kieth

Chapter 1
PROBLEM 1.1
The outer surface of a 0.2m-thick concrete wall is kept at a temperature of –5°C, while the inner surface is kept at 20°C. The thermal conductivity of the concrete is 1.2 W/(m
K). Determine the heat loss through a wall 10 m long and 3 m high.
GIVEN
10 m long, 3 m high, and 0.2 m thick concrete wall
Thermal conductivity of the concrete (k) = 1.2 W/(m K)
Temperature of the inner surface (Ti) = 20°C
Temperature of the outer surface (To) = –5°C
FIND
The heat loss through the wall (qk)
ASSUMPTIONS
One dimensional heat flow
The system has reached steady state
SKETCH

SOLUTION
The rate of heat loss through the wall is given by Equation (1.2) qk =

AK
( T)
L

qk =

(10 m) (3m) (1.2 W/(m K) )
(20°C – (–5°C))
0.2 m

qk = 4500 W
COMMENTS
Since the inside surface temperature is higher than the outside temperature heat is transferred from the inside of the wall to the outside of the wall.

1
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PROBLEM 1.2
The weight of the insulation in a spacecraft may be more important than the space required. Show analytically that the lightest insulation for a plane wall with a specified thermal resistance is that insulation which has the smallest product of density times thermal conductivity.
GIVEN
Insulating a plane wall, the weight of insulation is most significant
FIND
Show that lightest insulation for a given thermal resistance is that insulation which has the smallest product of density ( ) times thermal conductivity (k)
ASSUMPTIONS
One dimensional heat transfer through the wall
Steady state conditions
SOLUTION
The resistance of the wall (Rk), from Equation (1.13) is
Rk =

L
Ak

where
L = the thickness of the wall
A = the area of the wall
The weight of the wall (w) is w =

AL

Solving this for L
L =

w rA Substituting this expression for L into the equation for the resistance
Rk = w =
Therefore, when the product of

w r k A2

k Rk A2 k for a given resistance is smallest, the weight is also smallest.

COMMENTS
Since and k are physical properties of the insulation material they cannot be varied individually.
Hence in this type of design different materials must be tried to minimize the weight.
PROBLEM 1.3
A furnace wall is to be constructed of brick having standard dimensions 9 by 4.5 by
3 in. Two kinds of material are available. One has a maximum usable temperature of
1900°F and a thermal conductivity of 1 Btu/(h ft°F), and the other has a maximum temperature limit of 1600°F and a thermal conductivity of 0.5 Btu/(h ft°F). The bricks cost the same and can be laid in any manner, but we wish to design the most economical
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wall for a furnace with a temperature on the hot side of 1900°F and on the cold side of
400°F. If the maximum amount of heat transfer permissible is 300 Btu/h for each square foot of area, determine the most economical arrangements for the available bricks.
GIVEN
Furnace wall made of 9 4.5 3 inch bricks of two types
Type 1 bricks
Maximum useful temperature (T1,max) = 1900°F
Thermal conductivity (k1) = 1.0 Btu/(h ft°F)
Type 2 bricks
Maximum useful temperature (T2,max) = 1600°F
Thermal conductivity (k2) = 0.5 Btu/(h ft°F)
Bricks cost the same
Wall hot side (Thot) = 1900°F and cold side (Tcold) = 400°F
Maximum heat transfer permissible (qmax/A) = 300 Btu/(h ft2)
FIND
The most economical arrangement for the bricks
ASSUMPTIONS
One dimensional, steady state heat transfer conditions
Constant thermal conductivities
The contact resistance between the bricks is negligible
SKETCH

SOLUTION
Since the type 1 bricks have a higher thermal conductivity at the same cost as the type 2 bricks, the most economical wall would use as few type 1 bricks as possible. However, there must be a thick enough layer of type 1 bricks to keep the type 2 bricks at 1600°F or less.
For one dimensional conduction through the type 1 bricks (from Equation (1.2)) kA T qk =
L
qmax k = 1 (Thot – T12)
L1
A

where L1 = the minimum thickness of the type 1 bricks
Solving for L1 k1 L1 =
(Thot – T12)
Ê qmax ˆ
Á
Ë A ˜
¯

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L1 =

1.0 Btu/(h ft°F)
300 Btu/(h ft 2 )

(1900°F – 1600°F) = 1 ft

This thickness can be achieved with 4 layers of type 1 bricks using the 3 inch dimension.
Similarly, for one dimensional conduction through the type 2 bricks
L2 =

L2 =

k2
(T12 – Tcold)
Ê qmax ˆ
Á
Ë A ˜
¯
0.5 Btu/(h ft°F)
300 Btu/(h ft 2 )

(1600°F – 400°F) = 2 ft

This thickness can be achieved with 8 layers of type 2 brick using the 3 inch dimension. Therefore the most economical wall would be built using 4 layers of type 1 bricks and 8 layers of type 2 bricks with the three inch dimension of the bricks used as the thickness.
PROBLEM 1.4
To measure thermal conductivity, two similar 1-cm-thick specimens are placed in an apparatus shown in the accompanying sketch. Electric current is supplied to the
6-cm by 6-cm guarded heater, and a wattmeter shows that the power dissipation is 10 watts (W). Thermocouples attached to the warmer and to the cooler surfaces show temperatures of 322 and 300 K, respectively. Calculate the thermal conductivity of the material at the mean temperature in Btu/(h ft°F) and W/(m K).
GIVEN
Thermal conductivity measurement apparatus with two samples as shown
Sample thickness (L) = 1 cm = 0.01 cm
Area = 6 cm 6 cm = 36 cm2 = 0.0036 m2
Power dissipation rate of the heater (qh) = 10 W
Surface temperatures
Thot = 322 K and Tcold = 300 K
FIND
The thermal conductivity of the sample at the mean temperature in Btu/(h ft°F) and W/(m K)
ASSUMPTIONS
One dimensional, steady state conduction
No heat loss from the edges of the apparatus
SKETCH

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SOLUTION
By conservation of energy, the heat loss through the two specimens must equal the power dissipation of the heater. Therefore the heat transfer through one of the specimens is qh/2.
For one dimensional, steady state conduction (from Equation (1.2)) qk =

kA
L

T=

qh
2

Solving for the thermal conductivity
Ê qh ˆ
Á ˜L
Ë 2¯ k =
A

k =

T

(5 W)(0.01m)
(0.0036 m 2 ) (322 K - 300 K)

k = 0.63 W/(m K)
Converting the thermal conductivity in the English system of units using the conversion factor found on the inside front cover of the text book
Btu/(h ft°F) ˆ
Ê
k = 0.63 W/(m K) Á 0.057782
Ë
¯
W/(m K) ˜

k = 0.36 Btu/(h ft°F)
COMMENTS
In the construction of the apparatus care must be taken to avoid edge losses so all the heat generated will be conducted through the two specimens.
PROBLEM 1.5
To determine the thermal conductivity of a structural material, a large 6-in-thick slab of the material was subjected to a uniform heat flux of 800 Btu/(h ft2), while thermocouples embedded in the wall 2 in. apart were read over a period of time. After the system had reached equilibrium, an operator recorded the readings of the thermocouples as shown below for two different environmental conditions
Distance from the Surface (in.)

Temperature (°F)
Test 1

0
2
4
6

100
150
206
270
Test 2

0
2
4
6

200
265
335
406

From these data, determine an approximate expression for the thermal conductivity as a function of temperature between 100 and 400°F.
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GIVEN
Thermal conductivity test on a large, 6-in.-thick slab
Thermocouples are embedded in the wall 2 in. apart
Heat flux (q/A) = 800 Btu/(h ft2)
Two equilibrium conditions were recorded (shown above)
FIND
An approximate expression for thermal conductivity as a function of temperature between 100 and 400°F
ASSUMPTIONS
One dimensional conduction

SKETCH

SOLUTION
The thermal conductivity can be calculated for each pair of adjacent thermocouples using the equation for one dimensional conduction (Equation (1.2)) q =kA

DT
L

Solving for thermal conductivity k =

q L
A DT

This will yield a thermal conductivity for each pair of adjacent thermocouples which will then be assigned to the average temperature for that pair of thermocouples. As an example, for the first pair of thermocouples in Test 1, the thermal conductivity (ko) is
2
Ê
ˆ
ft
Á
˜
12
ko = 800 Btu/(h ft 2 ) Á
= 2.67 Btu/(h ft°F) o o ˜
Á 150 F - 100 F ˜
Ë
¯

(

)

The average temperature for this pair of thermocouples is
Tave =

100 o F + 150 o F
= 125 °F
2

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Thermal conductivities and average temperatures for the rest of the data can be calculated in a similar manner n

Temperature (°F)

Thermal conductivity Btu/(h ft°F)

1
2
3
4
5
6

125
178
238
232.5
300
370.5

2.67
2.38
2.08
2.05
1.90
1.88

These points are displayed graphically on the following page.

We will use the best fit quadratic function to represent the relationship between thermal conductivity and temperature k (T) = a + b T + c T 2
The constants a, b, and c can be found using a least squares fit.
Let the experimental thermal conductivity at data point n be designated as kn. A least squares fit of the data can be obtained as follows
The sum of the squares of the errors is
S =

 [kn - k (Tn )]2
N

S=

 kn2 - 2a  kn - N a 2 + 2ab Tn - 2b knTn + 2ac Tn2 + b2  Tn2 - 2 c

 k nTn2 + 2bc Tn3 + c2  Tn4
By setting the derivatives of S (with respect to a, b, and c) equal to zero, the following equations result Na+

Tnb +

Tn2 c =

kn

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Tn a +

Tn2 b +

Tn3 c =

2

3

4

Tn a +

Tn b +

kn Tn kn Tn2

Tn c =

For this problem
Tn = 1444
Tn2 = 3.853 105
Tn3 = 1.115 108
Tn4 = 3.432 1010 kn = 12.96 kn Tn = 2996 kn Tn2 = 7.748 105
Solving for a, b, and c a = 3.76 b = – 0.0106 c = 1.476 10–5
Therefore the expression for thermal conductivity as a function of temperature between 100 and 400°
F is k (T) = 3.76 – 0.0106 T + 1.476

10–5 T 2

This empirical expression for the thermal conductivity as a function of temperature is plotted with the thermal conductivities derived from the experimental data in the above graph.
COMMENTS
Note that the derived empirical expression is only valid within the temperature range of the experimental data.
PROBLEM 1.6
A square silicone chip 7 mm by 7 mm in size and 0.5 mm thick is mounted on a plastic substrate with its front surface cooled by a synthetic liquid flowing over it.
Electronic circuits in the back of the chip generate heat at a rate of 5 watts that have to be transferred through the chip. Estimate the steady state temperature difference between the front and back surfaces of the chip. The thermal conductivity of silicone is 150 W/(m K).
GIVEN
A 0.007 m by 0.007 m silicone chip
Thickness of the chip (L) = 0.5 mm = 0.0005 m
Heat generated at the back of the chip ( qG ) = 5 W
The thermal conductivity of silicon (k) = 150 W/(m K)
FIND
The steady state temperature difference ( T)
ASSUMPTIONS
One dimensional conduction (edge effects are negligible)
The thermal conductivity is constant
The heat lost through the plastic substrate is negligible

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SKETCH

SOLUTION
For steady state the rate of heat loss through the chip, given by Equation (1.2), must equal the rate of heat generation qk =

Ak
( T) = qG
L

Solving this for the temperature difference
T=

L qG kA T=

(0.0005) (5 W)
(150 W/(m K) ) (0.007 m) (0.007 m)

T = 0.34°C
PROBLEM 1.7
A warehouse is to be designed for keeping perishable foods cool prior to transportation to grocery stores. The warehouse has an effective surface area of 20,000 ft 2 exposed to an ambient air temperature of 90°F. The warehouse wall insulation (k = 0.1 Btu/(h ft°F)) is
3 in. thick. Determine the rate at which heat must be removed (Btu/h) from the warehouse to maintain the food at 40°F.
GIVEN
Cooled warehouse
Effective area (A) = 20,000 ft2
Temperatures
Outside air = 90°F and food inside = 40°F
Thickness of wall insulation (L) = 3 in. = 0.25 ft
Thermal conductivity of insulation (k) = 0.1 Btu/(h ft°F)
FIND
Rate at which heat must be removed (q)
ASSUMPTIONS
One dimensional, steady state heat flow
The food and the air inside the warehouse are at the same temperature
The thermal resistance of the wall is approximately equal to the thermal resistance of the wall insulation alone

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SKETCH

SOLUTION
The rate at which heat must be removed is equal to the rate at which heat flows into the warehouse.
There will be convective resistance to heat flow on the inside and outside of the wall. To estimate the upper limit of the rate at which heat must be removed these convective resistances will be neglected.
Therefore the inside and outside wall surfaces are assumed to be at the same temperature as the air inside and outside of the wall. Then the heat flow, from Equation (1.2), is q =

q =

kA
L

T

(0.1Btu/(h ft°F) ) (20,000 ft 2 )
0.25 ft

(90°F – 40°F)

q = 400,000 Btu/h

PROBLEM 1.8
With increasing emphasis on energy conservation, the heat loss from buildings has become a major concern. For a small tract house the typical exterior surface areas and
R-factors (area thermal resistance) are listed below
Element
Walls
Ceiling
Floor
Windows
Doors

Area (m2)

R-Factors = Area

150
120
120
20
5

Thermal Resistance [(m2 K/W)]
2.0
2.8
2.0
0.1
0.5

(a) Calculate the rate of heat loss from the house when the interior temperature is 22°C and the exterior is –5°C.
(b) Suggest ways and means to reduce the heat loss and show quantitatively the effect of doubling the wall insulation and the substitution of double glazed windows
(thermal resistance = 0.2 m2 K/W) for the single glazed type in the table above.
GIVEN
Small house
Areas and thermal resistances shown in the table above
Interior temperature = 22°C
Exterior temperature = –5°C
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FIND
(a) Heat loss from the house (qa)
(b) Heat loss from the house with doubled wall insulation and double glazed windows (qb). Suggest improvements. ASSUMPTIONS
All heat transfer can be treated as one dimensional
Steady state has been reached
The temperatures given are wall surface temperatures
Infiltration is negligible
The exterior temperature of the floor is the same as the rest of the house
SOLUTION
(a) The rate of heat transfer through each element of the house is given by Equations (1.33) and
(1.34)
q =

DT
Rth

The total rate of heat loss from the house is simply the sum of the loss through each element:
Ê
ˆ
1
1
1
1
1
Á
˜
+
+
+
+
Ê AR ˆ
Ê AR ˆ
Ê AR ˆ
Ê AR ˆ q = T Á Ê AR ˆ
˜
Á
˜
Á
˜
Á
˜
Á
˜
Á
˜
Á Ë A ¯ wall Ë A ¯ ceiling Ë A ¯ floor Ë A ¯ windows Ë A ¯ doors ˜
Á
˜
Ë
¯

q = (22°C – –5°C)
Ê
ˆ
Á
˜
1
1
1
1
1
Á
˜
+
+
+
+
Á Ê 2.0 (m 2 K)/W ˆ Ê 2.8 (m 2 K)/W ˆ Ê 2.0 (m2 K)/W ˆ Ê 0.5 (m 2 K)/W ˆ Ê 0.5 (m2 K)/W ˆ ˜
Á Á 150 m 2 ˜ Á 120 m2 ˜ Á 120 m 2 ˜ Á
˜ Á
˜˜
ËË
¯ Ë
¯ Ë
¯ Ë
¯ Ë
¯¯
20 m2
5 m2 q = (22°C – –5°C) (75 + 42.8 + 60 + 200 + 10) W/K q = 10,500 W
(b) Doubling the resistance of the walls and windows and recalculating the total heat loss: q = (22°C – –5°C)
Ê
ˆ
Á
˜
1
1
1
1
1
Á
˜
+
+
+
+
Á Ê 4.0 (m 2 K)/W ˆ Ê 2.8 (m 2 K)/W ˆ Ê 2.0 (m 2 K)/W ˆ Ê 0.2 (m 2 K)/W ˆ Ê 0.5 (m 2 K)/W ˆ ˜
ÁÁ
˜
˜ Ë
˜¯
Ë Ë 150 m 2 ˜ Á 120 m 2 ˜ Á 120 m 2 ˜ Á
¯ Ë
¯ Ë
¯ Ë
¯ Á
¯
20 m 2
5 m2 q = (22°C – –5°C) (37.5 + 42.8 + 60 + 100 + 10) W/K q = 6800 W
Doubling the wall and window insulation led to a 35% reduction in the total rate of heat loss.

11
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COMMENTS
Notice that the single glazed windows account for slightly over half of the total heat lost in case (a) and that the majority of the heat loss reduction in case (b) is due to the double glazed windows.
Therefore double glazed windows are strongly suggested.
PROBLEM 1.9
Heat is transferred at a rate of 0.1 kW through glass wool insulation (density = 100 kg/m3) of 5 cm thickness and 2 m2 area. If the hot surface is at 70°C, determine the temperature of the cooler surface.
GIVEN
Glass wool insulation with a density ( ) = 100 kg/m3
Thickness (L) = 5 cm = 0.05 m
Area (A) = 2 m2
Temperature of the hot surface (Th) = 70°C
Rate of heat transfer (qk) = 0.1 kW = 100 W
FIND
The temperature of the cooler surface (Tc)
ASSUMPTIONS
One dimensional, steady state conduction
Constant thermal conductivity
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 11
The thermal conductivity of glass wool at 20°C (k) = 0.036 W/(m K)
SOLUTION
For one dimensional, steady state conduction, the rate of heat transfer, from Equation (1.2), is qk =

Ak
(Th – Tc)
L

Solving this for Tc
Tc = Th –

qk L
Ak

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Tc = 70°C –

(100 W) (0.05 m)
(2 m 2 ) ( 0.036 W/m K )

Tc = 0.6°C
PROBLEM 1.10
A heat flux meter at the outer (cold) wall of a concrete building indicates that the heat loss through a wall of 10 cm thickness is 20 W/m2. If a thermocouple at the inner surface of the wall indicates a temperature of 22°C while another at the outer surface shows
6°C, calculate the thermal conductivity of the concrete and compare your result with the value in Appendix 2, Table 11.
GIVEN
Concrete wall
Thickness (L) = 100 cm = 0.1 m
Heat loss (q/A) = 20 W/m2
Surface temperature
Inner (Ti) = 22°C
Outer (To) = 6°C
FIND
The thermal conductivity (k) and compare it to the tabulated value
ASSUMPTIONS
One dimensional heat flow through the wall
Steady state conditions exist

SKETCH

SOLUTION
The rate of heat transfer for steady state, one dimensional conduction, from Equation (1.2), is qk =

kA
(Thot – Tcold)
L

Solving for the thermal conductivity
L
Êq ˆ k = Á k˜
Ë A ¯ (Ti - To )

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Ê 0.1m 2 ˆ k = (20 W/m 2 ) Á o
= 0.125 W/(m K)
Ë 22 C - 6oC ˜
¯

This result is very close to the tabulated value in Appendix 2, Table 11 where the thermal conductivity of concrete is given as 0.128 W/(m K).
PROBLEM 1.11
Calculate the heat loss through a 1-m by 3-m glass window 7 mm thick if the inner surface temperature is 20°C and the outer surface temperature is 17°C. Comment on the possible effect of radiation on your answer.
GIVEN
Window: 1 m by 3 m
Thickness (L) = 7 mm = 0.007 m
Surface temperature
Inner (Ti) = 20°C and outer (To) = 17°C
FIND
The rate of heat loss through the window (q)
ASSUMPTIONS
One dimensional, steady state conduction through the glass
Constant thermal conductivity
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 11
Thermal conductivity of glass (k) = 0.81 W/(m K)
SOLUTION
The heat loss by conduction through the window is given by Equation (1.2) qk = qk =

kA
(Thot – Tcold)
L

(0.81 W/(m K) ) (1m) (3m)
(0.007 m)

(20°C – 17°C)

qk = 1040 W
COMMENTS
Window glass is transparent to certain wavelengths of radiation, therefore some heat may be lost by radiation through the glass.
During the day sunlight may pass through the glass creating a net heat gain through the window.
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PROBLEM 1.12
If in Problem 1.11 the outer air temperature is –2°C, calculate the convective heat transfer coefficient between the outer surface of the window and the air assuming radiation is negligible.
Problem 1.11: Calculate the heat loss through a 1 m by 3 m glass window 7 mm thick if the inner surface temperature is 20°C and the outer surface temperature is 17°C.
Comment on the possible effect of radiation on your answer.
GIVEN
Window: 1 m by 3 m
Thickness (L) = 7 mm = 0.007 m
Surface temperatures
Inner (Ti) = 20°C and outer (To) = 17°C
The rate of heat loss = 1040 W (from the solution to Problem 1.11)
The outside air temperature = –2°C
FIND
The convective heat transfer coefficient at the outer surface of the window ( hc )
ASSUMPTIONS
The system is in steady state and radiative loss through the window is negligible
SKETCH

SOLUTION
For steady state the rate of heat transfer by convection (Equation (1.10)) from the outer surface must be the same as the rate of heat transfer by conduction through the glass qc = hc A T = qk

Solving for hc hc = hc =

qk
A (To - T• )

1040 W
(1m)(3m)(17 o C - - 2 o C)

hc = 18.2 W/(m2 K)

COMMENTS
This value for the convective heat transfer coefficient falls within the range given for the free convection of air in Table 1.4.
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PROBLEM 1.13
Using Table 1.4 as a guide, prepare a similar table showing the order of magnitudes of the thermal resistances of a unit area for convection between a surface and various fluids. GIVEN
Table 1.4— The order of magnitude of convective heat transfer coefficient ( hc )
FIND
The order of magnitudes of the thermal resistance of a unit area (A Rc)
SOLUTION
The thermal resistance for convection is defined by Equation (1.14) as
1
Rc = hc A
Therefore the thermal resistances of a unit area are simply the reciprocal of the convective heat transfer coefficient
1
A Rc = hc As an example, the first item in Table 1.4 is ‘air, free convection’ with a convective heat transfer coefficient of 6–30 W/(m2 K). Therefore the order of magnitude of the thermal resistances of a unit area for air, free convection is
1
2

30 W/(m K)

= 0.03 (m 2 K)/W to

1
2

6 W/(m K)

= 0.17 (m 2 K)/W

The rest of the table can be calculated in a similar manner
Order of Magnitude of Thermal Resistance of a Unit Area for Convection
Fluid
Air, free convection
Superheated steam or air, forced convection
Oil, forced convection
Water, forced convection
Water, boiling
Steam, condensing

W/(m2 K)
0.03–0.2
0.003–0.03
0.0006–0.02
0.0002–0.003
0.00002–0.0003
0.000008–0.0002

Btu/(h ft2 °F)
0.2–1.0
0.02–0.2
0.003–0.1
0.0005–0.02
0.0001–0.002
0.00005–0.001

COMMENTS
The extremely low thermal resistance in boiling and condensation suggests that these resistances can often be neglected in a series thermal network.
PROBLEM 1.14
A thermocouple (0.8-mm-OD wire) is used to measure the temperature of quiescent gas in a furnace. The thermocouple reading is 165°C. It is known, however, that the rate of radiant heat flow per meter length from the hotter furnace walls to the thermocouple wire is 1.1 W/m and the convective heat transfer coefficient between the wire and the gas is 6.8 W/(m2 K). With this information, estimate the true gas temperature. State your assumptions and indicate the equations used.
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GIVEN
Thermocouple (0.8 mm OD wire) in a furnace
Thermocouple reading (Tp) = 165°C
Radiant heat transfer to the wire (qr/L) = 1.1 W/m
Heat transfer coefficient ( hc ) = 6.8 W/(m2 K)
FIND
Estimate the true gas temperature (TG)
ASSUMPTIONS
The system is in equilibrium
Conduction along the thermocouple is negligible
Conduction between the thermocouple and the furnace wall is negligible
SKETCH

SOLUTION
Equilibrium and the conservation of energy require that the heat gain of the probe by radiation if equal to the heat lost by convection.
The rate of heat transfer by convection is given by Equation (1.10) qc = hc A

T = hc

D L (Tp – TG)

For steady state to exist the rate of heat transfer by convection must equal the rate of heat transfer by radiation qc = qr hc Êq ˆ
D L (Tp – TG) = Á r ˜ L
Ë L¯

Ê qr ˆ
Á ˜L
Ë L¯
TG = Tp – hc p D L

TG = 165°C –

(1.1W/m)

(6.8 W/(m2 K)) p (0.0008 m)

TG = 101°C
COMMENTS
This example illustrates that care must be taken in interpreting experimental measurements. In this case a significant correction must be applied to the thermocouple reading to obtain the true gas temperature. Can you suggest ways to reduce the correction?
17
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PROBLEM 1.15
Water at a temperature of 77°C is to be evaporated slowly in a vessel. The water is in a low pressure container which is surrounded by steam. The steam is condensing at 107°C.
The overall heat transfer coefficient between the water and the steam is 1100 W/(m2 K).
Calculate the surface area of the container which would be required to evaporate water at a rate of 0.01 kg/s.
GIVEN
Water evaporated slowly in a low pressure vessel surrounded by steam
Water temperature (Tw) = 77°C
Steam condensing temperature (Ts) = 107°C
Overall transfer coefficient between the water and the steam (U) = 1100 W/(m2 K)
Evaporation rate ( mw ) = 0.01 kg/s
FIND
The surface area (A) of the container required
ASSUMPTIONS
Steady state prevails
Vessel pressure is held constant at the saturation pressure corresponding to 77°C
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 13
The heat of vaporization of water at 77°C (hfg) = 2317 kJ/kg
SOLUTION
The heat transfer required to evaporate water at the given rate is q = mw hfg
For the heat transfer between the steam and the water q = U A T = mw hfg
Solving this for the transfer area mw h fg
A =
U DT
A =

(0.01kg/s) (2317 kJ/kg) (1000 J/kJ)

(1100 W/(m2 K)) (107 oC - 77o C)

A = 0.70 m2
18
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PROBLEM 1.16
The heat transfer rate from hot air at 100°C flowing over one side of a flat plate with dimensions 0.1 m by 0.5 m is determined to be 125 W when the surface of the plate is kept at 30°C. What is the average convective heat transfer coefficient between the plate and the air?
GIVEN
Flat plate, 0.1 m by 0.5 m, with hot air flowing over it
Temperature of plate surface (Ts) = 30°C
Air temperature (T ) = 100°C
Rate of heat transfer (q) = 125 W
FIND
The average convective heat transfer coefficient, hc, between the plate and the air
ASSUMPTION
Steady state conditions exist
SKETCH

SOLUTION
For convection the rate of heat transfer is given by Equation (1.10) qc = hc A T qc = hc A (T – Ts)
Solving this for the convective heat transfer coefficient yields hc = hc =

qc
A(T• - Ts )

125W
(0.1m)(0.5 m)(100 o C - 30o C)

hc = 35.7 W/(m2 K)

COMMENTS
One can see from Table 1.4 (order of magnitudes of convective heat transfer coefficients) that this result is reasonable for free convection in air.
Note that since T > Ts heat is transferred from the air to the plate.
PROBLEM 1.17
The heat transfer coefficient for a gas flowing over a thin flat plate 3 m long and
0.3 m wide varies with distance from the leading edge according to
19
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hc (x) = 10

–

1
4

W/(m 2 K)

If the plate temperature is 170°C and the gas temperature is 30°C, calculate (a) the average heat transfer coefficient, (b) the rate of heat transfer between the plate and the gas and (c) the local heat flux 2 m from the leading edge.
GIVEN
Gas flowing over a 3 m long by 0.3 m wide flat plate
Heat transfer coefficient (hc) is given by the equation above
The plate temperature (TP) = 170°C
The gas temperature (TG) = 30°C
FIND
(a) The average heat transfer coefficient ( hc )
(b) The rate of heat transfer (qc)
(c) The local heat flux at x = 2 m (qc (2)/A)
ASSUMPTIONS
Steady state prevails
SKETCH

SOLUTION
(a) The average heat transfer coefficient can be calculated by
1

1 L
10 4
1 L hc = Ú hc ( x) dx = Ú 10 ¥ 4 =
¥
L 0
L 0
L 3

3
4 L

|=

0

3

10 4 4
3
3 3

2

hc = 10.13 W/m K

(b) The total convective heat transfer is given by Equation (1.10) qc = hc A (TP – TG)

(

)

qc = 10.13 W/(m 2 K) (3 m) (0.3 m) (170°C – 30°C) qc = 1273 W
(c) The heat flux at x = 2 m is q ( x)
= hc(x) (TP – TG) = 10
A

-

1
4

(TP – TG)

20
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1

q (2)
= 10 (2) 4 (170°C – 30°C)
A

q (2)
= 1177 W/m2
A

COMMENTS
Note that the equation for hc does not apply near the leading edge of the plate since hc approaches infinity as x approaches zero. This behavior is discussed in more detail in Chapter 6.
PROBLEM 1.18
A cryogenic fluid is stored in a 0.3 m diameter spherical container in still air. If the convective heat transfer coefficient between the outer surface of the container and the air is 6.8 W/(m2 K), the temperature of the air is 27°C and the temperature of the surface of the sphere is –183°C, determine the rate of heat transfer by convection.
GIVEN
A sphere in still air
Sphere diameter (D) = 0.3 m
Convective heat transfer coefficient hc = 6.8 W/(m2 K)
Sphere surface temperature (Ts) = –183°C
Ambient air temperature (T ) = 27°C
FIND
Rate of heat transfer by convection (qc)
ASSUMPTIONS
Steady state heat flow
SKETCH

SOLUTION
The rate of heat transfer by convection is given by qc = hc A T qc = hc ( D2) (T – Ts)

(

qc = 6.8 W/(m 2 K)

)

(0.3 m)2 [27°C – (–183°C)]

qc = 404 W

21
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COMMENTS
Condensation would probably occur in this case due to the low surface temperature of the sphere. A calculation of the total rate of heat transfer to the sphere would have to take the rate on condensation and the rate of radiative heat transfer into account.
PROBLEM 1.19
A high-speed computer is located in a temperature controlled room of 26°C. When the machine is operating its internal heat generation rate is estimated to be 800 W. The external surface temperature is to be maintained below 85°C. The heat transfer coefficient for the surface of the computer is estimated to be 10 W/(m2 K). What surface area would be necessary to assure safe operation of this machine? Comment on ways to reduce this area.
GIVEN
A high-speed computer in a temperature controlled room
Temperature of the room (T ) = 26°C
Maximum surface temperature of the computer (Tc) = 85°C
Heat transfer coefficient (U) = 10 W/(m K)
Internal heat generation ( qG ) = 800 W
FIND
The surface area (A) required and comment on ways to reduce this area
ASSUMPTIONS
The system is in steady state
SKETCH

SOLUTION
For steady state the rate of heat transfer from the computer (given by Equation (1.33)) must equal the rate of internal heat generation q = U A T = qG
Solving this for the surface area
A =
A =

qG
U DT

800 W
= 1.4 m2 o o
10 W/(m K) (85 C - 26 K)

(

2

)

COMMENTS
Possibilities to reduce this surface area include
Increase the convective heat transfer from the computer by blowing air over it
Add fins to the outside of the computer
22
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PROBLEM 1.20
In order to prevent frostbite to skiers on chair lifts, the weather report at most ski areas gives both an air temperature and the wind chill temperature. The air temperature is measured with a thermometer that is not affected by the wind. However, the rate of heat loss from the skier increases with wind velocity, and the wind-chill temperature is the temperature that would result in the same rate of heat loss in still air as occurs at the measured air temperature with the existing wind.
Suppose that the inner temperature of a 3 mm thick layer of skin with a thermal conductivity of 0.35 W/(m K) is 35°C and the ambient air temperature is –20°C. Under calm ambient conditions the heat transfer coefficient at the outer skin surface is about 20
W/(m2 K) (see Table 1.4), but in a 40 mph wind it increases to 75 W/(m2 K).
(a) If frostbite can occur when the skin temperature drops to about 10°C, would you advise the skier to wear a face mask? (b) What is the skin temperature drop due to wind chill? GIVEN
Skier’s skin exposed to cold air
Skin thickness (L) = 3 mm = 0.003 m
Inner surface temperature of skin (Tsi) = 35°C
Thermal conductivity of skin (k) = 0.35 W/(m K)
Ambient air temperature (T ) = –20°C
Convective heat transfer coefficients
Still air (hc0) = 20 W/(m2 K)
40 mph air (hc40) = 75 W/(m2 K)
Frostbite occurs at an outer skin surface temperature (Tso) = 10°C
FIND
(a) Will frostbite occur under still or 40 mph wind conditions?
(b) Skin temperature drop due to wind chill.
ASSUMPTIONS
Steady state conditions prevail
One dimensional conduction occurs through the skin
Radiative loss (or gain from sunshine) is negligible
SKETCH

SOLUTION
The thermal circuit for this system is shown below

23
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(a) The rate of heat transfer is given by q =

Tsi - T•
DT
DT
=
=
Rtotal
Rk + Rc
Ê L ˆ Ê 1 ˆ
Á k A˜ + Á
Ë
¯ Ë hc A ˜
¯

T - T• q = si
L 1
A
+ k hc

The outer surface temperature of the skin in still air can be calculated by examining the conduction through the skin layer kA (Tsi – Tso) qk =
L
Solving for the outer skin surface temperature q L
Tso = Tsi – k
A k
The rate of heat transfer by conduction through the skin must be equal to the total rate of heat transfer, therefore Tso

È
˘
ÍT - T ˙ L
•
˙
= Tsi – Í si
Í L + 1 ˙ k
Í K hc ˙
Î
˚

Solving this for still air

(Tso)still air
(Tso)still air

È
˘
Í
˙
35o C - ( -20o C)
0.003m
˙
= 35°C – Í
1
Í 0.003m +
˙ 0.25 W/(m 2 K)
Í 0.25 W/(m K) 20 W/(m 2 K) ˙
Î
˚
= 24°C

For a 40 mph wind
È
˘
Í
˙
35o C - ( -20o C)
0.003m
˙
(Tso)40 mph = 35°C – Í
1
Í 0.003m +
˙ 0.25 W/(m 2 K)
Í 0.25 W/(m K) 75 W/(m 2 K) ˙
Î
˚
(Tso)40 mph = 9°C
Therefore, frostbite may occur under the windy conditions.
(b) Comparing the above results we see that the skin temperature drop due to the wind chill was
15°C.

PROBLEM 1.21
Using the information in Problem 1.20, estimate the ambient air temperature that could cause frostbite on a calm day on the ski slopes.
From Problem 1.20
Suppose that the inner temperature of a 3 mm thick layer of skin with a thermal conductivity of 0.35 W/(m K) is a temperature of 35°C. Under calm ambient conditions the heat transfer coefficient at the outer skin surface is about 20 W/(m2 K).
Frostbite can occur when the skin temperature drops to about 10°C.
24
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GIVEN
Skier’s skin exposed to cold air
Skin thickness (L) = 3 mm = 0.003 m
Inner surface temperature of skin (Tsi) = 35°C
Thermal conductivity of skin (k) = 0.35 W/(m K)
Convective heat transfer coefficient in still air ( hc ) = 20 W/(m2 K)
Frostbite occurs at an outer skin surface temperature (Tso) = 10°C
FIND
The ambient air temperature (T ) that could cause frostbite

ASSUMPTIONS
Steady state conditions prevail
One dimensional conduction occurs through the skin
Radiative loss (or gain from sunshine) is negligible
SKETCH

SOLUTION
The rate of conductive heat transfer through the skin at frostbite conditions is given by Equation (1.2) qk =

kA
(Tsi – Tso)
L

The rate of convective heat transfer from the surface of the skin, from equation (1.10), is qc = hc A (Tso – T )
These heat transfer rates must be equal qk = q c kA (Tsi – Tso) = hc A (Tso – T )
L

Solving for the ambient air temperature
Ê
Ê k ˆ k ˆ
T = Tso Á1 +
˜ – Tsi Á h L ˜
Ë hc L ¯
Ë c ¯

È
˘
0.25 W/(m K)
T = 10°C Í1 +
˙ – 35°C
2
È
˘
Í Î 20 W/(m K) ˚ (0.003m) ˚
˙
Î
25
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È
˘
0.25 W/(m K)
Í
˙
2
˘
Í Î 20 W/(m K) ˚ (0.003m) ˚
˙
ÎÈ

T = –94°C
PROBLEM 1.22
Two large parallel plates with surface conditions approximating those of a blackbody are maintained at 1500 and 500°F, respectively. Determine the rate of heat transfer by radiation between the plates in Btu/(h ft2) and the radiative heat transfer coefficient in
Btu/(h ft2 °F) and in W/(m2 K).
GIVEN
Two large parallel plates, approximately black bodies
Temperatures
T1 = 1500°F = 1960 R
T2 = 500°F = 960 R
FIND
(a) Rate of radiative heat transfer (qr/A) in Btu/(h ft2)
(b) Radiative heat transfer coefficient (hr) in Btu/(h ft2 °F) and W/(m2 K)
ASSUMPTIONS
Steady state prevails
Edge effects are negligible
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 5: Stefan-Boltzmann constant ( ) = 0.1714

10–8 Btu/(h ft2 R4)

SOLUTION
(a) The rate of heat transfer is given by Equation (1.16) qr = (T14 – T24)
A
qr
= 0.1714 ¥10-8 Btu/(h ft 2 R 4 ) (1960 R)4 - (960 R)4
A
qr
= 2.38 104 Btu/(h ft2)
A
(b) Let hr represent the radiative heat transfer coefficient

(

)(

)

qr = hr A T hr =

2.38 ¥ 104 È Btu /(h ft 2 ) ˘ qr 1
Î
˚
=
o o A DT
1500 F - 500 F
26

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hr = 23.8 Btu/(h ft 2 o F)
Converting this answer to SI units using the conversion factor found on the inside front cover of the text: Ê 5.678 W/(m 2 K) ˆ hr = 23.8 Btu/(h ft 2 o F) Á
Ë Btu/(h ft 2 o F) ˜
¯

hr = 135 W/(m2 K)
COMMENTS
Note that absolute temperatures must be used in the radiative heat transfer equation, whereas hr is based on the assumption that the rate of heat transfer is proportional to the temperature difference.
Hence hr cannot be applied to any other temperatures than those specified.
PROBLEM 1.23
A spherical vessel 0.3 m in diameter is located in a large room whose walls are at 27°C
(see sketch). If the vessel is used to store liquid oxygen at –183°C and the surface of the storage vessel as well as the walls of the room are black, calculate the rate of heat transfer by radiation to the liquid oxygen in watts and in Btu/h.
GIVEN
A black spherical vessel of liquid oxygen in a large black room
Liquid oxygen temperature (To) = –183°C = 90 K
Sphere diameter (D) = 0.3 m
Room wall temperature (Tw) = 27°C = 300 K
FIND
The rate of radiative heat transfer to the liquid oxygen in W and Btu/h
ASSUMPTIONS
Steady state prevails
The temperature of the vessel wall is the same as the temperature of the oxygen
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 5: The Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

27
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SOLUTION
The net radiative heat transfer to a black body in a black enclosure is given by Equation (1.16) qr = A

(T14 – T24)

qr =

D2

(Tw4 – To4)

qr =

(1 ft)2 0.1714 ¥10-8 Btu/(h ft 2 R 4 ) (540 R)4 – (163 R)4

(

)

qr = 454 Btu/h
Converting the net radiative heat transfer into SI units using the conversion factor given on the inside front cover of the text
Ê 0.2931W ˆ qr = 454 Btu/h Á
Ë Btu / h ˜
¯

qr = 133 W
COMMENTS
Note that absolute temperatures must be used in the radiative heat transfer equation.
PROBLEM 1.24
Repeat Problem 1.23 but assume that the surface of the storage vessel has an absorptance (equal to the emittance) of 0.1. Then determine the rate of evaporation of the liquid oxygen in kilograms per second and pounds per hour, assuming that convection can be neglected. The heat of vaporization of oxygen at –183°C is
213.3 kJ/kg.
From Problem 1.23: A spherical vessel of 0.3 m in diameter is located in a large room whose walls are at 27°C (see sketch). If the vessel is used to store liquid oxygen at –183°C and the surface of the storage vessel as well as the walls of the room are black, calculate the rate of heat transfer by radiation to the liquid oxygen in watts and in Btu/h.
GIVEN
A spherical vessel of liquid oxygen in a large black room
Emittance of vessel surface ( ) = 0.1
Liquid oxygen temperature (To) = –183°C = 90 K
Sphere diameter (D) = 0.3 m
Room wall temperature (Tw) = 27°C = 300 K
Heat of vaporization of oxygen (hfg) = 213.3 kJ/kg
FIND
(a) The rate of radiative heat transfer (qr) to the liquid oxygen in W and Btu/h
(b) The rate of evaporation of oxygen (mo) in kg/s and 1b/h
ASSUMPTIONS
Steady state prevails
The temperature of the vessel wall is equal to the temperature of the oxygen
Convective heat transfer is negligible

28
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SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 5: The Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
(a) The net radiative heat transfer from a gray body in a black enclosure, from Equation (1.17) is qr = A 1

1

(T14 – T24)

qr =

D2

(To4 – Tw4)

qr =

(0.3 m)2 (0.1) (5.67

10–8 [W/(m2 K4)] [(90 K)4 – (300 K)4)]

qk = –12.9 W
Converting this to English units using the conversion factor from the inside front cover of the text
Ê Btu / h ˆ qr = –12.9 W Á
Ë 0.2931W ˜
¯

qr = – 43.9 Btu/h
(b) The rate of evaporation of oxygen is given by mo =

qr h fg

mo =

(12.9 W) ( J/Ws )
(213.3 kJ/kg) (1000 J/kJ)

mo = 6.05

10–5 kg/s

mo = 6.05

Ê 7936.61b/h ˆ
10–5 kg/s Á
˜
Ë
¯
kg/s

In English units

mo = 0.48 lb/h

COMMENTS
Note that absolute temperatures must be used in the radiative heat transfer equation.
The negative sign in the rate of heat transfer indicates that the sphere is gaining heat from the surrounding wall.
29
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Note that the rate of heat transfer by radiation can be substantially reduced (see Problem 1.23) by applying a surface treatment, e.g., applying a metallic coating with low emissivity.
PROBLEM 1.25
Determine the rate of radiant heat emission in watts per square meter from a blackbody at (a) 150°C, (b) 600°C, (c) 5700°C.
GIVEN
A blackbody
FIND
The rate of radiant heat emission (qr) in W/m2 for a temperature of
(a) T = 150°C = 423 K
(b) T = 600°C = 873 K
(c) T = 5700°C = 5973 K
PROPERTIES AND CONSTANTS
From Appendix 2, Table 5: The Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
The rate of radiant heat emission from a blackbody is given by Equation (1.15) qr = qr =
A

A1 T14
T4

(a) For T = 423 K qr = [5.67
A

10–8 W/(m2 K4)] (423K)4

qr
= 1820 W/m2
A

(b) For T = 873 K qr = [5.67
A

0–8 W/(m2 K4)] (873 K)4

qr
= 32,900 W/m2
A

(c) For T = 5973 K qr = [(5.67
A
qr
= 7.2
A

10–8 W/(m2 K4)] (5974 K)4

107 W/m2

COMMENTS
Note that absolute temperatures must be used in radiative heat transfer equations.
30
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The rate of heat transfer is proportional to the absolute temperature to the fourth power, this results in a rapid increase in the rate of heat transfer with increasing temperature.
PROBLEM 1.26
108 m and approximates a blackbody with a surface
The sun has a radius of 7 temperature of about 5800 K. Calculate the total rate of radiation from the sun and the emitted radiation flux per square meter of surface area.
GIVEN
The sun approximates a blackbody
Surface temperature (Ts) = 5800 K
Radius (r) = 7 108 m
FIND
(a) The total rate of radiation from the sun (qr)
(b) The radiation flux per square meter of surface area (qr/A)
PROPERTIES AND CONSTANTS
From Appendix 2, Table 5: The Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
The rate of radiation from a blackbody, from Equation (1.15), is qr =

AT4
10–8 W/(m2 K4)] [4 (7

qr = [5.67 qr = 4.0

108 m)2] (5800 K)4

1026 W

The flux per square meter is given by qr =
A

T4

qr
= [5.67
A
qr
= 6.4
A

10–8 W/(m2 K4)] (5800 K)4
107 W/m2

COMMENTS
The solar radiation flux impinging in the earth’s atmosphere is only 1400 W/m2. Most of the radiation from the sun goes into space.
PROBLEM 1.27
A small gray sphere having an emissivity of 0.5 and a surface temperature of 1000°F is located in a blackbody enclosure having a temperature of 100°F. Calculate for this system: (a) the net rate of heat transfer by radiation per unit of surface area of the sphere, (b) the radiative thermal conductance in Btu/(h °F) if the surface area of the sphere is 0.1 ft2, (c) the thermal resistance for radiation between the sphere and its surroundings, (d) the ratio of thermal resistance for radiation to thermal resistance for convection if the convective heat transfer coefficient between the sphere and its surroundings is 2.0 Btu/(h ft2 °F), (e) the total rate of heat transfer from the sphere to the surroundings, and (f) the combined heat transfer coefficient for the sphere.
31
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GIVEN
Small gray sphere in a blackbody enclosure
Sphere emissivity ( s) = 0.5
Sphere surface temperature (T1) = 1000°F = 1460 R
Enclosure temperature (T2) = 100°F = 560 R
The surface area of the sphere (A) is 0.1 ft2
The convective transfer coefficient ( hc ) = 2.0 Btu/(h ft2 °F)
FIND
(a)
(b)
(c)
(d)
(e)
(f)

Rate of heat transfer by radiation per unit surface area
Radiative thermal conductance (Kr) in Btu/(h °F)
Thermal resistance for radiation (Rr)
Ratio of the radiative and conductive resistance
Total rate of heat transfer (qT) to the surroundings
Combined heat transfer coefficient ( hcr )

ASSUMPTIONS
Steady state prevails
The temperature of the fluid in the enclosure is equal to the enclosure temperature
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 5: The Stefan-Boltzmann constant ( ) = 0.1714

10–8 Btu/(h ft2 R4)

SOLUTION
(a) For a gray body radiating to a blackbody enclosure the net heat transfer is given by Equation
(1.17)
qr = A1 1 (T14 – T24) qr = (0.5) [0.1714
A

10–8 Btu/(h ft2 R4)] [(1460 R)4 – (560 R)4]

qr
= 3810 Btu/(h ft2)
A

(b) The radiative thermal conductance must be based on some reference temperature. Let the reference temperature be the enclosure temperature. Then, from Equation (1.21), the radiative thermal conductance is
Kr =

A1 f

1- 2

r (T14 - T14 )

T1 - T2¢

where f1–2 =

s

32
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Kr =

(0.1ft 2 ) (0.5)[0.1714 ¥ 10-8 Btu /(h ft 2 R 4 )](14604 - 5604 )R 4
1460 R - 560 R

Kr = 0.423 Btu/(h R)
(c) The thermal resistance for radiation is given by
1
1
Rr =
=
= 2.36 (h R)/Btu
0.423Btu /(h R)
Kr
(d) The convective thermal resistance is given by Equation (1.14)
Rc =

1
1
=
= 5.0 (h °F)/Btu
2
hc A
[2 Btu/(h ft °F)](0.1ft 2 )

Therefore the ratio of the radiative to the convective resistance is
Rr
2.36(h R)/Btu
=
= 0.47
5.0 (h R)/Btu
Rc

(e) The radiative and convective resistances are in parallel, therefore the total resistance, from Figure
1.18, is
Rtotal =

Rc Rr
Rc + Rr

=

(5.0) (2.36)
= 1.60 (h R)/Btu
5.0 + 2.36

The total heat transfer is given by: qT =

DT
1460 R - 560 R
=
= 561 Btu/h
1.60 (h R)/Btu
Rtotal

(f) The combined heat transfer coefficient can be calculated from qT = hcr A T hcr =

qT
561Btu/h
=
2
A DT
(0.1ft ) (1460 R - 560 R)

hcr = 6.23 Btu/(h ft2 °F)

COMMENTS
Note that absolute temperatures must be used in the radiative heat transfer equations.
Both heat transfer mechanisms are of the same order of magnitude in this situation.
PROBLEM 1.28
A spherical communications satellite 2 m in diameter is placed in orbit around the earth.
The satellite generates 1000 W of internal power from a small nuclear generator. If the surface of the satellite has an emittance of 0.3 and is shaded from solar radiation by the earth, estimate the surface temperature.
GIVEN
Spherical satellite
Diameter (D) = 2 m
Heat generation = 1000 W
Emittance ( ) = 0.3
33
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FIND
The surface temperature (Ts)
ASSUMPTIONS
The satellite radiates to space which behaves as a blackbody enclosure at 0 K
The system is in steady state
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 5: The Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
From Equation (1.17), the rate of the heat transfer from a gray body in a blackbody enclosure is qr = A 1

1

(T14 – T24)

Solving this for the surface temperature
1

1

Ê q ˆ 4 Ê qr ˆ 4
T1 = Á r ˜ = Á
Ë A1e1s ¯
Ë p D 2 e1s ˜
¯

For steady state the rate of heat transfer must equal the rate of internal generation, therefore the surface temperature is
1

Ê
ˆ4
1000 W
T1 = Á
= 262 K = –11°C
-8
2
2 4 ˜
Ë p (2 m) (0.3)5.67 ¥10 W/(m K ) ¯

PROBLEM 1.29
A long wire 0.03 inches in diameter with an emissivity of 0.9 is placed in a large quiescent air space at 20°F. If the wire is at 1000°F, calculate the net rate of heat loss.
Discuss your assumptions.
GIVEN
Long wire in still air
Wire diameter (D) = 0.03 in.
Wire temperature (Ts) = 1000°F = 1460 R
Emissivity ( ) = 0.9
Air temperature (T ) = 20°F = 480 R
FIND
The net rate of heat loss
34
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ASSUMPTIONS
The enclosure around the wire behaves as a blackbody enclosure at the temperature of the air
The natural convection heat transfer coefficient is 3 Btu/(h ft2 °F) (From Table 1.4)
Steady state conditions prevail
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 5: The Stefan-Boltmann constant ( ) = 0.1714

10–8 Btu/(h ft2 R4)

SOLUTION
The total rate of heat loss from the wire is the sum of the convective (Equation (1.10)) and radiative
(Equation (1.17)) losses
(Ts4 – T 4)

qtotal = hc A (Ts – T ) + A qtotal = [3 Btu/(h ft2 °F)]

(0.03 in) (1 ft/12 in) L (1460 R – 480 R)

0.03 ˆ
Ê
+ Áp L ft ˜ (0.9) [0.1714
Ë
12 ¯

10–8 Btu/(h ft2 R4)] [(1460 R)4 – (480 R)4]

qtotal
= 77 Btu/(h ft) = 77 Btu/h per foot of wire length
L

COMMENTS
The radiative heat transfer is about twice the magnitude of the convective transfer.
The enclosure is more likely a gray body, therefore the actual rate of loss will be smaller than we have calculated. The convective heat transfer coefficient may differ by a factor of two or three from our assumed value. PROBLEM 1.30
Wearing layers of clothing in cold weather is often recommended because dead-air spaces between the layers keep the body warm. The explanation for this is that the heat loss from the body is less. Compare the rate of heat loss for single 3/4-in-thick layer of wool [k = 0.020 Btu/(hr ft °F)] with three 1/4-in layers separated by 1/16-in air gaps. The thermal conductivity of air is 0.014 Btu/(hr ft °F).
GIVEN
Wool insulation
Thermal conductivities
Wool (kw) = 0.02 Btu/(h ft °F) and air (ka) = 0.014 Btu/(h ft °F)

35
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FIND
Compare the rate of heat loss for a single 0.75 in (.0625 ft) layer of wool to that of three 0.25 in.
(0.0208 ft) layers separated by 1/16 in (0.00521 ft) layers of air
ASSUMPTIONS
Heat transfer can be approximated as one dimensional, steady state conduction
SKETCH

SOLUTION
The thermal resistance for the single thick layer, from Equation (1.3), is
L
0.0625ft
1
Rka =
=
=
3.125 (h ft2°F)/Btu
2
K w A [0.02 Btu /(h ft °F)]A
A
Therefore the rate of conductive heat transfer is
DT
DT
=
= 0.32 A T Btu/(h ft2°F)
1
Rka
2
3.125(h ft °F)/Btu
A
The thermal resistance for the three thin layers is the sum of the resistance of the wool and the air between the layers

qka =

Rkb=

Lw
L
(2 layers) (0.00521ft/Layer )
(3layers) (0.0208ft/layer )
+
+ a =
2
k w A ka A
[0.02 Btu/(h ft °F)]A
[0.014 Btu/(h ft 2 °F)]A

Rkb =

1
3.86 (h ft2°F)/Btu
A

Therefore, the rate of conductive heat transfer for the three layer situation is
DT
DT qkb =
=
= 0.26 A T Btu/(h ft2°F)
1
k kb
2
3.86 (h ft °F)/Btu
A
Comparing the rate of heat loss for the two situations qkb 0.26
=
= 0.81 qka 0.32

Therefore, for the same temperature difference, the heat loss through the three layers of wool is only
81% of the heat loss through the single layer.

36
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PROBLEM 1.31
A section of a composite wall with the dimensions shown below has uniform temperatures of 200°C and 50°C over the left and right surfaces, respectively. If the thermal conductivities of the wall materials are: kA = 70 W/(m K), kB = 60 W/(m K), kc = 40 W/(m K) and kD = 20 W/(m K), determine the rate of heat transfer through this section of the wall and the temperatures at the interfaces.
GIVEN
A section of a composite wall
Thermal conductivities kA = 70 W/(m K) kB = 60 W/(m K) kC = 40 W/(m K) kD = 20 W/(m K)
Surface temperatures
Left side (TAs) = 200°C
Right side (TDs) = 50°C
FIND
(a) Rate of heat transfer through the wall (q)
(b) Temperature at the interfaces
ASSUMPTIONS
One dimensional conduction
The system is in steady state
The contact resistances between the materials is negligible
SKETCH

SOLUTION
The thermal circuit for the composite wall is shown below

(a) Each of these thermal resistances has a form given by Equation (1.3)
L
Rk =
Ak
Evaluating the thermal resistance for each component of the wall
37
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RA =

LA
0.02 m
=
= 0.0794 K/W
(0.06 m) (0.06 m)[70 W/(m K)]
AA k A

RB =

LB
0.025m
=
= 0.2315 K/W
(0.03m) (0.06 m)[60 W/(m K)]
AB k B

RC =

LC
0.025m
=
= 0.3472 K/W
(0.03m) (0.06 m)[40 W/(m K)]
AC kC

RD =

LD
0.04 m
=
= 0.5556 K/W
(0.06 m) (0.06 m)[20 W/(m K)]
AD k D

The total thermal resistance of the wall section, from Section 1.5.1, is
Rtotal = RA +

RB RC
+ RD
RB + RC

Rtotal = 0.0794 +

(0.2315) (0.3472)
+ 0.5556 K/W
0.2315 + 0.3472

Rtotal = 0.7738 K/W
The total rate of heat transfer through the composite wall is given by q =

200o C - 50o C
DT
=
= 194 W
Rtotal
0.7738 K/W

(b) The average temperature at the interface between material A and materials B and C (TABC) can be determined by examining the conduction through material A alone qka =

TAs - TABC
=q
RA

Solving for TABC
TABC = TAs – q RA = 200°C – (194 W) (0.0794 K/W) = 185°C
The average temperature at the interface between material D and materials B and C (TBCD) can be determined by examining the conduction through material D alone qkD =

TBCD - TDs
=q
RD

Solving for TBCD
TBCD = TDs + q RD = 50°C + (194 W) (0.5556 K/W) = 158°C
PROBLEM 1.32
Repeat the Problem 1.31 including a contact resistance of 0.1 K/W at each of the interfaces. Problem 1.31: A section of a composite wall with the dimensions shown in the schematic diagram below has uniform temperatures of 200°C and 50°C over the left and right surfaces, respectively. If the thermal conductivities of the wall materials are: kA = 70
W/(m K), kB = 60 W/(m K), kC = 40 W/(m K), and kD = 20 W/(m K), determine the rate of heat transfer through this section of the wall and the temperatures at the interfaces.
38
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GIVEN
Composite wall
Thermal conductivities: kA = 70 W/(m K) kB = 60 W/(m K) kC = 40 W/(m K) kD = 20 W/(m K)
Surface temperatures
Left side (TAs) = 200°C
Right side (TDs) = 50°C
Contact resistance at each interface (Ri) = 0.1 K/W
FIND
(a) Rate of heat transfer through the wall (q)
(b) Temperatures at the interfaces
ASSUMPTIONS
One dimensional conduction
The system is in steady state
SKETCH

SOLUTION
The thermal circuit for the composite wall with contact resistances is shown below

The values of the individual resistances, from Problem 1.31, are
RA = 0.0794 K/W

RB = 0.2315 K/W

RC = 0.3472 K/W

RD = 0.5556 K/W

(a) The total resistance for this system is
Rtotal = RA + Ri +

RB RC
+ Ri + RD
RB + RC

Rtotal = 0.0794 + 0.1 +

(0.2315) (0.3472)
+ 0.1 + 0.5556 K/W
0.2315 + 0.3472

Rtotal = 0.9738 K/W
39
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The total rate of heat transfer through the composite wall is given by q =

DT
200 ∞ C - 50 °C
=
= 154 W
0.9738 K/W
Rtotal

(b) The average temperature on the A side of the interface between material A and material B and C
(T1A) can be determined by examining the conduction through material A alone q =

TAs - T1A
RA

Solving for T1A
T1A = TAs – q RA = 200°C – (154 W) (0.0794 K/W) = 188°C
The average temperature on the B and C side of the interface between material A and materials B and
C (T1BC) can be determined by examining the heat transfer through the contact resistance
T -T q = 1A 1BC
Ri
Solving for T1BC
T1BC = T1A – q Ri = 188°C – (154 W) (0.1 K/W) = 172°C
The average temperature on the D side of the interface between material D and materials B and C
(T2D) can be determined by examining the conduction through material D alone
T - TDs q = 2D
RD
Solving for T2D
T2D = TDs + q RD = 50°C + (154 W) (0.5556 K/W) = 136°C
The average temperature on the B and C side of the interface between material D and materials B and
C (T2BC) can be determined by examining the heat transfer through the contact resistance
T
- T2 D q = 2 BC
Ri
Solving for T2BC
T2BC = T2D + q Ri = 136°C + (154 W) (0.1 K/W) = 151°C
COMMENTS
Note that the inclusion of the contact resistance lowers the calculated rate of heat transfer through the wall section by about 20%.
PROBLEM 1.33
Repeat the Problem 1.32 but assume that instead of surface temperatures, the given temperatures are those of air on the left and right sides of the wall and that the convective heat transfer coefficients on the left and right surfaces are 6 and 10 W/(m2
K), respectively.
Problem 1.32: Repeat the Problem 1.31 including a contact resistance of 0.1 K/W at each of the interfaces.
Problem 1.31: A section of a composite wall with the dimensions shown in the schematic diagram below has uniform temperatures of 200°C and 50°C over the left and right surfaces, respectively. If the thermal conductivities of the wall materials are: kA = 70
W/(m K), kB = 60 W/(m K), kC = 40 W/(m K), and kD = 20 W/(m K), determine the rate of heat transfer through this section of the wall and the temperatures at the interfaces.
40
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GIVEN
Composite wall
Thermal conductivities kA = 70 W/(m K) kB = 60 W/(m K) kC = 40 W/(m K) kD = 20 W/(m K)
Air temperatures
Left side (TA ) = 200°C
Right side (TD ) = 50°C
Contact resistance at each interface (Ri) = 0.1 K/W
Convective heat transfer coefficients
Left side ( hcA ) = 6 W/(m2 K)
Right side ( hcD ) = 10 W/(m2 K)
FIND
(a) Rate of heat transfer through the wall (q)
(b) Temperatures at the interfaces
ASSUMPTIONS
One dimensional, steady state conduction
SKETCH

SOLUTION
The thermal circuit for the composite wall with contact resistances and convection from the outer surfaces is shown below

The values of the individual conductive resistances, from Problem 1.31, are
RA = 0.0794 K/W
RB = 0.2315 K/W
RC = 0.3472 K/W
RD = 0.5556 K/W
The values of the convective resistances, using Equation (1.14), are
RcA =
RcD =

1 hcA A
1
hcD A

=
=

1
2

[6 W/(m K)](0.06 m) (0.06 m)

= 46.3 K/W

1
2

[10 W/(m K)](0.06 m) (0.06 m)

= 27.8 K/W

41
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(a) The total resistance for this system is
Rtotal = RcA + RA + Ri +

RB RC
+ Ri + RD + RcD
RB + RC

(0.2315) (0.3472)
0.2315 + 0.34472
+ 0.1 + 0.5556 + 27.8 K/W
Rtotal = 75.1 K/W
The total rate of heat transfer through the composite wall is given by

Rtotal = 46.3 + 0.0794 + 0.1 +

q =

DT
200 ∞C - 50 ∞C
=
= 2.0 W
Rtotal
75.1K/W

(b) The surface temperature on the left side of material A (TAs) can be determined by examining the convection from the surface of material A q =

TA• - TAs
RcA

Solving for TAs
TAs = TA – q RcA = 200°C – (2 W) (46.3 K/W) = 107.4°C
The average temperature on the A side of the interface between material A and material B and C (T1A) can be determined by examining the conduction through material A alone q =

TAs - T1A
RA

Solving for T1A
T1A = TAs – q RA = 107.4°C – (2 W) (0.0794 K/W) = 107.2°C
The average temperature on the B and C side of the interface between material A and materials B and
C (T1BC) can be determined by examining the heat transfer through the contact resistance q =

T1A - T1BC
Ri

Solving for T1BC
T1BC = T1A – q Ri = 107.2°C – (2 W) (0.1 K/W) = 107.0°C
The surface temperature on the D side of the wall (TDs) can be determined by examining the convection from that side of the wall q =

TDs - TD•
RcD

Solving for TDs
TDs = TD + q RcD = 50°C + (2 W) (27.8 K/W) = 105.6°C
The average temperature on the D side of the interface between material D and materials B and C
(T2D) can be determined by examining the conduction through material D alone q =

T2D - TDs
RD

Solving for T2D
T2D = TDs + q RD = 105.6°C + (2 W) (0.5556 K/W) = 106.7°C
The average temperature on the B and C side of the interface between material D and materials B and
C (T2BC) can be determined by examining the heat transfer through the contact resistance
T
- T2 D q = 2 BC
Ri
42
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Solving for T2BC
T2BC = T2D + q Ri = 106.7°C + (2 W) (0.1 K/W) = 106.9°C
COMMENTS
Note that the addition of the convective resistances reduced the rate of heat transfer through the wall section by a factor of 77.
PROBLEM 1.34
Mild steel nails were driven through a solid wood wall consisting of two layers, each 2.5 cm thick, for reinforcement. If the total cross-sectional area of the nails is 0.5% of the wall area, determine the unit thermal conductance of the composite wall and the percent of the total heat flow that passes through the nails when the temperature difference across the wall is 25°C. Neglect contact resistance between the wood layers.
GIVEN
Wood wall
Two layers 0.025 m thick each
Nail cross sectional area of nails = 0.5% of wall area
Temperature difference ( T) = 25°C
FIND
(a) The unit thermal conductance (k/L) of the wall
(b) Percent of total heat flow that passes through the wall
ASSUMPTIONS
One dimensional heat transfer through the wall
Steady state prevails
Contact resistance between the wall layers is negligible
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Tables 10 and 11
Thermal conductivities
Wood (Pine) (kw) = 0.15 W/(m K)
Mild steel (1% C) (ks) = 43 W/(m K)
SOLUTION
(a) The thermal circuit for the wall is

43
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The individual resistances are
Lw
0.05 m
1
Rw =
=
=
(0.995 Awall [0.15 W/(m K)]
Aw k w
Awall 0.335 (K m 2 )/W
Rs =

Ls
0.05 m
1
=
=
(0.005 Awall [43W/(m K)]
As k s
Awall 0.233 (K m 2 )/W

The total resistance of the wood and steel in parallel is
Rtotal =

Rw Rs
1 È (0.335) (0.233) ˘
1
(K m 2 )/W =
=
0.1374 (K m 2 )/W
Awall Í 0.335 + 0.233 ˙
Rw + Rs
Awall
Î
˚

The unit thermal conductance (k/L) is:
1
k
1
=
=
= 7.3 W/(K m2)
2
Rtotal Awall
L
0.1374(K m )/W
(b) The total heat flow through the wood and nails is given by
DT
25∞C qtotal =
=
1
Rtotal
0.1374(K m 2 )/W
Awall
qtotal
= 182 W/m2
Awall

The heat flow through the nails alone is
DT
qnails =
=
Rnails

25 ∞C
1
Awall

0.233(K m 2 )/W

qnails
= 107 W/m2
Awall

Therefore the percent of the total heat flow that passes through the nails is
107
Percent of heat flow through nails =
100 = 59%
182
PROBLEM 1.35
Calculate the rate of heat transfer through the composite wall in Problem 1.34 if the temperature difference is 25°C and the contact resistance between the sheets of wood is
0.005 m2 K/W.
Problem 1.34: To reinforce a solid wall consisting of two layers, each 2.5 cm thick, mild steel nails were driven through it. If the total cross sectional area of the nails is 0.5% of the wall area, determine the unit thermal conductance of the composite wall and the percent of the total heat flow that passes through the nails when the temperature difference across the wall is 25°C. Neglect contact resistance between the wood layers.
GIVEN
Wood wall
Two layers 0.025 m thick each, nailed together
Nail cross sectional area of nails = 0.5% of wall area
Temperature difference ( T) = 20°C
Contact resistance (A Ri) = 0.005 (m2 K)/W
FIND
The rate of heat transfer through the wall
44
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ASSUMPTIONS
One dimensional heat transfer through the wall
Steady state prevails
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Tables 10 and 11
Thermal conductivities
Wood (Pine) (kw) = 0.15 W/(m K)
Mild steel (1% C) (ks) = 43 W/(m K)
SOLUTION
The thermal circuit for the wall with contact resistance is shown below.

From Problem 1.34, the thermal resistance of the wood and the nails are
1
1
Rw =
0.335 (K m2)/W
Rs =
0.233 (K m2)/W
Awall
Awall
The combined resistance of the wood and the contact resistance in series is
1
1
È 0.355 (K m 2 )/W + 0.005 (K m 2 )/W ˘
Rwi = Rw + Ri = Rw +
(A Ri) =
˚
Awall Î
A
Rwi =

1
Awall

0.360 (K m2)/W

The total resistance equals the combined resistance of the wood and the contact resistance in parallel with the resistance of the nails
Rtotal =

Rwi Rs
1 È (0.360) (0.233) ˘
1
2
=
= 0.1415 (K m2)/W
Í 0.360 + 0.233 ˙ (K m )/W = A
Rwi + Rs
Awall Î
˚
wall

Therefore the rate of heat flow through the wall is:
25∞C
DT
=
q =
1
Rtotal
0.1415 (K m 2 )/W
Awall
q
Awall

= 172 W/m2

COMMENTS
In this case the inclusion of the contact resistance lowered the calculated rate of heat transfer by only
3% because most of the heat is transferred through the nails (see Problem 1.34).
45
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PROBLEM 1.36
Heat is transferred through a plane wall from the inside of a room at 22°C to the outside air at –2°C. The convective heat transfer coefficients at the inside and outside surfaces are 12 and 28 W/(m2 K), respectively. The thermal resistance of a unit area of the wall is
0.5 m2 K/W. Determining the temperature at the outer surface of the wall and the rate of heat flow through the wall per unit area.
GIVEN
Heat transfer through a plane wall
Air temperature
Inside wall (Ti) = 22°C and outside wall (To) = –2°C
Heat transfer coefficient
Inside wall ( hci ) = 12 W/(m2 K)
Outside wall ( hco ) = 28 W/(m2 K)
Thermal resistance of a unit area (A Rw) = 0.5 (m2 K)/W
FIND
(a) Temperature of the outer surface of the wall (Two)
(b) Rate of heat flow through the wall per unit area (q/A)
ASSUMPTIONS
One dimensional heat flow
Steady state has been reached
SKETCH

SOLUTION
The thermal circuit for the wall is shown below

The rate of heat transfer can be used to calculate the temperature of the outer surface of the wall, therefore part (b) will be solved first.
(b) The heat transfer situation can be visualized using the thermal circuit shown above. The total heat transfer through the wall, from Equations (1.33) and (1.34), is q =

DTtotal
Rtotal

The three thermal resistances are in series, therefore
Rtotal = Rci + Rw + R
Rtotal =

A Rw
1
1
+
+
A
Ahci
Ah•
46

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The heat flow through the wall is q =

q
=
A

Ti - To
1Ê 1
1ˆ
Á h + A Rw + h ˜
A Ë ci
•¯

22°C - ( -2°C)
1
1
+ 0.5(m 2 K)/W +
2
12 W/(m K)
28 W/(m 2 K)

q
= 38.8 W/m2
A

(a) The temperature of the outer surface of the wall can be calculated by examining the convective heat transfer from the outside of the wall (given by Equation (1.10)) qc = hco (Two – To)
A

Solving for Two
Two =

Ê
ˆ
q 1
1
+ To = (38.8 W/m2 Á
+ (–2°C) = – 0.6°C
2
A hco
Ë 28 W/(m K) ˜
¯

COMMENTS
Note that the conductive resistance of the wall is dominant compared to the convective resistance.
PROBLEM 1.37
How much fiberglass insulation [k = 0.035 W/(m K)] is needed to guarantee that the outside temperature of a kitchen oven will not exceed 43°C? The maximum oven temperature to be maintained by the convectional type of thermostatic control is 290°C, the kitchen temperature may vary from 15°C to 33°C and the average heat transfer coefficient between the oven surface and the kitchen is 12 W/(m2 K).
GIVEN
Kitchen oven wall insulated with fiberglass
Fiberglass thermal conductivity (k) = 0.035 W/(m K)
Convective transfer coefficient on the outside of wall ( hc ) = 12 W/(m2 K)
Maximum oven temperature (Ti) = 290°C
Kitchen temperature (T ) may vary: 15°C < T < 33°C
FIND
Thickness of fiberglass (L) to keep the temperature of the outer surface of the oven (Two) at 43°C or less
ASSUMPTIONS
One dimensional, steady state heat transfer prevails
The temperature of the inside of the wall (Twi) is the same as the oven temperature
The thermal resistance of the metal wall of the oven is negligible

47
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SKETCH

SOLUTION
For steady state conditions, the heat transfer by conduction through the wall, from Equation (1.2), must be equal to the heat transfer by convection from the outer surface of the wall, from Equation
(1.10)
kA qk =
(Twi – Two) = qc = hc A (Two – T )
L
Solving for L k (Twi - Two )
L = hc (Two - T• )
By examination of the above equation, the greatest thickness required for a given Two will occur when
Twi and T are at their maximum values
L =

0.035 W/(m K) (290o C - 43o C)
12 W/(m 2 K) (43o C - 33o C)

= 0.072 m = 7.2 cm

COMMENTS
In a real design a slightly thicker layer of insulation should be chosen to provide a margin of safety in case the convective heat transfer coefficient on the outside of the wall in some circumstances is less than expected due to the location of the oven in the kitchen or other unforseen factors.
PROBLEM 1.38
A heat exchanger wall consists of a copper plate 3/8 in. thick. The heat transfer coefficients on the two sides of the plate are 480 and 1250 Btu/(h ft2 °F), corresponding to fluid temperatures of 200 and 90°F, respectively. Assuming that the thermal conductivity of the wall is 220 Btu/(h ft °F), (a) compute the surface temperatures in °F, and (b) calculate the heat flux in Btu/(h ft2).
GIVEN
Heat exchanger wall, thickness (L) = 3/8 in = 0.03125 ft
Heat transfer coefficients hc1 = 480 Btu/(h ft2 °F) hc2 = 1250 Btu/(h ft2 °F)
Fluid temperatures
Tf 1 = 200°F
Tf 2 = 90°F
Thermal conductivity of the wall (k) = 220 Btu/(h ft °F)
FIND
(a) Surface temperatures (Tw1, Tw1) in °F
(b) The heat flux (q/A) in Btu/(h ft2)
48
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ASSUMPTIONS
One dimensional heat transfer prevails
The system has reached steady state
Radiative heat transfer is negligible
SKETCH

SOLUTION
The thermal circuit for the wall is shown below

The surface temperatures can only be calculated after the heat flux has been established, therefore part
(b) will be solved before part (a).
(b) The resistances are in series, therefore the total resistance is
3

Rtotal =

 Ri = Rc1 + Rw + Rc2 i =1

The total rate of heat transfer is given by Equation (1.33) and (1.34) q =

DT
DT
=
=
Rtotal
Rc1 + Rw + Rc 2

T1 - T2
1
L
1
+
+
hc1 A kA hc 2 A

Therefore the heat flux (q/A) is q =
A

200 oF - 90 oF
= 3.64
1
0.03125ft
1
+
+
180Btu/(h ft 2 oF) 220Btu/(h ft oF) 1250Btu/(h ft 2 oF)

104 Btu/(h ft2)

(a) Equation (1.10) can be applied to the convective heat transfer on the fluid 1 side qc = hc1 (Tf 1 – Tw1)
A

Solving for Tw1
Tw1 = Tf 1 –

q 1
= 200°F + [3.64
A hc1

Ê
ˆ
1
104 Btu/(h ft2)] Á
= 124°F
2o ˜
Ë 480 Btu/(h ft F) ¯

Similarly, on the fluid 2 side qc = hc 2 (Tw2 – Tf2)
A
Tw2 = Tf 2 –

q 1
= 90°F + [3.64
A hc 2

Ê
ˆ
1
104 Btu/(h ft2)] Á
= 119°F
2o ˜
Ë 1250 Btu/(h ft F) ¯
49

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PROBLEM 1.39
A submarine is to be designed to provide a comfortable temperature for the crew of no less than 70°F. The submarine can be idealized by a cylinder 30 ft in diameter and 200 ft in length. The combined heat transfer coefficient on the interior is about
2.5 Btu/(h ft2 °F), while on the outside the heat transfer coefficient is estimated to vary from about 10 Btu/(h ft2 °F) (not moving) to 150 Btu/(h ft2 °F) (top speed). For the following wall constructions, determine the minimum size in kilowatts of the heating unit required if the sea water temperatures vary from 34 to 55°F during operation. The
1
3 walls of the submarine are (a) in. aluminum (b) in. stainless steel with a 1 in. thick
2
4
3
layer fiberglass insulation on the inside and (c) of sandwich construction and a in. 4
1
in. thickness of stainless steel, a 1 in. thick layer of fiberglass insulation, and a
4
thickness of aluminum on the inside. What conclusions can you draw?
GIVEN
Submarine
Inside temperature (Ti) > 70°F
Can be idealized as a cylinder
Diameter (D) = 30 ft length (L) = 200 ft
Combined heat transfer coefficients
Inside ( hci ) = 2.5 Btu/(h ft2 °F)
Outside ( hco ): not moving = 10 Btu/(h ft2 °F) top speed: 150 Btu/(h ft2 °F)
Sea water temperature (To) varies: 34°F < To < 55°F
FIND
Minimum size of the heating unit (q) in kW for
(a) 0.5 inch thick aluminum walls
(b) 0.75 inch thick stainless steel with 1.0 inch of fiberglass insulation
(c) Sandwich of 0.75 inch stainless steel, 1.0 inch of fiberglass insulation, and 0.25 inch of aluminum
ASSUMPTIONS
Steady state prevails
Heat transfer can be approximated as heat transfer through a flat plate with the surface area of the cylinder Constant thermal conductivities
Contact resistance between the difference materials is negligible
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Tables 10, 11, and 12: The thermal conductivities are
50
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Aluminum (ka) = 136.3 Btu/(h ft °F) at 32°F
Stainless steel (ks) = 8.3 Btu/(h ft °F) at 68°F
Fiberglass insulation (kfg) = 0.20 Btu/(h ft °F) at 68°F
SOLUTION
The thermal circuits for the three cases are shown below

The total surface area of the idealized submarine (A) is
DL + 2

A =

p
D2
= (200 ft) (30 ft) +
(30 ft)2 = 20,260 ft2
4
2

(a) For case (a) the total resistance is
3

Rtotal = S Ri = Ri + Ra + Ro = i-1 1 hci A

+

L
1
+ ka A hco A

The heat transfer through the wall is
DT
=
Rtotal

q =

Ti - To
L
1
1
+ a + hci A ka A hco A

By examination of the above equation, the heater requirement will be the largest when To is at its minimum value and hco is at its maximum value q= 20,260 ft 2 (70°F - 34°F)
= 1.79
0.5
ft
1
1
12
+
+
2.5Btu/(h ft 2 °F) 136.3Btu/(h ft °F) 150 Btu/(h ft 2 °F)

106 Btu/h

Converting this result to kilowatts
1.79

Ê kW ˆ
106 Btu/h (0.2931 W/Btu/h ) Á
= 525 kW
Ë 1000 W ˜
¯

(b) Similarly, for case (b), the total resistance is
4

1

i=1

hci A

Rtotal = S Ri = Rs + Ra + Rfg + Ro =

+

L fg
Ls
1
+
+ k s A k fg A hco A

The size of heater needed is

51
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q=

20,260 ft 2 (70°F - 34°F)
0.75
1 ft ft
1
1
12
12
+
+
+
2
8.3Btu/(h ft °F) 0.02 Btu/(h ft °F) 150 Btu/(h ft 2 ∞F)
2.5Btu/(h ft °F)

q = 1.59

105 Btu/h

q = 46.6 kW
(c) The total resistance for case (c) is
5

1

i=1

hci A

Rtotal = S Ri = Rs + Ra + Rfg + Ra + Ro =

+

L fg
Ls
L
1
+
+ a + k s A k fg A ka A hco A

The size of heater needed is q= 20,260ft 2 (70°F - 34°F)
0.25
1 ft ft 1
1
12
12
+
+
+
+
2
2
2
2.5 Btu/(h ft ∞F) 8.3 Btu/(h ft ∞F) 0.02 Btu/(h ft ∞F) 136.3 Btu/(h ft 2 ∞F) 150 Btu/(h ft 2 ∞F)
0.75
ft
12

q = 1.59

105 Btu/h

q = 46.6 kW
COMMENTS
Neither the aluminum nor the stainless steel offers any appreciable resistance to heat loss.
Fiberglass or other low conductivity material is necessary to keep the heat loss down to a reasonable level. PROBLEM 1.40
A simple solar heater consists of a flat plate of glass below which is located a shallow pan filled with water, so that the water is in contact with the glass plate above it. Solar radiation is passing through the glass at the rate of 156 Btu/(h ft2). The water is at 200°F and the surrounding air is 80°F. If the heat transfer coefficients between the water and the glass and the glass and the air are 5 Btu/(h ft2 °F), and 1.2 Btu/(h ft2 °F), respectively, determine the time required to transfer 100 Btu per square foot of surface to the water in the pan. The lower surface of the pan may be assumed to be insulated.
GIVEN
A simple solar heater: shallow pan of water below glass, the water touches the glass
Solar radiation passing through glass (qr/A) = 156 Btu/(h ft2)
Water temperature (Tw) = 200°F
Surrounding air temperature (T ) = 80°F
Heat transfer coefficients
Between water and glass ( hcw ) = 5 Btu/(h ft2 °F)
Between glass and air ( hca ) = 1.2 Btu/(h ft2 °F)
FIND
The time (t) required to transfer 100 Btu/ft2 to the water
ASSUMPTIONS
One dimensional, steady state heat transfer prevails
The heat loss from the bottom of the pan is negligible
52
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The radiative loss from the top of the glass is negligible
The thermal resistance of the glass is negligible

SKETCH

SOLUTION
The total thermal resistance between the water and the surrounding air is the sum of the two convective thermal resistances
2

1

i=1

hcw A

Rtotal = S Ri = Rcw + Rca =
Rtotal =

1
2

A[5Btu/(h ft °F)]

+

+

1 hca A
1
2

A[1.2 Btu/(h ft °F)]

=

1
1.03 (h ft2°F)/Btu
A

The net rate of heat transfer to the water is qtotal q q q
DT
= r = c = r =
A
A
A A Rtotal
A
qtotal
200°F - 80°F
= 156 Btu/(h ft2) –
1
A
A Ê 1.03 (h ft 2 °F)/Btu ˆ
ËA
¯ qTotal = 40 Btu/(h ft2)
A
At this rate, the time required to transfer 100 Btu/h to the water is
100 Btu/ft 2
100 Btu/ft 2
=
qTotal
40 Btu/(h ft 2 )
A
t = 2.5 hours

t =

PROBLEM 1.41
A composite refrigerator wall is composed of 2 in. of corkboard sandwiched between a½
1
in. thick layer of oak and a in. thickness of aluminum lining on the inner surface.
32
The average convective heat transfer coefficients at the interior and exterior wall are 2 and 1.5 Btu/(h ft2 °F), respectively. (a) Draw the thermal circuit. (b) Calculate the individual resistances of the components of this composite wall and the resistances at the surfaces. (c) Calculate the overall heat transfer coefficient through the wall. (d) For an air temperature inside the refrigerator of 30°F and outside of 90°F, calculate the rate of heat transfer per unit area through the wall.
GIVEN
Refrigerator wall: oak, corkboard, and aluminum
Thicknesses
Oak (Lo) = 0.5 in
53
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Corkboard (Lc) = 2 in
1
Aluminum (La) = in 32
Convective heat transfer coefficients
Interior ( hci ) = 2 Btu/(h ft2 °F)
Exterior ( hco ) = 1.5 Btu/(h ft2 °F)
Air temperature
Inside (Ti) = 30°F and Outside (To) = 90°F
FIND
(a)
(b)
(c)
(d)

Draw the thermal circuit
The individual resistances
Overall heat transfer coefficient (U)
Rate of heat transfer per unit area (q/A)

ASSUMPTIONS
One dimensional, steady state heat transfer
Constant thermal conductivities
Contact resistance between the different materials is negligible
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Tables 11 and 12, the thermal conductivities are
Oak (ko) = 0.11 Btu/(h ft °F) at 68°F
Corkboard (kc) = 0.024 Btu/(h ft °F) at 68°F
Aluminum (ka) = 136 Btu/(h ft °F) at 32°F
SOLUTION
(a) The thermal circuit for the refrigerator wall is shown below

(b) The resistances to convection from the inner and outer surfaces is given by Equation (1.14)
1
Rc = hc A
Rci =
R =

1 hci A
1
hco A

=
=

1
2o

[2 Btu /(h ft F)]A

=

1
2o

[1.5Btu /(h ft F)]A

1
0.5 (h ft2 °F)/Btu
A

=

1
0.67 (h ft2 °F)/Btu
A

54
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The resistances to conduction through the components of the wall is given by Equation (1.3)
L
Rk =
Ak

Rka

Rkc

Rko

Ê 1 in ˆ Ê 1 ft/in ˆ
Ë 32 ¯ Ë 12
¯ 1
La
=
=
= 1.9
Ak a A[136 Btu/(ft °F)] A

10–5 (h ft2 °F)/Btu

2 ft Lc
1
12
=
=
=
6.9 (h ft2 °F)/Btu
A[0.024 Btu/(ft °F)]
Ak c
A
Ê 1 in ˆ Ê 1 ft/in ˆ
Ë 2 ¯ Ë 12
¯
Lo
1
=
=
=
0.38 (h ft2 °F)/Btu
A[0.11Btu/(ft °F)]
Ak o
A

(c) The overall heat transfer coefficient satisfies Equation (1.34)
1
UA =
Rtotal
1
1
=
U =
A Rtotal A ( Rci + Rka + Rkc + Rko + Rco )
1
U =
-5
(0.5 + 1.9 ¥ 10 + 6.9 + 0.38 + 0.67) (h ft 2 oF) /Btu
U = 0.12 Btu/(h ft2 °F)
(d) The rate of heat transfer through the wall is given by Equation (1.33) q = U T = 0.12 Btu/(h ft 2 °F) (90°F – 30°F) = 7.2 Btu/(h ft2)
A

(

)

COMMENTS
The thermal resistance of the corkboard is more than three times greater than the sum of the other resistances. The thermal resistance of the aluminum is negligible.
PROBLEM 1.42
An electronic device that internally generates 600 mW of heat has a maximum permissible operating temperature of 70°C. It is to be cooled in 25°C air by attaching aluminum fins with a total surface area of 12 cm2. The convective heat transfer coefficient between the fins and the air is 20 W/(m2 K). Estimate the operating temperature when the fins are attached in such a way that: (a) there exists a contact resistance between the surface of the device and the fin array of approximately
50 K/W, and (b) there is no contact resistance but the construction of the device is more expensive. Comment on the design options.
GIVEN
An electronic device with aluminum fin array
Device generates heat at a rate ( qG ) = 600 mW = 0.6 W
Surface area (A) = 12 cm2
Max temperature of device = 70°C
Air temperature (T ) = 25°C
Convective heat transfer coefficient ( hc ) = 20 W/(m2 K)

55
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FIND
Operating temperature (To) for
(a) contact resistance (Ri) = 50 K/W
(b) no contact resistance
ASSUMPTIONS
One dimensional heat transfer
Steady state has been reached
The temperature of the device is uniform
The temperature of the aluminum fins is uniform (the thermal resistance of the aluminum is negligible) The heat loss from the edges and back of the device is negligible
SKETCH

SOLUTION
(a) The thermal circuit for the case with contact resistance is shown below

The value of the convective resistance, from Equation (1.14), is
1
1
Rc =
=
= 41.7 K/W hc A
[20 W/(m 2 K)](0.0012 m 2 )
For steady state conditions, the heat loss from the device (q) must be equal to the heat generated by the device
T - T•
DT
q =
= o
= qG
Rtotal
Rc + Ri
Solving for To
To = T + qG (Rc + Ri) = 25°C + (0.6 W) (41.7 K/W + 50 K/W) = 80°C
(b) Similarly, the operating temperature of the device with no contact resistance is
To = T + qG Rc = 25°C + (0.6 W) (41.7 K/W) = 50°C
COMMENTS
The more expensive device with no contact resistance will have to be used to assure that the operating temperature does not exceed 70°C.
PROBLEM 1.43
To reduce the home heating requirements, modern building codes in many parts of the country require the use of double-glazed or double-pane windows, i.e., windows with two panes of glass. Some of these so called thermopane windows have an evacuated space between the two glass panes while others trap stagnant air in the space.
56
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(a) Consider a double-pane window with the dimensions shown in the following sketch.
If this window has stagnant air trapped between the two panes and the convective heat transfer coefficients on the inside and outside surfaces are 4 W/(m2 K) and 15 W/(m2 K), respectively, calculate the overall heat transfer coefficient for the system.
(b) If the inside air temperature is 22°C and the outside air temperature is –5°C, compare the heat loss through a 4 m2 double-pane window with the heat loss through a single-pane window. Comment on the effect of the window frame on this result.
(c) The total window area of a home heated by electric resistance heaters at a cost of
$.10/kWh is 80 m2. How much more cost can you justify for the double-pane windows if the average temperature difference during the six winter months when heating is required is about 15°C?
GIVEN
Double-pane window with stagnant air in gap
Convective heat transfer coefficients
Inside ( hci ) = 4 W/(m2 K)
Outside ( hco ) = 15 W/(m2 K)
Air temperatures
Inside (Ti) = 22°C
Outside (To) = –5°C
Single window area (Aw) = 4 m2
During the winter months, ( T) = 15°C
Heating cost = $.1.0/kWh
Total window area (AT) = 80 m2
FIND
(a) The overall heat transfer coefficient
(b) Compare heat loss of double- and single-pane window
ASSUMPTIONS
Steady state conditions prevail
Radiative heat transfer is negligible
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Tables 11 and 27, the thermal conductivities are window glass (kg) = 0.81 W/(m K) at 20°C; dry air (ka) = 0.0243 W/(m K) at 8.5°C
SOLUTION
The thermal circuit for the system is shown below

57
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The individual resistances are
Rco =

1
1
1
=
=
0.0667 (K m2)/W
2
hco A
A
[15 W/(m K)] A

Rk1 = Rk2 =
Rka =
Rci =

Lg
Ak g

=

0.007 m
1
= 0.00864 (K m2)/W
A[0.81W/(m K)] A

La
0.02 m
1
= 0.823 (K m2)/W
=
A[0.0243W/(m K)] A
Ak a

1 hci A

=

1
2

[4 W/(m K)] A

=

1
0.25 (K m2)/W
A

The total resistance for the double-pane window is
5

Rtotal = S Ri = Rco + Rk1 + Rka + Rk2 + Rci i=1 1
1
(0.0667 + 0.00864 + 0.823 + 0.00864 + 0.25) (m2 K)/W =
1.157 (K m2)/W
A
A
Therefore the overall heat transfer coefficient is
1
1
= 0.864 W/(m2 K)
Udouble =
=
A Rtotal 1.157 (m 2 K)/W

Rtotal =

(b) The rate of heat loss through the double-pane window is qdouble – U A T = [0.864 W/(m2 K)] (4 m2) [22°C – (–5°C)] = 93W
The thermal circuit for the single-pane window is

The total thermal resistance for the single-pane window is
3

Rtotal = S Ri = Rco + Rk1 + Rci = i=1 1
(0.0667 + 0.00864 + 0.25) (m2 K)/W
A

Rtotal = 0.325 (m2 K)/W
The overall heat transfer coefficient for the single-pane window is
1
1
=
= 3.08 W/(m2 K)
Usingle =
A Rtotal
0.325(m 2 K)/W
Therefore, the rate of heat loss through the single-pane window is qsingle = U A T = [3.07 W/(m2 K)] (4 m2) [22°C – (–5°C)] = 332 W
The heat loss through the double-pane window is only 28% of that through the single-pane window. (c) The average heat loss through double-pane windows during the winter months is qdouble = U AT T = [0.864 W/(m2 K)] (80 m2) 15°C = 1040 W
Therefore, the cost of the heat loss from the double-pane windows is
Costdouble = qdouble (heating cost)
Costdouble = (1040 W) ($0.10/kWh) (24 h/day) (182 heating days/year) (1 kW/1000 W)
Costdouble = $454/yr
58
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The average heat loss through the single-pane windows during the winter months is qsingle = U AT T = [3.07 W/(m2 K)] (80 m2) (15°C) = 3688 W
The cost of this heat loss is
Costsingle = qsingle (heat cost)
Costsingle = (3688 W) ($0.10/kWh) (24 h/day) (182 heating days/year) (1 kW/1000 W)
Costsingle = $1611/yr
The yearly savings of the double-pane windows is $1157. Therefore if we would like to have a payback period of two years, we would be willing to invest $2314 in double panes.
PROBLEM 1.44
A flat roof can be modeled as a flat plate insulated on the bottom and placed in the sunlight. If the radiant heat that the roof receives from the sun is 600 W/m2, the convection heat transfer coefficient between the roof and the air is 12 W/(m2 K), and the air temperature is 27°C, determine the roof temperature for the following two cases:
(a) Radiative heat loss to space is negligible. (b) The roof is black ( = 1.0) and radiates to space, which is assumed to be a black-body at 0 K.
GIVEN
A flat plate in the sunlight
Radiant heat received from the sun (qr/A) = 600 W/m2
Air temperature (T ) = 27°C
Convective heat transfer coefficient ( hc ) = 12 W/(m2 K)
FIND
The plate temperature (Tp)
ASSUMPTIONS
Steady state prevails
No heat is lost from the bottom of the plate
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 1, Table 5: Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
(a) For this case steady state and the conservation of energy require the heat lost by conduction, from
Equation (1.10), to be equal to the heat gained from the sun qc = hc A (Ts – T ) = qr
Solving for Ts

59
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Ts =

Ê
ˆ
qr 1
1
+ T = (600 W/m2) Á
+ (27°C) = 77°C
2
A hc
Ë 12 W/(m K) ˜
¯

(b) In this case, the solar gain must be equal to the sum of the convective loss, from Equation (1.10), and radiative loss, from equation (1.16) qr = hc (Tp – T ) +
A

(Tp4 – Tsp4)

600 W/m2 = 12 W/(m2 K) (Tp – 300K) + 5.67

10–8 W/(m2 K4) (Tp4 – 0)

By trial and error
Tp = 308 K = 35°C
COMMENTS
The addition of a second means of heat transfer from the plate in part (b) allows the plate to operate at a significantly lower temperature.
PROBLEM 1.45
A horizontal 3-mm-thick flat copper plate, 1 m long and 0.5 m wide, is exposed in air at
27°C to radiation from the sun. If the total rate of solar radiation absorbed is
300 W and the combined radiative and convective heat transfer coefficients on the upper and lower surfaces are 20 and 15 W/(m2 K), respectively, determine the equilibrium temperature of the plate.
GIVEN
Horizontal, 1 m long, 0.5 m wide, and 3 mm thick copper plate is exposed to air and solar radiation Air temperature (T ) = 27°C
Solar radiation absorbed (qsol) = 300 W
Combined transfer coefficients are upper surface ( h u ) = 20 W/(m2 K) lower surface ( h1 ) = 15 W/(m2 K)
FIND
The equilibrium temperature of the plate (Tp)
ASSUMPTIONS
Steady state prevails
The temperature of the plate is uniform
SKETCH

60
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SOLUTION
For equilibrium the heat gain from the solar radiation must equal the heat lost from the upper and lower surfaces qsol = hu A (Tp – T ) + h1 A (Tp – T )
Solving for Tp
Tp =

qsol
1
+T
A hu + h1

ˆ
1
Ê 300 W ˆ Ê
Tp = Á
˜Á
˜ + (27°C)
2
2
Ë (1m) (0.5 m) ¯ Ë 20 W/(m K) + 15 W/(m K) ¯

Tp = 44°C
PROBLEM 1.46
A small oven with a surface area of 3 ft2 is located in a room in which the walls and the air are at a temperature of 80°F. The exterior surface of the oven is at 300°F and the net heat transfer by radiation between the oven’s surface and the surroundings is 2000
Btu/h. If the average convective heat transfer coefficient between the oven and the surrounding air is 2.0 Btu/(h ft2 °F), calculate: (a) the net heat transfer between the oven and the surroundings in Btu/h, (b) the thermal resistance at the surface for radiation and convection, respectively, in (h °F)/Btu, and (c) the combined heat transfer coefficient in Btu/(h ft2 °F).
GIVEN
Small oven in a room
Oven surface area (A) = 3 ft2
Room wall and air temperature (T ) = 80°F
Surface temperature of the exterior of the oven (To) = 300°F
Net radiative heat transfer (qr) = 2000 Btu/h
Convective heat transfer coefficient ( hc ) = 2.0 Btu/(h ft2 °F)
FIND
(a) Net heat transfer (qT) in Btu/h
(b) Thermal resistance for radiation and convection (RT) in (h °F)/Btu
(c) The combined heat transfer coefficient ( hcr ) in Btu/(h ft2 °F)
ASSUMPTIONS
Steady state prevails
SKETCH

61
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SOLUTION
(a) The net heat transfer is the sum of the convective heat transfer, from Equation (1.10), and the net radiative heat transfer qT = qc + qr + hc A (To – T ) + qr qT = 2.0 Btu/(h ft2 °F) (3 ft2) (300°F – 80°F) + 2000 Btu/h qT = 1320 Btu/h + 2000 Btu/h = 3320 Btu/h
(b) The radiative resistance is
Rr =

To - T• 300°F - 80 ∞F
= 0.110 (h °F)/Btu
=
2000 Btu/h qr The convective resistance is
Rc =

To - T• 300 ∞F - 80 ∞F
= 0.167 (h °F)/Btu
=
1320 Btu / h qr These two resistances are in parallel, therefore the total resistance is given by
RT =

Rc Rr
Ê (0.167 (h °F)/Btu )(0.110 (h °F)/Btu ) ˆ
=Á
˜ = 0.0663 (h °F)/Btu
¯
(0.167 + 0.110) (h °F)/Btu
Rc + Rr Ë

(c) The combined heat transfer coefficient can be calculated from qT = hcr A T hcr =

qr
3320 Btu/h
=
= 5.0 Btu/(h ft2 °F)
2
A DT
(3ft ) (300 ∞F - 80°F)

COMMENTS
The thermal resistances for the convection and radiation modes are of the same order of magnitude.
Hence, neglecting either one would lead to a considerable error in the rate of heat transfer.
PROBLEM 1.47
A steam pipe 200 mm in diameter passes through a large basement room. The temperature of the pipe wall is 500°C, while that of the ambient air in the room is 20°C.
Determine the heat transfer rate by convection and radiation per unit length of steam pipe if the emissivity of the pipe surface is 0.8 and the natural convection heat transfer coefficient has been determined to be 10 W/(m2 K).
GIVEN
A steam pipe passing through a large basement room
Pipe diameter ( ) = 200 mm = 0.2 m
The temperature of the pipe wall (Tp) = 500°C = 773 K
Temperature of ambient air in the room (T ) = 20°C = 293 K
Emissivity of the pipe surface ( ) = 0.8
Natural convection heat transfer coefficient (hc) = 10 W/(m2 K)
FIND
Heat transfer rate by convection and radiation per unit length of the steam pipe (q/L)
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ASSUMPTIONS
Steady state prevails
The walls of the room are at the same temperature as the air in the room
The walls of the room are black ( = 1.0)

SKETCH

PROPERTIES AND CONSTANTS
From Appendix 1, Table 5, the Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
The net radiative heat transfer rate for a gray object in a blackbody enclosure is given by Equation
(1.17)
qr = A 1 qr =
L

1

(T14 – T24) =

(0.2 m) (0.8) [5.67

DL

(Tp4 – Ts4)

10–8 W/(m2 K4)] [(773 K)4 – (293 K)4]

qr
= 9970 W/m
L

The convective heat transfer rate is given by qc = hc A (Tp – T ) = hc ( D L) (Tp – T ) qc = [10 W/(m2 K)]
L

(0.2 m) (500°C – 20°C)

qc
= 3020 W/m
L

COMMENTS
Note that absolute temperatures must be used in the radiative heat transfer equation.
The radiation heat transfer dominates because of the high emissivity of the surface and the high surface temperature which enters to the fourth power in the rate of radiative heat loss.
PROBLEM 1.48
The inner wall of a rocket motor combustion chamber receives 50,000 Btu/(h ft2) by radiation from a gas at 5000° F. The convective heat transfer coefficient between the gas and the wall is 20 Btu/(h ft2 °F). If the inner wall of the combustion chamber is at a temperature of 1000° F, determine the total thermal resistance of a unit area of the wall in (h ft2 °F)/Btu and the heat flux. Also draw the thermal circuit.
63
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GIVEN
Wall of a rocket motor combustion chamber
Radiation to inner surface (qr/A) = 50,000 Btu/(h ft2)
Temperature of gas in chamber (Tg) = 5000°F
Convective heat transfer coefficient on inner wall (hc) = 20 Btu/(h ft2 °F)
Temperature of inner wall (Tw) = 1000°F
FIND
(a) Draw the thermal circuit
(b) The total thermal resistance of a unit area (A Rtotal) in (h ft2 °F)/Btu
ASSUMPTIONS
One dimensional heat transfer through the walls of the combustion chamber
Steady state heat flow
SKETCH

SOLUTION
(a) The thermal circuit for the chamber wall is shown below

(b) The total thermal resistance can be calculated from the total rate of heat transfer from the pipe qtotal =
A Rtotal =

DT
Rtotal
DT
Ê qtotal ˆ
Á
Ë A ˜
¯

The total rate of heat transfer is the sum of the radiative and convective heat transfer qtotal = qr + qc = qr + hc A T qtotal q
= r + hc
A
A

T

qtotal
= 50,000 Btu/(h ft2) + 20 Btu/(h ft2 °F) (5000°F – 1000°F) = 130,000 Btu/(h ft2)
A
64
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Therefore the thermal resistance of a unit area is
A Rtotal =

5000°F - 1000°F
2

130, 000 Btu /(h ft )

= 0.031 (h ft2 °F)/ Btu

An alternate method of solving part (b) is to calculate the radiative and convective resistances separately and then combine them in parallel as illustrated below.
The convective resistance is
Rc =

ˆ
1
1 Ê
1
1
=
0.05 (h ft2 °F)/ Btu
=
hc A A Á 20Btu /(h ft 2 oF) ˜
A
Ë
¯

Rr =

DT
DT
1 5000 o F - 1000 o F 1
=
=
0.08 (h ft2 °F)/ Btu
=
qr
A (qr / A)
A 50, 000 Btu /(h ft 2 ) A

The radiative resistance is

Combining these two resistances in parallel yields the total resistance
Rtotal =

A Rtotal =

Rr Rc
Rr + Rc

(0.08) (0.05)
(h ft 2 °F)/ Btu = 0.031 (h ft2 °F)/ Btu
0.08 + 0.05

PROBLEM 1.49
A flat roof of a house absorbs a solar radiation flux of 600 W/m2. The backside of the roof is well insulated, while the outside loses heat by radiation and convection to ambient air at 20°C. If the emittance of the roof is 0.80 and the convective heat transfer coefficient between the roof and the air is 12 W/(m2 K), calculate: (a) the equilibrium surface temperature of the roof, and (b) the ratio of convective to radiative heat loss.
Can one or the other of these be neglected? Explain your answer.
GIVEN
Flat roof of a house
Solar flux absorbed (qsol/A) = 600 W/m2
Back of roof is well insulated
Ambient air temperature (T ) = 20°C = 293 K
Emittance of the roof ( ) = 0.80
Convective heat transfer coefficient ( hc ) = 12 W/(m2 K)
FIND
(a) The equilibrium surface temperature (Ts)
(b) The ratio of the convective to radiative heat loss
ASSUMPTIONS
The heat transfer from the back surface of the roof is negligible
Steady state heat flow

65
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SKETCH

PROPERTIES AND CONSTANTS
From Appendix 1, Table 5, the Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
(a) For steady state the sum of the convective heat loss, from Equation (1.10), and the radiative heat loss, from Equation (1.15), must equal the solar gain qsol q q = c + r = hc (Ts – T ) +
A
A
A

Ts 4

(

)

600 W/m2 = 12 W/(m2 K) (Ts – 293K) + (0.8) 5.67 ¥10 -8 W/(m 2 K 4 ) Ts4
4.535

10–8 Ts4 + 12 Ts – 4116 = 0

By trial and error
Ts = 309 K = 36°C
(b) The ratio of the convective to radiative loss is

(

)

12 W/(m 2 K) (309 K - 293K ) qc h (T - T )
= c s 4• =
= 0.46
4
qr e s Ts
(0.8) 5.67 ¥ 10-8 W/(m 2 K 4 ) (309 K )

(

)

COMMENTS
Since the radiative and convective terms are of the same order of magnitude, neither one may be neglected without introducing significant error.
PROBLEM 1.50
Determine the power requirement of a soldering iron in which the tip is maintained at
400°C. The tip is a cylinder 3 mm in diameter and 10 mm long. Surrounding air temperature is 20°C and the average convective heat transfer coefficient over the tip is
20 W/(m2 K). Initially, the tip is highly polished giving it a very low emittance.
GIVEN
Soldering iron tip
Diameter (D) = 3 mm = 0.003 m
Length (D) = 10 mm = 0.01 m
Temperature of the tip (Tt) = 400°C
Temperature of the surrounding air (T ) = 20°C
Average convective heat transfer coefficient ( hc ) = 20 W/(m2 K)
Emittance is very low ( = 0)
FIND
The power requirement of the soldering iron ( q )
66
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ASSUMPTIONS
Steady state conditions exist
All power used by the soldering iron is used to heat the tip
Radiative heat transfer from the tip is negligible due to the low emittance
The end of the tip is flat
The tip is at a uniform temperature
SKETCH

SOLUTION
The power requirement of the soldering iron, q , is equal to the heat lost from the tip by convection qc = hco A T = hc ( D2/4 +

D L) (Tt – T ) = q

È p (0.003m)
˘
q = 20 W/(m 2 K) Í
+ p (0.003 m)(0.01m)˙ (400°C – 20°C)
4
Î
˚
q = 0.77 W
2

PROBLEM 1.51
The soldering iron tip in Problem 1.50 becomes oxidized with age and its gray-body emittance increases to 0.8. Assuming that the surroundings are at 20°C determine the power requirement for the soldering iron.
Problem 1.50:
Determine the power requirement of a soldering iron in which the tip is maintained at
400°C. The tip is a cylinder 3 mm in diameter and 10 mm long. Surrounding air temperature is 20°C and the average convective heat transfer coefficient over the tip is
20 W/(m2 K). Initially, the tip is highly polished giving it a very low emittance.
GIVEN
Soldering iron tip
Diameter (D) = 3 mm = 0.003 m
Length (D) = 10 mm = 0.01 m
Temperature of the tip (Tt) = 400°C
Temperature of the surrounding air (T ) = 20°C
Average convective heat transfer coefficient ( hc ) = 20 W/(m2 K)
Emittance of the tip ( ) = 0.8
FIND
The power requirement of the soldering iron ( q )
ASSUMPTIONS
Steady state conditions exist
All power used by soldering iron is used to heat the tip
The surroundings of the soldering iron behave as a blackbody enclosure
The end of the tip is flat
67
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SKETCH

PROPERTIES AND CONSTANTS
From Appendix 1, Table 5, the Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
The rate of heat loss by convection, from Problem 1.50, is 0.77 W.
The rate of heat loss by radiation is given by Equation (1.17) qr = A 1

1

Ê p D2
ˆ
(T14 – T24) = Á
+ p DL˜
Ë 4
¯

(Tt4 – Tw4)

È p (0.003m)2
˘
qr = Í
+ p (0.003 m)(0.01m)˙ (0.8) [5.67 10–8 W/(m2 K4)] [(673 K)4 – (293 K)4]
4
Î
˚
qr = 0.91 W
The power requirement of the soldering iron, q , is equal to the total rate of heat loss from the tip. The total heat loss is equal to the sum of the convective and radiative losses q = qc + qr = 0.77 W + 0.91 W = 1.68 W

COMMENTS
Note that the inclusion of the radiative term more than doubled the power requirement for the soldering iron.
The power required to maintain the desired temperature could be provided by electric resistance heating. PROBLEM 1.52
Some automobile manufacturers are currently working on a ceramic engine block that could operate without a cooling system. Idealize such an engine as a rectangular solid, 45 cm by 30 cm by 30 cm. Suppose that under maximum power output the engine consumes 5.7 liters of fuel per hour, the heat released by the fuel is 9.29 kWh per liter and the net engine efficiency (useful work output divided by the total heat input) is 0.33.
If the engine block is alumina with a gray-body emissivity of 0.9, the engine compartment operates at 150°C, and the convective heat transfer coefficient is 30 W/(m2 K), determine the average surface temperature of the engine block. Comment on the practicality of the concept.
GIVEN
Ceramic engine block, 0.45m by 0.3m by 0.3m
Engine gas consumption is 5.7 1/h
Heat released is 9.29 (kWh)/1
Net engine efficiency ( ) = 0.33
Emissivity ( = 0.9
Convective heat transfer coefficient (hc) = 30 W/(m2 K)
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Engine compartment temperature (Tc) = 150°C = 423 K
FIND
The surface temperature of the engine block (Ts)
Comment on the practicality
ASSUMPTIONS
Heat transfer has reached steady state
The engine compartment behaves as a blackbody enclosure
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 1, Table 5, the Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
The surface area of the idealized engine block is
A = 4 (0.45m) (0.3m) + 2(0.3m)2 = 0.72 m2
The rate of heat generation within the engine block is equal to the energy from the gasoline that is not transformed into useful work qG = (1 – ) mg hg = (1 – 0.33) (5.71/h) (9.29 (kWh)/1) = 35.5 kW

For steady state conditions, the net radiative and convective heat transfer from the engine block must be equal to the heat generation within the engine block qtotal = qr + qc = qG qG = A

(Ts4 – Tc4) + hc A (Ts – Tc)

(

)

35.5 kW = (0.72 m2) (0.9) 5.67 ¥ 10-8 W/(m 2 K 4 ) [Ts4 – (423 K)4] + (0.72 m2)

(30 W/(m2 K)) (Ts – 3.674

10–8 Ts4 + 21.6 Ts – 45656 = 0

By trial and error
Ts = 916 K = 643°C
COMMENTS
The engine operates at a temperature high enough to burn a careless motorist.
Note that absolute temperature must be used in radiation equations.
Hot spots due to the complex geometry of the actual engine may produce local temperatures much higher than 916 K.
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PROBLEM 1.53
A pipe carrying superheated steam in a basement at 10°C has a surface temperature of
150°C. Heat loss from the pipe occurs by radiation ( = 0.6) and natural convection
[ hc = 25 W/(m2 K)]. Determine the percentage of the total heat loss by these two mechanisms. GIVEN
Pipe in a basement
Pipe surface temperature (Ts) = 150°C = 423 K
Basement temperature (T ) = 10°C = 283 K
Pipe surface emissivity ( ) = 0.6
Convective heat transfer coefficient ( hc ) = 25 W/(m2 K)
FIND
The percentage of the total heat loss due to radiation and convection
ASSUMPTIONS
The system is in steady state
The basement behaves as a blackbody enclosure at 10°C
SKETCH

PROPERTIES AND CONSTANTS
From Appendix 1, Table 5: the Stefan-Boltzmann constant ( ) = 5.67

10–8 W/(m2 K4)

SOLUTION
The rate of heat transfer from a gray-body to a blackbody enclosure, from Equation (1.17), is qr = A 1

1

(T14 – T24) = A

qr
= (0.6) [5.67
A

(Ts4 – T 4)

10–8 W/(m2 K4)] [(423 K)4 – (283 K)4]

qr
= 870 W/m
L

The rate of heat transfer by convection, from Equation (1.10), is qc = hc A (Ts – T ) qc = 25 W/(m2 K) (423 K – 283 K) = 3500 W/m2
A

70
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The total rate of heat transfer is the sum of the radiative and convective rates q q qtotal = r + c = 870 W/m2 + 3500 W/m2 = 4370 W/m2
A A
A
The percentage of the total heat transfer due to radiation is qr / A qtotal / A

100 =

870
4370

100 = 20%

The percentage of the total heat transfer due to convection is qc / A qtotal / A

100 =

3500
4370

100 = 80%

COMMENTS
This pipe surface temperature and rate of heat loss are much too high to be acceptable. In practice, a layer of mineral wool insulation would be wrapped around the pipe. This would reduce the surface temperature as well as the rate of heat loss.
PROBLEM 1.54
For a furnace wall, draw the thermal circuit, determine the rate of heat flow per unit area, and estimate the exterior surface temperature under the following conditions: the convective heat transfer coefficient at the interior surface is 15 W/(m2 K); rate of heat flow by radiation from hot gases and soot particles at 2000°C to the interior wall surface is 45,000 W/m2; the unit thermal conductance of the wall (interior surface temperature is about 850°C) is 250 W/(m2 K); there is convection from the outer surface.
GIVEN
A furnace wall
Convective heat transfer coefficient ( hc ) = 15 W/(m2 K)
Temperature of hot gases inside furnace (Tg) = 2000°C
Rate of radiative heat flow to the interior of the wall (qr/A) = 45,000 W/m2
Unit thermal conductance of the wall (k/L) = 250 W/(m2 K)
Interior surface temperature (Twi) is about 850°C
Convection occurs from outer surface of the wall
FIND
(a) Draw the thermal circuit
(b) Rate of heat flow per unit area (q/A)
(c) The exterior surface temperature (Two)
ASSUMPTIONS
Heat flow through the wall is one dimensional
Steady state prevails
SKETCH

71
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SOLUTION
The thermal circuit for the furnace wall is shown below

The rate of heat flow per unit area through the wall is equal to the rate of convective and radiative heat flow to the interior wall q q q q
= r + c = r + hc (Tg – Twi)
A
A
A
A q = 45,000 W/m2 + 15 W/(m2 K) (2000°C – 850°C) = 62,250 W/m2
A

We can calculate the outer surface temperature of the wall by examining the conductive heat transfer through the wall given by Equation (1.2) qk =

KA
(Twi – Two)
L

Two = Twi –

Ê
ˆ
qk 1
1
= 850°C – (62,250 W/m2) Á
= 601°C
2
A k/L
Ë 250 W/(m K) ˜
¯

COMMENTS
The corner sections should be analyzed separately since the heat flow there is not one dimensional.
PROBLEM 1.55
Draw the thermal circuit for heat transfer through a double-glazed window. Include solar energy gain to the window and the interior space. Identify each of the circuit elements. Include solar radiation to the window and interior space.
GIVEN
Double-glazed window
FIND
The thermal circuit
SKETCH

72
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SOLUTION

where Rr1, Rr12, Rr2
Rk1, Rk2, Rk12
Rc1, Rc2
Trw, Tro
T
T1i, T1o, T2i qs1, qs2

= Radiative thermal resistances
= Conductive thermal resistances
= Convective thermal resistances
= Effective temperatures for radiative heat transfer
= Air temperatures
= Surface temperatures of the glass
= Solar energy incident on the window panes

PROBLEM 1.56
The ceiling of a tract house is constructed of wooden studs with fiberglass insulation between them. On the interior of the ceiling is plaster and on the exterior is a thin layer of sheet metal. A cross section of the ceiling with dimensions is shown below.
(a) The R-factor describes the thermal resistance of insulation and is defined by:
R-factor = L/keff = T/(q/A)
Calculate the R-factor for this type of ceiling and compare the value of this R-factor with that for a similar thickness of fiberglass. Why are the two different?
(b) Estimate the rate of heat transfer per square meter through the ceiling if the interior temperature is 22°C and the exterior temperature is –5°C.
GIVEN
Ceiling of a tract house, construction shown below
Inside temperature (Ti) = 22°C
Outside temperature (To) = –5°C
FIND
(a) R-factor for the ceiling (RFc). Compare this to the R-factor for the same thickness of fiberglass
(RFfg). Why do they differ?
(b) Rate of heat transfer (q/A)
ASSUMPTIONS
Steady state heat transfer
One dimensional conduction through the ceiling
Thermal resistance of the sheet metal is negligible

73
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SKETCH

PROPERTIES AND CONSTANTS
From Appendix 2, Table 11, the thermal conductivities of the ceiling materials are
Pine or fir wood studs (kw) = 0.15 W/(m K) at 20°C
Fiberglass (kfg) = 0.035 W/(m K) at 20°C
Plaster (kp) = 0.814 W/(m K) at 20°C
SOLUTION
The thermal circuit for the ceiling with studs is shown below

Rp = thermal resistance of the plaster
Rw = thermal resistance of the wood
Rfg = thermal resistance of the fiberglass
Each of these resistances can be evaluated using Equation (1.3) where Rp =

Rw =

Rfg =

LP
Awall k P

=

(0.5 in) (0.0254 m/in)

=

1

( Awall ) [0.814 W/(m K)] Awall

0.0156 K m2/W

Lw
(3.5in) (0.0254 m/in)
1
=
=
0.5927 K m2/W
Aw kw
( Aw ) [0.15 W/(m K)] Awall
L fg
A fg k fg

=

(3.5in) (0.0254 m/in)
1
=
2.54 K m2/W
Afg ) [0.035 W/(m K)] Awall
(

To convert these all to a wall area basis the fraction of the total wall area taken by the wood studs and the fiberglass must be calculated wood studs =

fiberglass =

Aw
1.5in
=
= 0.094
16 in
Awall

A fg
Awall

=

14.5in
= 0.906
16 in

Therefore the resistances of the studs and the fiberglass based on the wall area are
Rw =

1
1
0.5927 K m2/W =
6.31 K m2/W
0.094 Awall
Awall
74

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Rfg =

1
1
2.54 K m2/W =
2.80 K m2/W
0.906 Awall
Awall

The R-Factor of the wall is related to the total thermal resistance of the wall by
Rw R fg ˘
È
RFc = Awall Rtotal = Awall Í R p +
˙=
Rw + R fg ˚
Î

0.0156 +

(6.31) (2.8)
K m/W = 2.0 K m2/W
6.31 + 2.8

For 4 in. of fiberglass alone, the R-factor is
RFfg =

(4in) (0.0254 m/in)
L
=
= 2.9 K m2/W
0.035 W/(m K) k fg

The R-factor of the ceiling is only 69% that of the same thickness of fiberglass. This is mainly due to the fact that the wood studs act as a ‘thermal short’ conducting heat through the ceiling more quickly than the surrounding fiberglass.
(b) The rate of heat transfer through the ceiling is
DT
q
22°C - ( -5°C)
=
=
= 13.5 W/m2
RFc
A
2.0 K m 2 /W
COMMENTS
R-factors are given in handbooks. For example, Mark’s Standard Handbook for Mechanical
Engineers lists the R-factor of a multi-layer masonry wall as 6.36 Btu/(h ft2) = 20 W/m2.
PROBLEM 1.57
A homeowner wants to replace an electric hot-water heater. There are two models in the store. The inexpensive model costs $280 and has no insulation between the inner and outer walls. Due to natural convection, the space between the inner and outer walls has an effective conductivity of 3 times that of air. The more expensive model costs $310 and has fiberglass insulation in the gap between the walls. Both models are 3.0 m tall and have a cylindrical shape with an inner wall diameter of 0.60 m and a 5 cm gap. The surrounding air is at 25°C, and the convective heat transfer coefficient on the outside is
15 W/(m2 K). The hot water inside the tank results in an inside wall temperature of
60°C.
If energy costs 6 cents per kilowatt-hour, estimate how long it will take to pay back the extra investment in the more expensive hot-water heater. State your assumptions.
GIVEN
Two hot-water heaters
Height (H) = 3.0 m
Inner wall diameter (Di) = 0.60 m
Gap between walls (L) = 0.05 m
Water heater #1
Cost = $280.00
Insulation: none
Effective Conductivity between wall (keff) = 3(ka)
Water heater #2
Cost = $310.00
Insulation: Fiberglass
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Surrounding air temperature (T ) = 25°C
Convective heat transfer coefficient (hc) = 15 W/(m2 K)
Inside wall temperature (Twi) = 60°C
Energy cost = $0.06/kWh
FIND
The time it will take to pay back the extra investment in the more expensive hot-water heater
ASSUMPTIONS
Since the diameter is large compared to the wall thickness, one-dimensional heat transfer is assumed To simplify the analysis, we will assume there is no water drawn from the heater, therefore the inside wall is always at 60°C
Steady state conditions prevail
SKETCH

PROPERTIES AND CONSTANTS
From Appendix, Table 11 and 27: The thermal conductivities are fiberglass (ki) = 0.035 W/(m K) at 20°C dry air (ka) = 0.0279 W/(m K) at 60°C
SOLUTION
The areas of the inner and outer walls are
Ai = 2

p Di2
+
4

Di H = 2

p (0.6m) 2
+
4

(0.6 m) (3 m) = 6.22 m2

2 p Do p (0.7 m) 2
+ Do H = 2
+ (0.7 m) (3 m) = 7.37 m2
4
4
The average area for the air or insulation between the walls (Aa) = 6.8 m2.
The thermal circuit for water heater #1 is

Ao = 2

The rate of heat loss for water heater #1 is
DT
DT q1 =
=
=
Rtotal
Rk ,eff + Rco

Twi - T•
L
1
+
keff Aa h• Aco

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q1 =

60°C - 25°C
= 361 W = 0.361 kW
0.05 m
1
+
3[0.0279 W/(m K)](6.8 m 2 ) [15 W/(m 2 K)](7.37 m 2 )

Therefore the cost to operate water heater #1 is
Cost1 = q1 (energy cost) = 0.361 kW ($0.06/kWh) (24 h/day) = $0.52/day
The thermal circuit for water heater #2 is

The rate of heat loss from water heater #2 is q2 =

60°C - 25°C
= 160 W = 0.16 kW
0.05m
1
+
[0.035 W/(m K)](6.8 m 2 ) [15 W/(m 2 K)](7.37 m 2 )

Therefore the cost of operating water heater #2 is
Cost2 = q2 (energy cost) = 0.16 kW ($0.06/kWh) (24 h/day) = $0.23/day
The time to pay back the additional investment is the additional investment divided by the difference in operating costs
$310 - $280
Payback time =
$0.52 / day - $0.23 / day
Payback time = 103 days
COMMENTS
When water is periodically drawn from the water heater, energy must be supplied to heat the cold water entering the water heater. This would be the same for both water heaters. However, drawing water from the heater also temporarily lowers the temperature of the water in the heater thereby lowering the heat loss and lowering the cost savings of water heater #2. Therefore, the payback time calculated here is somewhat shorter than the actual payback time.
A more accurate, but much more complex estimate could be made by assuming a typical daily hot water usage pattern and power output of heaters. But since the payback time is so short, the increased complexity is not justified since it will not change the bottom line—buy the more expensive model and save money as well as energy!
PROBLEM 1.58
Liquid oxygen (LOX) for the Space Shuttle can be stored at 90 K prior to launch in a spherical container 4 m in diameter. To reduce the loss of oxygen, the sphere is insulated with superinsulation developed at the U.S. Institute of Standards and Technology’s
Cryogenic Division that has an effective thermal conductivity of 0.00012 W/(m K). If the outside temperature is 20°C on the average and the LOX has a heat of vaporization of
213 J/g, calculate the thickness of insulation required to keep the LOX evaporation rate below 200 g/h.
GIVEN
Spherical LOX tank with superinsulation
Tank diameter (D) = 4 m
LOX temperature (TLOX) = 90 K
Ambient temperature (T ) = 20°C = 293 K
Thermal conductivity of insulation (k) = 0.00012 W/(m K)
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Heat of vaporization of LOX (hfg) = 213 kJ/kg
Maximum evaporation rate ( mLox ) = 0.2 kg/h
FIND
The minimum thickness of the insulation (L) to keep evaporation rate below 0.2 kg/h
ASSUMPTIONS
The thickness is small compared to the sphere diameter so the problem can be considered one dimensional Steady state conditions prevail
Radiative heat loss is negligible
SKETCH

SOLUTION
The maximum permissible rate of heat transfer is the rate that will evaporate 0.2 kg/h of LOX
Ê h ˆ Ê 1000 J ˆ q = mLox h fg = (0.2 kg/h) (213 kJ/kg) Á
Á
˜ ( Ws/J ) = 11.8 W
Ë 3600 s ˜ Ë kJ ¯
¯

An upper limit can be put on the rate of heat transfer by assuming that the convective resistance on the outside of the insulation is negligible and therefore the outer surface temperature is the same as the ambient air temperature. With this assumption, heat transfer can be calculated using Equation (1.2), one dimensional steady state conduction qk =

kA k p D2
(Thot – Tcold) =
(T – TLOX)
L
L

Solving for the thickness of the insulation (L)
L =

[0.00012 W/(m K)] p (4 m)2 k p D2
(T – TLOX) =
(293 K – 90 K) = 0.10 m = 10 cm
11.8 W qk COMMENTS
The insulation thickness is small compared to the diameter of the tank. Therefore, the assumption of one dimensional conduction is reasonable.
PROBLEM 1.59
The heat transfer coefficient between a surface and a liquid is 10 Btu/(h ft2 °F). How many watts per square meter will be transferred in this system if the temperature difference is 10°C?
GIVEN
The heat transfer coefficient between a surface and a liquid (hc) = 10 Btu/(h ft2 °F)
Temperature difference ( T) = 10°C
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FIND
The rate of heat transfer in watts per square meter
ASSUMPTIONS
Steady state conditions
Surface temperature is higher than the liquid temperature
SKETCH

SOLUTION
The rate of convective heat transfer per unit area (qc/A) is qc = hc
A

Ê
ˆ
ft 2
Ê h ˆ Ê 1055 J ˆ
T = 10 Btu/(h ft2 °F) (10°C) (1.8 °F/°C) Á
˜ Á Btu ˜ (Ws/J) Á
2˜
Ë
¯
Ë 3600s ¯
Ë 0.0929 m ¯

qc
= 558 W/m2
A

COMMENTS
Note that the transfer coefficient is given in the English system of units but the answer is needed in SI units. Therefore, the conversion factors, found inside the front cover of the textbook,
1 Btu = 1055 J, 1 ft2 = 0.0929 m2, and 1°C = 1.8°F must be applied.
Also note that the units in the conversion factors can be cancelled just like a fraction. This is a good check. PROBLEM 1.60
The thermal conductivity of fiberglass insulation at 68°F is 0.02 Btu/(h ft °F). What is its value in SI units?
GIVEN
Thermal conductivity (k) = 0.02 Btu/(h ft °F)
FIND
Thermal conductivity is SI units: W/(m K)
SOLUTION
3.281ft ˆ
1055 J ˆ Ê h ˆ k = 0.02 Btu/(h ft 2 oF) Ê
(1.8 °F / °C)
(Ws/J ) Ê
Á
Á
Ë Btu ˜ Á 3600 s ˜
¯Ë
Ë m ˜
¯
¯

k = 0.035 W/ ( m °C ) = 0.035 W/(m K)
Comments
Note that 1°C = 1 K if a temperature difference is involved as in the units for thermal conductivity.

79
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PROBLEM 1.61
The thermal conductivity of silver at 212°F is 238 Btu/(h ft °F). What is the conductivity in SI units?
GIVEN
Thermal conductivity of silver (k) = 238 Btu/(h ft °F)
FIND
Thermal conductivity of silver in SI units: W/(m K)
SOLUTION
The conversion can be done one unit at a time ft 1055 J ˆ Ê h ˆ
Ê
ˆ k = 238 Btu/(h ft 2 oF) Ê
Á
˜ (Ws)/J Á 0.3048 m ˜ (1.8 °F/K )
Ë Btu ˜ Ë 3600s ¯
¯Á
Ë
¯

k = 412 W/(m K)
If the appropriate conversion factor is available, the whole group of units can be converted in one step
Ê 1.731W/(m K) ˆ k = 238 Btu/ ( h ft °F) Á
Ë Btu /(h ft °F) ˜
¯

k = 412 W/(m K)
COMMENTS
Although the single step conversion may be faster, it is important to understand the relationship between units of power, energy, length, etc. in the two systems. This understanding may be more effectively developed by converting each unit separately at first.
Also note that the names of the units can be canceled as a check on the final result.
PROBLEM 1.62
An ice chest (see sketch) is to be constructed from Styrofoam [k = 0.033 W/(m K)]. If the wall of the chest is 5 cm thick, calculate its R-value in (hr ft2 °F)/(Btu in).
GIVEN
Ice chest constructed of Styrofoam, k = 0.0333 W/(m K)
Wall thickness 5 cm
FIND
(a) R-value of the ice chest wall
ASSUMPTIONS
(a) One-dimensional, steady conduction
SKETCH

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SOLUTION
From Section 1.6 the R-value is defined as
R-value =

thickness thermal conductivity

The thermal conductivity in engineering units is k = (0.033 W/(m K))

[1Btu/(hr ft °F)]
= 0.019 Btu/(hr ft °F)
[1.731W/(m K)]

and the thickness is t = 5 cm

(1in)
= 1.97 in = 0.164 ft
(2.54 cm)

so
R-value =

(0.164ft)

(0.019 Btu/(hr ft °F))

= 8.634 (ft 2 hr °F)/Btu

From the problem statement, it is clear that we are asked to determine the R-value on a ‘per-inch’ basis. Dividing the above R-value by the thickness in inches, we get
R-value =

8.634
= 4.38 (ft 2 hr °F)/Btu in
1.97

PROBLEM 1.63
Estimate the R-values for a 2-inch-thick fiberglass board and a 1-inch-thick polyurethane foam layer. Then compare their respective conductivity-times-density products if the density for fiberglass is 50 kg/m3 and the density of polyurethane is 30 kg/m3. Use the units given in Figure 1.27.
GIVEN
2-inch-thick fiberglass board, density = 50 kg/m3
1-inch-thick polyurethane, density = 30 kg/m3
FIND
(a) R-values for both
(b) Conductivity-times-density products for both
ASSUMPTIONS
(a) One-dimensional, steady conduction
SOLUTION
Ranges of conductivity for both of these materials are given in Fig. 1.28. Using mean values we find: fiberglass board k = 0.04 W/(m K) polyurethane foam k = 0.025 W/(m K)
For the 2 inch fiberglass we have t = 2 inch = 0.051 m k = 0.04 W/(m K)
From section 1.6 the R-value is given by
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R-value =

thickness
0.051m
=
= 1.27 (m2 K)/W thermal conductivity
0.04 W/(m K)

and conductivity density = ( 0.04 W/(m K) ) (50 kg/m3 ) = 2 (Wkg)/(Km 4 )

For the 1 inch polyurethane we have t = 1 inch = 0.0254 m k = 0.025 W/(m K)
R-value = conductivity t
= 1 (m2 K)/W k (

density = (0.025 W/(m K)) 30 kg/m3

)

= 0.75 (Wkg)/(Km 4 )

Summarizing, we have

2 fiberglass board

R-value
[(m2 K)/W]
1.27

conductivity density
[(W kg)/(K m4)]
2

1 polyurethane foam

1

0.75

PROBLEM 1.64
A manufacturer in the U.S. wants to sell a refrigeration system to a customer in
Germany. The standard measure of refrigeration capacity used in the United States is the ‘ton’; a one-ton capacity means that the unit is capable of making about one ton of ice per day or has a heat removal rate of 12,000 Btu/hr. The capacity of the American system is to be guaranteed at three tons. What would this guarantee be in SI units?
GIVEN
A three-ton refrigeration unit to be sold in Germany
FIND
(a) The rating in SI units
SOLUTION
Converting the refrigeration capacity to SI units we have
Ê Whr ˆ
3 (12, 000 Btu/hr ) Á
= 10,548 W
Ë 3.413 Btu ˜
¯

Although the watt is a derived unit in the SI system, it would be used to express the capacity of the system rather than Newton meters per second.
PROBLEM 1.65
Referring to Problem 1.65, how many kilograms of ice can a 3-ton refrigeration unit produce in a 24-hour period? The heat of fusion of water is 330 kJ/kg.
From Problem 1.65: A manufacturer in the U.S. wants to sell a refrigeration system to a customer in Germany. The standard measure of refrigeration capacity used in the
United States is the ‘ton’; a one-ton capacity means that the unit is capable of making about one ton of ice per day or has a heat removal rate of 12,000 Btu/hr. The capacity of
82
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the American system is to be guaranteed at three tons. What would this guarantee be in
SI units?
GIVEN
A three-ton refrigeration unit
Heat of fusion of ice is 330 kJ/kg
FIND
(a) Kilograms of ice produced by the unit per 24 hour period
(b) The refrigeration unit capacity is the net value, i.e., it includes heat losses
ASSUMPTIONS
(a) Water is cooled to just above the freezing point before entering the unit
SOLUTION
The mass of ice produced in a given period of time t is given by mice =

q DT hf where hf is the heat of fusion and q is the rate of heat removal by the refrigeration unit. From Problem
1.65 we have q = 10,548 W. Inserting the given values we have mice =

(

(10,548 W) (24 hr)
= 2762 kg
Ê hr ˆ
5
3.30 ¥ 10 J / kg (Ws) / J
Ë 3600s ¯

)

PROBLEM 1.66
Explain a fundamental characteristic that differentiates conduction from convection and radiation. SOLUTION
Conduction is the only heat transfer mechanism that dominates in solid materials. Convection and radiation play important roles in fluids or, for radiation, in a vacuum. Under certain conditions, e.g., a transparent solid, radiation could be important in a solid.
PROBLEM 1.67
Explain in your own words: (a) what is the mode of heat transfer through a large steel plate that has its surfaces at specified temperatures? (b) what are the modes when the temperature on one surface of the steel plate is not specified, but the surface is exposed to a fluid at a specified temperature.
GIVEN
(a) Steel plate with specified surface temperatures
(b) Steel plate with one specified temperature and another surface exposed to a fluid
FIND
(a) Modes of heat transfer

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SKETCH

SOLUTION
(a) Since the surface temperatures are specified, the only mode of heat transfer of importance is conduction through the steel plate
(b) In addition to conduction to the steel plate, convection at the surface exposed to the fluid must be considered PROBLEM 1.68
What are the important modes of heat transfer for a person sitting quietly in a room?
What if the person is sitting near a roaring fireplace?
GIVEN
Person sitting quietly in a room
Person sitting in a room with a fireplace
FIND
(a) Modes of heat transfer for each situation
ASSUMPTIONS
The person is clothed
SOLUTION
(a) Since the person is clothed, we would need to consider conduction through the clothing, and convection and radiation from the exposed surface of the clothing.
(b) In addition to the modes identified in (a), we would need to consider that surfaces of the person oriented towards the fire would be absorbing radiation from the flames.
PROBLEM 1.69
Consider the cooling of (a) a personal computer with a separate CPU, and (b) a laptop computer. The reliable functioning of these machines depends upon their effective cooling. Identify and briefly explain all modes of heat transfer that are involved in the cooling process.
GIVEN
A personal computer with a separate CPU (the monitor, keyboard, and mouse are separate and not considered). A laptop computer.
FIND
Identify and describe modes of heat transfer involved in their cooling.
ASSUMPTIONS
The computers are turned on and in normal operation.
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SOLUTION
(a) The cooling would first involve conduction from microchips to heat sinks (finned structures) mounted on them as well as conduction to the surface of printed-circuit boards, and convection from heat sinks and printed-circuit boards to air flowing over them (most PCs have a fan that blows air through the computer compartment). From the printed circuit boards, which are mounted to the casing of the computer, heat would also be conducted to the casing. Furthermore, there would be some radiation from heat sinks and printed-circuit boards to the casing, and then from outer computer casing to the surroundings; from the outer casing there will also be convection (natural convection) heat loss to the room’s atmosphere.
(b) In a laptop computer, the heat produced in the microchips and other electrical circuitry would be conducted to the heat sinks mounted on them as well as through the circuit boards to the casing of the computer. From the outer casing the heat would then be dissipated by natural convection and radiation to the room’s atmosphere. Some laptop computers come mounted with a small fan, in which case heat removal by internal forced convection would also be part of the total thermal management (cooling strategy) of the device. Furthermore, some makers install heat pipes in the casing for heat removal. A heat pipe is a “wicking” device that involves evaporation of a thin liquid film inside the device and the condensation of vapor so generated (the student can learn more about a heat pipe in Chapter 10)
PROBLEM 1.70
Describe and compare the modes of heat loss through the single-pane and double-pane window assemblies shown in the sketch below.
GIVEN
A single-pane and a double-pane window assembly
FIND
(a) The modes of heat transfer for each
(b) Compare the modes of heat transfer for each
ASSUMPTIONS
The window assembly wood casing is a good insulator
SKETCH

SOLUTION
The thermal network for both cases is shown above and summarizes the situation. For the single-pane window, we have convection on both exterior surfaces of the glass, radiation from both exterior
85
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surfaces of the glass, and conduction through the glass. For the double-pane window, we would have these modes in addition to radiation and convection exchange between the facing surfaces of the glass panes. Since the overall thermal network for the double-pane assembly replaces the pane-conduction with two-pane conductions plus the convection/radiation between the two panes, the overall thermal resistance of the double-pane assembly should be larger. Therefore, we would expect lower heat loss through the double-pane window.
PROBLEM 1.71
A person wearing a heavy parka is standing in a cold wind. Describe the modes of heat transfer determining heat loss from the person’s body.
GIVEN
Person standing in a cold wind, wearing a heavy parka
FIND
(a) The modes of heat transfer
SKETCH

SOLUTION
The thermal circuit for the situation is shown above. Assume that the person is wearing one other garment, i.e., a shirt, under the parka. The modes of heat transfer include conduction through the shirt and the parka and convection from the outer surface of the parka to the cold wind. We expect that the largest thermal resistance will be the parka insulation. We have neglected radiation from the parka outer surface because its influence on the overall heat transfer will be small compared to the other terms. PROBLEM 1.72
Discuss the modes of heat transfer that determine the equilibrium temperature of the space shuttle Endeavor when it is in orbit. What happens when it reenters the earth’s atmosphere? GIVEN
Space shuttle Endeavor in orbit
Space shuttle Endeavor during reentry
FIND
(a) Modes of heat transfer
SKETCH

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SOLUTION
Heat generated internally will have to be rejected to the skin of the shuttle or to some type of radiator heat exchanger exposed to space. The internal loads that are not rejected actively, i.e., by a heat exchanger, will be transferred to the internal surface of the shuttle by radiation and convection, transferred by conduction through the skin, then radiated to space. These two paths of heat transfer must be sufficient to ensure that the interior is maintained at a comfortable working temperature.
During reentry, the exterior surface of the shuttle will be exposed to a heat flux that results from frictional heating by the atmosphere. In this case, it is likely that the net heat flow will be into the space shuttle. The thermal design must be such that during reentry the interior temperature does not exceed some safe value.

87
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