HAT IS THE BASIC DIFFERENCE BETWEEN A REFRIGERATION SYSTEM APPLIED IN AN AIR CONDITIONING AND IN THE HOUSEHOLD REFRIGERATOR
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WHAT IS THE BASIC DIFFERENCE BETWEEN A REFRIGERATION SYSTEM APPLIED IN AN AIR CONDITIONING AND IN THE HOUSEHOLD REFRIGERATOR?
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- (iii) What is the difference between a refrigerator and a heat pump?A two-stage refrigeration system operates with ammonia refrigerant flowing at the rate of 15 kg/min through the evaporator. The saturation temperature in the condenser and evaporator units have been noted to be 40⁰C and - 15⁰C respectively. If the system has intercooling by liquid refrigerant at 4.25 bar, determine (a) The capacity and COP of the system. (b) How will these parameters be affected if the compression is carried out in a single stage unit ; the operating temperature limits remaining the same? Use the p-h chart and property tables for saturated ammonia refrigerantA R-134a home refrigerator operates on a simple cycle at pressures 1.8 bar-abs and 7.5 bar-abs. The refrigerant mass flow-rate is 0.005 kg/s. Fill up Tables 3a and 36. Part 3. Calculation: Performance of a Simple Refrigeration Cycle. Table Ja. Selected Thermodynamic properties of R-134a refrigerant according to the given cycle. kJ/ kg a Specific Enthalpy at suction A Specific volume at suction m/ kg kJ /g-K e Specific entropy at suction d Specific enthalpy at discharge Specific enthalpy after expansion hs4 31 kJ/ kg kJ / kg Table 3b. Performance of the home refrigerator according to its cycle of operation. kWatt a. Evaporator cooling capacity b Condenser heat rejection rate Q Cond kWatt Power required by compressor W Comp C. kWatt d Volume displacement of compressor e. Coefficient of Performance-Ref VCamp Li/s COP Ref f Coefficient of Performance-Canot COP RC & Flash-gas formed after expansion X4 kg / kg %3D
- Choose correct answer: Refrigerators and heat pumps operate on the same cycle but differ in their objectives. Select one: True False5. A refrigerator with tetrafluoroethane as refrigerant operates with an evaporation temperature of 247.15 K (-26°C) and a condensation temperature of 300.15 K (27°C). Saturated liquid refrigerant from the condenser flows through an expansion valve into the evaporator, from which it emerges as saturated vapor. (a) For a cooling rate of 5.275 kW, what is the circulation rate of the refrigerant? (b) By how much would the circulation rate be reduced if the throttle valve were replaced by a turbine in which the refrigerant expands isentropically? (c) Suppose the cycle of (a) is modified by the inclusion of a countercurrent heat exchanger between the condenser and the throttle valve in which heat is transferred to vapor returning from the evaporator. If liquid from the condenser enters the exchanger at 300.15 K (27°C) and if vapor from the evaporator enters the exchanger at 247.15 K (-26°C) and leaves at 294.15 K (21°C), what is the circulation rate of the refrigerant?A parameter used to characterize the heat pump cycle performance is called the thermal efficiency. True or False?
- 3.) When a gas surrounded by air is compressed or expands adiabatically, its temperature rises or decreased even though there is no heat input or dissipated to the gas. Where does the energy come from to raise or lower the temperature? 4.) Why must a room air conditioner be placed in a window rather than just set on the floor and plugged in? Why can a refrigerator be set on the floor and plugged in?Modern (latest) innovation in domestic refrigerators and air conditioners.SUBJECT: THERMODYNAMIC COURSE: II ASSI.LACTURE: NATIQ ABBAS Example 2:- Refrigerant-134a enters the compressor of a refrigerator as superheated vapor at 0.14 MPa and -10°C at a rate of 0.05 kg/s and leaves at 0.8 MPa and 50°C. The refrigerant is cooled in the condenser to 26°C and 0.72 MPa and is throttled to 0.15 MPa. Disregarding any heat trans fer 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, and (b) the coefficient of performance of the refrigerator. Solution: - P 0.14 MPa T=-10 C dut 246.36 kJ/kg OF P 0.8 MPa h2 = 286.69 kJ/kg P 0.72 MPa T= 26 C h3 = 87.83 kJ/kg h3 = h = 87.83 KJ/kg h4 = h3 (throttling) h4 87.83 kJ/kg 0.8 MPa 0.72 MPa/ 26 C 0.15 MPa 0.14 MPa -10°C SUBJECT: THERMODYNAMIC COURSE: II ASSI.LACTURE: NATIQ ABBAS SAMARRA RING