study problems ch9

# study problems ch9 - CHAPTER 9 STUDY PROBLEMS SECOND-LAW...

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CHAPTER 9 STUDY PROBLEMS SECOND-LAW ANALYSIS FOR A CONTROL VOLUME The entropy equation for a control volume The steady state and the transient process The reversible steady state process Principle of the increase of entropy Efficiency FUNDAMENTALS of Thermodynamics Sixth Edition SONNTAG BORGNAKKE VAN WYLEN 87

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9.1 An ideal steam turbine A steam turbine receives 4 kg/s steam at 1 MPa 300 o C and there are two exit flows, 0.5 kg/s exits at 150 kPa and the rest exits at 15 kPa. Assume the turbine is ideal, adiabatic and that we can neglect kinetic energy everywhere. We want to determine the total power output. Solution: C.V. Turbine, steady operation adiabatic and reversible. Continuity Eq.6.9: m . 1 = m . + m . 3 ; 2 Energy Eq.6.10: m . 1 h 1 = m . 2 h 2 + m . 3 h 3 + W . T Entropy Eq.9.7: m . 1 s 1 + S . gen = m . 2 s 2 + m . 3 s 3 Process: S . gen = 0 W T 1 2 3 If we separately applied the second law to section 1 to 2 and section 2 to 3 we get Section 1 to 2: m . 1 s 1 = m . 2 s 2 + m . 3 s 2 = m . 1 s 2 s 1 = s 2 Section 2 to 3: m . 3 s 2 = m . 3 s 3 s 2 = s 3 s 1 = s 2 = s 3 that is constant s through turbine State 1 from Table B.1.3: h 1 = 3051.15 kJ/kg, s 1 = 7.1228 kJ/kg K, State 2 (P, s) : s 2 < s g so this state is two-phase x 2 = (s 2 – s f )/ s fg = (7.1228 – 1.4335) / 5.7897 = 0.98266, h 2 = 467.08 + x 2 × 2226.46 = 2654.9 kJ/kg State 3 (P, s) : s 3 < s g so this state is two-phase x 3 = (s 3 – s f )/ s fg = (7.1228 – 0.7548) / 7.2536 = 0.87791, h 3 = 225.91 + x 3 × 2373.14 = 2309.3 kJ/kg Now we can substitute into the energy equation W . T = m . 1 h 1 – m . 2 h 2 – m . 3 h 3 = 4 × 3051.15 - 0.5 × 2654.9 – 3.5 × 2309.3 = 2795 kW P v 1 2 3 1000 150 15 T s 1 2 3 300 120 111 54 15 kPa 150 kPa 200 kPa 1000 kPa 88
9.2 A cross-flowing heat exchanger with entropy generation An ordinary radiator receives 0.2 kg/s hot water at 80 o C, 110 kPa from a pipe and the water exits at 70 o C. The radiator heats the air outside of it so the air rises up (natural convection) pulling colder air towards the radiator behind it. This acts as a two fluid heat exchanger so assume the air comes in at 20 o C and leaves the radiator at 30 o C. We want to know the mass flow rate of air and the total entropy generation rate in the process. Solution: The schematic may look like this: 1 water 3 air 2 4 Energy Eq.6.10: m . H2O h 1 + m . AIR h 3 = m . H2O h + m . AIR h 2 4 Entropy Eq.9.7: m . H2O s 1 + m . AIR s 3 + S . gen = m . H2O s 2 + m . AIR s 4 Process: Constant pressure for air From B.1.1: h 1 = 334.88 kJ/kg; s 1 = 1.0752 kJ/kg K h 2 = 292.96 kJ/kg; s 2 = 0.9548 kJ/kg K Using A.5: h 4 - h 3 = C p (T 4 – T 3 ) = 1.004(30 – 20) = 10.04 kJ/kg s 4 - s 3 = C p ln ( T 4 T 3 ) – R ln ( P 4 P 3 ) = 1.004 ln 30 + 273 20 + 273 – 0 = 0.03368 kJ/kg K From energy equation m .

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study problems ch9 - CHAPTER 9 STUDY PROBLEMS SECOND-LAW...

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