Refrigeration/AC/Heat Pumps Dr. C. L. Jones Biosystems and Ag. Engineering

Dr. C. L. Jones Biosystems and Ag. Engineering

Enthalpy and Entropy Enthalpy Entropy, symbolized by h symbolized by S 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 / kg 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 Dr. C. L. Jones Biosystems and Ag. Engineering

Refrigeration Dr. C. L. Jones Biosystems and Ag. Engineering

Refrigeration Cooling = mr(ha – he) Energycompressor= mr(hb –ha) Dr. C. L. Jones Biosystems and Ag. Engineering

C.O.P. c.o.p.cooling c.o.p.heating c.o.p.heating = 1 + 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.cooling Dr. C. L. Jones Biosystems and Ag. Engineering

Dr. C. L. Jones Biosystems and Ag. Engineering

Refrigeration 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 systems Dr. C. L. Jones Biosystems and Ag. Engineering

Example 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) Dr. C. L. Jones Biosystems and Ag. Engineering

System 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) Dr. C. L. Jones Biosystems and Ag. Engineering

System 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. Dr. C. L. Jones Biosystems and Ag. Engineering

System Design: Compressor Equation 11.17 pg. 341 Compressor displacement = DNS mrvg = EvDNS Required compressor power per unit of refrigeration: Equation 11.18 pg 341 P’ = 100(hb – ha)/(Ec * (ha – he)) Dr. C. L. Jones Biosystems and Ag. Engineering

Examples: Refrigeration system with the following specs: Find: 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) COPcooling Dr. C. L. Jones Biosystems and Ag. Engineering

Examples: Refrigeration system with the following specs: Find: 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) COPcooling Dr. C. L. Jones Biosystems and Ag. Engineering