A closed rigid tank with a volume of 2 m^3 contains hydrogen gas initially at 32
ID: 1848984 • Letter: A
Question
A closed rigid tank with a volume of 2 m^3 contains hydrogen gas initially at 320 K and 180 kPa. Heat transfer from a reservoir at 500 K takes place until the gas temperature reaches 400 K.a. Calculate the entropy change and entropy production for the hydrogen gas during the process, both in KJ/K, if the boundary temperature for the gas is the same as the gas temperature throughout the process.
b. Determine the entropy production, in KJ/K, for an enlarged system which includes the tank and the reservoir.
c. Explain why the entropy production differ for parts (a) and (b).
d. Find the entropy production for the gas and the total value, in KJ/K, if the system boundary temperature is 450 K throughout the process.
e. The hydrogen gas process is repeated, but paddle-wheel work is used instead of heat transfer. Calculate the entropy production in this case.
f. Compare relative irreversibility of process (b.) (d.), or (e.)
Explanation / Answer
a) Entropy of the hot reservoir is assumed to have no irreversiblilites present so there is no entropy production involved. Therefore, entropy change is solely do to heat transfer. ie. Change in entropy = -Q/T(h) = -5100/(710) = -7.18309859 b) The same assumption as in part a) are made. Another assumption is that there is no work done for this "engine" which is weird but...i don't see any mention of work. Therefore: Change in entropy = Q/T(c) = 5100/207 = 24.6376812 c) The engine itself i am assuming is the total entropy change of the system. This is the total sum of the entropy in and the entropy out. What is flowing out for the hot reservoir is what flowing in to the system the signs will have to be flipped for the two answered found in pats a and b. Therefore the total change of the system (engine): = Q/T(ch) - Q/T(c) = -17.4545826 I guess that looks strange that the entropy change for the system is negative. The only evidence i can see that could possibly support that is that the process is reversible and it may be the case that the entropy change of the universe with no irreversibility what so ever is 0. That is the sum of parts a) , b) and c) is 0.
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