ME 200 L38 CARNOT REFRIGERATION CYCLE Read 10.1 Material not picked up this week may be recycled! ME 200 L38 CARNOT REFRIGERATION CYCLE Read 10.1 Kim Sees Office ME Gatewood Wing Room 2172 Examination and Quiz grades are available Blackboard Examinations and Quizzes can be picked up all of this week from Gatewood Room 2172 Material not picked up this week may be recycled! ThermoMentor © Program Spring 2014 MWF AM J. P. Gore Gatewood Wing 3166, Office Hours: MWF TAs: Robert Kapaku Dong Han
Features of Carnot Refrigeraiton Cycle Heat transfers between refrigerant and cold and warm regions are not reversible. Refrigerant temperature in evaporator is less than T C. Refrigerant temperature in condenser is greater than T H. Irreversible heat transfers have negative effect on performance. 2 sCarnot 3 Carnot 1 Carnot 4 Carnot 3 High-COP-Carnot 2 High-COP-Carnot 1 High-COP-Carnot
Carnot Refrigeration Cycle Compressor for a mixture is challenging to engineer Condenser with pressure gain is challenging. Expansion valve replaced by turbine generates only a small amount of work. Condenser input state is identical between the option cycles and may have mixture exit or saturated vapor exit. Hard to achieve refrigeration cycle but still the goal. Two-phase liquid-vapor mixture
Evaporator or Heat Exchanger that Cools The Carnot Refrigeration Cycle Applying mass and energy rate balances Adiabatic Compressor Condenser or Heat Exchanger that Heats Expansion Valve replaced by Turbine
Carnot Refrigeration and Heat Pump Cycle (a) compressor power, in kW, (b) refrigeration capacity, in tons, (c) coefficient of performance Example: Carnot refrigeration cycle using R-134a as the working fluid is shown below. For a refrigerant mass flow rate of 0.08 kg/s, determine State h (kJ/kg) 1 Carnot 2 Carnot 3 Carnot Carnot x1= h1=0.973(241.3)+(0.0267)(36.84)= =(1-x4) x4(0.9253) x4= h4=0.3033(204.46)+36.84=
The Carnot Refrigeration Cycle Wc= =-32.3 kJ/kg Qin= = Qout= = Wt= =6.44 COP_HP=162/( )=6.26 COP_Ref=136.92/( )=5.29 COP_HPT =TH/(TH-TL)=312.39/49.48=6.314 COP_Ref=TL/(TH-TL)=262.91/49.48=5.31 COP calculated based on temperature ratios and those calculated based On heat transfer and work done are identical within property tables accuracies.