Andrew Mason said:
Entropy describes a thermodynamic state, not a quantity of something physical. So it is very confusing to speak about entropy generation. One does not speak about generating temperature, for example.
Entropy generation is a well known term used in thermodynamic texts. It is not a term I coined by any means.
Andrew Mason said:
This is not exactly a thermodynamic system. You need some place to deliver the heat to. The system has to include the hot reservoir.
Of course is it a thermodynamic system. Thermodynamics is the study of energy. Let’s put some numbers in and see what happens to the entropy change of this system. Say the tank has 5kg of R134a initially at 20 °C and 140 kPa. The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa.
Since the tank is rigid v_2 = v_1. It is a closed system since no mass crosses the system boundary during the process. The entropy change of a substance is simply the difference between the entropy values at the final and initial states. Since the specific volume remains constant during this process the properties of the refrigerant at both states are:
State 1:
P_1 = 140 kPa
T_1 = 20 °C
Given these two intensive independent properties, one can see from a property table for R134a the following:
s_1 = 1.0624 \frac{kJ}{kg \cdot K}
v_1 = 0.16544 \frac{m^3}{kg}
State 2:
P_2 = 100 kPa
v_2 = v_1
Using the property table again one finds:
V_f = 0.0007259 \frac{m^3}{kg}
V_g = 0.19252 \frac{m^3}{kg}
From these the quality at state 2 is found to be 0.859.
Thus,
s_2 = s_f + x_2 \cdot s_{fg} = 0.07188 + (0.89)(0.87995) = 0.8278 \frac{kJ}{kg \cdot K}
The entropy change of the refrigerant during this process is:
\Delta{S} = m({s_2} - {s_1}) = (5)(0.8278 – 1.0624) \frac{kJ}{kg \cdot K}
EDIT: For some reason latex is not being displayed properly here. Delta S should equal m(s_2 - s_1).
= -1.173 kJ/K
Notice the negative sign. This indicates that the entropy of the system is decreasing during this process. This is not a violation of the second law, however, since it is the entropy generation that cannot be negative. This is why one must know the difference between the general term “entropy” and entropy generation. They are not the same. Equating the terms is more confusing, not less.
BTW, this is not a cyclic device as you seem to think (in my example).
Andrew Mason said:
The entropy of the cold reservoir is negative in a refrigeration cycle simply because heat flows out of it (dQ<0 so dQ/T < 0). This is the case whether the cycle is reversible or irreversible.
This was my point all along which you seemed to have been saying wasn’t possible. The stipulation about being reversible simply points out that the entropy generation is zero for a reversible process.
Andrew Mason said:
This is, again, confusing. If I take a Carnot engine system (engine + reservoirs) and I do work that is dissipated by friction externally to the system, is the
change in entropy of the system not still zero?
Friction is an irreversibility of a system. If friction is involved, the entropy generation is positive, not zero.
Andrew Mason said:
If entropy is a state function, how is it transferred or generated?
Entropy generation is a measure of the magnitudes of the irreversibilities present during a process.
Andrew Mason said:
There can be an entropy change associated with the transfer of work. Consider work done against friction force where the heat generated is transferred to a reservoir: the entropy of the reservoir will increase (by dQ/T = W/T).
AM
Being an organized form of energy, work is free of disorder or randomness and thus free of entropy. The work done against a frictional force results in an entropy change due to the irreversibility (i.e. the friction). It is not a result of the work itself, rather it is the fact that the organized energy (work) is being degraded to a less useful form (heat). Hence, the entropy generation is increased.
CS
