Refrigeration/AC/Heat Pumps

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

2. Refrigeration – Lecture 1 • Assignment: • Read Chapter 11 in Henderson/Perry • HW due 4/20 Dr. C. L. Jones Biosystems and Ag. Engineering

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

4. Enthalpy and Entropy • 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 / 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

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

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

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

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

9. 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

10. 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

11. 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

12. 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

13. 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

14. Examples: • 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) COPcooling Dr. C. L. Jones Biosystems and Ag. Engineering

15. Examples: • 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) COPcooling Dr. C. L. Jones Biosystems and Ag. Engineering