- Know basic of refrigeration - Able to analyze the efficiency of refrigeration system -

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Transcription:

Refrigeration cycle

Objectives - Know basic of refrigeration - Able to analyze the efficiency of refrigeration system - contents Ideal Vapor-Compression Refrigeration Cycle Actual Vapor-Compression Refrigeration Cycle Cascade refrigeration systems Multistage compression refrigeration systems

Refrigeration cycle Refrigeration is the transfer of heat from a lower temperature region to a higher temperature region Refrigeration cycle is the vapor-compression refrigeration cycle, where the refrigerant is vaporized and condenses alternately and is compressed in the vapor phase.

Refrigerator and Heat Pump Cyclic refrigeration device operating between two constant temperature reservoirs. In the Carnot cycle heat transfers take place at constant temperature. If our interest is the cooling load, the cycle is called the Carnot refrigerator. If our interest is the heat load, the cycle is called the Carnot heat pump. (Reversed Carnot cycle)

Refrigerator & Heat pump Coefficient of performance, COP COP R COP Desired output COP= Require input Refrigerator: is used to maintain the refrigerated space at a low temperature by removing heat from it Cooling effect QL = = Work input W HP net, in Heat pump: heat transfers from a low-temperature medium to a high temperature medium Heating effect QH = = Work input W COP = COP + 1 HP R net, in

Carnot refrigerator or a Carnot heat pump The reversed Carnot cycle is the most efficient refrigeration cycle operating between two specified temperature levels. A refrigerator or heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator or a Carnot heat pump Desired output COP= Require input T H T L COP COP R, Carnot HP, Carnot L ( 2 1) ( )( 2 1) H ( 2 1) ( )( ) T s s TL = = T T s s T T H L H L T s s TH = = T T s s T T H L 2 1 H L

The reversed Carnot cycle is not a suitable model for refrigeration cycle! Process 2 3 involves the compression of a liquid-vapor mixture, which requires a compressor that will handle two phase. Process 4 1 involves the expansion of high-moisture-content refrigerant in a turbine. (Reversed Carnot cycle)

Ideal Vapor-Compression Refrigeration Cycle T P s h Process Description 1-2 Isentropic compression 2-3 Constant pressure heat rejection in the condenser 3-4 Throttling in an expansion valve 4-1 Constant pressure heat addition in the evaporator

Expansion valve Condenser Q H Q L Evaporator Compressor

Energy analysis From 1 st and 2 nd Law analysis for steady flow P h COP COP R HP Q L = = W net, in Q H = = W net, in h h h h h h 1 4 2 1 h h 2 3 2 1

Example Refrigerant-134a is the working fluid in an ideal compression refrigeration cycle. The refrigerant leaves the evaporator at -20 o C and has a condenser pressure of 0.9 MPa. The mass flow rate is 3 kg/min. Find COP R and COP R, Carnot for the same T max and T min, and the tons of refrigeration. Use the Refrigerant-134a Tables State1 kj h1 238.41 Compressor inlet = kg o T1 = 20 C kj s1 = 0.9456 x1 = 1.0 kg K T s

State 2 Compressor exit kj h2s = 278.23 P2s = P2 = 900kPa kg o kj T2 s = 43.79 C s2s = s1 = 0.9456 kg K State3 kj h3 = 101.61 Condenser exit kg P3 = 900kPa kj s3 = 0.3738 x3 = 0.0 kg K State 4 x Throttle exit T T C s h4 = h 3 4 = 1 = 20 o 4 4 = 0.358 kj = 0.4053 kg K T s

COP R Q L m ( h h ) h h = = = W m ( h h) h h 1 4 1 4 net, in 2 1 2 1 T = (238.41 101.61) (278.23 238.41) kj kg kj kg = 3.44 s The tons of refrigeration (often called the cooling load or refrigeration effect) Q = m ( h h ) L 1 4 kg kj 1Ton = 3 (238.41 101.61) min kg kj 211 min = 1.94Ton

COP R, Carnot = T H TL T L ( 20 + 273) K = (43.79 ( 20)) K = 3.97 T s Another measure of the effectiveness of the refrigeration cycle is how much input power to the compressor, in horsepower, is required for each ton of cooling. The unit conversion is 4.715 hp per ton of cooling. W net, in 4.715 = Q COP L R 4.715 hp hp = = 1.37 3.44 Ton Ton

Actual Vapor-Compression Refrigeration Cycle Irreversibilities in various components - Pressure drop due to fluid friction - Heat transfer from or to surroundings

Example Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and -10 O C at a rate of 0.05 kg/s and laves at 0.8 MPa and 50 O C. The refrigerant is cooled in the condenser to 26 O C and 0.72 MPa and is throttled to 0.15 MPa. Disregarding any heat transfer and pressure drops in the connecting lines between the components, determine (a) the rate of heat removal from the refrigerated space and the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the coefficient of performance of the refrigerator.

Cascade refrigeration systems Due to very low temperature, the temperature range in a single vapor compression refrigeration cycle become very large, Resulting low COP COP R Cooling effect QL = = Work input W T net, in Increase COP R by decreasing work input or increasing heat remove s

Cascade refrigeration systems Very low temperatures can be achieved by operating two or more vaporcompression systems in series, called cascading. A( 5 8) = B( 2 m 3) ( h2 h A 3) = m B ( h5 h8) Q m B ( h1 h L 4) COPRcascade, = = W m ( h h ) + m ( h h ) m h h m h h net, in A 6 5 B 2 1

Air conditioning system Evaporator Expansion valve Pump Condenser Pump Cooling tower

Example Consider a two-stage cascade refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. Each stage operates on an ideal vapor compression refrigeration cycle with refrigerant-134a as the working fluid. Heat rejection from the lower cycle to the upper cycle takes place in an adiabatic counterflow heat exchanger where both streams enter at about 0.32 MPa. (In practice, the working fluid of the lower cycle is at a higher pressure and temperature in the heat exchanger for effective heat transfer). If the mass flow rate of the refrigeration through the upper cycle is 0.05 kg/s, determine (a) (b) (c) the mass flow rate of the refrigeration through the lower cycle, the rate of heat removal from the refrigerated space and the power input to the compressor, and the coefficient of performance of this cascade refrigerator.

Multistage compression refrigeration systems When the fluid used throughout the cascade refrigeration system is the same, the heat exchanger between the stages can be replace by a mixing chamber, called a flash chamber.

Example Consider a two-stage compression refrigeration system operating between the pressure limits of 0.8 and 0.14 MPa. The working fluid is refrigerant-134a. The refrigerant leaves the condenser as a saturated liquid and is throttled to a flash chamber operating at 0.32 MPa. Part of the refrigerant evaporates during this flashing process, and this vapor is mixed with the refrigerant leaving the low pressure compressor. The mixture is then compressed to the condenser pressure by the high pressure compressor. 0.8 MPa 0.8 MPa 0.32 MPa 0.32 MPa 0.14 MPa 0.14 MPa

The liquid in the flash chamber is throttled to the evaporator pressure and cools the refrigerated space as it vaporizes in the evaporator. Assuming the refrigerant leaves the evaporator as a saturated vapor and both compressors are isentropic, determine (a) (b) (c) the fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amount of heat removed from the refrigerated space and the compressor work per unit mass of refrigerant flowing through the condenser, and the coefficient of performance.