Application of Steady-State
Heat Transfer
Steady-state heat transfer
• Temperature in a system remains constant
with time.
• Temperature varies with location.
Conductive heat transfer in a
rectangular slab
dT
q x   kA dx
x qx
x
q
T1
T
dx    kdT
T2
A
T1
x
T
X
dx   k  dT

A x1
T1
1
T
q x  kA x
T1 > T2
Example
For the stainless steel plate 1 cm thick is
maintained at 110C, while the other face is at
90 C. Calculate temperature at 0.5 cm from the
110C-temperature face.
Given :
heat flux = 34,000 W/m2
thermal conductivity of stainless steel = 17 W/m C
Conductive Heat Transfer
through a Tubular Pipe
• Consider a long hollow cylinder
To
ro dr
ri
T
Ti
r
l
A( r )  2rl
Conductive Heat Transfer
through a Tubular Pipe
• Consider a long hollow cylinder
Example
A 2 cm thick steel pipe (k= 43
W/mC) with 6 cm inside
diameter is being used to convey
steam from a boiler to process
equipment for a distance of 40 m.
The
inside
pipe
surface
temperature is 115C, and the
outside pipe surface temperature
is 90C. Under steady state
conditions, calculate total heat
loss to the surrounding.
Heat conduction
in multilayered systems
Composite rectangular wall
(in series)
q
x1
x2
x3
k1
k2
k3
T1 T2 T3
R1
R2
R3
q
= composite thermal resistance
Composite rectangular wall
(in parallel)
q = A T k / x
= A T / (x/k)
 T1 + T2 + T3
q
= T
R = Resistance = x/k = 1/C
1/RT = 1/R1+1/ R2+1/ R3
= (1/(x1 / k 1))+
(1/(x2 / k 2))+ (1/(x3 / k 3))
x1 k1 T1
x2 k2 T2
x3 k3 T3
R1
R2
R3
q
and it is resistance which is additive when several
conducting layers lie between the hot and cool
regions, because A and Q are the same for all layers.
In a multilayer partition, the total conductance is
related to the conductance of its layers by:
So, when dealing with a multilayer partition, the
following formula is usually used:
Example
A cold storage wall (3m X 6m) is constructed of a 15 cm
thick concrete (k = 1.37 W/mC). Insulation must be
provided to maintain a heat transfer rate through the wall
at or below 500 W. If k of insulation is 0.04 W/mC. The
outside surface temperature of the wall is 38C and the
inside wall temperature is 5C.
Example
How many joules of thermal energy
flow through the wall per second?
------------------------------------------Heat is like a fluid: whatever flows
through the insulation must also
flow through the wood.
Across insulation:
Hins = (0.20)(40)(25 - T)/0.076
= 2631.6 -105.3 T
Across wood:
Hwood = (0.80)(40)(T - 4)/0.019
= 1684.2 T - 6736.8
Heat is like a fluid: whatever flows through the
insulation must also flow through the wood:
k (insulation) = 0.20 J/(s-m-C)
k (wood) = 0.80 J/(s-m-C)
Hwood = Hins
1684.2 T - 6736.8 = 2631.6 -105.3 T
1789.5 T = 9368.4
T = 5.235 C
H=Hwood=Hins
H= 1684.2 (5.235) - 6736.8 = 2080 J/s
H= 2631.6 - 105.3 (5.235) = 2080 J/s
Series and parallel one-dimensional heat
transfer through a composite wall and
electrical analog
B
F
C
A
E
G
D
1
RA
2
RB
RC
RD
3
4
RE
5
RF
RG
Composite cylindrical tube
(in series)
r1
r3
r2
Example
A stainless steel pipe (k= 17
W/mC) is being used to convey
heated oil. The inside surface
temperature is 130C. The pipe is
2 cm thick with an inside diameter
of 8 cm. The pipe is insulated
with 0.04 m thick insulation (k=
0.035
W/mC).
The
outer
insulation temperature is 25C.
Calculate the temperature of
interface between steel and
isulation. Assume steady-state
conditions.
THERMAL CONDUCTIVITY
CHANGE WITH TEMPERATURE
Heat transfer through a slab
k = k0(1+T)
dT
q x  kA
dx
qx 
km A
 X
(T1  T2 )
km is thermal conductivity at T =
T1 T 2
2
k0 A
 2
2
q 
((T2  T1 )  (T2  T1 ))
x
2
THERMAL CONDUCTIVITY
CHANGE WITH TEMPERATURE
Heat transfer through a cylindrical tube
kAdT
qr 
dr
qr  (k 0(1   T ))( 2rL )
dT
dr
dr
qr
 2k0 L(1  T )dT
r
2k0 L
qr 
(1   (Ti  T0 )(Ti  T0 ))
ln ro / ri
Problem
1. Find the heat transfer per unit area through the composite
wall. Assume one-dimensional heat flow.
Given:
kA = 150 W/mC
kB = 30 W/mC
kC = 50 W/mC
B
q
kD = 70 W/mC
A
AB = AD
AA = AC = 0.1 m2
C
D
T = 370C
T = 66C
2.5
cm
7.5
cm
5.0
cm
Problem
2. One side of a copper block 5 cm thick is maintained at
260C. The other side is covered with a layer of fiber
glass 2.5 cm thick. The outside of the fiber glass is
maintained at 38C, and the total heat flow through the
copper-fiber-glass combination is 44 kW. What is the
area of the slab?
3. A wall is constructed of 2.0 cm of copper, 3.0 mm of
asbestos, and 6.0 cm of fiber glass. Calculate the heat
flow per unit area for an overall temperature difference of
500C.
Problem
4. A certain material has a thickness of 30 cm and a
thermal conductivity of 0.04 W/mC. At a particular
instant in time the temperature distribution with x, the
distance from the left face, is T = 150x2 - 30x, where x is
in meters. Calculate the heat flow rates at x = 0 and x =
30 cm. Is the solid heating up or cooling down?
5. A certain material 2.5 cm thick, with a cross-sectional
area of 0.1 m2, has one side maintained at 35C and the
other at 95C. The temperature at the center plane of the
material is 62C, and the heat flow through the material
is 1 kW. Obtain an expression for the thermal
conductivity of the material as a function of temperature.