Table Of ContentProblem 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 = 5 W/m-K), each has thickness L = 1.0 cm. The surface of the wall at
B
x = 0 is perfectly insulated. A very thin heater is placed between the insulation and material A;
the heating element provides q(cid:5)′′=5000 W/m2 of heat. The surface of the wall at x = 2L is
exposed to fluid at T = 300 K with heat transfer coefficient h = 100 W/m2-K.
f,in in
materialA
q(cid:5)′′=5000W/m2 k =1W/m-K
A
insulated L= 1cm
x T =300K
f,in
h =100W/m2-K
in
L= 1cm materialB
k =5W/m-K
B
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:
L
R = (1)
A k A
A
L
R = (2)
B k A
B
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
R = (3)
conv,in h A
in
"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 R = 0.01 K/W, R = 0.002 K/W, and R = 0.01 K/W.
A B conv,in
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 (T ) is therefore:
htr
T =T +(R +R +R )q(cid:5)′′A (4)
htr f,in A B conv,in
T_htr=T_f_in+(R_A+R_B+R_conv_in)*q_flux*A "heater temperature"
which leads to T = 410 K.
htr
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:
T =T +(R +R )q(cid:5)′′A (5)
x=L f,in B conv,in
T =T +R q(cid:5)′′A (6)
x=2L f,in conv,in
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"
which leads to T = 360 K and T = 350 K. The temperature distribution is sketched on the
x=L x=2L
axes in Figure 3.
Figure 3: Sketch of temperature distribution.
Notice that the temperature drop through the two larger resistances (R and R ) are much larger
A B
than the temperature drop across the small resistance, R .
B
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 k = 2.0 W/m-K and L = 1.0 cm.
C
q(cid:5)′′=5000W/m2 materialA
materialC
k =1W/m-K
k =2W/m-K A
C
L= 1cm
insulated
x T =300K
f,in
h =100W/m2-K
in
materialB
L= 1cm L= 1cm k =5W/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, R , as shown
C
below:
Notice that the boundary condition at the end of R corresponds to the insulated wall; that is, no
C
heat can be transferred through this resistance. The resistance to conduction through material C
is:
L
R = (7)
C k A
C
"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 R = 0.005 K/W.
C
e.) Use your resistance network from (d) to determine the temperature of the heating element.
Because there is no heat transferred through R , all of the heat must still go through materials A
C
and B and be convected from the inner surface of the wall. Therefore, the answer is not changed
from part (b), T = 410 K.
htr
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.
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, R′′=0.01 K-m2/W, and the surface of the wall at
c
x = -L is exposed to fluid at T = 400 K with a heat transfer coefficient h = 10 W/m2-K.
f,out out
q(cid:5)′′=5000W/m2 materialA
materialC
k =1W/m-K
k =2W/m-K A
C
L= 1cm
T =400K
hf,ou=t 10W/m2-K x Tf,in =300K
out h =100W/m2-K
in
materialB
L= 1cm k =5W/m-K
B
L= 1cm
R′′=0.01K-m2/W
c
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 (q(cid:5) and q(cid:5) ) and so the problem is not as easy to solve.
out in
The additional resistances are computed according to:
1
R = (8)
conv,out h A
out
R′′
R = c (9)
contact A
"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 R = 0.01 K/W and R = 0.1 K/W.
contact conv,out
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:
q(cid:5)′′A=q(cid:5) +q(cid:5) (10)
in out
The heat transfer rates can be related to T according to:
htr
( )
T −T
q(cid:5) = htr f,in (11)
in R +R +R +R
A contact B conv,in
( )
T −T
q(cid:5) = htr f,out (12)
out R +R
C conv,out
These are 3 equations in 3 unknowns, T , q(cid:5) and q(cid:5) , and therefore can be solved
htr out in
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"
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 T = 446 K. The other intermediate temperatures shown on the resistance
htr
diagram can be computed:
T =T −R q(cid:5) (13)
x=L− htr A in
T =T −(R +R )q(cid:5) (14)
x=L+ htr A contact in
T =T −(R +R +R )q(cid:5) (15)
x=2L htr A contact B in
T =T −R q(cid:5) (16)
x=−L htr C out
"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 T = 400.4 K, T = 354.7 K, T = 345.6 K, and T = 443.8 K.
x=L- x=L+ x=2L x=-L
i.) Sketch the temperature distribution on the axes provided below.
P1.1-1: Viscosity of a dilute gas
Momentum transfer occurs in a fluid due to interactions between molecules that results in a
transfer of momentum. This process is characterized by viscosity, which relates the shear stress
to a velocity gradient in the same way Fourier’s Law relates heat flux to a temperature gradient.
It is not surprising, then, that the viscosity and thermal conductivity of an ideal gas are analogous
transport properties.
a.) Using reasoning similar to that provided in Section 1.1.2 for thermal conductivity, show that
the viscosity of an ideal gas can be estimated according to μ∝ T MW /σ2.
Consider momentum transfer through a fluid in which a velocity gradient has been established in
the x-direction, as shown in Figure 1. We can evaluate the net rate of momentum transferred
through a plane that is located at position x. The flux of molecules passing through the plane
from left-to-right (i.e., in the positive x-direction) is proportional to the number density of the
molecules (n ) and their mean velocity (v ). The molecules that are moving in the positive x-
ms ms
direction experienced their last interaction at x–L (on average), where L is the distance
ms ms
between molecular interactions. The rate of momentum associated with these molecules per unit
area is the product of the rate of molecules passing through the plane (n v ) momentum and
ms ms
the momentum per molecule; the momentum per molecule is the product of the mass of the
molecule (M) and its x-velocity at the point where it experienced its last collision, x-L
ms
(M u ). Therefore, the rate of momentum passing through the plane from left-to-right (M(cid:5)′′ )
x−L x+
ms
is given approximately by:
M(cid:5) ′′ ≈n v M u (1)
x+ ms ms ms,x−L
ms
Similarly, the momentum per unit area passing through the plane from right-to-left (M(cid:5)′′ ) is
x−
given by:
M(cid:5) ′′ ≈n v M u (2)
x+ ms ms x+L
ms
Velocity
M(cid:5)′′
x+
M(cid:5)′′
x−
L
ms
x-L x+L
ms ms
Position
x
Figure 1: Momentum flows through a plane in a material.
The net rate of momentum flux passing through the plane per unit area in the positive x-direction
(M(cid:5) ′′) is the difference between M(cid:5) ′′ and M(cid:5) ′′ ,
x+ x−
M(cid:5) ′′≈n v M (u −u ) (3)
ms ms x−L x+L
ms ms
which can be rearranged to yield:
( )
u −u ∂u
M(cid:5) ′′≈−2n v M L x+Lms x−Lms ≈−2n v M L (4)
ms ms ms L (cid:8)(cid:11)ms(cid:11)(cid:9)ms (cid:11)(cid:11)(cid:10)ms ∂x
(cid:8)(cid:11)(cid:11)(cid:9)ms(cid:11)(cid:11)(cid:10)
∝μ
∂u
∂x
Comparing Eq. (4) with the definition of viscosity shows that the viscosity is proportional to the
product of the number of molecules per unit volume, their average velocity, the mass of each
molecule, and the mean distance between their interactions.
μ∝n v M L (5)
ms ms ms
The mass of a molecule is the molecular weight, MW. As noted in Eq. (1-14), kinetic theory
indicates that
R T
v ∝ univ (6)
ms MW
where R is the universal gas constant and T is the absolute temperature. The distance between
univ
molecular interactions was derived in Eq. (1-17) is
1
L = (7)
ms n πσ2
ms
where σ is the equivalent radius of the molecule. Substituting Eqs. (6) and (7) into Eq. (5) shows
that:
1
μ∝ T MW (8)
σ2
which is identical to Eq. (1-18) for conductivity if the specific heat capacity is removed.
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:
c T
k ∝ v (1)
σ2 MW
Equation (1) is written as an equality by including a constant of proportionality (C ):
k
c T
k =C v (2)
kσ2 MW
Solving for C leads to:
k
kσ2 MW
C = (3)
k c T
v
which indicates that C has units m-kg1.5/s-kgmol05-K0.5.
k
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 , σ, and MW) are evaluated using EES' built-in funcions. Equation (3) is used to
v
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"
