Modern Engineering Thermodynamics

7.11 Heat Transport of Entropy 229
and comparing Eqs. (7.48) and (7.49) allows us to identify the following terms: 10
ðdS T Þ Q = d Q
T
act
(7.50)
and
ðdS T Þ W
= 0 (7.51)
ðdS P Þ Q
= Q
T dT 2 act
ðdS P Þ W
= dW T
irr
(7.52)
(7.53)
Integrating Eq. (7.49) and then differentiating it with respect to time produces an entropy rate balance equation
for a system as
_S = dS
dt = d Z
d Q
+ d Z
dt T
act
dt
= _S T Q +
_S T W +
_S P Q + _S P
W
Q
T dT + d Z
2
act
dt
dW
T
irr
(7.54)
so that
_S T Q = d Z
d Q
dt T
act
(7.55)
_S T
W = 0 (7.56)
and
_S P Q = d Z
dt
_S P W = d Z
dt
Q
T 2 dT
act
dW
T
irr
(7.57)
(7.58)
7.11 HEAT TRANSPORT OF ENTROPY
The integration of Eq. (7.50) gives the heat transport of entropy as
Z
ðS T Þ Q
= d
Σ Q T b
= ∑ Q T b
act
Σ
(7.59)
where Σ is the surface area of the system and T b is the local absolute temperature of the system boundary corresponding
to the value of the local heat transfer at the boundary, Q. However,ifQ and T b vary continuously
along the boundary, then Eq. (7.59) is not easy to evaluate. To produce a more useful version of this equation,
let q be the heat transfer per unit area and let _q be the heat transfer rate per unit area (i.e., the heat “flux”).
Then, define the heat transport of entropy per unit area as q/T b = ðdQ/dAÞ/T b and define the heat transport rate
of entropy as _q /T = ðd 2 Q/dA dtÞ/T b so that Eq. (7.59) becomes
Z Z Z
ðS T Þ Q
=
Σ
q
T b
dA =
act
Σ
τ
_q
T b
dA dt (7.60)
act
10 Note that we could also attempt to use the identity of Eq. (5.44) on the irreversible work term and decompose it into
dW irr /T = dW/T ð Þ irr
+ ðW irr /T 2 Þ, then we would be tempted to equate d(S T ) W = d(W/T) irr . This would be incorrect because the
irreversible work always occurs inside the system boundary and therefore cannot be associated with a transport term that measures
quantities crossing the system boundary.