study problems ch11 - SONNTAG/BORGNAKKE STUDY PROBLEM 11-1...

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SONNTAG/BORGNAKKE STUDY PROBLEM 11-1 CHAPTER 11 STUDY PROBLEMS POWER AND REFRIGERATION SYSTEMS The steam power plant, Rankine cycle The reheat cycle and feedwater heaters The gas turbine, Brayton cycle, jet engine Reciprocating engine cycles, Otto, Diesel and Stirling cycles Vapor-compression refrigeration cycle Air standard refrigeration cycle and combined cycles FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN
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11.1 A simple Rankine cycle A power plant on the North Pole uses R-22 with a boiler outlet of 3000 kPa, 110 C. The condenser operates at –30 C with the ambient being at –40 C. a) Find the specific energy transfer in all components. b) Find the cycle thermal efficiency. Q W T 3 2 4 1 Condenser Boiler Turbine W P Q B P v 1 2 3 4 T s 1 2 3 4 State out of boiler 3 (P, T) Table B.4.2: h 3 = 306.74 kJ/kg, s 3 = 0.9555 kJ/kg K State 1 (T 1 , x 1 = 0) Tbl. B.4.1: P 1 = 163.5 kPa, h 1 = 10.73 kJ/kg, v 1 = 0.000725 m 3 /kg C.V. Pump. Reversible adiabatic: s 2 = s 1 implemented from Eq.9.18 with constant v. w P1 = v 1 (P 2 – P 1 ) = 0.000725 m 3 /kg (3000 – 163.5) kPa = 2.06 kJ/kg State out of boiler 3 (P, T) Table B.4.2: h 3 = 306.74 kJ/kg, s 3 = 0.9555 kJ/kg K C.V. Turbine reversible, adiabatic: s 4 = s 3 x 4 = (s 4 – s f )/s fg = (0.9555 – 0.0449)/0.9335 = 0.97547 h 4 = h f + x 4 h fg = 10.73 + 0. 97547 × 227.0 = 232.16 kJ/kg w T = h 3 – h 4 = 306.74 – 232.16 = 74.58 kJ/kg C.V. Steam generator (boiler) q H = h 3 – h 2 = 306.74 – 12.79 = 293.95 kJ/kg C.V. Condenser q L = h 4 – h 1 = 232.16 – 10.73 = 221.43 kJ/kg Net work w net = w T – w P1 = 74.58 – 2.06 = 72.52 kJ/kg (= q net = q H – q L ) Cycle efficiency: η thermal = w net q H = 72.52 293.95 = 0.247 2
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11.2 A Rankine cycle with reheat Consider an ideal steam reheat cycle where steam enters the high-pressure turbine at 3.0 MPa, 400 o C, and then expands to 0.8 MPa. It is then reheated at the 800 kPa and expands to 10 kPa in the low-pressure turbine. To what temperature should it be reheated to have a minimum quality of 90.116% in the turbine? Calculate the cycle thermal efficiency. Solution: C.V. LP Turbine section State 6: 10 kPa, x = 0.90116 => h 6 = 191.81 + 0.90116 × 2392.82 = 2348.12 kJ/kg s 6 = 0.6492 + 0.90116 × 7.501 = 7.4088 kJ/kg K State 5: 800 kPa, s 5 = s 6 => T 5 = 350 o C , h 5 = 3161.68 kJ/kg C.V. HP Turbine section P 3 = 3 MPa, T 3 = 400 o C => h 3 = 3230.82 kJ/kg, s 3 = 6.9211 kJ/kg K s 4 = s 3 => h 4 = 2891.6 kJ/kg; C.V. Pump reversible, adiabatic and assume incompressible flow w P = v 1 (P 2 - P 1 ) = 0.00101(3000 - 10) = 3.02 kJ/kg, h 2 = h 1 + w P = 191.81 + 3.02 = 194.83 kJ/kg Q W T 3 2 4 1 Condenser Boiler Turbine W P Q H 5 6 L cb s 3 MPa 10 kPa 1 4 6 2 T 3 5 w T,tot = h 3 - h 4 + h 5 - h 6 = 3230.82 - 2891.6 + 3161.68 – 2348.12 = 1152.78 kJ/kg q H1 = h 3 - h 2 = 3230.82 - 194.83 = 3036 kJ/kg q H = q H1 + h 5 - h 4 = 3036 + 3161.68 - 2891.6 = 3306.08 kJ/kg η CYCLE = (w T,tot - w P )/q H = (1152.78 - 3.02)/3306.08 = 0.348 3
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11.3 An open FWH Rankine cycle
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study problems ch11 - SONNTAG/BORGNAKKE STUDY PROBLEM 11-1...

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