Modern Engineering Thermodynamics

7.13 Entropy Production Mechanisms 231
and since
_W = 0 here (no work modes are present), we have
_Q = _mðh 2 − h 1 Þ
= _m f½h f ð100:°CÞ + x 2 h fg ð100:°CÞŠ − h f ð100:°CÞg
= _mx 2 h fg ð100:°CÞ = ð3:00 kg/minÞð0:750Þð2257 kJ/kgÞ = 5078 kJ/min
Then, Eq. (7.61a) gives
ð _S T Þ Q =
5078 kJ/min
100: + 273:15 K
= 13:6 kJ/min.K
Exercises
13. Suppose the system in Example 7.7 is a closed, rigid vessel containing 3.00 kg of water, and the water is heated in the
same manner from a saturated liquid at 100.°C to a liquid-vapor mixture at 100.°C with quality of 75.0%. Determine
the heat transport of entropy for this process. Answer: (S T ) Q = 12.6 kJ/K.
14. A new heat exchanger has been designed, where the local heat flux (heat “flux” is heat transfer per unit area) is directly
proportional to the local surface temperature T b of the heat transfer area A. Then, _q = KT b over the surface area A of the
heat exchanger. Determine an expression for the heat transport rate of entropy for this system. Answer: ð _S T Þ Q
= KA.
15. An electric motor draws 800. W of electrical power and is 95.0% efficient. This means that 0.950(800.) = 760. W leaves
the motor as mechanical shaft power and (1 − 0.950)(800.) = 40.0 W leaves as heat generated by the electrical and
mechanical losses in the motor. Determine the heat transport rate of entropy of the motor if it has a uniform surface
temperature of 30.0°C. Answer: ð _S T Þ Q
= 0:132 W/K.
7.12 WORK MODE TRANSPORT OF ENTROPY
Integration of Eq. (7.51) for all possible work modes clearly gives
and, from Eq. (7.56), we also have
Work mode transport of entropy
ðS T Þ W
= 0 (7.62)
Work mode transport rate of entropy
_S T
W = 0 (7.63)
This produces the surprising result that none of the work modes discussed in Chapter 4 transports entropy into
or out of a system. However, as is shown later, the irreversibilities of these work modes always contribute to the
production of entropy within the system.
7.13 ENTROPY PRODUCTION MECHANISMS
One of the main problems with Eq. (7.22) is that it applies only to reversible processes. All of our auxiliary
formulae for heat transfer have been developed on an empirical basis and therefore always give the actual or irreversible
rather than the reversible heat transport of energy. Also, the process inefficiencies are mostly due to losses
WHY DO WORK MODES NOT TRANSPORT ENTROPY?
Let us suppose you have an electric ceiling fan in your room. When you turn it on, you have an electric work “mode” coming
into the room. The fan blows your papers around and makes a mess (disorder). So why is this not a “work mode”
transport of entropy (disorder)?
Well, “transports” have to be able to go both ways, in and out of the system, like heat transfer. If we run the fan backward,
it does not put all your papers back where they were. So, in this case, your ceiling fan does not “transport” disorder
(entropy) into your room, it “produces” disorder (entropy) inside your room. It is an entropy production process, not an
entropy transport process. Work modes can only produce entropy; they cannot transport it across the system boundary like
heat transfer.