solution of PS5 (2)(1)
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Transcript of solution of PS5 (2)(1)
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University of Houston
Department of Mechanical Engineering
MECE 2334: Thermodynamics I
Solution of problem set #5: Fall Semester 2014
Problem1
i) Yes. The entropy change of a closed system during an irreversible process can be positive,
negative or zero. Note that a closed system does not mean an isolated system. We could still
have entropy and heat transfer.
> 0According to 2ndlaw of thermodynamics,
= = Although is positive, the can be positive, negativeor or zero.ii) While the heat transfer is the same for the two processes, the entropy generation in
process (a) is larger since the same amount of heat is transferred across a larger
temperature difference. We can quantify the entropy generation of both cases by applyingthe Second Law in turn to the rod connecting the two reservoirs, noting that since the
temperature gradients are confined to the rod, all the entropy generation occurs in the rod
in both cases (the heat reservoirs are at a uniform temperature and do not have any
entropy generation).
While the following is not required by the problem, we can to formalize the above
conclusions by applying the Second Law to the rod
=
0 =
( ) ( )
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, = ( 2000
1000
) (2000
800
) = 0 . 5
, > ,It means that (a) is more irriversible.
iii) According to fist law we have
=
And,If > 0, we can have = 0. A net heat transfer need not always occur. In fact, wehave already seen in class examples of closed systems undergoing reversible processes
with and without heat transfer, namely quasi-static, isothermal compression of a gas and
quasi-static, adiabatic compression of a gas. In both cases, work is being done on the
system to compress the gas.
iv) For the closed system:
= a) = 0 , > 0 , > 0 It is inpossible process.b) = 0 , = 0 , > 0impossible process.c)
= 0 , < 0 , > 0it is possible if = 0.d) > 0 , = 0 , if > 0it is possible.e) > 0 , > 0 , < 0It is impossible. cannot be negative.f) < 0 , , > 0 it is possible if
< 0.
Problem 2
A System can be defined as an oil and spring. We have three states in this process.
State 1: Initial condition x=0. Spring is at zero force, and the oil and spring are at 20.State 2: x=0.1 m.
State 3: Spring is cut and returns to the unextended length.
=
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l n (
) = 0
= = = 12 1
2 10000.1 =4000
4000 = 5 =293.00125
l n () 4000 293.00125
293 =0.01706 =0.01706
Problem 3
a) Gas properties:
m=1 kg
R: Gas constant = 287.058
:
: Gas expands adiabatically, and with no dissipation (reversibly), the entropy of the gas does not
change:
= l n() l n () = 0
= ()
/+
Gas experiences no heat transfers, the first law for the gas can be written as:
= = = 1
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0
0
Finally, the maximum work is equal:
=
b) The gas expands isothermally.
= = /=ln/For an ideal gas in an isothermal condition,
= As
0 And
Thus,
C) Yes, they are different. In the first case, we have no heat transfer to the system. The work
done by the system is related to internal energy of the gas, and the maximum work is difference
between the initial internal energy of the gas and the final internal energy (at zero pressure).
In the second case, there is heat transfer. As our process is isothermal, during expansion from Pto P dP, system requires a heat transfer to remain in the same temperature. Therefore,
according to the first law, the work is equal to the heat transfer into the system.
Problem 4
a)
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b)
12 _ _ + 0 0 + _
23 + 0 + + + + +
34 + + _ 0 0 _ +
41 0 + _ _ _ _ 0
c)
12:
= 0 = ln () l n (
) = 0
741297 ln ( 300) = 2 9 7 l n (0.11 ) = 155.20 =155.20300=144.8
= 0.1 1 = 0.9 = m R = 0.178200
= = 0.921888
= 0.743688
= = 214.5 = = 0
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=280155.20=124.8
= 0 = m R = 2 2 9 7 2800.110 = 1.663200
= 1663200921888 10 = 0.741312
= = 0 . 1 1 0 74131210 = 741.312 = = 185 = = 259
= ln () l n () = 2 741297 ln ( 280155.20) 0 = 1.22 /
34
= =300
= 3 0 0 2 8 0 = 2 0 = ln () l n (
) = 0
2 741297 ln (300
280) = 2 2 9 7 l n (
0.1) =0.13
= 0.03 = m R = 1.37
=0.29 = = 29.6
= = 29 6 = 0
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41
= 0Thus, = 0
= 1 0.13 = 0.87 = 0.1781.37 =1.1925
= ln () ln () = 0 2 2 9 7 l n ( 10.13) = 1.22 /To determine the heat transfer, apply the Second Law for reversible isothermal processes:
= = 0
= = 367 = 0, = = 367 These results make sense! The First Law is satisfied for the cycles net work and heat
transfer, and the net change in all the state properties (T, P, V, S, and U) is zero as it
should be since a cycle by definition ends at the same state as it started.
Lastly, calculating the coefficient of performance of the cycle (assuming it is used as arefrigerator and therefore using the cooling heat transfer, = into the cycle, asdesired quantity):
= =2.4
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