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

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

3. 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

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

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

6. 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

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

8. 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

9. 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

10. 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

11. 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

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

13. 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

14. 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