Refrigeration/AC/Heat PumpsRefrigeration – Lecture 1Slide 3Enthalpy and EntropyRefrigerationSlide 6C.O.P.Slide 8Slide 9Example 11.1 pg. 332 for R-12System Design: EvaporatorSlide 12System Design: CompressorExamples:Slide 151Dr. C. L. JonesBiosystems and Ag. EngineeringRefrigeration/AC/Heat Pumps2Dr. C. L. JonesBiosystems and Ag. EngineeringRefrigeration – Lecture 1•Assignment: –Read Chapter 11 in Henderson/Perry–HW due 4/203Dr. C. L. JonesBiosystems and Ag. Engineering4Dr. C. L. JonesBiosystems and Ag. EngineeringEnthalpy and Entropy•Entropy, –symbolized by S–a measure of the energy in a system or process that is unavailable to do work. –IOW…how much energy is spread out in a process, or how widely spread out it becomes at a specific temperature –how much energy was spread out in raising T from 0 K to xxxK–kJ / kg K •Enthalpy –symbolized by h–A thermodynamic function of a system,equivalent to the sum of the internal energy of the system plus the product of its volume multiplied by the pressure exerted on it by its surroundings.–IOW…heat content of a system–kJ / kg5Dr. C. L. JonesBiosystems and Ag. EngineeringRefrigeration6Dr. C. L. JonesBiosystems and Ag. EngineeringRefrigeration•Cooling = mr(ha – he)•Energycompressor= mr(hb –ha)7Dr. C. L. JonesBiosystems and Ag. EngineeringC.O.P.•c.o.p.cooling–Factor that designates the useful cooling capacity per unit of energy supplied by the compressor•c.o.p.heating–Heating capacity at the condenser per unit of energy supplied by the compressor•c.o.p.heating = 1 + c.o.p.cooling8Dr. C. L. JonesBiosystems and Ag. Engineering9Dr. C. L. JonesBiosystems and Ag. EngineeringRefrigeration•Definition of refrigeration: process of removing heat from a body having a temperature below the temperature of its surroundings…transferring heat energy from a lower to a higher temperature•Natural refrigeration: produced by using natural ice•Mechanical refrigeration: accomplished by refrigerating engines operated on thermo principles–Aborption system–Vapor compression systems10Dr. C. L. JonesBiosystems and Ag. EngineeringExample 11.1 pg. 332 for R-12•Determine COP for cooling for R12 when operating at an evaporator T of -15C and a condenser T of 30C if there is no superheating in the evaporator and no subcooling in the condenser. (use table 11-4 pg 331)11Dr. C. L. JonesBiosystems and Ag. EngineeringSystem Design: Evaporator•The lower the evaporator temperature, the lower the low-side-pressure required and a compressor with greater vol. capacity is required.•q (kW) = mmc(t1 - t2) = mr(ha- he)12Dr. C. L. JonesBiosystems and Ag. EngineeringSystem Design: Evaporator•Determining refrigerant mass flow rate: Example 11.2 using R12 pg 340 first part•Determine mr for an R12 refrig. System to produce 3.5 kW of cooling. Evap. T = -15C(same conditions as ex. 11.1)•Determine the evap. surface area needed to cool air from 30 to 10 C and mass flow rate of the air being cooled.13Dr. C. L. JonesBiosystems and Ag. EngineeringSystem Design: Compressor•Equation 11.17 pg. 341–Compressor displacement = DNSmrvg = EvDNS•Required compressor power per unit of refrigeration:–Equation 11.18 pg 341P’ = 100(hb – ha)/(Ec * (ha – he))14Dr. C. L. JonesBiosystems and Ag. EngineeringExamples:•Refrigeration system with the following specs:–Requires 15 kw–Uses R134a for refrigerant–Evaporator Temp = -18C–Condenser is watercooled w/ 22C water–Compressor vol. eff = 83%, thermal eff = 89%•Find:–A) high and low side pressures–B) compressor displacement–C) compressor power per unit of refrigeration required–D) Refrigerant rate–E) COPcooling15Dr. C. L. JonesBiosystems and Ag. EngineeringExamples:•Refrigeration system with the following specs:–Requires 15 kw–Uses R134a for refrigerant Ammonia–Evaporator Temp = -18C–Condenser is watercooled w/ 22C water–Compressor vol. eff = 83%, thermal eff = 89%•Find:–A) high and low side pressures–B) compressor displacement–C) compressor power per unit of refrigeration required–D) Refrigerant rate–E)
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