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PROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal conductivity k and inner temperature, T1. FIND: The...

A thick slab of polymer composite at 40C is immersed in a large stirred oil bath kept at 4C. A thermocouple 1.3cm below the slab's surface reds 26.2C after 3 minutes. What is the thermal diffusivity of the slab?
Heat Transfer solutions.pdf

PROBLEM 1.1
KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal
conductivity k and inner temperature, T1.
FIND: The outer temperature of the wall, T2.
SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions,
(3) Constant properties.
ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law, q cond = q x = q ′′ ⋅ A = -k
x T −T
dT
⋅ A = kA 1 2 .
dx
L Solving for T2 gives T2 = T1 − q cond L
.
kA Substituting numerical values, find T2 = 415 C - 3000W × 0.025m
0.2W / m ⋅ K × 10m2 T2 = 415 C - 37.5 C
T2 = 378 C.
COMMENTS: Note direction of heat flow and fact that T2 must be less than T1. < PROBLEM 1.2
KNOWN: Inner surface temperature and thermal conductivity of a concrete wall.
FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from
-15 to 38°C.
SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3)
Constant properties, (4) Outside wall temperature is that of the ambient air.
ANALYSIS: From Fourier’s law, it is evident that the gradient, dT dx = − q′′ k , is a constant, and
x hence the temperature distribution is linear, if q′′ and k are each constant. The heat flux must be
x
constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends
only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15°C
are ) ( 25 C − −15 C
dT
T1 − T2
q′′ = − k
=k
= 1W m ⋅ K
= 133.3W m 2 .
x
dx
L
0.30 m
q x = q′′ × A = 133.3 W m 2 × 20 m 2 = 2667 W .
x (1)
(2) < Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15
≤ T2 ≤ 38°C, with different wall thermal conductivities, k.
3500 Heat loss, qx (W) 2500 1500 500 -500 -1500
-20 -10 0 10 20 30 40 Ambient air temperature, T2 (C)
Wall thermal conductivity, k = 1.25 W/m.K
k = 1 W/m.K, concrete wall
k = 0.75 W/m.K For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearily from +2667 W to -867 W and is zero
when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with
increasing thermal conductivity.
COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane
wall would not be linear. PROBLEM 1.3
KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency
of gas furnace and cost of natural gas.
FIND: Daily cost of heat loss.
SCHEMATIC: ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties.
ANALYSIS: The rate of heat loss by conduction through the slab is T −T
7°C
q = k ( LW ) 1 2 = 1.4 W / m ⋅ K (11m × 8 m )
= 4312 W
t
0.20 m < The daily cost of natural gas that must be combusted to compensate for the heat loss is Cd = q Cg ηf ( ∆t ) = 4312 W × $0.01/ MJ
0.9 ×106 J / MJ ( 24 h / d × 3600s / h ) = $4.14 / d < COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation
between it and the concrete. PROBLEM 1.4
KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed
thickness.
FIND: Thermal conductivity, k, of the wood.
SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be
determined from Fourier’s law, Eq. 1.2. Rearranging,
k=q′′
x L
W
= 40
T − T2
m2
1 k = 0.10 W / m ⋅ K. 0.05m ( 40-20 ) C < COMMENTS: Note that the °C or K temperature units may be used interchangeably when
evaluating a temperature difference. PROBLEM 1.5
KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions.
FIND: Heat loss through window.
SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from
Fourier’s law, Eq. 1.2.
T −T
q′′ = k 1 2
x
L

W (15-5 ) C
q′′ = 1.4
x
m ⋅ K 0.005m
q′′ = 2800 W/m 2 .
x
Since the heat flux is uniform over the surface, the heat loss (rate) is
q = q ′′ × A
x
q = 2800 W / m2 × 3m2
q = 8400 W.
COMMENTS: A linear temperature distribution exists in the glass for the prescribed
conditions. < PROBLEM 1.6
KNOWN: Width, height, thickness and thermal conductivity of a single pane window and
the air space of a double pane window. Representative winter surface temperatures of single
pane and air space.
FIND: Heat loss through single and double pane windows.
SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-state
conditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy induced
motion).
ANALYSIS: From Fourier’s law, the heat losses are Single Pane: $
T1 − T2
2 35 C = 19, 600 W
qg = k g A
= 1.4 W/m ⋅ K 2m
L
0.005m () () T −T
25 $C
Double Pane: qa = k a A 1 2 = 0.024 2m2
= 120 W
L
0.010 m
COMMENTS: Losses associated with a single pane are unacceptable and would remain
excessive, even if the thickness of the glass were doubled to match that of the air space. The
principal advantage of the double pane construction resides with the low thermal conductivity
of air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, use
of the double pane construction would also increase the surface temperature of the glass
exposed to the room (inside) air. PROBLEM 1.7
KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures.
FIND: Thickness of styrofoam insulation needed to maintain heat load below prescribed
value.
SCHEMATIC: ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through 5
2
walls of area A = 4m , (3) Steady-state conditions, (4) Constant properties.
ANALYSIS: Using Fourier’s law, Eq. 1.2, the heat rate is
q = q ′′ ⋅ A = k ∆T
A total
L
2 Solving for L and recognizing that Atotal = 5×W , find
5 k ∆ T W2
L=
q  L= () 5 × 0.03 W/m ⋅ K 35 - (-10 ) C 4m 2 500 W L = 0.054m = 54mm. < COMMENTS: The corners will cause local departures from one-dimensional conduction
and a slightly larger heat loss. PROBLEM 1.8
KNOWN: Dimensions and thermal conductivity of food/beverage container. Inner and outer
surface temperatures.
FIND: Heat flux through container wall and total heat load.
SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through bottom
wall, (3) Uniform surface temperatures and one-dimensional conduction through remaining
walls.
ANALYSIS: From Fourier’s law, Eq. 1.2, the heat flux is
$
T − T 0.023 W/m ⋅ K ( 20 − 2 ) C
′′ = k 2 1 =
q
= 16.6 W/m 2
L
0.025 m < Since the flux is uniform over each of the five walls through which heat is transferred, the
heat load is
q = q′′ × A total = q′′ H ( 2W1 + 2W2 ) + W1 × W2 q = 16.6 W/m2 0.6m (1.6m + 1.2m ) + ( 0.8m × 0.6m ) = 35.9 W < COMMENTS: The corners and edges of the container create local departures from onedimensional conduction, which increase the heat load. However, for H, W1, W2 >> L, the
effect is negligible. PROBLEM 1.9
KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of that
through a composite wall of prescribed thermal conductivity and thickness.
FIND: Thickness of masonry wall.
SCHEMATIC: ASSUMPTIONS: (1) Both walls subjected to same surface temperatures, (2) Onedimensional conduction, (3) Steady-state conditions, (4) Constant properties.
ANALYSIS: For steady-state conditions, the conduction heat flux through a one-dimensional
wall follows from Fourier’s law, Eq. 1.2,
q ′′ = k ∆T
L where ∆T represents the difference in surface temperatures. Since ∆T is the same for both
walls, it follows that
L1 = L2 k1
q ′′
⋅ 2.
k2
q1
′′ With the heat fluxes related as
q1′ = 0.8 q ′′

2
L1 = 100mm 0.75 W / m ⋅ K
1
×
= 375mm.
0.25 W / m ⋅ K
0.8 < COMMENTS: Not knowing the temperature difference across the walls, we cannot find the
value of the heat rate. PROBLEM 1.10
KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boil
water. Rate of heat transfer to the pan.
FIND: Outer surface temperature of pan for an aluminum and a copper bottom.
SCHEMATIC: ASSUMPTIONS: (1) One-dimensional, steady-state conduction through bottom of pan.
ANALYSIS: From Fourier’s law, the rate of heat transfer by conduction through the bottom
of the pan is
T −T
q = kA 1 2
L
Hence,
T1 = T2 + qL
kA 2
where A = π D2 / 4 = π (0.2m ) / 4 = 0.0314 m 2 .
Aluminum: T1 = 110 $C + Copper: T1 = 110 $C + 600W ( 0.005 m ) ( 240 W/m ⋅ K 0.0314 m 2
600W (0.005 m ) ( 390 W/m ⋅ K 0.0314 m2 )
) = 110.40 $C = 110.25 $C COMMENTS: Although the temperature drop across the bottom is slightly larger for
aluminum (due to its smaller thermal conductivity), it is sufficiently small to be negligible for
both materials. To a good approximation, the bottom may be considered isothermal at T ≈
110 °C, which is a desirable feature of pots and pans. PROBLEM 1.11
KNOWN: Dimensions and thermal conductivity of a chip. Power dissipated on one surface.
FIND: Temperature drop across the chip.
SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heat
dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction in
chip.
ANALYSIS: All of the electrical power dissipated at the back surface of the chip is
transferred by conduction through the chip. Hence, from Fourier’s law,
P = q = kA ∆T
t or
∆T = t ⋅P
kW 2 = ∆T = 1.1 C. 0.001 m × 4 W 2 150 W/m ⋅ K ( 0.005 m ) < COMMENTS: For fixed P, the temperature drop across the chip decreases with increasing k
and W, as well as with decreasing t. PROBLEM 1.12
KNOWN: Heat flux gage with thin-film thermocouples on upper and lower surfaces; output
voltage, calibration constant, thickness and thermal conductivity of gage.
FIND: (a) Heat flux, (b) Precaution when sandwiching gage between two materials.
SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat conduction in gage,
(3) Constant properties.
ANALYSIS: (a) Fourier’s law applied to the gage can be written as
q ′′ = k ∆T
∆x and the gradient can be expressed as
∆T
∆E / N
=
∆x
SABt
where N is the number of differentially connected thermocouple junctions, SAB is the Seebeck
coefficient for type K thermocouples (A-chromel and B-alumel), and ∆x = t is the gage
thickness. Hence,
q ′′ = k∆E
NSABt q ′′ = 1.4 W / m ⋅ K × 350 × 10-6 V
= 9800 W / m2 .
-6 V / $ C × 0.25 × 10-3 m
5 × 40 × 10 < (b) The major precaution to be taken with this type of gage is to match its thermal
conductivity with that of the material on which it is installed. If the gage is bonded
between laminates (see sketch above) and its thermal conductivity is significantly different
from that of the laminates, one dimensional heat flow will be disturbed and the gage will
read incorrectly.
COMMENTS: If the thermal conductivity of the gage is lower than that of the laminates,
will it indicate heat fluxes that are systematically high or low? PROBLEM 1.13
KNOWN: Hand experiencing convection heat transfer with moving air and water.
FIND: Determine which condition feels colder. Contrast these results with a heat loss of 30 W/m2 under
normal room conditions.
SCHEMATIC: ASSUMPTIONS: (1) Temperature is uniform over the hand’s surface, (2) Convection coefficient is
uniform over the hand, and (3) Negligible radiation exchange between hand and surroundings in the case
of air flow.
ANALYSIS: The hand will feel colder for the condition which results in the larger heat loss. The heat
loss can be determined from Newton’s law of cooling, Eq. 1.3a, written as q′′ = h ( Ts − T∞ )
For the air stream: q′′ = 40 W m 2 ⋅ K 30 − ( −5) K = 1, 400 W m 2
air < For the water stream:
2
2
q′′
water = 900 W m ⋅ K (30 − 10 ) K = 18,000 W m < COMMENTS: The heat loss for the hand in the water stream is an order of magnitude larger than when
in the air stream for the given temperature and convection coefficient conditions. In contrast, the heat
loss in a normal room environment is only 30 W/m2 which is a factor of 400 times less than the loss in
the air stream. In the room environment, the hand would feel comfortable; in the air and water streams,
as you probably know from experience, the hand would feel uncomfortably cold since the heat loss is
excessively high. PROBLEM 1.14
KNOWN: Power required to maintain the surface temperature of a long, 25-mm diameter cylinder
with an imbedded electrical heater for different air velocities.
FIND: (a) Determine the convection coefficient for each of the air velocity conditions and display
the results graphically, and (b) Assuming that the convection coefficient depends upon air velocity as
h = CVn, determine the parameters C and n.
SCHEMATIC:
V(m/s)
Pe′ (W/m)
h (W/m2⋅K) 1
450
22.0 2
658
32.2 4
983
48.1 8
1507
73.8 12
1963
96.1 ASSUMPTIONS: (1) Temperature is uniform over the cylinder surface, (2) Negligible radiation
exchange between the cylinder surface and the surroundings, (3) Steady-state conditions.
ANALYSIS: (a) From an overall energy balance on the cylinder, the power dissipated by the
electrical heater is transferred by convection to the air stream. Using Newtons law of cooling on a per
unit length basis, ′
Pe = h (π D )(Ts − T∞ )

where Pe is the electrical power dissipated per unit length of the cylinder. For the V = 1 m/s
condition, using the data from the table above, find

h = 450 W m π × 0.025 m 300 − 40 C = 22.0 W m 2⋅K ( ) < Repeating the calculations, find the convection coefficients for the remaining conditions which are
tabulated above and plotted below. Note that h is not linear with respect to the air velocity.
(b) To determine the (C,n) parameters, we plotted h vs. V on log-log coordinates. Choosing C =
22.12 W/m2⋅K(s/m)n, assuring a match at V = 1, we can readily find the exponent n from the slope of
the h vs. V curve. From the trials with n = 0.8, 0.6 and 0.5, we recognize that n = 0.6 is a reasonable < 100
80
60
40
20
0 2 4 6 8 10 12 Air velocity, V (m/s)
Data, smooth curve, 5-points Coefficient, h (W/m^2.K) Coefficient, h (W/m^2.K) choice. Hence, C = 22.12 and n = 0.6. 100
80
60
40
20
10
1 2 4 6 Air velocity, V (m/s)
Data , smooth curve, 5 points
h = C * V^n, C = 22.1, n = 0.5
n = 0.6
n = 0.8 8 10 PROBLEM 1.15
KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required
to maintain a specified surface temperature for water and air flows.
FIND: Convection coefficients for the water and air flow convection processes, hw and ha,
respectively.
SCHEMATIC: ASSUMPTIONS: (1) Flow is cross-wise over cylinder which is very long in the direction
normal to flow.
ANALYSIS: The convection heat rate from the cylinder per unit length of the cylinder has
the form
q′ = h (π D ) ( Ts − T∞ )
and solving for the heat transfer convection coefficient, find
h= q′
.
π D (Ts − T∞ ) Substituting numerical values for the water and air situations:
Water hw = Air ha = 28 × 103 W/m 
π × 0.030m (90-25 ) C
400 W/m π × 0.030m (90-25 ) C = 4,570 W/m 2 ⋅ K = 65 W/m 2 ⋅ K. <
< COMMENTS: Note that the air velocity is 10 times that of the water flow, yet
hw ≈ 70 × ha.
These values for the convection coefficient are typical for forced convection heat transfer with
liquids and gases. See Table 1.1. PROBLEM 1.16
KNOWN: Dimensions of a cartridge heater. Heater power. Convection coefficients in air
and water at a prescribed temperature.
FIND: Heater surface temperatures in water and air.
SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) All of the electrical power is transferred
to the fluid by convection, (3) Negligible heat transfer from ends.
ANALYSIS: With P = qconv, Newton’s law of cooling yields
P=hA (Ts − T∞ ) = hπ DL ( Ts − T∞ )
P
Ts = T∞ +
.
hπ DL
In water,
Ts = 20 C + 2000 W
5000 W / m ⋅ K × π × 0.02 m × 0.200 m
2 Ts = 20 C + 31.8 C = 51.8 C. < In air,
Ts = 20 C + 2000 W
50 W / m ⋅ K × π × 0.02 m × 0.200 m
2 Ts = 20 C + 3183 C = 3203 C. < COMMENTS: (1) Air is much less effective than water as a heat transfer fluid. Hence, the
cartridge temperature is much higher in air, so high, in fact, that the cartridge would melt.
(2) In air, the high cartridge temperature would render radiation significant. PROBLEM 1.17
KNOWN: Length, diameter and calibration of a hot wire anemometer. Temperature of air
stream. Current, voltage drop and surface temperature of wire for a particular application.
FIND: Air velocity
SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from the wire by
natural convection or radiation.
ANALYSIS: If all of the electric energy is transferred by convection to the air, the following
equality must be satisfied
Pelec = EI = hA (Ts − T∞ )
where A = π DL = π ( 0.0005m × 0.02m ) = 3.14 × 10−5 m 2 .
Hence,
h= EI
5V × 0.1A
=
= 318 W/m 2 ⋅ K
A (Ts − T∞ ) 3.14 × 10−5m 2 50 $C ( ( ) V = 6.25 × 10−5 h 2 = 6.25 ×10−5 318 W/m 2 ⋅ K ) 2 = 6.3 m/s < COMMENTS: The convection coefficient is sufficiently large to render buoyancy (natural
convection) and radiation effects negligible. PROBLEM 1.18
KNOWN: Chip width and maximum allowable temperature. Coolant conditions.
FIND: Maximum allowable chip power for air and liquid coolants.
SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides and
bottom, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat transfer by
radiation in air.
ANALYSIS: All of the electrical power dissipated in the chip is transferred by convection to
the coolant. Hence,
P=q
and from Newton’s law of cooling,
2 P = hA(T - T∞) = h W (T - T∞).
In air,
2 2 Pmax = 200 W/m ⋅K(0.005 m) (85 - 15) ° C = 0.35 W. < In the dielectric liquid
2 2 Pmax = 3000 W/m ⋅K(0.005 m) (85-15) ° C = 5.25 W. < COMMENTS: Relative to liquids, air is a poor heat transfer fluid. Hence, in air the chip can
dissipate far less energy than in the dielectric liquid. PROBLEM 1.19
KNOWN: Length, diameter and maximum allowable surface temperature of a power
transistor. Temperature and convection coefficient for air cooling.
FIND: Maximum allowable power dissipation.
SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through base of
transistor, (3) Negligible heat transfer by radiation from surface of transistor.
ANALYSIS: Subject to the foregoing assumptions, the power dissipated by the transistor is
equivalent to the rate at which heat is transferred by convection to the air. Hence,
Pelec = q conv = hA (Ts − T∞ ) ( ) 2
where A = π DL + D2 / 4 = π 0.012m × 0.01m + ( 0.012m ) / 4 = 4.90 ×10−4 m 2 . For a maximum allowable surface temperature of 85°C, the power is ( Pelec = 100 W/m2 ⋅ K 4.90 × 10−4 m 2 ) (85 − 25)$ C = 2.94 W < COMMENTS: (1) For the prescribed surface temperature and convection coefficient,
radiation will be negligible relative to convection. However, conduction through the base
could be significant, thereby permitting operation at a larger power.
(2) The local convection coefficient varies over the surface, and hot spots could exist if there
are locations at which the local value of h is substantially smaller than the prescribed average
value. PROBLEM 1.20
KNOWN: Air jet impingement is an effective means of cooling logic chips.
FIND: Procedure for measuring convection coefficients associated with a 10 mm × 10 mm chip.
SCHEMATIC: ASSUMPTIONS: Steady-state conditions.
ANALYSIS: One approach would be to use the actual chip-substrate system, Case (a), to perform the
measurements. In this case, the electric power dissipated in the chip would be transferred from the chip
by radiation and conduction (to the substrate), as well as by convection to the jet. An energy balance for
the chip yields q elec = q conv + q cond + q rad . Hence, with q conv = hA ( Ts − T∞ ) , where A = 100
mm2 is the surface area of the chip, q
− q cond − q rad
h = elec
A (Ts − T∞ ) (1) While the electric power ( q elec ) and the jet ( T∞ ) and surface ( Ts ) temperatures may be measured, losses
from the chip by conduction and radiation would have to be estimated. Unless the losses are negligible
(an unlikely condition), the accuracy of the procedure could be compromised by uncertainties associated
with determining the conduction and radiation losses.
A second approach, Case (b), could involve fabrication of a heater assembly for which the
conduction and radiation losses are controlled and minimized. A 10 mm × 10 mm copper block (k ~ 400
W/m⋅K) could be inserted in a poorly conducting substrate (k < 0.1 W/m⋅K) and a patch heater could be
applied to the back of the block and insulated from below. If conduction to both the substrate and
insulation could thereby be rendered negligible, heat would be transferred almost exclusively through the
block. If radiation were rendered negligible by applying a low emissivity coating (ε < 0.1) to the surface
of the copper block, virtually all of the heat would be transferred by convection to the jet. Hence, q cond
and q rad may be neglected in equation (1), and the expression may be used to accurately determine h
from the known (A) and measured ( q elec , Ts , T∞ ) quantities.
COMMENTS: Since convection coefficients associated with gas flows are generally small, concurrent
heat transfer by radiation and/or conduction must often be considered. However, jet impingement is one
of the more effective means of transferring heat by convection and convection coefficients well in excess
of 100 W/m2⋅K may be achieved. PROBLEM 1.21
KNOWN: Upper temperature set point, Tset, of a bimetallic switch and convection heat
transfer coefficient between clothes dryer air and exposed surface of switch.
FIND: Electrical power for heater to maintain Tset when air temperature is T∞ = 50°C.
SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Electrical heater is perfectly insulated
from dryer wall, (3) Heater and switch are isothermal at Tset, (4) Negligible heat transfer from
sides of heater or switch, (5) Switch surface, As, loses heat only by convection.
ANALYSIS: Define a control volume around the bimetallic switch which experiences heat
input from the heater and convection heat transfer to the dryer air. That is,


Ein - Eout = 0
qelec - hAs ( Tset − T∞ ) = 0.
The electrical power required is,
qelec = hAs ( Tset − T∞ )
qelec = 25 W/m2 ⋅ K × 30 × 10-6 m2 ( 70 − 50 ) K=15 mW. < COMMENTS: (1) This type of controller can achieve variable operating air temperatures
with a single set-point, inexpensive, bimetallic-thermostatic switch by adjusting power levels
to the heater.
(2) Will the heater power requirement increase or decrease if the insulation pad is other than
perfect? PROBLEM 1.22
KNOWN: Hot vertical plate suspended in cool, still air. Change in plate temperature with time at
the instant when the plate temperature is 225°C.
FIND: Convection heat transfer coefficient for this condition.
SCHEMATIC: ASSUMPTIONS: (1) Plate is isothermal and of uniform temperature, (2) Negligible radiation
exchange with surroundings, (3) Negligible heat lost through suspension wires.
ANALYSIS: As shown in the cooling curve above, the plate temperature decreases with time. The
condition of interest is for time to. For a control surface about the plate, the conservation of energy
requirement is 


E in - E out = E st
dT
− 2hA s ( Ts − T∞ ) = M c p
dt
where As is the surface area of one side of the plate. Solving for h, find h= h= Mcp dT
2As (Ts − T∞ ) dt
3.75 kg × 2770 J/kg ⋅ K
2 × ( 0.3 × 0.3) m 2 ( 225 − 25 ) K × 0.022 K/s=6.4 W/m 2 ⋅ K < COMMENTS: (1) Assuming the plate is very highly polished with emissivity of 0.08, determine
whether radiation exchange with the surroundings at 25°C is negligible compared to convection.
(2) We will later consider the criterion for determining whether the isothermal plate assumption is
reasonable. If the thermal conductivity of the present plate were high (such as aluminum or copper),
the criterion would be satisfied. PROBLEM 1.23
KNOWN: Width, input power and efficiency of a transmission. Temperature and convection
coefficient associated with air flow over the casing.
FIND: Surface temperature of casing.
SCHEMATIC: ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3)
Negligible radiation.
ANALYSIS: From Newton’s law of cooling, q = hAs ( Ts − T∞ ) = 6 hW 2 ( Ts − T∞ )
where the output power is η Pi and the heat rate is q = Pi − Po = Pi (1 − η ) = 150 hp × 746 W / hp × 0.07 = 7833 W
Hence, Ts = T∞ + q
6 hW 2 = 30°C + 7833 W
6 × 200 W / m 2 ⋅ K × (0.3m ) 2 = 102.5°C COMMENTS: There will, in fact, be considerable variability of the local convection coefficient
over the transmission case and the prescribed value represents an average over the surface. < PROBLEM 1.24
KNOWN: Air and wall temperatures of a room. Surface temperature, convection coefficient
and emissivity of a person in the room.
FIND: Basis for difference in comfort level between sum...

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