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Modern Engineering Thermodynamics

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9.9 Unsteady State Processes in Open Systems 301<br />

a. From Eq. (9.42) for adiabatic filling, we have<br />

½ 1 ðS P Þ 2 Š adiabatic<br />

filling<br />

b. From Eq. (9.46) for isothermal filling, we have<br />

½ 1 ðS P Þ 2 Š isothermal<br />

filling<br />

= ðm 2 Þ adiabatic c p lnðkÞ = ð24:3 lbmÞ½0:219 Btu/ðlbm.RÞŠ ln ð1:39Þ = 1:75 Btu/R<br />

filling<br />

which is about 20% greater than the entropy produced in part a.<br />

<br />

= ðm 2 Þ isothermal R = ð33:8 lbmÞ 48:29 ft <br />

.lbf/ðlbm.RÞ<br />

= 2:10 Btu/R<br />

filling<br />

778:16 ft.lbf/Btu<br />

Exercises<br />

22. If the volume of the rigid container in Example 9.8 is reduced from 3.00 ft 3 to 2.00 ft 3 , what is the new entropy<br />

production for the filling processes described in parts a and b? Keep the values of all the other variables the same as<br />

they are in Example 9.8. Answer: (a) [ 1 (S P ) 2 ] adiabatic = 1.17 Btu/R, (b) [ 1 (S P ) 2 ] isothermal = 1.40 Btu/R.<br />

23. The filling pressure in Example 9.8 is to be reduced from 2000. psia to 1500. psia. Determine the new entropy<br />

production for the filling processes described in parts a and b. Keep the values of all the other variables the same as they<br />

are in Example 9.8. Answer: (a) [ 1 (S P ) 2 ] adiabatic = 1.31 Btu/R, (b) [ 1 (S P ) 2 ] isothermal = 1.57 Btu/R.<br />

24. If the rigid container discussed in Example 9.8 is filled with air instead of oxygen, determine the new entropy<br />

production for the filling described in parts a and b. Keep the values of all the other variables the same as they are in<br />

Example 9.8. Answer: (a) [ 1 (S P ) 2 ] adiabatic = 1.76 Btu/R, (b) [ 1 (S P ) 2 ] isothermal = 2.10 Btu/R.<br />

25. Determine the amount of entropy produced as 15.7 kg of argon gas is isothermally compressed into a 1.75 m 3 rigid<br />

container at 25.0°C. Answer: [ 1 (S P ) 2 ] isothermal = 3.27 kJ/K.<br />

Equations (9.42), (9.46), and (9.48) can be combined to give<br />

<br />

½ 1 ðS P Þ 2 Š isothermal<br />

½ 1<br />

ðS P Þ 2<br />

Š adiabatic<br />

<br />

<br />

ideal<br />

gas<br />

= k − 1<br />

ln k<br />

which is greater than 1.0 for k > 1.0. This result is shown in Figure 9.17.<br />

However, the specific entropy production ratio on a per unit mass of charge basis, that is, as 1 (S P ) 2 /m 2 = 1 (s P ) 2 ,is<br />

<br />

½ 1 ðs P Þ 2 Š isothermal<br />

½ 1<br />

ðs P Þ 2<br />

Š adiabatic<br />

<br />

<br />

ideal<br />

gas<br />

= k − 1<br />

k ln k<br />

whichislessthan1.0fork > 1.0, because the two processes do not take on the same total charge of gas (see<br />

Eq. (9.48)).<br />

1.5<br />

( 1 (S P ) 2 ) iso<br />

( 1 (S P ) 2 ) adiab ideal gas<br />

1.4<br />

1.3<br />

1.2<br />

1.1<br />

FIGURE 9.17<br />

Entropy production ratio for an ideal gas.<br />

1.0<br />

1.0 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0<br />

Specific heat ratio, k = c p /c v

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