (814) The variation of local Nusselt number along a tube in turbulent flow for
both uniform surface temperature and uniform surface heat flux is given in
Figure 8–9 for the range of Reynolds numbers encountered in heat transfer
equipment. We make these important observations from this figure:
• The Nusselt numbers and thus the convection heat transfer coefficients
are much higher in the entrance region. FIGURE 8–8
Variation of the friction
factor and the convection
heat transfer coefficient in the flow
direction for flow in a tube (Pr 1). cen58933_ch08.qxd 9/4/2002 11:29 AM Page 426 426
HEAT TRANSFER
800
700
Nux, T (Ts = constant)
·
Nux, H (q s = constant) Nux, T Nux, H 600
500 D 400 105 200 FIGURE 8–9
Variation of local Nusselt number
along a tube in turbulent flow for both
uniform surface temperature and
uniform surface heat flux
[Deissler (1953), Ref. 4]. 105 Re = 2 300 6 0 2 4 6 8 10 104 3 100 104
104 12 14 16 18 20 x/D • The Nusselt number reaches a constant value at a distance of less than
10 diameters, and thus the flow can be assumed to be fully developed for
x 10D.
• The Nusselt numbers for the uniform surface temperature and uniform
surface heat flux conditions are identical in the fully developed regions,
and nearly identical in the entrance regions. Therefore, Nusselt number
is insensitive to the type of thermal boundary condition, and the turbulent
flow correlations can be used for either type of boundary condition.
Precise correlations for the friction and heat transfer coefficients for the entrance regions are available in the literature. However, the tubes used in practice in forced convection are usually several times the length of either entrance
region, and thus the flow through the tubes is often assumed to be fully developed for the entire length of the tube. This simplistic approach gives reasonable results for long tubes and conservative results for short ones. 8–4
. GENERAL THERMAL ANALYSIS You will recall that in the absence of any work interactions (such as electric
resistance heating), the conservation of energy equation for the steady flow of
a fluid in a tube can be expressed as (Fig. 8–10) Q
Ti
·CT
m pi I Te
·
m Cp Te Energy balance:
·
·
Q = m Cp(Te – Ti ) FIGURE 8–10
The heat transfer to a fluid flowing in
a tube is equal to the increase in
the energy of the fluid. ·
Q ·
m Cp(Te Ti) (W) (815) where Ti and Te are the mean fluid temperatures at the inlet and exit of the
·
tube, respectively, and Q is the rate of heat transfer to or from the fluid. Note
that the temperature of a fluid flowing in a tube remains constant in the absence of any energy interactions through the wall of the tube.
The thermal conditions at the surface can usually be approximated with
constant) or
reasonable accuracy to be constant surface temperature (Ts
·
constant surface heat flux (qs constant). For example, the constant surface cen58933_ch08.qxd 9/4/2002 11:29 AM Page 427 427
CHAPTER 8 temperature condition is realized when a phase change process such as boiling or condensation occurs at the outer surface of a tube. The constant surface
heat flux condition is realized when the tube is subjected to radiation or electric resistance heating uniformly from all directions.
Surface heat flux is expressed as
·
qs hx (Ts (W/m2) Tm) (816) where hx is the local heat transfer coefficient and Ts and Tm are the surface and
the mean fluid temperatures at that location. Note that the mean fluid temperature Tm of a fluid flowing in a tube must change during heating or cooling.
Therefore, when hx h constant, the surface temperature Ts must change
·
·
when qs constant, and the surface heat flux qs must change when Ts con·
stant. Thus we may have either Ts constant or qs constant at the surface
of a tube, but not both. Next we consider convection heat transfer for these
two common cases. ·
Constant Surface Heat Flux (qs
·
In the case of qs constant) T constant, the rate of heat transfer can also be expressed as
·
Q q·s As ·
m Cp(Te Ti) (W) Entrance
region Fully developed
region Ts (817)
Te Then the mean fluid temperature at the tube exit becomes
Te Ti ·
qs As
·
m Cp Ti Note that the mean fluid temperature increases linearly in the flow direction
in the case of constant surface heat flux, since the surface area increases linearly in the flow direction (As is equal to the perimeter, which is constant,
times the tube length).
·
The surface temperature in the case of constant surface heat flux qs can be
determined from
·
qs h(Ts Tm) → Ts Tm ·
qs
h dTm
·
q s(pdx) →
dx ·
qs p
·
m Cp dTs
dx ·
qs = constant constant L Ti x Te FIGURE 8–11
Variation of the tube surface
and the mean fluid temperatures
along the tube for the case of
constant surface heat flux.
·
δ Q = h (Ts – Tm ) dA
Tm (820) where p is the perimeter of the tube.
·
Noting that both qs and h are constants, the differentiation of Eq. 8–19 with
respect to x gives
dTm
dx 0 (819) In the fully developed region, the surface temperature Ts will also increase linearly in the flow direction since h is constant and thus Ts Tm constant
(Fig. 8–11). Of course this is true when the fluid properties remain constant
during flow.
The slope of the mean fluid temperature Tm on a Tx diagram can be determined by applying the steadyflow energy balance to a tube slice of thickness
dx shown in Figure 8–12. It gives
·
m Cp dTm Tm ·
qs
∆T = Ts – Tm = ––
h (818) (821) Tm + dTm . ·
m Cp(Tm + d Tm) m CpTm Ts
dx FIGURE 8–12
Energy interactions for a differential
control volume in a tube. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 428 428
HEAT TRANSFER Also, the requirement that the dimensionless temperature profile remains
unchanged in the fully developed region gives
Ts
x Ts since Ts T
Tm Tm 0 → T (r)
Ts 2 Ts1 x ·
qs T
x 0 dTs
dx T
x → (822) dTs
dx ·
qs p
· Cp
m dTm
dx constant (823) Then we conclude that in fully developed flow in a tube subjected to constant
surface heat flux, the temperature gradient is independent of x and thus the
shape of the temperature profile does not change along the tube (Fig. 8–13).
·
2
For a circular tube, p 2 R and m
m Ac
m( R ), and Eq. 8–23
becomes
T
x Circular tube:
x1 Tm constant. Combining Eqs. 8–20, 8–21, and 8–22 gives
T
x T (r) Ts
x 1
Ts dTs
dx ·
2qs
mCp R dTm
dx constant (824) x2 FIGURE 8–13
The shape of the temperature
profile remains unchanged in the
fully developed region of a tube
subjected to constant surface heat flux. where m is the mean velocity of the fluid. Constant Surface Temperature (Ts constant) From Newton’s law of cooling, the rate of heat transfer to or from a fluid
flowing in a tube can be expressed as
·
Q hAs Tave hAs(Ts Tm)ave (W) (825) where h is the average convection heat transfer coefficient, As is the heat transfer surface area (it is equal to DL for a circular pipe of length L), and Tave
is some appropriate average temperature difference between the fluid and the
surface. Below we discuss two suitable ways of expressing Tave.
constant) case, Tave can be
In the constant surface temperature (Ts
expressed approximately by the arithmetic mean temperature difference
Tam as
Tave Tam Ti
Ts Te
2
Tb (Ts Ti) (Ts
2 Te) Ts Ti Te
2
(826) where Tb (Ti Te)/2 is the bulk mean fluid temperature, which is the arithmetic average of the mean fluid temperatures at the inlet and the exit of
the tube.
Note that the arithmetic mean temperature difference Tam is simply the average of the temperature differences between the surface and the fluid at the
inlet and the exit of the tube. Inherent in this definition is the assumption that
the mean fluid temperature varies linearly along the tube, which is hardly ever
the case when Ts constant. This simple approximation often gives acceptable results, but not always. Therefore, we need a better way to evaluate Tave.
Consider the heating of a fluid in a tube of constant cross section whose
inner surface is maintained at a constant temperature of Ts. We know that the cen58933_ch08.qxd 9/4/2002 11:29 AM Page 429 429
CHAPTER 8 mean temperature of the fluid Tm will increase in the flow direction as a result
of heat transfer. The energy balance on a differential control volume shown in
Figure 8–12 gives
·
m Cp dTm h(Ts Tm)dAs (827) That is, the increase in the energy of the fluid (represented by an increase in
its mean temperature by dTm) is equal to the heat transferred to the fluid from
the tube surface by convection. Noting that the differential surface area is
d(Ts Tm),
dAs pdx, where p is the perimeter of the tube, and that dTm
since Ts is constant, the relation above can be rearranged as
d(Ts
Ts Integrating from x
Tm Te) gives Tm)
Tm 0 (tube inlet where Tm
Ts
ln
Ts Te
Ti T hp
· dx
mCp (828) Ti) to x Ts = constant Ts Tm ∆Ti L (tube exit where ∆T
e ∆T = Ts – Tm Ti hAs
·
mCp (829) (Tm approaches Ts asymptotically)
0 where As pL is the surface area of the tube and h is the constant average
convection heat transfer coefficient. Taking the exponential of both sides and
solving for Te gives the following relation which is very useful for the determination of the mean fluid temperature at the tube exit:
Te Ts (Ts ·
Ti) exp( hAs /m Cp) ln[(Ts hAs
Te)/(Ts Ti)] x
Te Ti Ts = constant
(830) This relation can also be used to determine the mean fluid temperature Tm(x)
at any x by replacing As pL by px.
Note that the temperature difference between the fluid and the surface decays exponentially in the flow direction, and the rate of decay depends on the
·
magnitude of the exponent hAx /m Cp, as shown in Figure 8–14. This dimensionless parameter is called the number of transfer units, denoted by
NTU, and is a measure of the effectiveness of the heat transfer systems. For
NUT 5, the exit temperature of the fluid becomes almost equal to the surface temperature, Te Ts (Fig. 8–15). Noting that the fluid temperature can
approach the surface temperature but cannot cross it, an NTU of about 5 indicates that the limit is reached for heat transfer, and the heat transfer will not increase no matter how much we extend the length of the tube. A small value of
NTU, on the other hand, indicates more opportunities for heat transfer, and the
heat transfer will continue increasing as the tube length is increased. A large
NTU and thus a large heat transfer surface area (which means a large tube)
may be desirable from a heat transfer point of view, but it may be unacceptable from an economic point of view. The selection of heat transfer equipment
usually reflects a compromise between heat transfer performance and cost.
·
Solving Eq. 8–29 for m Cp gives
·
m Cp L (831) FIGURE 8–14
The variation of the mean fluid
temperature along the tube for the
case of constant temperature.
Ts = 100°C
Ti =
20°C ·
m , Cp Te As, h
·
NTU = h As / mCp Te , °C 0.01
0.05
0.10
0.50
1.00
5.00
10.00 20.8
23.9
27.6
51.5
70.6
99.5
100.0 FIGURE 8–15
An NTU greater than 5 indicates that
the fluid flowing in a tube will reach
the surface temperature at the exit
regardless of the inlet temperature. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 430 430
HEAT TRANSFER Substituting this into Eq. 8–17, we obtain
·
Q hAs Tln (832) where
Tln ln[(Ts Ti Te
Te)/(Ts Ti)] Te
Ti
ln( Te / Ti) (833) is the logarithmic mean temperature difference. Note that Ti Ts Ti
and Te Ts Te are the temperature differences between the surface and
the fluid at the inlet and the exit of the tube, respectively. This Tln relation
appears to be prone to misuse, but it is practically failsafe, since using Ti in
place of Te and vice versa in the numerator and/or the denominator will, at
most, affect the sign, not the magnitude. Also, it can be used for both heating
(Ts Ti and Te) and cooling (Ts Ti and Te) of a fluid in a tube.
The logarithmic mean temperature difference Tln is obtained by tracing the
actual temperature profile of the fluid along the tube, and is an exact representation of the average temperature difference between the fluid and the surface. It truly reflects the exponential decay of the local temperature difference.
When Te differs from Ti by no more than 40 percent, the error in using the
arithmetic mean temperature difference is less than 1 percent. But the error increases to undesirable levels when Te differs from Ti by greater amounts.
Therefore, we should always use the logarithmic mean temperature difference
when determining the convection heat transfer in a tube whose surface is
maintained at a constant temperature Ts. EXAMPLE 8–1 Water enters a 2.5cminternaldiameter thin copper tube of a heat exchanger
at 15°C at a rate of 0.3 kg/s, and is heated by steam condensing outside at
120°C. If the average heat transfer coefficient is 800 W/m2 C, determine the
length of the tube required in order to heat the water to 115°C (Fig. 8–16). Steam
Ts = 120°C
Water
15°C
0.3 kg/s 115°C
D = 2.5 cm FIGURE 8–16
Schematic for Example 8–1. Heating of Water in a Tube by Steam SOLUTION Water is heated by steam in a circular tube. The tube length
required to heat the water to a specified temperature is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Fluid properties are constant. 3 The convection heat transfer coefficient is constant. 4 The conduction
resistance of copper tube is negligible so that the inner surface temperature of
the tube is equal to the condensation temperature of steam.
Properties The specific heat of water at the bulk mean temperature of
(15 115)/2 65°C is 4187 J/kg °C. The heat of condensation of steam at
120°C is 2203 kJ/kg (Table A9).
Analysis Knowing the inlet and exit temperatures of water, the rate of heat
transfer is determined to be ·
Q ·
m Cp(Te Ti) (0.3 kg/s)(4.187 kJ/kg °C)(115°C The logarithmic mean temperature difference is 15°C) 125.6 kW cen58933_ch08.qxd 9/4/2002 11:29 AM Page 431 431
CHAPTER 8 Te
Ti Ts
Ts 120°C 115°C 5°C
120°C 15°C 105°C
Te
Ti
5 105
32.85°C
ln( Te / Ti) ln(5/105) Tln Te
Ti The heat transfer surface area is ·
Q hAs Tln → As ·
Q
h Tln 125.6 kW
(0.8 kW/m2 · °C)(32.85°C) 4.78 m2 Then the required length of tube becomes As DL → L As
D 4.78 m2
(0.025 m) 61 m Discussion The bulk mean temperature of water during this heating process
is 65°C, and thus the arithmetic mean temperature difference is Tam
120 65 55°C. Using Tam instead of Tln would give L 36 m, which is
grossly in error. This shows the importance of using the logarithmic mean temperature in calculations. 8–5 I LAMINAR FLOW IN TUBES We mentioned earlier that flow in tubes is laminar for Re 2300, and that the
flow is fully developed if the tube is sufficiently long (relative to the entry
length) so that the entrance effects are negligible. In this section we consider
the steady laminar flow of an incompressible fluid with constant properties in
the fully developed region of a straight circular tube. We obtain the momentum and energy equations by applying momentum and energy balances to a
differential volume element, and obtain the velocity and temperature profiles
by solving them. Then we will use them to obtain relations for the friction factor and the Nusselt number. An important aspect of the analysis below is that
it is one of the few available for viscous flow and forced convection.
In fully developed laminar flow, each fluid particle moves at a constant
axial velocity along a streamline and the velocity profile (r) remains unchanged in the flow direction. There is no motion in the radial direction, and
thus the velocity component v in the direction normal to flow is everywhere
zero. There is no acceleration since the flow is steady.
Now consider a ringshaped differential volume element of radius r, thickness dr, and length dx oriented coaxially with the tube, as shown in Figure
8–17. The pressure force acting on a submerged plane surface is the product
of the pressure at the centroid of the surface and the surface area.
The volume element involves only pressure and viscous effects, and thus
the pressure and shear forces must balance each other. A force balance on the
volume element in the flow direction gives
(2 rdrP)x (2 rdrP)x dx (2 rdx )r (2 rdx )r dr 0 (834) τr dr Px Px dx τr (r) dr R r
x dx max FIGURE 8–17
Free body diagram of a cylindrical
fluid element of radius r, thickness dr,
and length dx oriented coaxially with a
horizontal tube in fully developed
steady flow. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 432 432
HEAT TRANSFER which indicates that in fully developed flow in a tube, the viscous and pressure forces balance each other. Dividing by 2 drdx and rearranging,
r Px (r )x Px dx (r )r dr dx dr 0 (835) Taking the limit as dr, dx → 0 gives Substituting d(r )
dr dP
dx r 0 (836) (d /dr) and rearranging gives the desired equation,
d
d
r dr r dr dP
dx (837) The quantity d /dr is negative in tube flow, and the negative sign is included
to obtain positive values for . (Or, d /dr
d /dy since y R r.) The
left side of this equation is a function of r and the right side is a function of x.
The equality must hold for any value of r and x, and an equality of the form
f(r) g(x) can happen only if both f(r) and g(x) are equal to constants. Thus
we conclude that dP/dx constant. This can be verified by writing a force
balance on a volume element of radius R and thickness dx (a slice of the tube),
which gives dP/dx
2 s/R. Here s is constant since the viscosity and
the velocity profile are constants in the fully developed region. Therefore,
dP/dx constant.
Equation 8–37 can be solved by rearranging and integrating it twice to give
dP
dx 1
4 (r) C1lnr C2 (838) The velocity profile (r) is obtained by applying the boundary conditions
/ r 0 at r 0 (because of symmetry about the centerline) and
0 at
r R (the noslip condition at the tube surface). We get
R2 dP
1
4 dx (r) r2
R2 (839) Therefore, the velocity profile in fully developed laminar flow in a tube is
parabolic with a maximum at the centerline and minimum at the tube surface.
Also, the axial velocity is positive for any r, and thus the axial pressure gradient dP/dx must be negative (i.e., pressure must decrease in the flow direction because of viscous effects).
The mean velocity is determined from its definition by substituting
Eq. 8–39 into Eq. 8–2, and performing the integration. It gives
m 2
R2 R rdr
0 2
R2 R
0 R2 dP
1
4 dx r2
rdr
R2 R2 dP
8 dx (840) Combining the last two equations, the velocity profile is obtained to be
(r) 2 m 1 r2
R2 (841) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 433 433
CHAPTER 8 This is a convenient form for the velocity profile since m can be determined
easily from the flow rate information.
The maximum velocity occurs at the centerline, and is determined from
Eq. 8–39 by substituting r 0,
2 max (842) m Therefore, the mean velocity is onehalf of the maximum velocity. Pressure Drop
A quantity of interest in the analysis of tube flow is the pressure drop P since
it is directly related to the power requirements of the fan or pump to maintain
flow. We note that dP/dx constant, and integrate it from x 0 where the
pressure is P1 to x L where the pressure is P2. We get
P1 P2 dP
dx P
L L (843) Note that in fluid mechanics, the pressure drop P is a positive quantity, and
is defined as P P1 P2. Substituting Eq. 8–43 into the m expression in
Eq. 8–40, the pressure drop can be expressed as
Laminar flow: 8L
R2 P 32 L
D2 m m (844) In practice, it is found convenient to express the pressure drop for all types of
internal flows (laminar or turbulent flows, circular or noncircular tubes,
smooth or rough surfaces) as (Fig. 8–18) ∆P 2 P f m
L
D2 (845) D m L where the dimensionless quantity f is the friction factor (also called the
Darcy friction factor after French engineer Henry Darcy, 1803–1858, who
first studied experimentally the effects of roughness on tube resistance). It
should not be confused with the friction coefficient Cf (also called the Fanning
2
f/4.
friction factor), which is defined as Cf
s(
m /2)
Equation 8–45 gives the pressure drop for a flow section of length L provided that (1) the flow section is horizontal so that there are no hydrostatic or
gravity effects, (2) the flow section does not involve any work devices such as
a pump or a turbine since they change the fluid pressure, and (3) the cross sectional area of the flow section is constant and thus the mean flow velocity is
constant.
Setting Eqs. 8–44 and 8–45 equal to each other and solving for f gives the
friction factor for the fully developed laminar flow in a circular tube to be
Circular tube, laminar: f 64
D m 64
Re (846) This equation shows that in laminar flow, the friction factor is a function of
the Reynolds number only and is independent of the roughness of the tube ρm
Pressure drop: ∆P = f L
D2 2 FIGURE 8–18
The relation for pressure
drop is one of the most general
relations in fluid mechanics, and it is
valid for laminar or turbulent flows,
circular or noncircular pipes, and
smooth or rough surfaces. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 434 434
HEAT TRANSFER surface. Once the pressure drop is available, the required pumping power is
determined from
·
Wpump ·
VP (847) ·
where V is the volume flow rate of flow, which is expressed as
·
V D4 P
128 L (848) This equation is known as the Poiseuille’s Law, and this flow is called the
Hagen–Poiseuille flow in honor of the works of G. Hagen (1797–1839) and
J. Poiseuille (1799–1869) on the subject. Note from Eq. 8–48 that for a specified flow rate, the pressure drop and thus the required pumping power is proportional to the length of the tube and the viscosity of the fluid, but it is
inversely proportional to the fourth power of the radius (or diameter) of the
tube. Therefore, the pumping power requirement for a piping system can be
reduced by a factor of 16 by doubling the tube diameter (Fig. 8–19). Of course
the benefits of the reduction in the energy costs must be weighed against the
increased cost of construction due to using a larger diameter tube.
The pressure drop is caused by viscosity, and it is directly related to the wall
shear stress. For the ideal inviscid flow, the pressure drop is zero since there
are no viscous effects. Again, Eq. 8–47 is valid for both laminar and turbulent
flows in circular and noncircular tubes. ⋅
Wpump = 16 hp
D ⋅
Wpump = 1 hp 2D FIGURE 8–19
The pumping power requirement for
a laminar flow piping system can
be reduced by a factor of 16 by
doubling the pipe diameter. dx mCpTx R4 P
8L PR2
R2
8L ave Ac dx r
dr Temperature Profile and the Nusselt Number
In the analysis above, we have obtained the velocity profile for fully developed flow in a circular tube from a momentum balance applied on a volume
element, determined the friction factor and the pressure drop. Below we obtain the energy equation by applying the energy balance to a differential volume element, and solve it to obtain the temperature profile for the constant
surface temperature and the constant surface heat flux cases.
Reconsider steady laminar flow of a fluid in a circular tube of radius R. The
fluid properties , k, and Cp are constant, and the work done by viscous
stresses is negligible. The fluid flows along the xaxis with velocity . The
flow is fully developed so that is independent of x and thus
(r). Noting that energy is transferred by mass in the xdirection, and by conduction in
the rdirection (heat conduction in the xdirection is assumed to be negligible),
the steadyflow energy balance for a cylindrical shell element of thickness dr
and length dx can be expressed as (Fig. 8–20) Qr ·
m CpTx mCpTx
Qr FIGURE 8–20
The differential volume element
used in the derivation of
energy balance relation. dr ·
where m
Ac
after rearranging, ·
m CpTx dx ·
Qr ·
Qr dr 0 (849) (2 rdr). Substituting and dividing by 2 rdrdx gives, Cp Tx dx dx Tx ·
·
1 Q r dr Q r
2 rdx
dr (850) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 435 435
CHAPTER 8 or
·
Q
2 Cp rdx r T
x 1 (851) where we used the definition of derivative. But
·
Q
r k2 rdx r Substituting and using T
r 2 kdx r r T
r (852) k/ Cp gives
α
T
r dr r r T
x (853) which states that the rate of net energy transfer to the control volume by mass
flow is equal to the net rate of heat conduction in the radial direction. Constant Surface Heat Flux
For fully developed flow in a circular pipe subjected to constant surface heat
flux, we have, from Eq. 8–24,
dTs
dx T
x ·
2qs
mCpR dTm
dx constant (854) If heat conduction in the xdirection were considered in the derivation of
Eq. 8–53, it would give an additional term 2T/ x2, which would be equal to
zero since T/ x constant and thus T T(r). Therefore, the assumption that
there is no axial heat conduction is satisfied exactly in this case.
Substituting Eq. 8–54 and the relation for velocity profile (Eq. 8–41) into
Eq. 8–53 gives
·
4qs
1
kR r2
R2 dT
1d
r dr r dr (855) which is a secondorder ordinary differential equation. Its general solution is
obtained by separating the variables and integrating twice to be
T ·
qs 2
r
kR r2
4R2 C1 r C2 (856) The desired solution to the problem is obtained by applying the boundary
conditions T/ x 0 at r 0 (because of symmetry) and T Ts at r R.
We get
T Ts ·
qs R 3
k4 r2
R2 r4
4R4 (857) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 436 436
HEAT TRANSFER The bulk mean temperature Tm is determined by substituting the velocity and
temperature profile relations (Eqs. 8–41 and 8–57) into Eq. 8–4 and performing the integration. It gives
Tm ·
Combining this relation with qs
h Ts ·
11 qs R
24 k h(Ts Tm) gives 48 k
11 D 24 k
11 R (858) 4.36 k
D (859) or
Circular tube, laminar (q·x constant): Nu hD
k 4.36 (860) Therefore, for fully developed laminar flow in a circular tube subjected to
constant surface heat flux, the Nusselt number is a constant. There is no dependence on the Reynolds or the Prandtl numbers. Constant Surface Temperature
A similar analysis can be performed for fully developed laminar flow in a circular tube for the case of constant surface temperature Ts. The solution procedure in this case is more complex as it requires iterations, but the Nusselt
number relation obtained is equally simple (Fig. 8–21): Ts = constant
64
f = –––
Re D Nu = 3.66 max
m Fully developed
laminar flow FIGURE 8–21
In laminar flow in a tube with
constant surface temperature,
both the friction factor and
the heat transfer coefficient
remain constant in the fully
developed region. Circular tube, laminar (Ts constant): Nu hD
k 3.66 (861) The thermal conductivity k for use in the Nu relations above should be evaluated at the bulk mean fluid temperature, which is the arithmetic average of the
mean fluid temperatures at the inlet and the exit of the tube. For laminar flow,
the effect of surface roughness on the friction factor and the heat transfer coefficient is negligible. Laminar Flow in Noncircular Tubes
The friction factor f and the Nusselt number relations are given in Table 8–1
for fully developed laminar flow in tubes of various cross sections. The
Reynolds and Nusselt numbers for flow in these tubes are based on the hydraulic diameter Dh 4Ac /p, where Ac is the cross sectional area of the tube
and p is its perimeter. Once the Nusselt number is available, the convection
heat transfer coefficient is determined from h kNu/Dh. Developing Laminar Flow in the Entrance Region
For a circular tube of length L subjected to constant surface temperature, the
average Nusselt number for the thermal entrance region can be determined
from (Edwards et al., 1979)
Entry region, laminar: Nu 3.66 1 0.065 (D/L) Re Pr
0.04[(D/L) Re Pr]2/3 (862) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 437 437
CHAPTER 8 TABLE 8–1
Nusselt number and friction factor for fully developed laminar flow in tubes of
various cross sections (Dh 4Ac /p, Re
hDh /k)
m Dh /v, and Nu Tube Geometry
Circle a/b
or ° Ts — Nusselt Number
·
Const.
q s Const. Friction Factor
f 3.66 4.36 64.00/Re 2.98
3.39
3.96
4.44
5.14
5.60
7.54 3.61
4.12
4.79
5.33
6.05
6.49
8.24 56.92/Re
62.20/Re
68.36/Re
72.92/Re
78.80/Re
82.32/Re
96.00/Re a/b
1
2
4
8
16 3.66
3.74
3.79
3.72
3.65 4.36
4.56
4.88
5.09
5.18 64.00/Re
67.28/Re
72.96/Re
76.60/Re
78.16/Re 10°
30°
60°
90°
120° 1.61
2.26
2.47
2.34
2.00 2.45
2.91
3.11
2.98
2.68 50.80/Re
52.28/Re
53.32/Re
52.60/Re
50.96/Re D Rectangle b
a Ellipse b
a a/b
1
2
3
4
6
8 Triangle θ Note that the average Nusselt number is larger at the entrance region, as expected, and it approaches asymptotically to the fully developed value of 3.66
as L → . This relation assumes that the flow is hydrodynamically developed
when the fluid enters the heating section, but it can also be used approximately for flow developing hydrodynamically.
When the difference between the surface and the fluid temperatures is large,
it may be necessary to account for the variation of viscosity with temperature.
The average Nusselt number for developing laminar flow in a circular tube in
that case can be determined from [Sieder and Tate (1936), Ref. 26]
Nu 1.86 Re Pr D
L 1/3 b
s 0.14 (863) All properties are evaluated at the bulk mean fluid temperature, except for
which is evaluated at the surface temperature. s, cen58933_ch08.qxd 9/4/2002 11:29 AM Page 438 438
HEAT TRANSFER The average Nusselt number for the thermal entrance region of flow
between isothermal parallel plates of length L is expressed as (Edwards
et al., 1979)
Entry region, laminar: Nu 7.54 1 0.03 (Dh /L) Re Pr
0.016[(Dh /L) Re Pr]2/3 (864) where Dh is the hydraulic diameter, which is twice the spacing of the plates.
This relation can be used for Re 2800.
EXAMPLE 8–2 3 ft/s 0.15 in. Pressure Drop in a Pipe Water at 40°F (
62.42 lbm/ft3 and
3.74 lbm/ft h) is flowing in a 0.15in.diameter 30ftlong pipe steadily at an average velocity of 3 ft/s (Fig. 8–22).
Determine the pressure drop and the pumping power requirement to overcome
this pressure drop. 30 ft FIGURE 8–22
Schematic for Example 8–2. SOLUTION The average flow velocity in a pipe is given. The pressure drop and
the required pumping power are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects
are negligible, and thus the flow is fully developed. 3 The pipe involves no components such as bends, valves, and connectors.
Properties The density and dynamic viscosity of water are given to be
62.42 lbm/ft3 and
3.74 lbm/ft h 0.00104 lbm/ft s.
Analysis First we need to determine the flow regime. The Reynolds number is
mD Re (62.42 lbm/ft3)(3 ft/s)(0.12/12 ft) 3600 s
3.74 lbm/ft · h
1h 1803 which is less than 2300. Therefore, the flow is laminar. Then the friction factor
and the pressure drop become f 64
Re 64
1803 0.0355 2 P 30 ft (62.42 lbm/ft3)(3 ft/s)2
1 lbf
m
L
0.0355
D2
2
0.12/12 ft
32.174 lbm · ft/s2
930 lbf/ft2 6.46 psi
f The volume flow rate and the pumping power requirements are
·
2
V
(3 ft/s)[ (0.12/12 ft)2/4] 0.000236 ft3/s
m Ac
m ( D /4)
·
Wpump ·
VP (0.000236 ft3/s)(930 lbf/ft2) 1W
0.737 lbf · ft/s 0.30 W Therefore, mechanical power input in the amount of 0.30 W is needed to overcome the frictional losses in the flow due to viscosity. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 439 439
CHAPTER 8 EXAMPLE 8–3 Flow of Oil in a Pipeline through a Lake Consider the flow of oil at 20°C in a 30cmdiameter pipeline at an average
velocity of 2 m/s (Fig. 8–23). A 200mlong section of the pipeline passes
through icy waters of a lake at 0°C. Measurements indicate that the surface
temperature of the pipe is very nearly 0°C. Disregarding the thermal resistance
of the pipe material, determine (a) the temperature of the oil when the pipe
leaves the lake, (b) the rate of heat transfer from the oil, and (c) the pumping
power required to overcome the pressure losses and to maintain the flow of the
oil in the pipe. Icy lake, 0°C
20°C Oil
2 m/s D = 0.3 m Te 0°C
200 m FIGURE 8–23
Schematic for Example 8–3. SOLUTION Oil flows in a pipeline that passes through icy waters of a lake at
0°C. The exit temperature of the oil, the rate of heat loss, and the pumping
power needed to overcome pressure losses are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The surface temperature of
the pipe is very nearly 0°C. 3 The thermal resistance of the pipe is negligible.
4 The inner surfaces of the pipeline are smooth. 5 The flow is hydrodynamically
developed when the pipeline reaches the lake.
Properties We do not know the exit temperature of the oil, and thus we cannot
determine the bulk mean temperature, which is the temperature at which the
properties of oil are to be evaluated. The mean temperature of the oil at the
inlet is 20°C, and we expect this temperature to drop somewhat as a result
of heat loss to the icy waters of the lake. We evaluate the properties of the oil
at the inlet temperature, but we will repeat the calculations, if necessary,
using properties at the evaluated bulk mean temperature. At 20°C we read
(Table A14)
k 888 kg/m3
0.145 W/m °C Cp
Pr 901 10 6 m2/s
1880 J/kg °C
10,400 Analysis (a) The Reynolds number is
m Dh Re (2 m/s)(0.3 m)
901 10 6 m2/s 666 which is less than the critical Reynolds number of 2300. Therefore, the flow is
laminar, and the thermal entry length in this case is roughly Lt 0.05 Re Pr D 0.05 666 10,400 (0.3 m) 104,000 m which is much greater than the total length of the pipe. This is typical of fluids
with high Prandtl numbers. Therefore, we assume thermally developing flow
and determine the Nusselt number from Nu hD
k
3.66
37.3 3.66 1 1 0.065 (D/L) Re Pr
0.04 [(D/L) Re Pr]2/3 0.065(0.3/200)
0.04[(0.3/200) 666
666 10,400
10,400]2/3 cen58933_ch08.qxd 9/4/2002 11:29 AM Page 440 440
HEAT TRANSFER Note that this Nusselt number is considerably higher than the fully developed
value of 3.66. Then, h k
Nu
D 0.145 W/m
(37.3)
0.3 m 18.0 W/m2 °C Also, As
·
m pL
Ac m DL
(0.3 m)(200 m) 188.5 m2
(888 kg/m3)[1 (0.3 m)2](2 m/s) 125.5 kg/s
4 Next we determine the exit temperature of oil from ·
Ti) exp ( hAs /m Cp)
(18.0 W/m2 · °C)(188.5 m2)
0°C [(0 20)°C] exp
(125.5 kg/s)(1880 J/kg · °C)
19.71°C Te Ts (Ts Thus, the mean temperature of oil drops by a mere 0.29°C as it crosses the
lake. This makes the bulk mean oil temperature 19.86°C, which is practically
identical to the inlet temperature of 20°C. Therefore, we do not need to reevaluate the properties.
(b) The logarithmic mean temperature difference and the rate of heat loss from
the oil are Tln
·
Q Ti Te
20 19.71
19.85°C
Ts Te
0 19.71
ln
ln
0 20
Ts Ti
hAs Tln (18.0 W/m2 °C)(188.5 m2)( 19.85°C) 6.74 104 Therefore, the oil will lose heat at a rate of 67.4 kW as it flows through the pipe
in the icy waters of the lake. Note that Tln is identical to the arithmetic mean
temperature in this case, since Ti
Te.
(c) The laminar flow of oil is hydrodynamically developed. Therefore, the friction
factor can be determined from f 64
Re 64
666 0.0961 Then the pressure drop in the pipe and the required pumping power become P
·
Wpump 3
2
2
200 m (888 kg/m )(2 m/s)
m
L
0.0961
1.14
D2
0.3 m
2
5
2
·
m P (125.5 kg/s)(1.14 10 N/m )
16.1 kW
3
888 kg/m f 105 N/m2 Discussion We will need a 16.1kW pump just to overcome the friction in the
pipe as the oil flows in the 200mlong pipe through the lake. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 441 441
CHAPTER 8 8–6 I TURBULENT FLOW IN TUBES We mentioned earlier that flow in smooth tubes is fully turbulent for Re
10,000. Turbulent flow is commonly utilized in practice because of the higher
heat transfer coefficients associated with it. Most correlations for the friction
and heat transfer coefficients in turbulent flow are based on experimental
studies because of the difficulty in dealing with turbulent flow theoretically.
For smooth tubes, the friction factor in turbulent flow can be determined
from the explicit first Petukhov equation [Petukhov (1970), Ref. 21] given as
Smooth tubes: f (0.790 ln Re 1.64) 2 104 Re 106 (865) The Nusselt number in turbulent flow is related to the friction factor through
the Chilton–Colburn analogy expressed as
Nu 0.125 f RePr1/3 (866) Once the friction factor is available, this equation can be used conveniently to
evaluate the Nusselt number for both smooth and rough tubes.
For fully developed turbulent flow in smooth tubes, a simple relation for the
Nusselt number can be obtained by substituting the simple power law relation
f 0.184 Re 0.2 for the friction factor into Eq. 8–66. It gives
Nu 0.7
Re 0.023 Re0.8 Pr1/3 Pr 160
10,000 (867) which is known as the Colburn equation. The accuracy of this equation can be
improved by modifying it as
Nu 0.023 Re0.8 Pr n (868) where n 0.4 for heating and 0.3 for cooling of the fluid flowing through
the tube. This equation is known as the Dittus–Boelter equation [Dittus and
Boelter (1930), Ref. 6] and it is preferred to the Colburn equation.
The fluid properties are evaluated at the bulk mean fluid temperature Tb
(Ti Te)/2. When the temperature difference between the fluid and the wall is
very large, it may be necessary to use a correction factor to account for the different viscosities near the wall and at the tube center.
The Nusselt number relations above are fairly simple, but they may give
errors as large as 25 percent. This error can be reduced considerably to less
than 10 percent by using more complex but accurate relations such as the second Petukhov equation expressed as
Nu 1.07 ( f /8) Re Pr
12.7( f /8)0.5 (Pr2/3 1) 0.5
104 Pr
Re 2000
5 106 (869) The accuracy of this relation at lower Reynolds numbers is improved by modifying it as [Gnielinski (1976), Ref. 8]
Nu ( f /8)(Re 1000) Pr
1 12.7( f /8)0.5 (Pr2/3 1) 0.5 Pr
3 103 2000
Re 5 106 (870) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 442 442
HEAT TRANSFER Relative
Roughness,
/L Friction
Factor,
f 0.0*
0.00001
0.0001
0.0005
0.001
0.005
0.01
0.05 where the friction factor f can be determined from an appropriate relation such
as the first Petukhov equation. Gnielinski’s equation should be preferred
in calculations. Again properties should be evaluated at the bulk mean fluid
temperature.
The relations above are not very sensitive to the thermal conditions at the
·
tube surfaces and can be used for both Ts constant and qs constant cases.
Despite their simplicity, the correlations already presented give sufficiently
accurate results for most engineering purposes. They can also be used to obtain rough estimates of the friction factor and the heat transfer coefficients in
the transition region 2300
Re
10,000, especially when the Reynolds
number is closer to 10,000 than it is to 2300.
The relations given so far do not apply to liquid metals because of their
very low Prandtl numbers. For liquid metals (0.004 Pr 0.01), the following relations are recommended by Sleicher and Rouse (1975, Ref. 27) for
104 Re 106: 0.0119
0.0119
0.0134
0.0172
0.0199
0.0305
0.0380
0.0716 *Smooth surface. All values are for Re
and are calculated from Eq. 8–73. FIGURE 8–24
The friction factor is
minimum for a smooth pipe
and increases with roughness. Standard sizes for Schedule 40
steel pipes
Nominal
Size, in. Actual Inside
Diameter, in. ⁄8
⁄4
3
⁄8
1
⁄2
3
⁄4
1
11⁄2
2
21⁄2
3
5
10 0.269
0.364
0.493
0.622
0.824
1.049
1.610
2.067
2.469
3.068
5.047
10.02 1 constant:
constant: Nu
Nu 4.8
6.3 0.0156 Re0.85 Pr0.93
s
0.0167 Re0.85 Pr0.93
s (871)
(872) where the subscript s indicates that the Prandtl number is to be evaluated at
the surface temperature. Rough Surfaces
6 10 , Any irregularity or roughness on the surface disturbs the laminar sublayer,
and affects the flow. Therefore, unlike laminar flow, the friction factor and
the convection coefficient in turbulent flow are strong functions of surface
roughness.
The friction factor in fully developed turbulent flow depends on the
Reynolds number and the relative roughness /D. In 1939, C. F. Colebrook
(Ref. 3) combined all the friction factor data for transition and turbulent flow
in smooth as well as rough pipes into the following implicit relation known as
the Colebrook equation.
1 TABLE 8–2 1 Liquid metals, Ts
Liquid metals, q·s f 2.0 log /D
3.7 2.51
Re f (turbulent flow) (873) In 1944, L. F. Moody (Ref. 17) plotted this formula into the famous Moody
chart given in the Appendix. It presents the friction factors for pipe flow as a
function of the Reynolds number and /D over a wide range. For smooth
tubes, the agreement between the Petukhov and Colebrook equations is very
good. The friction factor is minimum for a smooth pipe (but still not zero because of the noslip condition), and increases with roughness (Fig. 8–24).
Although the Moody chart is developed for circular pipes, it can also be
used for noncircular pipes by replacing the diameter by the hydraulic diameter. At very large Reynolds numbers (to the right of the dashed line on the
chart) the friction factor curves corresponding to specified relative roughness
curves are nearly horizontal, and thus the friction factors are independent of
the Reynolds number. In calculations, we should make sure that we use the internal diameter of the pipe, which may be different than the nominal diameter.
For example, the internal diameter of a steel pipe whose nominal diameter is
1 in. is 1.049 in. (Table 8–2). cen58933_ch08.qxd 9/4/2002 11:29 AM Page 443 443
CHAPTER 8 Commercially available pipes differ from those used in the experiments in
that the roughness of pipes in the market is not uniform, and it is difficult to
give a precise description of it. Equivalent roughness values for some commercial pipes are given in Table 8–3, as well as on the Moody chart. But it
should be kept in mind that these values are for new pipes, and the relative
roughness of pipes may increase with use as a result of corrosion, scale
buildup, and precipitation. As a result, the friction factor may increase by a
factor of 5 to 10. Actual operating conditions must be considered in the design
of piping systems. Also, the Moody chart and its equivalent Colebrook equation involve several uncertainties (the roughness size, experimental error,
curve fitting of data, etc.), and thus the results obtained should not be treated
as “exact.” It is usually considered to be accurate to 15 percent over the entire range in the figure.
The Colebrook equation is implicit in f, and thus the determination of the
friction factor requires tedious iteration unless an equation solver is used.
An approximate explicit relation for f is given by S. E. Haaland in 1983
(Ref. 9) as
1
f 1.8 log 6.9
Re /D
3.7 1.11 (874) The results obtained from this relation are within 2 percent of those obtained
from Colebrook equation, and we recommend using this relation rather than
the Moody chart to avoid reading errors.
In turbulent flow, wall roughness increases the heat transfer coefficient h by
a factor of 2 or more [Dipprey and Sabersky (1963), Ref. 5]. The convection
heat transfer coefficient for rough tubes can be calculated approximately from
the Nusselt number relations such as Eq. 8–70 by using the friction factor
determined from the Moody chart or the Colebrook equation. However, this
approach is not very accurate since there is no further increase in h with f for
f 4fsmooth [Norris (1970), Ref. 20] and correlations developed specifically for
rough tubes should be used when more accuracy is desired. TABLE 8–3
Equivalent roughness values for
new commercial pipes*
Roughness,
Material
Glass, plastic
Concrete
Wood stave
Rubber,
smoothed
Copper or
brass tubing
Cast iron
Galvanized
iron
Wrought iron
Stainless steel
Commercial
steel ft mm 0 (smooth)
0.003–0.03 0.9–9
0.0016
0.5
0.000033 0.01 0.000005
0.00085 0.0015
0.26 0.0005
0.00015
0.000007 0.15
0.046
0.002 0.00015 0.045 *The uncertainty in these values can be as much
as 60 percent. Developing Turbulent Flow in the Entrance Region
The entry lengths for turbulent flow are typically short, often just 10 tube
diameters long, and thus the Nusselt number determined for fully developed
turbulent flow can be used approximately for the entire tube. This simple approach gives reasonable results for pressure drop and heat transfer for long
tubes and conservative results for short ones. Correlations for the friction and
heat transfer coefficients for the entrance regions are available in the literature
for better accuracy. r
0 Turbulent Flow in Noncircular Tubes
The velocity and temperature profiles in turbulent flow are nearly straight
lines in the core region, and any significant velocity and temperature gradients
occur in the viscous sublayer (Fig. 8–25). Despite the small thickness of
laminar sublayer (usually much less than 1 percent of the pipe diameter), the
characteristics of the flow in this layer are very important since they set the
stage for flow in the rest of the pipe. Therefore, pressure drop and heat transfer characteristics of turbulent flow in tubes are dominated by the very thin (r)
Turbulent layer
Overlap layer
Laminar sublayer FIGURE 8–25
In turbulent flow, the velocity
profile is nearly a straight line in the
core region, and any significant
velocity gradients occur in the
viscous sublayer. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 444 444
HEAT TRANSFER viscous sublayer next to the wall surface, and the shape of the core region is
not of much significance. Consequently, the turbulent flow relations given
above for circular tubes can also be used for noncircular tubes with reasonable
accuracy by replacing the diameter D in the evaluation of the Reynolds number by the hydraulic diameter Dh 4Ac /p. Flow through Tube Annulus
Tube Di Do Annulus Some simple heat transfer equipments consist of two concentric tubes, and are
properly called doubletube heat exchangers (Fig. 8–26). In such devices, one
fluid flows through the tube while the other flows through the annular space.
The governing differential equations for both flows are identical. Therefore,
steady laminar flow through an annulus can be studied analytically by using
suitable boundary conditions.
Consider a concentric annulus of inner diameter Di and outer diameter Do.
The hydraulic diameter of annulus is FIGURE 8–26
A doubletube heat exchanger that
consists of two concentric tubes. TABLE 8–4
Nusselt number for fully developed
laminar flow in an annulus with
one surface isothermal and the
other adiabatic (Kays and Perkins,
Ref. 14)
Di /Do Nui
—
17.46
11.56
7.37
5.74
4.86 3.66
4.06
4.11
4.23
4.43
4.86 4 (D2
o
(Do 4Ac
p Nui hi Dh
k (b) Roughened surface Roughness FIGURE 8–27
Tube surfaces are often roughened,
corrugated, or finned in order to
enhance convection heat transfer. Do Di (875) and Nuo ho Dh
k (876) For fully developed turbulent flow, the inner and outer convection coefficients are approximately equal to each other, and the tube annulus can be
treated as a noncircular duct with a hydraulic diameter of Dh Do Di. The
Nusselt number in this case can be determined from a suitable turbulent flow
relation such as the Gnielinski equation. To improve the accuracy of Nusselt
numbers obtained from these relations for annular flow, Petukhov and Roizen
(1964, Ref. 22) recommend multiplying them by the following correction factors when one of the tube walls is adiabatic and heat transfer is through the
other wall:
0.86 Di
Do 0.16 Fi 0.86 Di
Do 0.16 Fo (a) Finned surface Fin D2)/4
i
Di) Annular flow is associated with two Nusselt numbers—Nui on the inner
tube surface and Nuo on the outer tube surface—since it may involve heat
transfer on both surfaces. The Nusselt numbers for fully developed laminar
flow with one surface isothermal and the other adiabatic are given in
Table 8–4. When Nusselt numbers are known, the convection coefficients for
the inner and the outer surfaces are determined from Nuo 0
0.05
0.10
0.25
0.50
1.00 Dh (outer wall adiabatic) (877) (inner wall adiabatic) (878) Heat Transfer Enhancement
Tubes with rough surfaces have much higher heat transfer coefficients than
tubes with smooth surfaces. Therefore, tube surfaces are often intentionally roughened, corrugated, or finned in order to enhance the convection
heat transfer coefficient and thus the convection heat transfer rate (Fig. 8–27).
Heat transfer in turbulent flow in a tube has been increased by as much as cen58933_ch08.qxd 9/4/2002 11:29 AM Page 445 445
CHAPTER 8 400 percent by roughening the surface. Roughening the surface, of course,
also increases the friction factor and thus the power requirement for the pump
or the fan.
The convection heat transfer coefficient can also be increased by inducing
pulsating flow by pulse generators, by inducing swirl by inserting a twisted
tape into the tube, or by inducing secondary flows by coiling the tube.
EXAMPLE 8–4 Pressure Drop in a Water Pipe Water at 60°F (
62.36 lbm/ft3 and
2.713 lbm/ft h) is flowing steadily
in a 2in.diameter horizontal pipe made of stainless steel at a rate of 0.2 ft3/s
(Fig. 8–28). Determine the pressure drop and the required pumping power input for flow through a 200ftlong section of the pipe. 0.2 ft3/s
water 2 in. 200 ft FIGURE 8–28
Schematic for Example 8–4. SOLUTION The flow rate through a specified water pipe is given. The pressure
drop and the pumping power requirements are to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects
are negligible, and thus the flow is fully developed. 3 The pipe involves no components such as bends, valves, and connectors. 4 The piping section involves
no work devices such as a pump or a turbine.
Properties The density and dynamic viscosity of water are given by
62.36
lbm/ft3 and
2.713 lbm/ft h 0.0007536 lbm/ft s, respectively.
Analysis First we calculate the mean velocity and the Reynolds number to
determine the flow regime:
·
V
Ac
Re ·
V
0.2 ft3/s
9.17 ft/s
2
D /4
(2/12 ft)2/4
D (62.36 lbm/ft3)(9.17 ft/s)(2/12 ft) 3600 s
2.713 lbm/ft · h
1h 126,400 which is greater than 10,000. Therefore, the flow is turbulent. The relative
roughness of the pipe is /D 0.000007 ft
2/12 ft 0.000042 The friction factor corresponding to this relative roughness and the Reynolds
number can simply be determined from the Moody chart. To avoid the reading
error, we determine it from the Colebrook equation: 1 2.0 log f /D
3.7 2.51
Re f → 1
f 2.0 log 0.000042
3.7 2.51
126,400 f Using an equation solver or an iterative scheme, the friction factor is determined to be f 0.0174. Then the pressure drop and the required power input
become
2 P ·
Wpump 200 ft (62.36 lbm/ft3)(9.17 ft/s)2
1 lbf
L
0.0174
D2
2
2/12 ft
32.2 lbm · ft/s2
1700 lbf/ft2 11.8 psi
·
1W
V P (0.2 ft3/s)(1700 lbf/ft2)
461 W
0.737 lbf · ft/s
f cen58933_ch08.qxd 9/4/2002 11:29 AM Page 446 446
HEAT TRANSFER Therefore, power input in the amount of 461 W is needed to overcome the frictional losses in the pipe.
Discussion The friction factor also could be determined easily from the explicit
Haaland relation. It would give f
0.0172, which is sufficiently close to
0.0174. Also, the friction factor corresponding to
0 in this case is 0.0171,
which indicates that stainless steel pipes can be assumed to be smooth with
negligible error. EXAMPLE 8–5
·
qs = constant
15°C Water
D = 3 cm 65°C Heating of Water by Resistance Heaters in a Tube Water is to be heated from 15°C to 65°C as it flows through a 3cminternaldiameter 5mlong tube (Fig. 8–29). The tube is equipped with an electric resistance heater that provides uniform heating throughout the surface of the
tube. The outer surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to the water in the tube.
If the system is to provide hot water at a rate of 10 L/min, determine the power
rating of the resistance heater. Also, estimate the inner surface temperature of
the pipe at the exit. 5m FIGURE 8–29
Schematic for Example 8–5. SOLUTION Water is to be heated in a tube equipped with an electric resistance heater on its surface. The power rating of the heater and the inner surface
temperature are to be determined.
Assumptions 1 Steady flow conditions exist. 2 The surface heat flux is uniform.
3 The inner surfaces of the tube are smooth.
Properties The properties of water at the bulk mean temperature of Tb
(Ti Te)/2 (15 65)/2 40°C are (Table A9).
k 992.1 kg/m3
0.631 W/m °C
/
0.658 10 Cp
Pr
6 4179 J/kg °C
4.32 m2/s Analysis The cross sectional and heat transfer surface areas are Ac
As D2 1 (0.03 m)2 7.069
4
pL
DL
(0.03 m)(5 m) 1
4 ·
The volume flow rate of water is given as V
the mass flow rate becomes ·
m ·
V (992.1 kg/m3)(0.01 m3/min) 10 4 m2
0.471 m2
0.01 m3/min. Then 10 L/min 9.921 kg/min 0.1654 kg/s To heat the water at this mass flow rate from 15°C to 65°C, heat must be supplied to the water at a rate of ·
Q ·
m Cp(Te Ti)
(0.1654 kg/s)(4.179 kJ/kg °C)(65
34.6 kJ/s 34.6 kW 15)°C cen58933_ch08.qxd 9/4/2002 11:29 AM Page 447 447
CHAPTER 8 All of this energy must come from the resistance heater. Therefore, the power
rating of the heater must be 34.6 kW.
The surface temperature Ts of the tube at any location can be determined
from ·
qs h(Ts Tm) → Ts Tm ·
qs
h where h is the heat transfer coefficient and Tm is the mean temperature of the
fluid at that location. The surface heat flux is constant in this case, and its
value can be determined from ·
qs ·
Q
As 34.6 kW
0.471 m2 73.46 kW/m2 To determine the heat transfer coefficient, we first need to find the mean velocity of water and the Reynolds number: ·
V
Ac 0.010 m3/min
14.15 m/min
7.069 10 4 m2
(0.236 m/s)(0.03 m)
mD
10,760
0.658 10 6 m2/s m Re 0.236 m/s which is greater than 10,000. Therefore, the flow is turbulent and the entry
length is roughly Lh Lt 10D 10 0.03 0.3 m which is much shorter than the total length of the pipe. Therefore, we can assume fully developed turbulent flow in the entire pipe and determine the Nusselt number from hD
k Nu 0.023 Re0.8 Pr0.4 0.023(10,760)0.8 (4.34)0.4 69.5 Then, k
Nu
D h 0.631 W/m · °C
(69.5)
0.03 m 1462 W/m2 °C and the surface temperature of the pipe at the exit becomes Ts Tm ·
qs
h 65°C 73,460 W/m2
1462 W/m2 · °C 115°C Discussion Note that the inner surface temperature of the pipe will be 50°C
higher than the mean water temperature at the pipe exit. This temperature difference of 50°C between the water and the surface will remain constant
throughout the fully developed flow region. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 448 448
HEAT TRANSFER EXAMPLE 8–6 Ts = 60°C
0.2 m Te Air
1 atm
80°C Heat Loss from the Ducts of a Heating System Hot air at atmospheric pressure and 80°C enters an 8–mlong uninsulated
square duct of cross section 0.2 m 0.2 m that passes through the attic of a
house at a rate of 0.15 m3/s (Fig. 8–30). The duct is observed to be nearly
isothermal at 60°C. Determine the exit temperature of the air and the rate of
heat loss from the duct to the attic space. 0.2 m
8m FIGURE 8–30 SOLUTION Heat loss from uninsulated square ducts of a heating system in
the attic is considered. The exit temperature and the rate of heat loss are to be
determined. Schematic for Example 8–6.
Assumptions 1 Steady operating conditions exist. 2 The inner surfaces of the
duct are smooth. 3 Air is an ideal gas. Properties We do not know the exit temperature of the air in the duct, and thus
we cannot determine the bulk mean temperature of air, which is the temperature at which the properties are to be determined. The temperature of air at the
inlet is 80°C and we expect this temperature to drop somewhat as a result of
heat loss through the duct whose surface is at 60°C. At 80°C and 1 atm we
read (Table A15) k 0.9994 kg/m3
0.02953 W/m °C
2.097 10 5 m2/s Cp
Pr 1008 J/kg °C
0.7154 Analysis The characteristic length (which is the hydraulic diameter), the mean
velocity, and the Reynolds number in this case are Dh
m Re 4Ac 4a2
a 0.2 m
p
4a
·
V
0.15 m3/s
3.75 m/s
Ac
(0.2 m)2
(3.75 m/s)(0.2 m)
m Dh
2.097 10 5 m2/s 35,765 which is greater than 10,000. Therefore, the flow is turbulent and the entry
lengths in this case are roughly Lh Lt 10D 10 0.2 m 2m which is much shorter than the total length of the duct. Therefore, we can
assume fully developed turbulent flow in the entire duct and determine the
Nusselt number from Nu hDh
k 0.023 Re0.8 Pr0.3 0.023(35,765)0.8 (0.7154)0.3 91.4 cen58933_ch08.qxd 9/4/2002 11:29 AM Page 449 449
CHAPTER 8 Then, 0.02953 W/m · °C
k
Nu
(91.4) 13.5 W/m2 °C
Dh
0.2 m
pL 4aL 4 (0.2 m)(8 m) 6.4 m2
·
V (1.009 kg/m3)(0.15 m3/s) 0.151 kg/s h
As
·
m Next, we determine the exit temperature of air from Te Ts ·
Ti) exp ( hAs /m Cp) (Ts 60°C [(60 (13.5 W/m2 · °C)(6.4 m2)
(0.151 kg/s)(1008 J/kg · °C) 80)°C] exp 71.3°C
Then the logarithmic mean temperature difference and the rate of heat loss
from the air become Ti Te
80 71.3
15.2°C
Ts Te
60 71.3
ln
ln
60 80
Ts Ti
hAs Tln (13.5 W/m2 °C)(6.4 m2)( 15.2°C) Tln
·
Q 1313 W Therefore, air will lose heat at a rate of 1313 W as it flows through the duct in
the attic.
Discussion The average fluid temperature is (80
71.3)/2
75.7°C, which
is sufficiently close to 80°C at which we evaluated the properties of air. Therefore, it is not necessary to reevaluate the properties at this temperature and to
repeat the calculations. SUMMARY
Internal flow is characterized by the fluid being completely
confined by the inner surfaces of the tube. The mean velocity
and mean temperature for a circular tube of radius R are expressed as
m 2
R2 R (r, x)rdr and 0 Tm 2
2
mR R Trdr
0 dynamic entry length Lh. The region beyond the entrance
region in which the velocity profile is fully developed is the
hydrodynamically fully developed region. The length of the region of flow over which the thermal boundary layer develops
and reaches the tube center is the thermal entry length Lt. The
region in which the flow is both hydrodynamically and thermally developed is the fully developed flow region. The entry
lengths are given by The Reynolds number for internal flow and the hydraulic diameter are defined as
Re mD mD and Dh Lh, laminar 0.05 Re D
Lt, laminar 0.05 Re Pr D Pr Lh, laminar
Lh, turbulent Lt, turbulent 10D 4Ac
p The flow in a tube is laminar for Re
2300, turbulent for
Re 10,000, and transitional in between.
The length of the region from the tube inlet to the point at
which the boundary layer merges at the centerline is the hydro ·
For q s constant, the rate of heat transfer is expressed as
·
·
Q q·s As m Cp(Te Ti) cen58933_ch08.qxd 9/4/2002 11:29 AM Page 450 450
HEAT TRANSFER For Ts For fully developed turbulent flow with smooth surfaces,
we have constant, we have
·
·
Q hAs Tln m Cp(Te Ti)
·
Te Ts (Ts Ti)exp( hAs /m Cp)
Ti Te
Ti
Te
Tln
ln[(Ts Te)/(Ts Ti)] ln( Te / Ti) f
Nu
Nu The pressure drop and required pumping power for a volume
·
flow rate of V are
2 m
L
D2 P ·
Wpump and Nu ·
VP For fully developed laminar flow in a circular pipe, we have:
(r)
f
·
V 2 m 64
D 1 m ave Ac r2
R2 max 64
Re
PR2
R2
8L r2
R2 1 R4 P
8L ·
Circular tube, laminar (q s constant): Nu Circular tube, laminar (Ts constant): Nu R4 P
128 L
hD
k
hD
k 3.66 Nu Circular tube: Nu Parallel plates: Nu 0.065(D/L) Re Pr
1 0.04[(D/L) Re Pr]2/3
Re Pr D 1/3 b 0.14
1.86
s
L
0.03(Dh /L) Re Pr
7.54
1 0.016[(Dh /L) Re Pr]2/3
3.66 104 Re 106 0.7 Pr 160
Re 10,000
0.023 Re0.8 Prn with n 0.4 for heating and 0.3 for
cooling of fluid
( f/8)(Re 1000) Pr
0.5 Pr 2000
1 12.7( f/8)0.5 (Pr2/3 1) 3 103 Re 5 106
0.023 Re0.8 Pr1/3 constant:
constant: Nu
Nu 4.8
6.3 0.0156 Re0.85 Pr0.93
s
0.0167 Re0.85 Pr0.93
s For fully developed turbulent flow with rough surfaces, the
friction factor f is determined from the Moody chart or
1 4.36 2 The fluid properties are evaluated at the bulk mean fluid
temperature Tb
(Ti
Te)/2. For liquid metal flow in the
range of 104 Re 106 we have:
Ts
q·s For developing laminar flow in the entrance region with constant surface temperature, we have
Circular tube: Nu (0.790 ln Re 1.64)
0.125f Re Pr1/3 f 2.0 log /D
3.7 2.51
Re f 1.8 log 6.9
Re /D
3.7 1.11 For a concentric annulus, the hydraulic diameter is Dh
Do Di, and the Nusselt numbers are expressed as
Nui hi Dh
k and Nuo ho Dh
k where the values for the Nusselt numbers are given in
Table 8–4. REFERENCES AND SUGGESTED READING
1. M. S. Bhatti and R. K. Shah. “Turbulent and Transition
Flow Convective Heat Transfer in Ducts.” In Handbook
of SinglePhase Convective Heat Transfer, ed. S. Kakaç,
R. K. Shah, and W. Aung. New York: Wiley
Interscience, 1987. 2. A. P. Colburn. Transactions of the AIChE 26 (1933),
p. 174.
3. C. F. Colebrook. “Turbulent flow in Pipes, with Particular
Reference to the Transition between the Smooth and cen58933_ch08.qxd 9/4/2002 11:29 AM Page 451 451
CHAPTER 8 Rough Pipe Laws.” Journal of the Institute of Civil
Engineers London. 11 (1939), pp. 133–156. Surfaces.” In Augmentation of Convective Heat Transfer,
ed. A. E. Bergles and R. L. Webb. New York:
ASME, 1970. 4. R. G. Deissler. “Analysis of Turbulent Heat Transfer and
Flow in the Entrance Regions of Smooth Passages.” 1953.
Referred to in Handbook of SinglePhase Convective
Heat Transfer, ed. S. Kakaç, R. K. Shah, and W. Aung.
New York: Wiley Interscience, 1987. 21. B. S. Petukhov. “Heat Transfer and Friction in Turbulent
Pipe Flow with Variable Physical Properties.” In
Advances in Heat Transfer, ed. T. F. Irvine and J. P.
Hartnett, Vol. 6. New York: Academic Press, 1970. 5. D. F. Dipprey and D. H. Sabersky. “Heat and Momentum
Transfer in Smooth and Rough Tubes at Various Prandtl
Numbers.” International Journal of Heat Mass Transfer 6
(1963), pp. 329–353. 22. B. S. Petukhov and L. I. Roizen. “Generalized
Relationships for Heat Transfer in a Turbulent Flow of
a Gas in Tubes of Annular Section.” High Temperature
(USSR) 2 (1964), pp. 65–68. 6. F. W. Dittus and L. M. K. Boelter. University of
California Publications on Engineering 2 (1930), p. 433. 23. O. Reynolds. “On the Experimental Investigation of the
Circumstances Which Determine Whether the Motion
of Water Shall Be Direct or Sinuous, and the Law of
Resistance in Parallel Channels.” Philosophical
Transactions of the Royal Society of London 174 (1883),
pp. 935–982. 7. D. K. Edwards, V. E. Denny, and A. F. Mills. Transfer
Processes. 2nd ed. Washington, DC: Hemisphere, 1979.
8. V. Gnielinski. “New Equations for Heat and Mass
Transfer in Turbulent Pipe and Channel Flow.”
International Chemical Engineering 16 (1976),
pp. 359–368.
9. S. E. Haaland. “Simple and Explicit Formulas for the
Friction Factor in Turbulent Pipe Flow.” Journal of Fluids
Engineering (March 1983), pp. 89–90.
10. J. P. Holman. Heat Transfer. 8th ed. New York:
McGrawHill, 1997.
11. F. P. Incropera and D. P. DeWitt. Introduction to Heat
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New York: McGrawHill, 1979.
25. R. K. Shah and M. S. Bhatti. “Laminar Convective
Heat Transfer in Ducts.” In Handbook of SinglePhase
Convective Heat Transfer, ed. S. Kakaç, R. K. Shah, and
W. Aung. New York: Wiley Interscience, 1987.
26. E. N. Sieder and G. E. Tate. “Heat Transfer and Pressure
Drop of Liquids in Tubes.” Industrial Engineering
Chemistry 28 (1936), pp. 1429–1435. 12. S. Kakaç, R. K. Shah, and W. Aung, eds. Handbook of
SinglePhase Convective Heat Transfer. New York:
Wiley Interscience, 1987. 27. C. A. Sleicher and M. W. Rouse. “A Convenient
Correlation for Heat Transfer to Constant and Variable
Property Fluids in Turbulent Pipe Flow.” International
Journal of Heat Mass Transfer 18 (1975), pp. 1429–1435. 13. W. M. Kays and M. E. Crawford. Convective Heat and
Mass Transfer. 3rd ed. New York: McGrawHill, 1993. 28. N. V. Suryanarayana. Engineering Heat Transfer. St. Paul,
MN: West, 1995. 14. W. M. Kays and H. C. Perkins. Chapter 7. In Handbook of
Heat Transfer, ed. W. M. Rohsenow and J. P. Hartnett.
New York: McGrawHill, 1972. 29. F. M. White. Heat and Mass Transfer. Reading, MA:
AddisonWesley, 1988. 15. F. Kreith and M. S. Bohn. Principles of Heat Transfer.
6th ed. Pacific Grove, CA: Brooks/Cole, 2001.
16. A. F. Mills. Basic Heat and Mass Transfer. 2nd ed.
Upper Saddle River, NJ: Prentice Hall, 1999.
17. L. F. Moody. “Friction Factors for Pipe Flows.”
Transactions of the ASME 66 (1944), pp. 671–684.
18. M. Molki and E. M. Sparrow. “An Empirical Correlation
for the Average Heat Transfer Coefficient in Circular
Tubes.” Journal of Heat Transfer 108 (1986),
pp. 482–484.
19. B. R. Munson, D. F. Young, and T. Okiishi. Fundamentals
of Fluid Mechanics. 4th ed. New York: Wiley, 2002.
20. R. H. Norris. “Some Simple Approximate Heat Transfer
Correlations for Turbulent Flow in Ducts with Rough 30. S. Whitaker. “Forced Convection Heat Transfer
Correlations for Flow in Pipes, Past Flat Plates, Single
Cylinders, and for Flow in Packed Beds and Tube
Bundles.” AIChE Journal 18 (1972), pp. 361–371.
31. W. Zhiqing. “Study on Correction Coefficients of
Laminar and Turbulent Entrance Region Effects in
Round Pipes.” Applied Mathematical Mechanics 3
(1982), p. 433. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 452 452
HEAT TRANSFER PROBLEMS*
General Flow Analysis
8–1C Why are liquids usually transported in circular pipes? 8–2C Show that the Reynolds number for flow in a circular
·
tube of diameter D can be expressed as Re 4m /( D ).
8–3C Which fluid at room temperature requires a larger
pump to move at a specified velocity in a given tube: water or
engine oil? Why?
8–4C What is the generally accepted value of the Reynolds
number above which the flow in smooth pipes is turbulent?
8–5C What is hydraulic diameter? How is it defined? What
is it equal to for a circular tube of diameter?
8–6C How is the hydrodynamic entry length defined for flow
in a tube? Is the entry length longer in laminar or turbulent
flow?
8–7C Consider laminar flow in a circular tube. Will the
friction factor be higher near the inlet of the tube or near the
exit? Why? What would your response be if the flow were
turbulent?
8–8C How does surface roughness affect the pressure drop in
a tube if the flow is turbulent? What would your response be if
the flow were laminar?
8–9C How does the friction factor f vary along the flow direction in the fully developed region in (a) laminar flow and
(b) turbulent flow?
8–10C What fluid property is responsible for the development of the velocity boundary layer? For what kinds of fluids
will there be no velocity boundary layer in a pipe?
8–11C What is the physical significance of the number of
·
transfer units NTU hA/m Cp? What do small and large NTU
values tell about a heat transfer system?
8–12C What does the logarithmic mean temperature difference represent for flow in a tube whose surface temperature is
constant? Why do we use the logarithmic mean temperature
instead of the arithmetic mean temperature?
8–13C How is the thermal entry length defined for flow in a
tube? In what region is the flow in a tube fully developed? *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. 8–14C Consider laminar forced convection in a circular tube.
Will the heat flux be higher near the inlet of the tube or near the
exit? Why?
8–15C Consider turbulent forced convection in a circular
tube. Will the heat flux be higher near the inlet of the tube or
near the exit? Why?
8–16C In the fully developed region of flow in a circular
tube, will the velocity profile change in the flow direction?
How about the temperature profile?
8–17C Consider the flow of oil in a tube. How will the
hydrodynamic and thermal entry lengths compare if the flow is
laminar? How would they compare if the flow were turbulent?
8–18C Consider the flow of mercury (a liquid metal) in a
tube. How will the hydrodynamic and thermal entry lengths
compare if the flow is laminar? How would they compare if the
flow were turbulent?
8–19C What do the mean velocity m and the mean temperature Tm represent in flow through circular tubes of constant
diameter?
8–20C Consider fluid flow in a tube whose surface temperature remains constant. What is the appropriate temperature difference for use in Newton’s law of cooling with an average
heat transfer coefficient?
8–21 Air enters a 20cmdiameter 12mlong underwater
duct at 50°C and 1 atm at a mean velocity of 7 m/s, and is
cooled by the water outside. If the average heat transfer coefficient is 85 W/m2 °C and the tube temperature is nearly equal
to the water temperature of 5°C, determine the exit temperature
of air and the rate of heat transfer.
8–22 Cooling water available at 10°C is used to condense
steam at 30°C in the condenser of a power plant at a rate of
0.15 kg/s by circulating the cooling water through a bank of
5mlong 1.2cminternaldiameter thin copper tubes. Water
enters the tubes at a mean velocity of 4 m/s, and leaves at a
temperature of 24C. The tubes are nearly isothermal at 30°C.
Determine the average heat transfer coefficient between the
water and the tubes, and the number of tubes needed to achieve
the indicated heat transfer rate in the condenser.
8–23 Repeat Problem 8–22 for steam condensing at a rate of
0.60 kg/s.
8–24 Combustion gases passing through a 3cminternaldiameter circular tube are used to vaporize waste water at atmospheric pressure. Hot gases enter the tube at 115 kPa and
250°C at a mean velocity of 5 m/s, and leave at 150°C. If the
average heat transfer coefficient is 120 W/m2 °C and the inner surface temperature of the tube is 110°C, determine (a) the
tube length and (b) the rate of evaporation of water.
8–25 Repeat Problem 8–24 for a heat transfer coefficient of
60 W/m2 °C. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 453 453
CHAPTER 8 Laminar and Turbulent Flow in Tubes
8–26C How is the friction factor for flow in a tube related to
the pressure drop? How is the pressure drop related to the
pumping power requirement for a given mass flow rate?
8–27C Someone claims that the shear stress at the center of
a circular pipe during fully developed laminar flow is zero.
Do you agree with this claim? Explain. terline) is measured to be 6 m/s. Determine the velocity at the
Answer: 8 m/s
center of the pipe.
8–37 The velocity profile in fully developed laminar flow in
a circular pipe of inner radius R 2 cm, in m/s, is given by
(r) 4(1 r2/R2). Determine the mean and maximum velocities in the pipe, and the volume flow rate. ( r2
(r) = 4 1 – ––
R2 8–28C Someone claims that in fully developed turbulent
flow in a tube, the shear stress is a maximum at the tube surface. Do you agree with this claim? Explain. ) R = 2 cm 8–29C Consider fully developed flow in a circular pipe with
negligible entrance effects. If the length of the pipe is doubled,
the pressure drop will (a) double, (b) more than double, (c) less
than double, (d) reduce by half, or (e) remain constant. FIGURE P8–37 8–30C Someone claims that the volume flow rate in a circular pipe with laminar flow can be determined by measuring the
velocity at the centerline in the fully developed region, multiplying it by the cross sectional area, and dividing the result by
2. Do you agree? Explain. 8–39 Water at 10°C (
999.7 kg/m3 and
1.307 10 3
kg/m s) is flowing in a 0.20cmdiameter 15mlong pipe
steadily at an average velocity of 1.2 m/s. Determine (a) the
pressure drop and (b) the pumping power requirement to overcome this pressure drop. 8–31C Someone claims that the average velocity in a circular pipe in fully developed laminar flow can be determined by
simply measuring the velocity at R/2 (midway between the
wall surface and the centerline). Do you agree? Explain.
8–32C Consider fully developed laminar flow in a circular
pipe. If the diameter of the pipe is reduced by half while the
flow rate and the pipe length are held constant, the pressure
drop will (a) double, (b) triple, (c) quadruple, (d) increase by a
factor of 8, or (e) increase by a factor of 16.
8–33C Consider fully developed laminar flow in a circular
pipe. If the viscosity of the fluid is reduced by half by heating
while the flow rate is held constant, how will the pressure drop
change?
8–34C How does surface roughness affect the heat transfer
in a tube if the fluid flow is turbulent? What would your response be if the flow in the tube were laminar?
8–35 Water at 15°C (
999.1 kg/m3 and
1.138 10 3
kg/m s) is flowing in a 4cmdiameter and 30m long horizontal pipe made of stainless steel steadily at a rate of 5 L/s.
Determine (a) the pressure drop and (b) the pumping power
requirement to overcome this pressure drop.
5 L/s 8–38 Answers: (a) 188 kPa,(b) 0.71 W 8–40 Water is to be heated from 10°C to 80°C as it flows
through a 2cminternaldiameter, 7mlong tube. The tube is
equipped with an electric resistance heater, which provides
uniform heating throughout the surface of the tube. The outer
surface of the heater is well insulated, so that in steady operation all the heat generated in the heater is transferred to
the water in the tube. If the system is to provide hot water at a
rate of 8 L/min, determine the power rating of the resistance
heater. Also, estimate the inner surface temperature of the pipe
at the exit.
8–41 Hot air at atmospheric pressure and 85°C enters a
10mlong uninsulated square duct of cross section 0.15 m
0.15 m that passes through the attic of a house at a rate of
0.10 m3/s. The duct is observed to be nearly isothermal at
70°C. Determine the exit temperature of the air and the rate of
heat loss from the duct to the air space in the attic.
Answers: 75.7°C, 941 W
Attic
space Air
85°C 70°C
0.1 m3/s 4 cm 30 m FIGURE P8–35
8–36 In fully developed laminar flow in a circular pipe, the
velocity at R/2 (midway between the wall surface and the cen Repeat Problem 8–37 for a pipe of inner radius 5 cm. FIGURE P8–41
8–42 Reconsider Problem 8–41. Using EES (or other)
software, investigate the effect of the volume
flow rate of air on the exit temperature of air and the rate of
heat loss. Let the flow rate vary from 0.05 m3/s to 0.15 m3/s. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 454 454
HEAT TRANSFER Plot the exit temperature and the rate of heat loss as a function
of flow rate, and discuss the results. Parabolic
solar collector 8–43 Consider an air solar collector that is 1 m wide and 5 m
long and has a constant spacing of 3 cm between the glass
cover and the collector plate. Air enters the collector at 30°C at
a rate of 0.15 m3/s through the 1mwide edge and flows along
the 5mlong passage way. If the average temperatures of the
glass cover and the collector plate are 20°C and 60°C, respectively, determine (a) the net rate of heat transfer to the air in the
collector and (b) the temperature rise of air as it flows through
the collector. Air
30°C
0.15 m3/s 5m Insulation Collector
plate, 60°C Glass
cover
20°C FIGURE P8–43
8–44 Consider the flow of oil at 10°C in a 40cmdiameter
pipeline at an average velocity of 0.5 m/s. A 300mlong section of the pipeline passes through icy waters of a lake at 0°C.
Measurements indicate that the surface temperature of the pipe
is very nearly 0°C. Disregarding the thermal resistance of the
pipe material, determine (a) the temperature of the oil when the
pipe leaves the lake, (b) the rate of heat transfer from the oil,
and (c) the pumping power required to overcome the pressure
losses and to maintain the flow oil in the pipe.
8–45 Consider laminar flow of a fluid through a square channel maintained at a constant temperature. Now the mean velocity of the fluid is doubled. Determine the change in the
pressure drop and the change in the rate of heat transfer between the fluid and the walls of the channel. Assume the flow
regime remains unchanged.
8–46 Repeat Problem 8–45 for turbulent flow. 8–47E The hot water needs of a household are to be met by
heating water at 55°F to 200°F by a parabolic solar collector at
a rate of 4 lbm/s. Water flows through a 1.25in.diameter thin
aluminum tube whose outer surface is blackanodized in order
to maximize its solar absorption ability. The centerline of the
tube coincides with the focal line of the collector, and a glass Water
200°F
4 lbm/s Glass tube
Water tube FIGURE P8–47E
sleeve is placed outside the tube to minimize the heat losses. If
solar energy is transferred to water at a net rate of 350 Btu/h
per ft length of the tube, determine the required length of the
parabolic collector to meet the hot water requirements of this
house. Also, determine the surface temperature of the tube at
the exit.
8–48 A 15cm 20cm printed circuit board whose components are not allowed to come into direct contact with air for
reliability reasons is to be cooled by passing cool air through a
20cmlong channel of rectangular cross section 0.2 cm 14
cm drilled into the board. The heat generated by the electronic
components is conducted across the thin layer of the board to
the channel, where it is removed by air that enters the channel
at 15°C. The heat flux at the top surface of the channel can be
considered to be uniform, and heat transfer through other surfaces is negligible. If the velocity of the air at the inlet of the
channel is not to exceed 4 m/s and the surface temperature of
the channel is to remain under 50°C, determine the maximum
total power of the electronic components that can safely be
mounted on this circuit board. Air
15°C
Air channel
0.2 cm × 14 cm Electronic
components FIGURE P8–48
8–49 Repeat Problem 8–48 by replacing air with helium,
which has six times the thermal conductivity of air.
8–50 Reconsider Problem 8–48. Using EES (or other)
software, investigate the effects of air velocity at
the inlet of the channel and the maximum surface temperature
on the maximum total power dissipation of electronic components. Let the air velocity vary from 1 m/s to 10 m/s and the
surface temperature from 30°C to 90°C. Plot the power dissipation as functions of air velocity and surface temperature, and
discuss the results. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 455 455
CHAPTER 8 8–51 Air enters a 7mlong section of a rectangular duct of
cross section 15 cm 20 cm at 50°C at an average velocity of
7 m/s. If the walls of the duct are maintained at 10°C, determine (a) the outlet temperature of the air, (b) the rate of heat
transfer from the air, and (c) the fan power needed to overcome
the pressure losses in this section of the duct.
Answers: (a) 32.8°C, (b) 3674 W, (c) 4.2 W 8–52 Reconsider Problem 8–51. Using EES (or other)
software, investigate the effect of air velocity on
the exit temperature of air, the rate of heat transfer, and the fan
power. Let the air velocity vary from 1 m/s to 10 m/s. Plot the
exit temperature, the rate of heat transfer, and the fan power as
a function of the air velocity, and discuss the results.
8–53 Hot air at 60°C leaving the furnace of a house enters a
12mlong section of a sheet metal duct of rectangular cross
section 20 cm 20 cm at an average velocity of 4 m/s. The
thermal resistance of the duct is negligible, and the outer surface of the duct, whose emissivity is 0.3, is exposed to the cold
air at 10°C in the basement, with a convection heat transfer coefficient of 10 W/m2 °C. Taking the walls of the basement to
be at 10°C also, determine (a) the temperature at which the hot
air will leave the basement and (b) the rate of heat loss from the
hot air in the duct to the basement.
10°C
ho = 10 W/ m2·°C
12 m Hot air
60°C
4 m /s Air duct
20 cm × 20 cm
ε = 0.3 mine (a) the exit temperature of air and (b) the highest component surface temperature in the duct.
8–56 Repeat Problem 8–55 for a circular horizontal duct of
15cm diameter.
8–57 Consider a hollowcore printed circuit board 12 cm
high and 18 cm long, dissipating a total of 20 W. The width of
the air gap in the middle of the PCB is 0.25 cm. The cooling air
enters the 12cmwide core at 32°C at a rate of 0.8 L/s. Assuming the heat generated to be uniformly distributed over the two
side surfaces of the PCB, determine (a) the temperature at
which the air leaves the hollow core and (b) the highest temperature on the inner surface of the core.
Answers: (a) 54.0°C, (b) 72.8°C 8–58 Repeat Problem 8–57 for a hollowcore PCB dissipating 35 W.
8–59E Water at 54°F is heated by passing it through 0.75in.internaldiameter thinwalled copper tubes. Heat is supplied to
the water by steam that condenses outside the copper tubes at
250°F. If water is to be heated to 140°F at a rate of 0.7 lbm/s,
determine (a) the length of the copper tube that needs to be
used and (b) the pumping power required to overcome pressure
losses. Assume the entire copper tube to be at the steam temperature of 250°F.
8–60 A computer cooled by a fan contains eight PCBs, each
dissipating 10 W of power. The height of the PCBs is 12 cm
and the length is 18 cm. The clearance between the tips of the
components on the PCB and the back surface of the adjacent
PCB is 0.3 cm. The cooling air is supplied by a 10W fan
mounted at the inlet. If the temperature rise of air as it flows
through the case of the computer is not to exceed 10°C, determine (a) the flow rate of the air that the fan needs to deliver,
(b) the fraction of the temperature rise of air that is due to the
heat generated by the fan and its motor, and (c) the highest
allowable inlet air temperature if the surface temperature of the
Air
outlet FIGURE P8–53
0.3 cm 8–54 Reconsider Problem 8–53. Using EES (or other)
software, investigate the effects of air velocity
and the surface emissivity on the exit temperature of air and the
rate of heat loss. Let the air velocity vary from 1 m/s to 10 m/s
and the emissivity from 0.1 to 1.0. Plot the exit temperature
and the rate of heat loss as functions of air velocity and emissivity, and discuss the results.
8–55 The components of an electronic system dissipating
90 W are located in a 1mlong horizontal duct whose cross
section is 16 cm
16 cm. The components in the duct are
cooled by forced air, which enters at 32°C at a rate of
0.65 m3/min. Assuming 85 percent of the heat generated inside
is transferred to air flowing through the duct and the remaining
15 percent is lost through the outer surfaces of the duct, deter 18 cm Air
inlet FIGURE P8–60 PCB, 10 W cen58933_ch08.qxd 9/4/2002 11:29 AM Page 456 456
HEAT TRANSFER components is not to exceed 70°C anywhere in the system. Use
air properties at 25°C. Air, 0.27 m3/s
10°C, 95 kPa Review Problems
8–61 A geothermal district heating system involves the transport of geothermal water at 110°C from a geothermal well to a
city at about the same elevation for a distance of 12 km at a rate
of 1.5 m3/s in 60cmdiameter stainless steel pipes. The fluid
pressures at the wellhead and the arrival point in the city are to
be the same. The minor losses are negligible because of the
large lengthtodiameter ratio and the relatively small number
of components that cause minor losses. (a) Assuming the
pumpmotor efficiency to be 65 percent, determine the electric
power consumption of the system for pumping. (b) Determine
the daily cost of power consumption of the system if the unit
cost of electricity is $0.06/kWh. (c) The temperature of geothermal water is estimated to drop 0.5°C during this long flow.
Determine if the frictional heating during flow can make up for
this drop in temperature.
8–62 Repeat Problem 8–61 for cast iron pipes of the same
diameter. 20
cm Air
Compressor
150 hp FIGURE P8–65
water, determine the outlet temperature of air as it leaves the
underwater portion of the duct. Also, for an overall fan efficiency of 55 percent, determine the fan power input needed to
overcome the flow resistance in this section of the duct. 8–63 The velocity profile in fully developed laminar flow in
a circular pipe, in m/s, is given by (r) 6(1 100r2) where
r is the radial distance from the centerline of the pipe in m. Determine (a) the radius of the pipe, (b) the mean velocity
through the pipe, and (c) the maximum velocity in the pipe. Air
25°C, 3 m/s 8–64E The velocity profile in fully developed laminar flow
of water at 40°F in a 80ftlong horizontal circular pipe, in ft/s,
is given by (r) 0.8(1 625r2) where r is the radial distance from the centerline of the pipe in ft. Determine (a) the
volume flow rate of water through the pipe, (b) the pressure
drop across the pipe, and (c) the useful pumping power required to overcome this pressure drop.
8–65 The compressed air requirements of a manufacturing
facility are met by a 150hp compressor located in a room that
is maintained at 20°C. In order to minimize the compressor
work, the intake port of the compressor is connected to the
outside through an 11mlong, 20cmdiameter duct made of
thin aluminum sheet. The compressor takes in air at a rate of
0.27 m3/s at the outdoor conditions of 10°C and 95 kPa. Disregarding the thermal resistance of the duct and taking the heat
transfer coefficient on the outer surface of the duct to be 10
W/m2 °C, determine (a) the power used by the compressor to
overcome the pressure drop in this duct, (b) the rate of heat
transfer to the incoming cooler air, and (c) the temperature rise
of air as it flows through the duct.
8–66 A house built on a riverside is to be cooled in summer
by utilizing the cool water of the river, which flows at an average temperature of 15°C. A 15mlong section of a circular
duct of 20cm diameter passes through the water. Air enters the
underwater section of the duct at 25°C at a velocity of 3 m/s.
Assuming the surface of the duct to be at the temperature of the 11 m 15°C
Air
River, 15°C FIGURE P8–66
8–67 Repeat Problem 8–66 assuming that a 0.15mmthick
layer of mineral deposit (k 3 W/m °C) formed on the inner
surface of the pipe.
8–68E The exhaust gases of an automotive engine
leave the combustion chamber and enter a
8ftlong and 3.5in.diameter thinwalled steel exhaust pipe at
800°F and 15.5 psia at a rate of 0.2 lbm/s. The surrounding
ambient air is at a temperature of 80°F, and the heat transfer
coefficient on the outer surface of the exhaust pipe is
3 Btu/h ft2 °F. Assuming the exhaust gases to have the properties of air, determine (a) the velocity of the exhaust gases at
the inlet of the exhaust pipe and (b) the temperature at which
the exhaust gases will leave the pipe and enter the air.
8–69 Hot water at 90°C enters a 15m section of a cast iron
pipe (k 52 W/m °C) whose inner and outer diameters are 4
and 4.6 cm, respectively, at an average velocity of 0.8 m/s. The
outer surface of the pipe, whose emissivity is 0.7, is exposed to
the cold air at 10°C in a basement, with a convection heat cen58933_ch08.qxd 9/4/2002 11:29 AM Page 457 457
CHAPTER 8
Tambient = 10°C 8–72 Liquidcooled systems have high heat transfer coefficients associated with them, but they have the inherent disadvantage that they present potential leakage problems.
Therefore, air is proposed to be used as the microchannel
coolant. Repeat Problem 8–71 using air as the cooling fluid instead of water, entering at a rate of 0.5 L/s. ε = 0.7
Hot
water
90°C
0.8 m/s 15 m FIGURE P8–69
transfer coefficient of l5 W/m2 °C. Taking the walls of the
basement to be at 10°C also, determine (a) the rate of heat loss
from the water and (b) the temperature at which the water
leaves the basement.
8–70 Repeat Problem 8–69 for a pipe made of copper (k
386 W/m °C) instead of cast iron.
8–71 D. B. Tuckerman and R. F. Pease of Stanford University demonstrated in the early 1980s that integrated circuits can
be cooled very effectively by fabricating a series of microscopic channels 0.3 mm high and 0.05 mm wide in the back of
the substrate and covering them with a plate to confine the
fluid flow within the channels. They were able to dissipate
790 W of power generated in a 1cm2 silicon chip at a junctiontoambient temperature difference of 71°C using water as the
coolant flowing at a rate of 0.01 L/s through 100 such channels
under a 1cm 1cm silicon chip. Heat is transferred primarily through the base area of the channel, and it was found that
the increased surface area and thus the fin effect are of lesser
importance. Disregarding the entrance effects and ignoring any
heat transfer from the side and cover surfaces, determine
(a) the temperature rise of water as it flows through the microchannels and (b) the average surface temperature of the base of
the microchannels for a power dissipation of 50 W. Assume the
water enters the channels at 20°C. Cover
plate
1 cm 0.3
mm 0.05 mm Silicon
substrate Electronic
circuits
on this side FIGURE P8–71 Microscopic
channels 8–73 Hot exhaust gases leaving a stationary diesel engine at
450°C enter a l5cmdiameter pipe at an average velocity of
3.6 m/s. The surface temperature of the pipe is 180°C. Determine the pipe length if the exhaust gases are to leave the pipe
at 250°C after transferring heat to water in a heat recovery unit.
Use properties of air for exhaust gases.
8–74 Geothermal steam at 165°C condenses in the shell side
of a heat exchanger over the tubes through which water flows.
Water enters the 4cmdiameter, 14mlong tubes at 20°C at a
rate of 0.8 kg/s. Determine the exit temperature of water and
the rate of condensation of geothermal steam.
8–75 Cold air at 5°C enters a l2cmdiameter 20mlong
isothermal pipe at a velocity of 2.5 m/s and leaves at 19°C.
Estimate the surface temperature of the pipe.
8–76 Oil at 10°C is to be heated by saturated steam at 1 atm
in a doublepipe heat exchanger to a temperature of 30°C. The
inner and outer diameters of the annular space are 3 cm and
5 cm, respectively, and oil enters at with a mean velocity of 0.8
m/s. The inner tube may be assumed to be isothermal at 100°C,
and the outer tube is well insulated. Assuming fully developed
flow for oil, determine the tube length required to heat the oil
to the indicated temperature. In reality, will you need a shorter
or longer tube? Explain. Design and Essay Problems
8–77 Electronic boxes such as computers are commonly
cooled by a fan. Write an essay on forced air cooling of electronic boxes and on the selection of the fan for electronic
devices.
8–78 Design a heat exchanger to pasteurize milk by steam in
a dairy plant. Milk is to flow through a bank of 1.2cm internal
diameter tubes while steam condenses outside the tubes at
1 atm. Milk is to enter the tubes at 4°C, and it is to be heated to
72°C at a rate of 15 L/s. Making reasonable assumptions, you
are to specify the tube length and the number of tubes, and the
pump for the heat exchanger.
8–79 A desktop computer is to be cooled by a fan. The electronic components of the computer consume 80 W of power
under fullload conditions. The computer is to operate in environments at temperatures up to 50°C and at elevations up to
3000 m where the atmospheric pressure is 70.12 kPa. The exit
temperature of air is not to exceed 60°C to meet the reliability
requirements. Also, the average velocity of air is not to exceed
120 m/min at the exit of the computer case, where the fan is installed to keep the noise level down. Specify the flow rate of
the fan that needs to be installed and the diameter of the casing
of the fan. cen58933_ch08.qxd 9/4/2002 11:29 AM Page 458 ...
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This note was uploaded on 01/28/2010 for the course HEAT ENG taught by Professor Ghaz during the Spring '10 term at University of Guelph.
 Spring '10
 Ghaz

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