Solution Manual for Fundamentals of Heat and Mass Transfer, 7th Edition

Name: Solution Manual for Fundamentals of Heat and Mass Transfer, 7th Edition
Brand: Academic Documents Platform
Price: 20 USD
Availability: InStock
Author: Jack Murphy

Simplify your textbook learning with Solution Manual for Fundamentals of Heat and Mass Transfer, 7th Edition, providing step-by-step solutions to every chapter.

Jack Murphy

Contributor

4.9

4 months ago

Preview (16 of 2638)

PROBLEM 1.1
KNOWN: Thermal conductivity, thickness and temperature difference across a sheet of rigid
extruded insulation.
FIND: (a) The heat flux through a 2 m × 2 m sheet of the insulation, and (b) The heat rate
through the sheet.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: From Equation 1.2 the heat flux is
1 2
x
T - TdT
q = -k = k
dx L
′′
Solving,
"
x
W 10 K
q = 0.029 ×
m K 0.02 m⋅
x 2
W
q = 14.5 m
′′ <
The heat rate is
2
x x 2
W
q = q A = 14.5 × 4 m = 58 W
m
′′ ⋅ <
COMMENTS: (1) Be sure to keep in mind the important distinction between the heat flux
(W/m2 ) and the heat rate (W). (2) The direction of heat flow is from hot to cold. (3) Note that
a temperature difference may be expressed in kelvins or degrees Celsius.
qcond
A = 4 m 2
T 2T 1
k = 0.029 ⋅
W
m K
x
L = 20 mm
T 1 – T 2 = 10˚C
qcond
A = 4 m 2
T 2T 1
k = 0.029 ⋅
W
m K
x
L = 20 mm
T 1 – T 2 = 10˚C

PROBLEM 1.2
KNOWN: Thickness and thermal conductivity of a wall. Heat flux applied to one face and
temperatures of both surfaces.
FIND: Whether steady-state conditions exist.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal energy
generation.
ANALYSIS: Under steady-state conditions an energy balance on the control volume shown is
2
in out cond 1 2( ) / 12 W/m K(50 C 30 C) / 0.01 m 24,000 W/mq q q k T T L′′ ′′ ′′= = = − = ⋅ ° − ° =
Since the heat flux in at the left face is only 20 W/m2 , the conditions are not steady state. <
COMMENTS: If the same heat flux is maintained until steady-state conditions are reached, the
steady-state temperature difference across the wall will be
ΔT = 2
/ 20 W/m 0.01 m /12 W/m K 0.0167 Kq L k′′ = × ⋅ =
which is much smaller than the specified temperature difference of 20°C.
q” = 20 W/m2
L = 10 mm
k = 12 W/m·KT1 = 50°C
T2 = 30°C
q″cond

PROBLEM 1.3
KNOWN: Inner surface temperature and thermal conductivity of a concrete wall.
FIND: Heat loss by conduction through the wall as a function of outer surface temperatures ranging from
-15 to 38°C.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3)
Constant properties.
ANALYSIS: From Fourier’s law, if xq′′ and k are each constant it is evident that the gradient,
xdT dx q k′′= − , is a constant, and hence the temperature distribution is linear. The heat flux must be
constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends
only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15°C
are
( ) 21 2
x
25 C 15 CdT T T
q k k 1W m K 133.3 W m
dx L 0.30 m
− −−
′′ = − = = ⋅ =
o o
. (1)
2 2
x xq q A 133.3 W m 20 m 2667 W′′= × = × = . (2) <
Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of outer surface temperature,
-15 ≤ T2 ≤ 38°C, with different wall thermal conductivities, k.
For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearly from +2667 W to -867 W and is zero
when the inside and outer surface temperatures are the same. The magnitude of the heat rate increases
with increasing thermal conductivity.
COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane
wall would not be linear.
-20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)
-1500
-500
500
1500
2500
3500
Heat loss, qx (W)
Wall thermal conductivity, k = 1.25 W/m.K
k = 1 W/m.K, concrete wall
k = 0.75 W/m.K
Outside surface
-20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)
-1500
-500
500
1500
2500
3500
Heat loss, qx (W)
Wall thermal conductivity, k = 1.25 W/m.K
k = 1 W/m.K, concrete wall
k = 0.75 W/m.K
Outside surface

PROBLEM 1.4
KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency
of gas furnace and cost of natural gas.
FIND: Daily cost of heat loss.
SCHEMATIC:
ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties.
ANALYSIS: The rate of heat loss by conduction through the slab is
( ) ( )1 2T T 7 C
q k LW 1.4 W / m K 11m 8 m 4312 W
t 0.20 m
− °
= = ⋅ × = <
The daily cost of natural gas that must be combusted to compensate for the heat loss is
( ) ( )g
d 6
f
q C 4312 W $0.02 / MJ
C t 24 h / d 3600s / h $8.28 / d
0.9 10 J / MJ
η
×
= Δ = × =
× <
COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation
between it and the concrete.

PROBLEM 1.5
KNOWN: Thermal conductivity and thickness of a wall. Heat flux through wall. Steady-state
conditions.
FIND: Value of temperature gradient in K/m and in °C/m.
SCHEMATIC:
L = 20 mm
k = 2.3 W/m·K
q”x = 10 W/m 2
x
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.
ANALYSIS: Under steady-state conditions,
" 2
10 W/m 4.35 K/m 4.35 C/m
2.3 W/m K
xdT q
dx k
= − = − = − = − °
⋅ <
Since the K units here represent a temperature difference, and since the temperature difference is the
same in K and °C units, the temperature gradient value is the same in either units.
COMMENTS: A negative value of temperature gradient means that temperature is decreasing with
increasing x, corresponding to a positive heat flux in the x-direction.

Loading page 6...

PROBLEM 1.6
KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed
thickness.
FIND: Thermal conductivity, k, of the wood.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be
determined from Fourier’s law, Eq. 1.2. Rearranging,
( )
L W 0.05m
k=q 40
T T m 40-20 C
x 21 2
′′ =
− o
k = 0.10 W / m K.⋅ <
COMMENTS: Note that the °C or K temperature units may be used interchangeably when
evaluating a temperature difference.

Loading page 7...

PROBLEM 1.7
KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions.
FIND: Heat loss through window.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from
Fourier’s law, Eq. 1.2.
( )
T T
q k L
15-5 CW
q 1.4 m K 0.005m
q 2800 W/m .
1 2
x
x
2
x
−
′′ =
′′ = ⋅
′′ =
o
Since the heat flux is uniform over the surface, the heat loss (rate) is
q = qx A
q = 2800 W / m2 3m2
′′ ×
×
q = 8400 W. <
COMMENTS: A linear temperature distribution exists in the glass for the prescribed
conditions.

Loading page 8...

PROBLEM 1.8
KNOWN: Net power output, average compressor and turbine temperatures, shaft dimensions and
thermal conductivity.
FIND: (a) Comparison of the conduction rate through the shaft to the predicted net power output of
the device, (b) Plot of the ratio of the shaft conduction heat rate to the anticipated net power output of
the device over the range 0.005 m ≤ L ≤ 1 m and feasibility of a L = 0.005 m device.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Net power output is
proportional to the volume of the gas turbine.
PROPERTIES: Shaft (given): k = 40 W/m⋅K.
ANALYSIS: (a) The conduction through the shaft may be evaluated using Fourier’s law, yielding
( ) ( )2 3 2( ) 40W/m K(1000 400) C
" / 4 (70 10 m) / 4 92.4W
1m
h c
c
k T T
q q A d
L
π
π −− ⋅ − °
= = = × =
The ratio of the conduction heat rate to the net power output is
6
6
92.4W 18.5 10
5 10 W
q
r P
−
= = = ×
× <
(b) The volume of the turbine is proportional to L3 . Designating L a = 1 m, d a = 70 mm and Pa as the
shaft length, shaft diameter, and net power output, respectively, in part (a),
d = d a × (L/La); P = P a × (L/L a)3
and the ratio of the conduction heat rate to the net power output is
( ) ( )( )22 2
3 2
3 2 6 6 2
2 2
( ) ( ) ( )
/ 4 / / 4 /
" 4
( / )
40W/m K(1000 400) C (70 10 m) 1m / 5 10 W 18.5 10 m4=
h c h c h c
a a a a a
c
a a
k T T k T T k T T
d d L L d L P
q A L Lr P P P L L L
L L
π
π
π
π − −
− − −
= = = =
⋅ − ° × × × ×
=
Continued…
L = 1m
d = 70 mm
P = 5 MW
TurbineCompressor
Shaft T h = 1000°C
T c = 400°C
Combustion
chamber
L = 1m
d = 70 mm
P = 5 MW
TurbineCompressor
Shaft T h = 1000°C
T c = 400°C
Combustion
chamber
k = 40 W/m·K

Loading page 9...

PROBLEM 1.8 (Cont.)
The ratio of the shaft conduction to net power is shown below. At L = 0.005 m = 5 mm, the shaft
conduction to net power output ratio is 0.74. The concept of the very small turbine is not feasible since
it will be unlikely that the large temperature difference between the compressor and turbine can be
maintained. <
COMMENTS: (1) The thermodynamics analysis does not account for heat transfer effects and is
therefore meaningful only when heat transfer can be safely ignored, as is the case for the shaft in part
(a). (2) Successful miniaturization of thermal devices is often hindered by heat transfer effects that
must be overcome with innovative design.
Ratio of shaft conduction to net power
0 0.2 0.4 0.6 0.8 1
L (m)
0.0001
0.001
0.01
0.1
1
r

Loading page 10...

PROBLEM 1.9
KNOWN: Width, height, thickness and thermal conductivity of a single pane window and
the air space of a double pane window. Representative winter surface temperatures of single
pane and air space.
FIND: Heat loss through single and double pane windows.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-state
conditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy induced
motion).
ANALYSIS: From Fourier’s law, the heat losses are
Single Pane: ( )T T 35 C21 2q k A 1.4 W/m K 2m 19, 600 Wg g L 0.005m
−
= = ⋅ =
o
<
Double Pane: ( )T T 25 C21 2q k A 0.024 2m 120 Wa a L 0.010 m
−
= = =
o
<
COMMENTS: Losses associated with a single pane are unacceptable and would remain
excessive, even if the thickness of the glass were doubled to match that of the air space. The
principal advantage of the double pane construction resides with the low thermal conductivity
of air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, use
of the double pane construction would also increase the surface temperature of the glass
exposed to the room (inside) air.

Loading page 11...

PROBLEM 1.10
KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures.
FIND: Thickness of styrofoam insulation needed to maintain heat load below prescribed
value.
SCHEMATIC:
ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through 5
walls of area A = 4m
2 , (3) Steady-state conditions, (4) Constant properties.
ANALYSIS: Using Fourier’s law, Eq. 1.2, the heat rate is
q = q A = k T
L A total′′ ⋅ Δ
Solving for L and recognizing that Atotal = 5×W
2 , find
L = 5 k T W
q
2
Δ
( ) ( )5 0.03 W/m K 35 - -10 C 4m
L = 500 W
2⎡ ⎤× ⋅ ⎣ ⎦o
L = 0.054m = 54mm. <
COMMENTS: The corners will cause local departures from one-dimensional conduction
and a slightly larger heat loss.

Loading page 12...

PROBLEM 1.11
KNOWN: Heat flux at one face and air temperature and convection coefficient at other face of plane
wall. Temperature of surface exposed to convection.
FIND: If steady-state conditions exist. If not, whether the temperature is increasing or decreasing.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal energy generation.
ANALYSIS: Conservation of energy for a control volume around the wall gives
st
in out g
dE E E E
dt = − +& & &
[ ]st
in in
2 2 2
( ) ( )
20 W/m 20 W/m K(50 C 30 C) 380 W/m
s s
dE q A hA T T q h T T A
dt
A A
∞ ∞
′′ ′′= − − = − −
= − ⋅ ° − ° = −⎡ ⎤⎣ ⎦
Since dEst /dt ≠ 0, the system is not at steady-state. <
Since dEst /dt < 0, the stored energy is decreasing, therefore the wall temperature is decreasing. <
COMMENTS: When the surface temperature of the face exposed to convection cools to 31°C, qin =
qout and dEst/dt = 0 and the wall will have reached steady-state conditions.
q” = 20 W/m2
Ts = 50°C
h = 20 W/m2 ·K
T∞ = 30°C
Air
q” conv

Loading page 13...

PROBLEM 1.12
KNOWN: Dimensions and thermal conductivity of food/beverage container. Inner and outer
surface temperatures.
FIND: Heat flux through container wall and total heat load.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through bottom
wall, (3) Uniform surface temperatures and one-dimensional conduction through remaining
walls.
ANALYSIS: From Fourier’s law, Eq. 1.2, the heat flux is
( )0.023 W/m K 20 2 CT T 22 1q k 16.6 W/m
L 0.025 m
⋅ −−
′′ = = =
o
<
Since the flux is uniform over each of the five walls through which heat is transferred, the
heat load is
( )q q A q H 2W 2W W Wtotal 1 2 1 2′′ ′′ ⎡ ⎤= × = + + ×⎣ ⎦
( ) ( )2q 16.6 W/m 0.6m 1.6m 1.2m 0.8m 0.6m 35.9 W⎡ ⎤= + + × =⎣ ⎦ <
COMMENTS: The corners and edges of the container create local departures from one-
dimensional conduction, which increase the heat load. However, for H, W1 , W2 >> L, the
effect is negligible.

Loading page 14...

PROBLEM 1.13
KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of that
through a composite wall of prescribed thermal conductivity and thickness.
FIND: Thickness of masonry wall.
SCHEMATIC:
ASSUMPTIONS: (1) Both walls subjected to same surface temperatures, (2) One-
dimensional conduction, (3) Steady-state conditions, (4) Constant properties.
ANALYSIS: For steady-state conditions, the conduction heat flux through a one-
dimensional wall follows from Fourier’s law, Eq. 1.2,
′′q = k T
L
Δ
where ΔT represents the difference in surface temperatures. Since ΔT is the same for both
walls, it follows that
L = L k
k
q
q1 2 1
2
2
1
⋅ ′′
′′ .
With the heat fluxes related as
′′ = ′′q 0.8 q1 2
L = 100mm 0.75 W / m K
0.25 W / m K
1
0.8 = 375mm.1
⋅
⋅ × <
COMMENTS: Not knowing the temperature difference across the walls, we cannot find the
value of the heat rate.

Loading page 15...

PROBLEM 1.14
KNOWN: Expression for variable thermal conductivity of a wall. Constant heat flux.
Temperature at x = 0.
FIND: Expression for temperature gradient and temperature distribution.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction.
ANALYSIS: The heat flux is given by Fourier’s law, and is known to be constant, therefore
x
dT
q k constant
dx
′′ = − =
Solving for the temperature gradient and substituting the expression for k yields
x xq qdT
dx k ax b
′′ ′′
= − = − + <
This expression can be integrated to find the temperature distribution, as follows:
xqdT dx dx
dx ax b
′′
= − +∫ ∫
Since xq constant′′ = , we can integrate the right hand side to find
( )xq
T ln ax b c
a
′′
= − + +
where c is a constant of integration. Applying the known condition that T = T1 at x = 0,
we can solve for c.
Continued…
q”
x
k = ax + b
T1

Loading page 16...

13 more pages available. Scroll down to load them.

Preview Mode

100%

Study Now!

XY-Copilot AI

Unlimited Access

Secure Payment

Instant Access

24/7 Support

Document Chat

Document Details

Subject

Mechanical Engineering

Solution Manual for Fundamentals of Heat and Mass Transfer, 7th Edition

Study Now!

Document Details

Related Documents

Solution Manual For Fundamentals of Heat and Mass Transfer, 8th Edition

A system is a group of parts that work together to

Solution Manual for Introduction to Robotics: Mechanics and Control, 4th Edition

Solution Manual For Thermal Radiation Heat Transfer, 6th Edition

Solution Manual For Thermodynamics: An Interactive Approach, 1st Edition

Solution Manual For System Dynamics, 4th Edition

Solution Manual for Fluid Mechanics for Engineers, 2017 Edition

Solution Manual For Design Of Machine Elements, 8th Edition

Solution Manual for Engineering Materials: Properties and Selection, 9th Edition

Solution Manual For Theory Of Vibrations With Applications, 5th Edition

Company

Explore

Study Tools