handout09 - 3.6 Power and Refrigeration Cycles AIM Show...

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3.6 Power and Refrigeration Cycles AIM: Show common ways of assembling some of the components previously discussed. Power Cycle We can extract energy in the form of work from a fluid that has energy in the form of heat. Pump Boiler Turbine Condenser Qb Wt Qc Wp 1 2 3 4 NOTE: Boiler and condenser are heat exchangers. NOTE: There is a pressure loss in the turbine. Need a pump to compensate for this. 3–41
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NOTE: The thermal efficiency , symbol η (eta), is defined as η = work we get out heat we put in (3.103) = - d W p d t + d W t d t d Q b d t . (3.104) NOTE: This formula will depend on how the terms are defined. In our sign convention, W t will be negative and W p positive. NOTE: We will see (second law of thermodynamics) that there is an upper limit to the thermal efficiency. EXAMPLE: (Schaum) Steam leaves the boiler at P 2,3 = 4000kPa and T 3 = 600 C It exits the turbine at P 1,4 = 20kPa as saturated steam ( x = 1 ). It exits the condensor as saturated water ( x = 0 ). QUESTION: What is the thermal efficiency, ignoring pressure loss in boiler and condenser. ANSWER: We need to calculate d W t d t , d W p d t and d Q b d t . We choose a mass of 1kg and mass flux of ˙ m = 1 kg s . The pump: We know the enthalpy of saturated water from the steam tables (Schaum p. 355) H spec 1 = 251.4 kJ kg . The pump increases the pressure from P 1 to P 2 : d W p d t = - ˙ m P 1 - P 2 ρ (3.105) 3–42
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= - 1 kg s " 20kPa - 4000kPa 1000 kg m 3 # = 3.98 kJ s . (3.106) We can now calculate H spec 2 because we know in general that d W p d t = - ˙ m h H spec 1 - H spec 2 i . (3.107) For ˙ m = 1 kg s , we have H spec 2 = d W p d t + H spec 1 (3.108) = 3.98 kJ kg + 251.4 kJ kg (3.109) = 255.38 kJ kg . (3.110) The boiler: From the steam tables, we find H spec 3 = 3674 kJ kg . Then the boiler heating rate d Q b d t = - ˙ m h H spec 2 - H spec 3 i (3.111) = - 1 kg s 255.38 kJ kg - 3674 kJ kg = 3418.6 kJ s . (3.112) 3–43
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fi The turbine : From the steam tables, we find H spec 4 = 2610 kJ kg . d W t d t = - ˙ m h H spec 3 - H spec 4 i (3.113) = - 1 kg s 3674 kJ kg - 2610 kJ kg = - 1064 kJ s . (3.114) Now, finally, we can calculate η = - d W p d t + d W t d t d Q b d t (3.115) = - 3.98 kJ s + - 1064 kJ s 3418.6 kJ s (3.116) = 0.31 . (3.117) 3–44
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Refrigeration Cycle We can transfer heat from one place to another. Evaporator Expansion Valve Condenser Compressor Wcmp Qe Qc NOTE: Two names for the same cycle, depending on the main purpose of the application: cool something (ice cream) down: refrigeration cycle heat something (building) up: heat pump QUESTION: How do we measure efficiency in this case? ANSWER: As the aim here is to transfer heat, not perform work, we do not use η in this case. 3–45
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Instead, introduce coefficient of performance (COP) : For a refrigerator: COP f ridge = heat we extract work we put in (3.118) = d Q e d t d W cmp d t . (3.119) For a heat pump: COP hp = heat we inject work we put in (3.120) = d Q c d t d W cmp d t .
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