Modern Engineering Thermodynamics

7.14 Heat Transfer Production of Entropy 233
EXAMPLE 7.9
Determine an equation for the steady state entropy production rate due to pure heat conduction in an insulated horizontal
rod connecting a high-temperature (T 1 ) thermal reservoir with a low-temperature (T 2 ) thermal reservoir.
Solution
First, draw a sketch of the system (Figure 7.18).
Thermal source at
temperature
T 1 > T 2
Area A 1
Q
System boundary
temperature T 1
Insulation
Rod
System
boundary
x
Rod
L
Heat
conducting
rod
Thermal reservoir
at temperature
T 2 < T 1
Problem schematic
Area A 2 = A 1
Q
System sketch
System boundary
temperature T 2
FIGURE 7.18
Example 7.9.
The unknown is an equation for the steady state entropy production rate for the system. The entropy production rate due to
heat transfer is given by Eq. (7.66) as
_S P
Q = − Z
V
_q
dT
T 2 dx
dV
actual
For steady state conditions, _q = constant across areas A 1 and A 2 , but _q = 0 on the remaining surfaces of the system boundary.
Since A l = A 2 = A and _Q = _qA is a constant, then using dV = Adx, we can write
_q
dT
0 T 2 dx
= _Q 1 − 1 T 2 T 1
_S P
Q = − Z L
Z L
Adx= − _Q
0
1
T 2
Q
= _ ðT 1 − T 2 Þ> 0
T 1 T 2
dT
dx
dx = − _Q
For a pure one-dimensional, steady state conduction heat transfer, Fourier’s law gives
Z T2
T1
dT
T 2
or
dT
dx = − T 1 − T 2 Q
= − _ L k t A = constant
_Q = k t AðT 1 − T 2 Þ/L
Putting this information into the preceding entropy production rate formula gives
_S P
Q = Q _ ðT 1 − T 2 Þ = k tA
ð
T 1 T 2 T 1 T 2 L T 1 − T 2 Þ 2 > 0
Notice that, in Example 7.9, the entropy production rate becomes zero only when T 1 = T 2 (i.e., when _Q = 0), so
that a reversible conduction heat transfer is actually impossible. Also note that, the larger the temperature difference
T 1 − T 2 through which a given heat transfer _Q occurs, the larger the associated heat transfer entropy production rate
becomes.