Modern Engineering Thermodynamics

232 CHAPTER 7: Second Law of Thermodynamics and Entropy Transport and Production Mechanisms
produced by entropy production, so it is useful to have a tool to investigate these losses in the design and
analysis process.
From the work of Carnot, Thomson, and Clausius discussed earlier, it seems clear that energy and entropy are
related in some way. Therefore, in this section, we investigate the possibility that the energy transport mechanisms
of heat, work, and mass flow are also mechanisms for entropy production. First, we investigate heat and
work production of entropy by again restricting our analysis to closed systems. In Chapter 9, we expand this
investigation to include mass flow entropy production.
7.14 HEAT TRANSFER PRODUCTION OF ENTROPY
To integrate Eq. (7.52) properly, we define a one-dimensional thermal entropy production rate per unit volume,
σ 0 , which is the rate of entropy production per unit volume due to the actual heat transfer, as
d 2 ðS P Þ Q
dt dV
= d _S P
Q
dV
= σ Q = − _q
dT
T 2 dx
actual
(7.64)
where the minus sign appears because dT/dx always is negative when _q is positive. The temperature T in this
equation is the local absolute temperature inside the system boundary evaluated at the point where the local
internal heat flux _q occurs. It is generally not the same as the local system boundary temperature T b except in
the case of an isothermal system. Equation (7.64) can then be integrated to give
dðS P Þ Q
= Q T 2 dT = − Z
and, with a second integration, we have
τ
_q
dT
T 2 dx
dt dV =
actual
Heat transfer entropy production
Z Z
_q
Z Z
ðS
dT
P Þ Q
= −
dt dV =
V τ T 2 dx
actual
V τ
Z
τ
σ Q dt dV
Differentiation of Eq. (7.65) yields the heat transfer entropy production rate term as
σ Q dt dV (7.65)
Heat transfer entropy production rate
Z
_S P Q = − _q
Z
dT
V T 2 dV = σQdV (7.66)
dx actual V
EXAMPLE 7.8
An electric motor has a volume of 2.50 × 10 −3 m 3 and operates with a constant internal thermal entropy production rate
per unit volume of σ Q = 53:7W=K.m 3 . Determine the heat production of entropy inside this motor for the time period τ =
30.0 min of operation.
Solution
First, draw a sketch of the system (Figure 7.17).
Theunknownistheheatproductionofentropy inside this motor for the time period
τ = 30.0 min of operation. The heat production of entropy for this system is given be
Eq. (7.65) as
Z Z " _q
ðS P Þ Q
= −
dT # Z Z
V τ T dt dV = σ 2 Q dt dV
dx
V τ
actual
Since σ Q is a constant for τ = 30.0 min of operation, this equation reduces to
ðS P Þ Q = σ Q τV
and we can calculate
wherewehaveusedthefactthat1W= 1J/s.
ðS P Þ Q = ð53:7W/K.m 3 Þð2:50 × 10 − 3 m 3 Þð30:0minÞð60 sec/minÞ = 242 J/K
Electric
motor
∀ = 2.50 × 10 −3 m 3
σ Q = 53.7 W/K . m
FIGURE 7.17
Example 7.8.
(S P ) Q = ?