Find: a) Mass of the air, in kg, b) final temperature of the air, in K, and c) amount of entropy produced, in kj/k.

PROBLEM 6.25 Three m 3 of air in a rigid, insulated container fitted with a paddle wheel is initially at 295 K, 200 kpa. The air receives 1546 kj of work from the paddle wheel. Assuming the ideal gas model, determine for the air (a) the mass, in kg, (b) final temperature, in K, and (c) the amount of entropy produced, in KJ/K. Schematic and Given Data: Air V = 3 m 3 T 1 = 295 K p 1 = 200 kpa W 12 shaft = -1546 kj Given: Three m 3 of air is in a closed system with a paddle wheel. See above figure for given values. Find: a) Mass of the air, in kg, b) final temperature of the air, in K, and c) amount of entropy produced, in kj/k. Engineering Model: 1. Air is a closed system. 2. For the air, Q = 0, and there are no overall changes in kinetic or potential energy. 3. The air is modeled as an ideal gas, taking into account variations in specific heat. Analysis: a) From Table A-22 at T 1 = 295 K: u 1 = 210.49 kj/kg Using the ideal gas equation of state: m = p 1V = (200 103 N m 2 ) (3m 3 ) RT 1 ( 8314 N m = ) (295 K) 28.97 Kg K 7.087 kg b) An energy balance reduces to ΔU 12 = W 12. Therefore, W 12 = -m (u 2 u 1 ). Rearranging, u 2 = -W 12 /m + u 1 = ( 1546 ) + (210.49) = 428.64 kj/kg 7.087 Find corresponding T 2 from Table A-22 with interpolation, T 2 = 311 K c) For an adiabatic process involving an ideal gas taking into account variations in specific volume, Eqs. 6.20a and 6.24 simplify to

σ = m (s 2 s 1 R M ln (p 2 p 1 )) Using ideal gas model for the process with fixed m and v, restate as σ = m (s 2 s 1 R M ln (T 2 T 1 )) From Table A-22 at T 1, s 1 = 1.68515 kj/kg K Using Table A-22 with interpolation at T 2, s 2 = 1.73812 kj/kg K σ = (7.087 kg)[(1.73812 kj/kg K 1.68515kJ/kg K) ( 8.314 ) kj/kg K(ln (311 28.97 295 ))] = 0.268 kj/k

PROBLEM 6.51

PROBLEM 6.51 (CONTINUED)

PROBLEM 6.80

PROBLEM 6.60 Air at 400 kpa, 980 K enters a turbine operating at steady state and exits at 100 kpa, 670 K. Heat transfer from the turbine occurs at an average outer surface temperature of 315 K at the rate of 30 kj per kg of air flowing. Kinetic and potential energy effects are negligible. Assuming the air is modeled as an ideal gas with variations in specific heat, determine (a) the rate power is developed, in kj per kg of air flowing, and (b) the rate of entropy production within the turbine, in kj/k per kg of air flowing. Schematic and Given Data: 2 T 2 = 670 K p 2 = 100 kpa Turbine T b = 315 K T 1 = 980 K p 1 = 400 kpa 1 Given: Air enters and exits a turbine operating at steady state with heat transfer. See above figure for values. Find: (a) Rate power is developed, in kj/kg of air flowing through the turbine (b) Rate of entropy production within the turbine in kj/kg K of air flowing through the turbine Engineering Model: 1. The control volume shown in the sketch is at steady state. 2. Heat transfer occurs at T b = 315K. 3. Kinetic and potential energy effects are negligible. 4. The air is modeled as an ideal taking into account variations in specific heat. Analysis: (a) An energy rate balance reads, 0 = Q cv W cv + (h 1 h 2 ) Simplified, W cv = Q cv + (h 1 h 2 ) (1) For states 1 and 2 at 980K and 670K, respectively, and using Table A-22:

h 1 = 1023.25 kj/kg, s 1 = 2.94468 kj/kg K h 2 = 681.14 kj/kg, s 2 = 2.52589 kj/kg K Substituting these values into equation (1), W cv = 30 kj/kg + (1023.25 681.14)kJ/kg = 312.11 kj/kg (b) An entropy rate balance reads, 0 = Q cv T b + (s 1 s 2 ) + σ cv Considering variations in specific heat, Eq. 6.20a states, s 2 s 1 = s 2 s 1 R M ln (p 2 p 1 ) Simplified, σ cv σ cv σ cv = ( 30 kj/kg 315K = 0.074 kj/kg K = Q cv/ T b + (s 2 s 1 R M ln (p 2 p 1 )) kj ) + (2.52589 2.94468 kj kg K kg K (8.314 28.97 ) ln (100 kpa 400 kpa ))

PROBLEM 6.100 Refrigerant 22 in a refrigeration system enters one side of a counter-flow heat exchanger at 12 bar, 28 o C. The refrigerant exits at 22 bar, 20 o C. A separate stream of R-22 enters the other side of the heat exchanger as saturated vapor at 2 bar and exits as superheated vapor at 2 bar. The mass flow rates of the two streams are equal. Stray heat transfer from the heat exchanger to its surroundings and kinetic and potential energy effects are negligible. Determine the entropy production in the heat exchanger, in kj/k per kg of refrigerant flowing. What gives rise to the entropy production in this application? KNOWN: Two streams of R-22 pass though opposite sides of a counter-flow heat exchanger operating at steady state with equal mass flow rates. Data are known for each stream. FIND: Determine the entropy production for the heat exchanger per unit mass of refrigerant flowing. 12 bar T SCHEMATIC AND GIVEN DATA: 12 bar (1) (2) 12 bar (1) 28 o C 20 o 28 o C. C 20 o C (4) (3) 2 bar 2 bar sup. vapor sat. vapor ENGINEERING MODEL: (1) The control volume is at -25.8 o C steady state. (2). (3) Kinetic and potential energy effects are negligible. ANALYSIS: To fix state 4, we write mass and energy rate balances. The mass balances reduce at steady state to and. Further, (2). (3) 2 bar.. (4) s T 4 0 = + [(h 1 h 2 ) + (h 3 h 4 )] h 4 = h 1 h 2 + h 3 From Table A-7: h 1 h f (28 o C) = 79.05 kj/kg and s 1 s f (28 o C) = 0.2936 kj/kg K h 2 h f (20 o C) = 69.09 kj/kg and s 2 s f (20 o C) = 0.2607 kj/kg K From Table A-8: h 3 = h g (2 bar) = 239.88 kj/kg and s 3 = s g (2 bar) = 0.9691 kj/kg K h 4 = h 1 h 2 + h 3 = 79.05 69.09 + 239.88 = 249.84 kj/kg Interpolating in Table A-9: T 4-9.815 o C and s 4 1.0081 kj/kg K The entropy rate balance reduces as follows: 0 = + [(s 1 s 2 ) + (s 3 s 4 )] + Thus = (0.2607 0.2936) + (1.0081 0.9691) = 0.0061 kj/ kg K The entropy production is due to irreversible heat transfer between the two streams. There would be a small effect of frictional pressure drop, but pressure drops have been ignored.