THERMODYNAMICS PRACTICE PROBLEMS 1.

1 THERMODYNAMICS PRACTICE PROBLEMS . 1. A Carnot refrigerator has a coefficient of performance of 10. If the refrigerator's interior is to be kept at 45 C, the temperature of the refrigerator's high temperature reservoir is most nearly (A) 250K. (B) 270K. (C) 300K. (D) 350K. Solution For a refrigerator, Tlow COP =. Thigh Tlow Solve for the hot side temperature. Tlow 45 C + 273. Thigh = + Tlow = + ( 45 C + 273). COP 10. = Answer is (A). 2. Helium is compressed isentropically from 1 atmosphere and 5 C to a pressure of 8. atmospheres. The ratio of specific heats for helium is 5/3. What is the final temperature of the helium?

2 (A) 290 C. (B) 340 C. (C) 370 C. (D) 650 C. Solution 1 k T2 P1 . k = . T1 P2 . Copyright 2001 Professional Publications, Inc. THERMODYNAMICS PRACTICE PROBLEMS - 1. 5. 1 . 1 k 3 = 3 5 = =. k 5 5. 3. 1 k P1 k 1 atm . T2 = T1 = (5D C + 273) . P2 8 atm . = (366 C). Answer is (C). 3. The thermal efficiency of a Carnot cycle operating between 170 C and 620 C is closest to (A) 44%. (B) 50%. (C) 63%. (D) 73%. Solution Thigh = 620 C + 273 = 893K. Tlow = 170 C + 273 = 443K. Thigh Tlow 893K 443K. Carnot = = = ( ). Thigh 893K. Answer is (B). 4. Superheated steam at MPa and 275 C expands isentropically to MPa.

3 What is the quality factor of the resulting vapor? The data for the steam are as follows. For MPa, 275 C: h = kJ kg ; s = kJ kg K. For MPa, dry saturated vapor: hg = kJ kg ; sg = kg K. For MPa, saturated liquid: h f = kg ; s f = kJ kg K. (A) 91%. (B) 92%. (C) 93%. (D) 94%. Copyright 2001 Professional Publications, Inc. THERMODYNAMICS PRACTICE PROBLEMS - 2. Solution The entropy is unchanged in an isentropic process. s = xsg + (1 x ) s f kJ kJ kJ . = x + (1 x ) . kg K kg K kg K . = x = Answer is (D). 5. A compressor takes atmospheric air (molecular weight of 29 kg/kmol) at kPa and 20 C.

4 And delivers it at MPa and 175 C. The compression process is polytropic. The work required to compress one unit mass of air is most nearly (A) 50 kJ/kg (B) 100 kJ/kg (C) 150 kJ/kg (D) 200 kJ/kg Solution For a process with polytropic exponent n, n 1. T1 P1 n = . T2 P2 . n 1. 20 C + 273 kPa n = . 175 C + 273 1034 kPa . n 1. = ( ) n Take the base-10 logarithm of both sides. n 1 . log ( ) = log ( ). n . n 1. = n Copyright 2001 Professional Publications, Inc. THERMODYNAMICS PRACTICE PROBLEMS - 3. n = If air's specific gas constant is not known, it can be calculated. kJ. R kmol K = kJ kg K.

5 R= =. molecular weight kg 29. kmol P2 v2 Pv R (T2 T1 ). w= 1 1. =. 1 n 1 n kJ . kg K (175 C 20 C ). = = kJ kg 1 Answer is (D). 6. When kg of an ideal gas ( specific heat at constant volume = kJ kg K ) is heated at constant volume to a final temperature of 425 C, the total entropy increase is kJ/K. The initial temperature of the gas is most nearly (A) 200 C. (B) 210 C. (C) 220 C. (D) 240 C. Solution The total entropy increase for an ideal gas is T v . S = m cv ln 2 + R ln 2 . T1 v1 . v2 = v1. kJ kJ 425 C + 273 . = ( kg) ln . K kg K T1 . 698 K . ln = T1 . Take the antilogarithm of both sides and solve for T1.

6 Copyright 2001 Professional Publications, Inc. THERMODYNAMICS PRACTICE PROBLEMS - 4. T1 = 489K (489K 273 = 216 C). Answer is (C). 7. Steam enters a turbine with a velocity of 40 m/s and an enthalpy of kJ/kg. At the outlet, 2 meters lower than the inlet, the velocity is 162 m/s, and the enthalpy is kJ/kg. A heat loss of 1 kJ/kg is experienced from the turbine casing. The work output per unit mass is closest to (A) 650 kJ/kg (B) 700 kJ/kg (C) 720 kJ/kg (D) 750 kJ/kg Solution The steady flow energy equation is V2 V2 . W out = m hin + in + Z in g hexit + exit + Z exit g + Q in 2 2.

7 M . 2. m . 40 (2 m) . W out kJ s s2 . = + + . m kg J J . (2) 1000 1000 . kJ kJ . m . 2.. 162 . kJ s kJ. + + 0 1. kg J kg (2) 1000 . kJ .. = kJ kg Answer is (D). Copyright 2001 Professional Publications, Inc. THERMODYNAMICS PRACTICE PROBLEMS - 5. 8. Compressed carbon dioxide (molecular weight = 44) is kept in a full m3 tank at 100 C and 500 kPa. The mass of the carbon dioxide in the tank is most nearly (A) kg (B) kg (C) kg (D) kg Solution The specific gas constant is kJ. R kmol K = kJ kg K. R= =. molecular weight kg 44. kmol Use the ideal gas law. PV = mRT. pV (500 kPa)( m3 ). m= =.

8 RT kJ . kg K (100 C + 273).. Answer is (C). 9. A Carnot refrigeration cycle is used to keep a freezer at 5 C. heat is rejected at 20 C. If the heat removal rate is 30 kW, the COP of the refrigeration cycle is most nearly (A) 9. (B) 10. (C) 11. (D) 12. Solution Thigh = 20 C + 273 = 293K. Tlow = 5 C + 273 = 268K. Copyright 2001 Professional Publications, Inc. THERMODYNAMICS PRACTICE PROBLEMS - 6. Tlow 268K. COP = = = Thigh Tlow 293K 268K. Answer is (C). 10. kg of air is heated in a constant volume process from 20 C to 100 C. The specific heat at constant volume is kJ kg K . The change in entropy for the heating process is most nearly (A) kJ/K.

9 (B) kJ/K. (C) kJ/K. (D) kJ/K. Solution T v . S = m cv ln 2 + R ln 2 . T1 v1 . v2 = v1. kJ 100 C + 273 . S = kg ln . kg K 20 C + 273 . = Answer is (C). Copyright 2001 Professional Publications, Inc. THERMODYNAMICS PRACTICE PROBLEMS - 7.