8/9/2019 solution of PS5 (2)(1)

1/14

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 =

( ) ( )

8/9/2019 solution of PS5 (2)(1)

2/14

, = ( 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.

=

8/9/2019 solution of PS5 (2)(1)

3/14

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

8/9/2019 solution of PS5 (2)(1)

4/14

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)

8/9/2019 solution of PS5 (2)(1)

5/14

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

8/9/2019 solution of PS5 (2)(1)

6/14

=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

8/9/2019 solution of PS5 (2)(1)

7/14

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

8/9/2019 solution of PS5 (2)(1)

8/14

8/9/2019 solution of PS5 (2)(1)

9/14

8/9/2019 solution of PS5 (2)(1)

10/14

8/9/2019 solution of PS5 (2)(1)

11/14

8/9/2019 solution of PS5 (2)(1)

12/14

8/9/2019 solution of PS5 (2)(1)

13/14

8/9/2019 solution of PS5 (2)(1)

14/14