1

HVAC317 - Refrigeration

Refrigeration Theory

2

Terms

• Heat: A form of energy.• Refrigeration: The process of removing

heat from a space.• British Thermal Unit (Btu): The

conventional unit of measurement of heat.

3

Terms

−A Btu is defined as the amount of heat required to raise or lower the temperature of one pound of water (or any substance) one degree Fahrenheit.

– Substances other than water require different amounts of Btu.

4

Terms

• Thermodynamics: The laws of thermodynamics state that heat will always travel from warmer to cooler areas.

• Pressure: Standard atmospheric pressure is 14.7 psia (pounds per square inch absolute pressure).

5

Terms

• Refrigeration gauges are calibrated to read zero psig (pounds per square inch gauge pressure) at standard atmospheric pressure.

• Vacuum: Pressures below atmospheric pressure.

6

Terms

• Compound Gauge: A gauge that can read above and below atmospheric pressure (this is always the low side gauge on your manifolds).– Above atmospheric pressure these

gauges will read psig.– Below atmospheric pressure these gauges

will read inches of mercury (Hg).

7

Terms

• Sensible Heat: A change in temperature measured with an ordinary thermometer.

• Latent Heat: Heat that causes a change in state.

• Change in State: Conversion of a substance from one form to another. This is NOT a temperature change, it is a molecular structure change.– For example:

• Water to Ice, or Ice to Water• Water to Steam, or Steam to Water

8

Terms

• Specific Heat: The amount of Btu it takes to raise the temperature of one pound (1 lb) of any substance one degree (1)

• Ambient Temperature: The temperature of air that is around an object or device.

9

Terms

• Enthalpy: The total amount of heat that is around a substance.

• Ton: A refrigeration rating that is calculated when Btu are known.– 1 ton = 12,000 Btu per hour– Compressor tonnage and horsepower are

proportionally equal.– 1 ton = 1 hp = 12,000 Btu/h

10

Terms

• Temperature Pressure Chart (TP Chart): A cross reference between the temperatures, pressures and the boiling points of refrigerant.

11

Terms

• Methods of heat transfer– Convection: Heat transfer by fluid or air

movement. (Air Conditioning / Forced air systems)

– Conduction: Particle to Particle transfer of heat. (Radiant Heat)

– Radiation: Transfer of heat by rays. (sunlight)

12

Classification of Systems

• High Temperature– Start at about 45°F– End at about 60°F– Used in flower shops and candy storage.

13

Classification of Systems

• Medium Temperature– Start at about 30°F – End at about 45°F– Used in products such as dairy goods, beer,

wine.– Many products will not freeze at 32°F.

14

Classification of Systems

• Low Temperature Systems– Start at 32°F, and decrease from there– Used in anything frozen such as ice, ice cream,

frozen foods, meat storage.– Must have a defrost cycle.

15

Temp / Pressure Relationship

• The temperature at which a substance changes state. It is dependent on pressure.– Lowering the pressure lowers the

saturation or boiling point.– Raising the pressure raises the saturation

or boiling point.

16

Refrigeration Calculations

• To find the total Btu needed to cool a substance from one temperature to another (without a change in state):– Find the net weight of the substance.– Find the degrees of temperature change

(Delta T or T).– Find the specific heat of the substance.– Btu/hr = weight * T * Spec. Heat

17

Example 1

• Cooling 500 lbs of Cranberries from 70F to 35F

T = 70 - 35 = 35Btu/hr = weight * T * Spec. HeatBtu/hr = 500 * 35 * .91Btu/hr = 15,925

18

Example 2

A grocery store takes in 400 lbs of cucumbers and needs to cool them from the 93°F truck to the long term storage temperature of 45°F. Calculate the Btu.

T = .93 - 45 = 48Specific Heat of Cucumbers = .93

LBS T SpecHeat

LatBtu

TotalBtu

A 400 48 .93 ---- 17,856

19

Example 3

Need to chill 50 gallons of water from 75°F degrees to 35°F.

– Specific heat water = 1.0– T = 75 - 35 = 40– 1 gallon water = 8.345 lbs– Weight = 8.345 * 50 = 417.25

LBS T SpecHeat

LatBtu

TotalBtu

A 417.25 40 1.0 - 16,690

20

Example 3, cont’d:

So, it takes 16,690 Btu to cool 50 gallons of water from 75°F to 35°F.

The only substance with a specific heat of 1.0 is water.

21

Refrigeration Calculations

• To find the total Btu needed to cool a substance from one temperature to another with a change of state:– Find the net weight of the substance.– Find the degree of temperature change

(delta T or T).– Find the specific heat of the substance.– Find the latent heat of the substance.

22

Refrigeration Calculations

– Find the temperature at which the substance changes state (32°F for water).

– Calculate the number of Btu prior to the change in state.

– Calculate the numer of Btu after the change in state.

– Calculate the number of Btu for the state change.

– Add them all together for total Btu needed.

23

Example 1

• A warehouse gets 100 pounds of celery. It is delivered at 70°F. It needs to be cooled to -10°F for long term storage. How many Btu are needed?

24

Example 1, cont’d

• Find the latent heat changes:– Celery freezes at 29.7° F

70

29.729.7

-10Latent

Sensible

25

Example 1, cont’d

• Two areas of sensible heat change:– 70 - 29.7 degrees = 40.3°F– 29.7 - -10 degrees = 39.7°F

• Latent Heat = 135 Btu

26

Example 1, cont’d

LBS T SpecHeat

LatBTUs

TotalBTUs

A 100 40.3 .95 3828.5

B 100 135 13500

C 100 39.7 .48 1905.6

TOTAL 19234.1

Conclusion: It takes 19,234.1 Btu to cool celery from 70°F to -10°F.

27

Example 1, cont’d

• Note that some substances have a different sensible heat above and below their freezing points.

28

LBS T SpecHeat

LatBtu

TotalBtu

A

B

C

TOTAL