Modern Engineering Thermodynamics

230 CHAPTER 7: Second Law of Thermodynamics and Entropy Transport and Production Mechanisms
HOW DOES HEAT TRANSPORT ENTROPY?
Take a glass of water, for example. If we heat it up by “transporting” heat into it, the water molecules move faster and
become more disordered. If we add enough heat, the water boils and becomes steam, and the water molecules become
even more disordered. If we cool the water by transporting heat out of it, the molecules slow down, and when it freezes
they form a very ordered crystalline structure. However, if you cool down your bedroom it does not become more
organized in the sense that all the stuff on the floor ends up in the closet. It only gets more organized at the molecular
level as everything freezes.
where τ is the time over which the heat transport occurs. Differentiating this equation with respect to time gives
the corresponding transport rate term as
Z
_S T Q = _q
dA (7.61)
Σ T b act
In the case where _q and T b are constant for time τ over the surface Σ of area A, Eq. (7.60) reduces to
Heat transport of entropy when _q and T b are constant:
ðS T Þ Q
=
_q
ðτAÞ = 1 Q 2
T b act
T b
and Eq. (7.61) reduces to
Heat transport rate of entropy when _q and T b are constant:
_S T
Q = Q _
T b act
act
(7.60a)
(7.61a)
Otherwise, the exact relations between ð_q /T b Þ act , time, and surface area must be known before the integral in
Eqs. (7.60) and (7.61) can be evaluated.
EXAMPLE 7.7
A dishwashing process in a restaurant has 3.00 kg/min of saturated liquid water heated in a steady flow process at 100.°C
until it has a quality of 75.0%. Determine the heat transport rate of entropy for this process.
Solution
First, draw a sketch of the system (Figure 7.16).
m = 3.00 kg/min
x in = 0
T in = 100.°C
Dish-washer
Heat
(S T ) Q = ?
m = 3.00 kg/min
x out = 75.0%
T out = 100.°C
FIGURE 7.16
Example 7.7.
The unknown is the heat transport rate of entropy for this process, and the material is water. The heat transport rate of
entropy is given by Eq. (7.61a) as
_S T
Q = Q _
T b act
Neglecting any changes in flow stream kinetic and potential energy, an energy rate balance for this steady state, steady flow
system is
_Q − _W + _m ðh 1 − h 2 + 0 + 0Þ = 0