Solutions Manual
Heat Transfer
By
Gregory Nellis,
Sanford Klein
( All Chapters Included - 100% Verified Solutions )
1
,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 kA
= 1 W/m-K and B with kB = 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 q ′′ = 5000 W/m 2 of heat. The surface of the wall at x = 2L is
exposed to fluid at Tf,in = 300 K with heat transfer coefficient hin = 100 W/m2-K.
material A
q ′′ = 5000 W/m
2
kA = 1 W/m-K
insulated L = 1 cm
x T f ,in = 300 K
hin = 100 W/m -K
2
L = 1 cm material B
kB = 5 W/m-K
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:
2
, L
RA = (1)
kA A
L
RB = (2)
kB A
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
Rconv ,in = (3)
hin A
"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:
Thtr = T f ,in + ( RA + RB + Rconv ,in ) 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:
Tx = L = T f ,in + ( RB + Rconv ,in ) q ′′ A (5)
Tx = 2 L = T f ,in + Rconv ,in 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"
3
, 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.
q ′′ = 5000 W/m
2
material A
material C kA = 1 W/m-K
kC = 2 W/m-K
L = 1 cm
insulated
x T f ,in = 300 K
hin = 100 W/m -K
2
material B
L = 1 cm L = 1 cm k = 5 W/m-K
B
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:
4
Heat Transfer
By
Gregory Nellis,
Sanford Klein
( All Chapters Included - 100% Verified Solutions )
1
,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 kA
= 1 W/m-K and B with kB = 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 q ′′ = 5000 W/m 2 of heat. The surface of the wall at x = 2L is
exposed to fluid at Tf,in = 300 K with heat transfer coefficient hin = 100 W/m2-K.
material A
q ′′ = 5000 W/m
2
kA = 1 W/m-K
insulated L = 1 cm
x T f ,in = 300 K
hin = 100 W/m -K
2
L = 1 cm material B
kB = 5 W/m-K
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:
2
, L
RA = (1)
kA A
L
RB = (2)
kB A
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
Rconv ,in = (3)
hin A
"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:
Thtr = T f ,in + ( RA + RB + Rconv ,in ) 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:
Tx = L = T f ,in + ( RB + Rconv ,in ) q ′′ A (5)
Tx = 2 L = T f ,in + Rconv ,in 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"
3
, 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.
q ′′ = 5000 W/m
2
material A
material C kA = 1 W/m-K
kC = 2 W/m-K
L = 1 cm
insulated
x T f ,in = 300 K
hin = 100 W/m -K
2
material B
L = 1 cm L = 1 cm k = 5 W/m-K
B
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:
4
