Solution Manual for Heat Transfer , 1st Edition

Name: Solution Manual for Heat Transfer , 1st Edition
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P1.1-2 (1-1 in text): Conductivity of a dilute gas
Section 1.1.2 provides an approximation for the thermal conductivity of a monatomic gas at ideal
gas conditions. Test the validity of this approximation by comparing the conductivity estimated
using Eq. (1-18) to the value of thermal conductivity for a monotonic ideal gas (e.g., low
pressure argon) provided by the internal function in EES. Note that the molecular radius, σ, is
provided in EES by the Lennard-Jones potential using the function sigma_LJ.
a.) What is the value and units of the proportionality constant required to make Eq. (1-18) an
equality?
Equation (1-18) is repeated below:
2
vc T
k MW
σ
∝ (1)
Equation (1) is written as an equality by including a constant of proportionality (Ck):
2
v
k
c T
k C MW
σ
= (2)
Solving for Ck leads to:
2
k
v
k MW
C c T
σ
= (3)
which indicates that Ck has units m-kg 1.5 /s-kgmol 05
-K0.5 .
The inputs are entered in EES for Argon at relatively low pressure (0.1 MPa) and 300 K.
"Problem 1.1-2"
$UnitSystem SI MASS RAD PA K J
$TABSTOPS 0.2 0.4 0.6 0.8 3.5 in
T=300 [K] "temperature"
F$='Argon' "fluid"
P_MPa=0.1 [MPa] "pressure, in MPa"
P=P_MPa*convert(MPa, Pa) "pressure"
The conductivity, specific heat capacity, Lennard-Jones potential, and molecular weight of
Argon (k, c v,
σ, and MW) are evaluated using EES' built-in funcions. Equation (3) is used to
evaluate the proportionality constant.
k=conductivity(F$,T=T,P=P) "conductivity"
cv=cv(F$,T=T,P=P) "specific heat capacity at constant volume"
MW=molarMass(F$) "molecular weight"
sigma=sigma_LJ(F$) "Lennard-Jones potential"
C_k=k*sigma^2*sqrt(MW/T)/cv "constant of proportionality"

which leads to Ck = 2.619x10-24 m-kg1.5 /s-kgmol 0.5 -K0.5 .
b.) Plot the value of the proportionality constant for 300 K argon at pressures between 0.01 and
100 MPa on a semi-log plot with pressure on the log scale. At what pressure does the
approximation given in Eq. (1-18) begin to fail at 300 K for argon?
Figure 1 illustrates the constant of proportionality as a function of pressure for argon at 300 K.
The approximation provided by Eq. (1-18) breaks down at approximately 1 MPa.
0.001 0.01 0.1 1 10 100
0 x10 0
10 -24
2 x10 -24
3 x10 -24
4 x10 -24
5 x10 -24
6 x10 -24
7 x10 -24
8 x10 -24
Pressure (MPa)
C k (m-kg
1.5 /s-kgmol
0.5
-K
0.5
)
Figure 1: Constant of proportionality in Eq. (3) as a function of pressure for argon at 300 K.

Problem 1.2-3 (1-2 in text): Conduction through a Wall
Figure P1.2-3 illustrates a plane wall made of a very thin (thw = 0.001 m) and conductive (k =
100 W/m-K) material that separates two fluids, A and fluid B. Fluid A is at TA = 100°C and the
heat transfer coefficient between the fluid and the wall is Ah = 10 W/m2 -K while fluid B is at TB
= 0°C with Bh = 100 W/m2 -K.
th w = 0.001 m
k = 100 W/m-K
2
100 C
10 W/m -K
A
A
T
h
= °
= 2
0 C
100 W/m -K
B
B
T
h
= °
=
Figure P1.2-3: Plane wall separating two fluids
a.) Draw a resistance network that represents this situation and calculate the value of each
resistor (assuming a unit area for the wall, A = 1 m2 ).
Heat flowing from fluid A to fluid B must pass through a fluid A-to-wall convective resistance
(Rconv,A), a resistance to conduction through the wall (Rcond), and a wall-to-fluid B convective
resistance (Rconv,B). These resistors are in series. The network and values of the resistors are
shown in Figure 2.
TA = 100°C
,
1
conv A
A
R h A
= w
cond
t
R k A
=
K
0.1 W
K
0.0001 W
,
1
cond B
B
R h A
=
K
0.01 W
TB = 0°C
Figure 2: Thermal resistance network representing the wall.
b.) If you wanted to predict the heat transfer rate from fluid A to B very accurately, then which
parameter (e.g., thw, k, etc.) would you try to understand/measure very carefully and which
parameters are not very important? Justify your answer.
The largest resistance in a series network will control the heat transfer. For the wall above, the
largest resistance is Rconv,A. Therefore, I would focus on predicting this resistance accurately.
This would suggest that Ah is the most important parameter and the others do not matter much.

Problem 1.2-8 (1-3 in text): Frozen Gutters
You have a problem with your house. Every spring at some point the snow immediately
adjacent to your roof melts and runs along the roof line until it reaches the gutter. The water in
the gutter is exposed to air at temperature less than 0°C and therefore freezes, blocking the gutter
and causing water to run into your attic. The situation is shown in Figure P1.2-8.
snow melts at this surface
2
, 15 W/m -Kout outT h =
snow, ks = 0.08 W/m-K
2
22 C, 10 W/m -Kin inT h= ° =
Lins = 3 inch
insulation, kins = 0.05 W/m-K
plywood,
0.5 inch, 0.2 W/m-Kp pL k= =
Ls = 2.5 inch
Figure P1.2-8: Roof of your house.
The air in the attic is at Tin = 22°C and the heat transfer coefficient between the inside air and the
inner surface of the roof is inh = 10 W/m2 -K. The roof is composed of a Lins = 3.0 inch thick
piece of insulation with conductivity k ins = 0.05 W/m-K that is sandwiched between two Lp = 0.5
inch thick pieces of plywood with conductivity k p = 0.2 W/m-K. There is an Ls = 2.5 inch thick
layer of snow on the roof with conductivity k s = 0.08 W/m-K. The heat transfer coefficient
between the outside air at temperature Tout and the surface of the snow is outh = 15 W/m2 -K.
Neglect radiation and contact resistances for part (a) of this problem.
a.) What is the range of outdoor air temperatures where you should be concerned that your
gutters will become blocked by ice?
The input parameters are entered in EES and converted to base SI units (N, m, J, K) in order to
eliminate any unit conversion errors; note that units should still be checked as you work the
problem but that this is actually a check on the unit consistency of the equations.
"P1.2-8: Frozen Gutters"
$UnitSystem SI MASS RAD PA K J
$TABSTOPS 0.2 0.4 0.6 0.8 3.5 in
T_in=converttemp(C,K,22) "temperature in your attic"
L_ins=3 [inch]*convert(inch,m) "insulation thickness"
L_p=0.5 [inch]*convert(inch,m) "plywood thickness"
k_ins=0.05 [W/m-K] "insulation conductivity"
k_p=0.2 [W/m-K] "plywood conductivity"
k_s=0.08 [W/m-K] "snow conductivity"
L_s=2.5 [inch]*convert(inch,m) "snow thickness"
h_in=10 [W/m^2-K]
"heat transfer coefficient between attic air and inner surface of roof"
h_out=15 [W/m^2-K]
"heat transfer coefficient between outside air and snow"
A=1 [m^2] "per unit area"

The problem may be represented by the resistance network shown in Figure 2.
Figure 2: Resistance network representing the roof of your house.
The network includes resistances that correspond to convection with the inside and outside air:
,
1
conv out
out
R h A
= (1)
,
1
conv in
in
R h A
= (2)
where A is 1 m2 in order to accomplish the problem on a per unit area basis. There are also
conduction resistances associated with the insulation, plywood and snow:
ins
ins
ins
L
R k A
= (3)
p
p
p
L
R k A
= (4)
s
s
s
L
R k A
= (5)
R_conv_out=1/(h_out*A) "outer convection resistance"
R_s=L_s/(k_s*A) "snow resistance"
R_p=L_p/(k_p*A) "plywood resistance"
R_ins=L_ins/(k_ins*A) "insulation resistance"
R_conv_in=1/(h_in*A) "inner convection resistance"
Which leads to Rconv,out = 0.07 K/W, Rs = 0.79 K/W, Rp = 0.06 K/W, Rins = 1.52 K/W and Rconv,in
= 0.10 K/W. Therefore, the dominant effects for this problem are conduction through the
insulation and the snow; the other effects (convection and the plywood conduction) are not
terribly important since the largest resistances will dominate in a series network.
If the snow at the surface of the room is melting then the temperature at the connection between
Rs and Rp must be Ts = 0°C (see Figure 2). Therefore, the heat transferred through the roof ( q in
Figure 2) must be:

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( )
, 2
in s
conv in p ins
T T
q R R R
−
= + +
(6)
The temperature of the outside air must therefore be:
( ),out s s conv outT T q R R= − + (7)
T_s=converttemp(C,K,0)
"roof-to-snow interface temperature must be melting point of water"
q_dot=(T_in-T_s)/(R_conv_in+2*R_p+R_ins)
"heat transfer from the attic to the snow when melting point is reached"
T_out=T_s-q_dot*(R_s+R_conv_out)
"outside temperature required to reach melting point at roof surface"
T_out_C=converttemp(K,C,T_out) "outside temperature in C"
which leads to Tout = -10.8°C. If the temperature is below this then the roof temperature will be
below freezing and the snow will not melt. If the temperature is above 0°C then the water will
not refreeze upon hitting the gutter. Therefore, the range of temperatures of concern are -10.8°C
< Tout < 0°C.
b.) Would your answer change much if you considered radiation from the outside surface of the
snow to surroundings at Tout? Assume that the emissivity of snow is
εs = 0.82.
The modified resistance network that includes radiation is shown in Figure 3.
Figure 3: Resistance network representing the roof of your house and including radiation.
The additional resistance for radiation is in parallel with convection from the surface of the snow
as heat is transferred from the surface by both mechanisms. The radiation resistance can be
calculated approximately according to:
3
1
4
rad
s
R T A
ε σ
= (8)
where T is the average temperature of the surroundings and the snow surface. In order to get a
quick idea of the magnitude of this resistance we can approximate T with its largest possible
value (which will result in the largest possible amount of radiation); the maximum temperature
of the snow is 0°C:
e_s=0.82 [-] "emissivity of snow"

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Problem P1.2-11 (1-4 in text)
Figure P1.2-11(a) illustrates a composite wall. The wall is composed of two materials (A with k A
= 1 W/m-K and B with k B = 5 W/m-K), each has thickness L = 1.0 cm. The surface of the wall at
x = 0 is perfectly insulated. A very thin heater is placed between the insulation and material A;
the heating element provides 2
5000 W/mq′′ = of heat. The surface of the wall at x = 2L is
exposed to fluid at Tf,in = 300 K with heat transfer coefficient inh = 100 W/m2 -K.
L = 1 cm
L = 1 cm
x
2
5000 W/mq′′ =
insulated
material A
kA = 1 W/m-K
material B
kB = 5 W/m-K
,
2
300 K
100 W/m -K
f in
in
T
h
=
=
Figure P1.2-11(a): Composite wall with a heater.
You may neglect radiation and contact resistance for parts (a) through (c) of this problem.
a.) Draw a resistance network to represent this problem; clearly indicate what each resistance
represents and calculate the value of each resistance.
The input parameters are entered in EES:
“P1.2-11: Heater"
$UnitSystem SI MASS RAD PA K J
$TABSTOPS 0.2 0.4 0.6 0.8 3.5 in
"Inputs"
q_flux=100 [W/m^2] "heat flux provided by the heater"
L = 1.0 [cm]*convert(cm,m) "thickness of each layer"
k_A=1.0 [W/m-K] "conductivity of material A"
k_B=5.0 [W/m-K] "conductivity of material B"
T_f_in=300 [K] "fluid temperature at inside surface"
h_in=100 [W/m^2-K] "heat transfer on inside surface"
A=1 [m^2] "per unit area"
The resistance network that represents the problem shown in Figure 2 is:
Figure 2: Resistance network.
The resistances due to conduction through materials A and B are:

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A
A
L
R k A
= (1)
B
B
L
R k A
= (2)
where A is the area of the wall, taken to be 1 m2 in order to carry out the analysis on a per unit
area basis. The resistance due to convection is:
,
1
conv in
in
R h A
= (3)
"part (a)"
R_A=L/(k_A*A) "resistance to conduction through A"
R_B=L/(k_B*A) "resistance to conduction through B"
R_conv_in=1/(h_in*A)
"resistance to convection on inner surface"
which leads to RA = 0.01 K/W, RB = 0.002 K/W, and Rconv,in = 0.01 K/W.
b.) Use your resistance network from (a) to determine the temperature of the heating element.
The resistance network for this problem is simple; the temperature drop across each resistor is
equal to the product of the heat transferred through the resistor and its resistance. In this simple
case, all of the heat provided by the heater must pass through materials A, B, and into the fluid
by convection so these resistances are in series. The heater temperature (Thtr) is therefore:
( ), ,htr f in A B conv inT T R R R q A′′= + + + (4)
T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A "heater temperature"
which leads to Thtr = 410 K.
c.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch is
consistent with your solution from (b).
The temperatures at x = L and x = 2L can be computed according to:
( ), ,x L f in B conv inT T R R q A= ′′= + + (5)
2 , ,x L f in conv inT T R q A= ′′= + (6)
T_L=T_f_in+(R_B+R_conv_in)*q_flux*A "temperature at x=L"
T_2L=T_f_in+R_conv_in*q_flux*A "temperature at x=2L"

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which leads to Tx=L = 360 K and Tx=2L = 350 K. The temperature distribution is sketched on the
axes in Figure 3.
Figure 3: Sketch of temperature distribution.
Notice that the temperature drop through the two larger resistances (RA and RB) are much larger
than the temperature drop across the small resistance, RB.
Figure P1.2-11(b) illustrates the same composite wall shown in Figure P1.2-11(a), but there is an
additional layer added to the wall, material C with kC = 2.0 W/m-K and L = 1.0 cm.
L = 1 cm
L = 1 cm
x
2
5000 W/mq′′ =
insulated
material A
kA = 1 W/m-K
material B
k B = 5 W/m-K
,
2
300 K
100 W/m -K
f in
in
T
h
=
=
L = 1 cm
material C
kC = 2 W/m-K
Figure P1.2-11(b): Composite wall with Material C.
Neglect radiation and contact resistance for parts (d) through (f) of this problem.
d.) Draw a resistance network to represent the problem shown in Figure P1.2-11(b); clearly
indicate what each resistance represents and calculate the value of each resistance.
There is an additional resistor corresponding to conduction through material C, RC, as shown
below:

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Notice that the boundary condition at the end of RC corresponds to the insulated wall; that is, no
heat can be transferred through this resistance. The resistance to conduction through material C
is:
C
C
L
R k A
= (7)
"part (b)"
k_C=2.0 [W/m-K] "conductivity of material C"
R_C=L/(k_C*A) "resistance to conduction through C"
which leads to RC = 0.005 K/W.
e.) Use your resistance network from (d) to determine the temperature of the heating element.
Because there is no heat transferred through RC, all of the heat must still go through materials A
and B and be convected from the inner surface of the wall. Therefore, the answer is not changed
from part (b), Thtr = 410 K.
f.) Sketch the temperature distribution on the axes provided below. Make sure that the sketch is
consistent with your solution from (e).
The answer is unchanged from part (c) except that there is material to the left of the heater.
However, no heat is transferred through material C and therefore there is no temperature gradient
in the material.

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Figure P1.2-11(c) illustrates the same composite wall shown in Figure P1.2-11(b), but there is a
contact resistance between materials A and B, 2
0.01 K-m /WcR′′ = , and the surface of the wall at
x = -L is exposed to fluid at Tf,out = 400 K with a heat transfer coefficient outh = 10 W/m2 -K.
L = 1 cm
L = 1 cm
x
2
5000 W/mq′′ = material A
kA = 1 W/m-K
material B
kB = 5 W/m-K
,
2
300 K
100 W/m -K
f in
in
T
h
=
=
L = 1 cm
material C
k C = 2 W/m-K
2
0.01 K-m /WcR′′ =
,
2
400 K
10 W/m -K
f out
out
T
h
=
=
Figure P1.2-11(c): Composite wall with convection at the outer surface and contact resistance.
Neglect radiation for parts (g) through (i) of this problem.
g.) Draw a resistance network to represent the problem shown in Figure P1.2-11(c); clearly
indicate what each resistance represents and calculate the value of each resistance.
The additional resistances associated with contact resistance and convection to the fluid at the
outer surface are indicated. Notice that the boundary condition has changed; heat provided by
the heater has two paths ( outq and inq ) and so the problem is not as easy to solve.

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The additional resistances are computed according to:
,
1
conv out
out
R h A
= (8)
c
contact
R
R A
′′
= (9)
"part (c)"
R``_c=0.01 [K-m^2/W] "area specific contact resistance"
h_out=10 [W/m^2-K] "heat transfer coefficient"
T_f_out=400 [K] "fluid temperature on outside surface"
R_contact=R``_c/A "contact resistance"
R_conv_out=1/(h_out*A)
"convection resistance on outer surface"
which leads to Rcontact = 0.01 K/W and Rconv,out = 0.1 K/W.
h.) Use your resistance network from (j) to determine the temperature of the heating element.
It is necessary to carry out an energy balance on the heater:
in outq A q q′′ = + (10)
The heat transfer rates can be related to Thtr according to:
( ),
,
htr f in
in
A contact B conv in
T T
q R R R R
−
= + + +
(11)
( ),
,
htr f out
out
C conv out
T T
q R R
−
= +
(12)
These are 3 equations in 3 unknowns, Thtr, outq and inq , and therefore can be solved
simultaneously in EES (note that the previous temperature calculations from part (b) must be
commented out):
{T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A "heater temperature"
T_L=T_f_in+(R_B+R_conv_in)*q_flux*A "temperature at x=L"

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T_2L=T_f_in+R_conv_in*q_flux*A "temperature at x=2L"}
q_flux*A=q_dot_in+q_dot_out "energy balance on the heater"
q_dot_in=(T_htr-T_f_in)/(R_A+R_contact+R_B+R_conv_in) "heat flow to inner fluid"
q_dot_out=(T_htr-T_f_out)/(R_C+R_conv_out) "heat flow to outer fluid"
which leads to Thtr = 446 K. The other intermediate temperatures shown on the resistance
diagram can be computed:
x L htr A inT T R q= − = − (13)
( )x L htr A contact inT T R R q= + = − + (14)
( )2x L htr A contact B inT T R R R q= = − + + (15)
x L htr C outT T R q=− = − (16)
"intermediate temperatures"
T_Lm=T_htr-R_A*q_dot_in
T_Lp=T_htr-(R_A+R_contact)*q_dot_in
T_2L=T_htr-(R_A+R_contact+R_B)*q_dot_in
T_mL=T_htr-R_C*q_dot_out
which leads to Tx=L- = 400.4 K, Tx=L+ = 354.7 K, Tx=2L = 345.6 K, and Tx=-L = 443.8 K.
i.) Sketch the temperature distribution on the axes provided below.

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Problems 1.2-12 (1-5 in text): Floor Heater
You have decided to install a strip heater under the linoleum in your bathroom in order to keep
your feet warm on cold winter mornings. Figure P1.2-12 illustrates a cross-section of the
bathroom floor. The bathroom is located on the first story of your house and is W = 2.5 m wide
by L = 2.5 m long. The linoleum thickness is thL = 5 mm and has conductivity k L = 0.05 W/m-K.
The strip heater under the linoleum is negligibly thin. Beneath the heater is a piece of plywood
with thickness thp = 5 mm and conductivity k p = 0.4 W/m-K. The plywood is supported by ths =
6 cm thick studs that are W s = 4 cm wide with thermal conductivity k s = 0.4 W/m-K. The center-
to-center distance between studs is ps = 25.0 cm. Between each stud are pockets of air that can
be considered to be stagnant with conductivity k air = 0.025 W/m-K. A sheet of drywall is nailed
to the bottom of the studs. The thickness of the drywall is thd = 9.0 mm and the conductivity of
drywall is k d = 0.1 W/m-K. The air above in the bathroom is at Tair,1 = 15°C while the air in the
basement is at Tair,2 = 5°C. The heat transfer coefficient on both sides of the floor is h = 15
W/m2 -K. You may neglect radiation and contact resistance for this problem.
strip heater
linoleum, kL = 0.05 W/m-K
thL = 5 mm
thp = 5 mm
plywood, kp = 0.4 W/m-K
ths = 6 cm
Ws = 4 cm
studs, ks = 0.4 W/m-K
ps = 25 cm
thd = 9 mm
drywall, k d = 0.1 W/m-K
air, ka = 0.025 W/m-K
2
,1 15 C, 15 W/m -KairT h= ° =
2
,2 5 C, 15 W/m -KairT h= ° =
Figure P1.2-12: Bathroom floor with heater.
a.) Draw a thermal resistance network that can be used to represent this situation. Be sure to
label the temperatures of the air above and below the floor (Tair,1 and Tair,2 ), the temperature
at the surface of the linoleum (TL), the temperature of the strip heater (Th), and the heat input
to the strip heater ( hq ) on your diagram.
The resistance diagram corresponding to this problem is shown in Figure 2.

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Solution Manual for Heat Transfer , 1st Edition

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