BASICS OF REFRIGERATION & AIR CONDITIONING

TEMPERATURE

Temperature is the level of heat or the measure of the intensity of heat of a

substance. Temperature is directly proportional to heat. As heat is added to

a substance, its temperature increases. As heat is removed, its temperature

decreases.

The English system uses the Fahrenheit to measure temperature. The metric

or the SI unit of temperature is the Celsius. A thermometer is an instrument

that is used to measure temperature.

At standard atmospheric conditions, the pressure is 29.92 in. Hg or 14.7 psia

at sea level. At these given conditions:

a. water boils at 212 °F

b. water freezes and becomes ice at 32 °F

Absolute zero is the temperature at which there is zero heat and molecules have

zero movement.

The absolute zero temperature in the Rankine scale is 0 °R. This is equal

to -460 °F.

°R = °F + 460

The absolute zero temperature in the Kelvin scale is 0 °K. This is equal

to -273 °C.

°K = °C + 273

To convert from °C to °F:

°F = (1.8 x °C) + 32

Example: Convert 100 °C, the boiling point of water to °F

°F = (1.8 x 100) + 32

°F = (180) + 32

°F = 212 °F

To convert from °F to °C:

Example: Convert 32 °F, the freezing point of water to °C

°C = (°F - 32)/1.8

°C = (32 - 32)/1.8

°C = 0 °C

TEMPERATURE AND PRESSURE RELATIONSHIP

Charle's Law:

If the pressure of a gas is constant, its absolute temperature is directly

proportional to its volume,

(V1/T1) = (V2/T2)

Example:

Find the absolute temperature at which 50 liters of a gas at 21 C

must be raised at in order to have a final volume of 100 liters.

Solution:

V1 = 50 liters

T1 = 21 + 273 = 294 K

V2 = 100 liters

(V1/T1) = (V2/T2)

50/294 = 100/T2

T2 = 294 x 100/50

T2 = 588 K

TEMPERATURE AND VOLUME RELATIONSHIP

Charle's Law:

If the volume of a gas is constant, its absolute temperature is directly

proportional to its pressure,

(P1/T1) = (P2/T2)

Example:

A certain gas has an initial pressure of 500 Kpaa at 21 C. Determine its final

pressure when the temperature is increased to 35 C.

Solution:

P1 = 500 Kpaa

T1 = 21 + 273 = 294 K

T2 = 35 + 273 = 308 K

(P1/T1) = (P2/T2)

500/294 = P2/308

P2 = 500 x 308/294

P2 = 524 Kpaa

PRESSURE

Pressure is the force acting on a unit area. Pressure is inversely proportional

to the area of contact. The smaller the area of contact, the greater the

pressure exerted. Pressure is directly proportional to the force.

Pressure flows from a higher pressure to a lower pressure until it equalizes.

P = F/A

Example:

Find the pressure on one sole of the foot of a 180 lb person. What is the

pressure if the same person stands on both feet?

Solution:

a. The area of one sole of a foot of an average person is 30 sq in.

P = F/A

P = 180/30

P = 6 psi

b. The area of both soles of the feet of an average person is 60 sq. in.

P = 180/60

P = 3 psi

Thus, the pressure exerted is twice greater standing on one foot than the

pressure when standing on both feet.

PRESSURE AND DEPTH

Pressure is directly proportional to the depth and the relative density

of the liquid.

P = 9.8 x d x RD

where:

P = pressure in Kpa

d = depth of liquid in meters

RD = relative density of the liquid

Relative density is also known as Specific Gravity. It is the ratio of the density of

a substance to the density of a reference substance (water). It is also equal to

the ratio of the mass of a substance to the mass of an equal volume of water.

SG = Density of substance/Density of water

SG = mass of substance/mass of equal volume of water

The specific gravity of an object which will float on water is equal to the

fraction of the submerged volume. It is also equal to the ratio of the

submerged height to the object's total height.

If the specific gravity of a substance is lesser than 1, it will float on water. If

it is greater than 1, it will be submerged in water.

Example:

The relative density of a certain oil is 0.82. Find the pressure at the bottom

of the oil storage tank if the oil level is 2 meters.

Solution:

d = 2 m

RD = 0.82

P = 9.8 x d x RD

P = 9.8 x 2 x 0.82

P = 16 Kpa

Example:

An empty container has a mass of 50 g. When filled with water, the total

mass is 200 g. When filled with a liquid with an equal volume as that of

the water, the total mass is 300 g. Calculate the specific gravity of the

liquid.

Solution:

mass of liquid = 300 - 50 = 250 g

mass of water = 200 - 50 = 150 g

SG = mass of substance/mass of equal volume of water

SG = 250/150

SG = 1.67

Example:

A block of wood is 10 inches in length. It was dropped in a container filled

with water. The submerged height was 8 inches. Find the relative density of

the given wood.

Solution:

submerged height = 8 in

total height = 10 in

RD = submerged height/total height

RD = 8/10

RD = 0.8

PRESSURE AND VOLUME RELATIONSHIP

Boyle’s law:

If the temperature is constant, the absolute pressure of a gas is inversely

proportional to its volume.

P1 x V1 = P2 x V2

Example:

The piston compressed a gas from 12 liters at 100 Kpaa to a final volume of 1 liter.

Determine the final pressure.

Solution:

P1 = 100 Kpaa

V1 = 12 liters

V2 = 1 liter

P1 x V1 = P2 x V2

P2 = 100 x 12/1

P2 = 1200 Kpa

PARTIAL PRESSURES

Dalton’s law:

The total pressure of a confined mixture of gases is the sum of the individual

pressures of the gases in the mixture.

Pm x Vm = (P1 x V1) + (P2 x V2) + (P3 x V3) + ...

Example:

Tank A contains Nitrogen at a pressure 2400 Kpaa and a volume of 100 liters.

Tank B contains gas at a pressure of 1000 Kpaa and a volume of 200 liters. If

the tanks are connected and the valves of both tanks are opened to allow

the gases to mix, calculate the total pressure of the gas mixture.

Solution:

P1 = 2400 Kpaa

V1 = 100 L

P2 = 1000 Kpaa

V2 = 200 L

Vm = 100 + 200 = 300 L

Pm x Vm = (P1 x V1) + (P2 x V2)

Pm x 300 = (2400 x 100) + (1000 x 200)

Pm = 1467 Kpaa

IDEAL GAS LAW (PERFECT GAS LAW)

P1 x V1/T1 = P2 x V2/T2

Example:

An air compressor compresses air from 20 liters at 100 Kpaa and 27 C to 2 liters

and 1200 Kpaa. Find the final temperature of air.

Solution:

P1 = 100 Kpaa

V1 = 20 liters

T1 = 27 + 273 = 300 K

P2 = 1200 Kpaa

V2 = 2 liters

P1 x V1/T1 = P2 x V2/T2

100 x 20/300 = 1200 x 2/T2

T2 = 360 K

4 LAWS OF THERMODYNAMICS

1. Zeroth law

If two systems are in thermal equilibrium with a third system, all three systems

are in thermal equilibrium.

2. First law

Energy cannot be created nor destroyed but can be transformed from one form

into another. Perpetual motion machines are impossible.

3. Second law

The entropy (disorder) increases and the energy decreases whenever energy is

transformed from one form to another.

4. Third law

At absolute zero temperature, all atoms of a substance will stop moving.

The entropy of a system is zero at absolute zero temperature.

3 LAWS OF NEWTON

1. Inertia

A body will stay at rest or in motion unless acted upon by external force.

2. Acceleration

Acceleration is directly proportional to the force and inversely proportional

to the mass. Force is the product of an object's mass and acceleration.

a = F/m

F = m x a

3. For every action, there is an equal and opposite reaction.

DENSITY

Density is the mass per unit volume of a substance.

D = m/V

m = D x V

density of water (4 C/39 F) = 62 lb/cu ft (approx 8 lb/gal)

density of air (sea level, 15 C/59 F) = 1.22 kg/m^3

density of air (sea level, 15 C/59 F) = 0.076 lb/cu ft (approx 1 oz/cu ft)

1 ounce (oz) = 28 grams

1 cubic foot (cu ft) = 7.5 gallons (US)

1 cubic meter = 35 cu ft (264 US gal)

Example:

Calculate the mass of 1 gallon of water.

Solution:

V = 1 gallon

m = D x V

m = 8 x 1

m = 8 lb

SPECIFIC VOLUME

Specific volume is the volume of a substance per unit mass.

Specific volume is the inverse (reciprocal) of density.

SV = V/m

SV = 1/D

The specific volume of dry air at standard atmospheric conditions is 13.33 cu ft/lb.

WEIGHT

Weight is the force caused by Earth's gravity.

Weight is the number (value) you see on a weighing scale.

Weight is a form of Force.

W = m x g

F = m x g

W = F

where:

W = weight of object, Newtons

m = mass of object, kilograms

g = acceleration due to gravity, equal to 9.8 m/s^2

F = force, Newtons

Example:

What is the weight of an object having a mass of 8 kg?

Solution:

m = 8 kg

W = m x g

W = 8 x 9.8

W = 78 N

WORK

Work is the product of the force and the distance travelled.

Work is also the product of Pressure and Volume (flow work).

Work = F x d

Work = P x V

Example:

A sand bag with a mass of 20 kg is raised to a height of 1 meter. Find the

work required. How much is the work done by the sand bag if it was dropped

1 meter?

Solution:

Force = mass x acceleration due to gravity

F = m x g

F = 20 x 9.8

F = 196 N

d = 1 m

Work = F x d

Work = 196 x 1

Work = 196 Nm (Joules)

POWER

Power is the rate of doing work. Power is work per unit of time.

Power is also the product of Force and Velocity.

Power = Work/time

Power = F x d/t

Power = Force x velocity

1 Horsepower (HP) = 746 watts

1 Kilowatt (KW) = 1.34 HP

Example:

How much power is required by a pump to raise 1 drum of water (approx.

200 liters, 55 US gallons) to a height of 10 meters in 2 minutes?

Solution:

m = 200 kg (the mass of 1 Liter of water is 1 Kilogram)

d = 10 m

t = 2 x 60 = 120 sec

F = m x g

F = 200 x 9.8

F = 1960 N

Work = F x d

Work = 1960 x 10

Work = 19,600 J

Power = Work/time

Power = 19,600/120

Power = 163 J/s (watts)

Horsepower = 163/746

Horsepower = 0.22 HP

EFFICIENCY

Efficiency is the ratio of Power output to Power input. It is expressed in %.

Efficiency = Power output/Power input

% Efficiency = (Power output/Power input) x 100%

Example:

If the pump in the previous example is 90% efficient, calculate the required

power input to the pump.

Solution:

Power input = Power output/Efficiency

Power input = 0.22/0.90

Power input = 0.24 HP (approx. 1/4 HP)

ENERGY

Energy is the ability to do work.

Potential Energy

Potential Energy (PE) is the energy of a body due to its position (height).

PE is equal to the work done in lifting that given body to that height.

PE = F x d

PE = Work

Example:

Calculate the potential energy of 4 liters (approx 1 gallon) of water at a

height of 3 meters (approx 10 feet).

Solution:

F = m x g

F = 4 x 9.8 = 39 N

d = 3 m

PE = F x d

PE = 39 x 3

PE = 117 J

Note:

The power required by a 100 watt light bulb is approximately equal to the

work required to lift 1 gallon of water to a height of 10 feet in 1 second.

Kinetic Energy

Kinetic Energy (KE) is the energy of a body due to its motion.

Kinetic Energy is equal to 1/2 times the product of mass and the square of

velocity.

KE = 0.5 x m x v^2

Example:

A car travelling at 90 km/hr (55 mph) have a speed of 25 meters per second.

If the car has a mass of 2000 kg, determine its kinetic energy at this speed.

Solution:

m = 2000 kg

v = 25 m/s

KE = 0.5 x m x v^2

KE = 0.5 x 2000 x 25 x 25

KE = 625,000 J

KE = 625 KJ

HEAT

Heat is a form of energy.

Heat naturally flows from a hotter substance to a colder substance.

Heat is directly proportional to mass, specific heat and temperature difference.

Q = m x C x TD

where:

Q = heat

m = mass

C = specific heat

TD = temperature difference

1 Btu = 1.05 KJ

1 KJ = 0.95 Btu

Specific Heat

Specific heat is the amount of heat required to raise 1 pound of a substance

1 degree Fahrenheit.

specific heat of water = 1 Btu/lb-F (4.2 KJ/kg-C)

specific heat of ice = 0.5 Btu/lb-F (2.1 KJ/kg-C)

specific heat of steam = 0.5 Btu/lb-F (2.1 KJ/kg-C)

specific heat of air = 0.24 Btu/lb-F (1 KJ/kg-C)

Example:

A hot water tank is to be heated to a temperature of 140 °F (60 C). If the

temperature of city cold water is 45 °F (7 C), find the heat required to heat

a 50 gallon residential hot water tank.

Solution:

m = D x V

m = 8 lb/gal x 50 gal = 400 lb

T1 = 45 F

T2 = 140 F

Q = m x C x (T2 - T1)

Q = 400 x 1 x (140 - 45)

Q = 38,000 Btu

Note:

Residential water heaters (boilers) range in size from 50,000 - 150,000 Btu/hr.

Sensible heat

Sensible heat is a form of heat with a change in temperature without a change

of state. It can be measured by a thermometer.

Latent heat

Latent is a form of heat with a change of state without change in temperature.

This cannot be measured by a thermometer.

latent heat of vaporization of water = 970 Btu/lb (2260 KJ/kg)

latent heat of fusion of ice = 144 Btu/lb (334 KJ/kg)

Example:

The temperature of a domestic refrigerator freezer compartment is 0 °F to 5 °F

(-18 C to -15 C). Calculate the total heat required to melt 1 lb of ice at 0 F and

turn it to steam at 222 F.

Solution:

m = 1 lb

T ice = 0 F

T ice melting point = 32 F

T water boiling point = 212 F

T steam = 222 F

a. Sensible heat of ice

specific heat of ice = 0.5 Btu/lb-F

Q = m x C x (T ice melting point - T ice)

Q = 1 x 0.5 x (32 - 0)

Q = 16 Btu

b. Latent heat of ice

latent heat of fusion (LHF) of ice = 144 Btu/lb

Q = m x LHF

Q = 1 x 144

Q = 144 Btu

c. Sensible heat of water (melted ice to boiling water)

specific heat of water = 1 Btu/lb-F

Q = m x C x (T water boiling point - T ice melting point)

Q = 1 x 1 x (212 - 32)

Q = 180 Btu

d. Latent heat of vapor

latent heat of vaporization (LHV) of water = 970 Btu/lb

Q = m x LHV

Q = 1 x 970

Q = 970 Btu

e. Sensible heat of steam (superheated vapor at 222 F)

specific heat of steam = 0.5 Btu/lb-F

Q = m x C x (T steam - T water boiling point)

Q = 1 x 0.5 x (222 - 212)

Q = 5 Btu

f. Total heat

Qtotal = 16 + 144 + 180 + 970 + 5

Qtotal = 1315 Btu

REFRIGERATION

Refrigeration is the process of removing heat from a substance and transferring

that heat to another substance.

Refrigeration Capacity (Rating)

One ton of refrigeration is the amount of heat required to melt 1 ton (2000 lb)

of ice in 24 hours. Refrigeration equipment are rated in tons (tonnage) or KW.

1 Ton of Refrigeration (TOR) = 288,000 Btu/24 hr

1 Ton of Refrigeration (TOR) = 12,000 Btu/hr

1 Ton of Refrigeration (TOR) = 200 Btu/min

1 Ton of Refrigeration (TOR) = 3.5 KW

1 Ton of Refrigeration (TOR) = 4.7 HP

1 KW = 3412 Btu/hr

1 KW-hr = 3412 Btu

1 watt = 3.4 Btu/hr

1 HP = 2544 Btu/hr

Example:

A house has a 75,000 Btu/hr furnace for winter heating. Calculate its power

in KW. The furnace runs an average of 10 hours per day. If the cost of energy

is 7 cents/KW-hr, find the daily and monthly power consumption and energy

cost of heating the house during the winter season.

Solution:

Power = 75,000 Btu/hr

Energy cost = 7 cents/KW-hr = $0.07/KW-hr

Furnace run time = 10 hr/day

a. Power in KW

Power = 75,000 Btu/hr x 1 KW/3412 Btu/hr

Power = 75,000/3412

Power = 22 KW

b. Daily power consumption for heating

Power per day = 22 KW x 10 hr/day

Power per day = 220 KW-hr/day

c. Daily energy cost for heating in the winter season

Energy cost per day = 220 KW-hr/day x $0.07/KW-hr

Energy cost per day = $15

d. Monthly power consumption for heating

Power per month = 220 KW-hr/day x 30 days/month

Power per month = 6600 KW-hr/month

e. Monthly energy cost for heating in the winter season

Energy cost per month = 6600 KW-hr/month x $0.07/KW-hr

Energy cost per month = $462

AIR CONDITIONING

Air-conditioning is a branch of refrigeration that deals with comfort cooling.

Air conditioning provides comfortable:

(1) temperature

(2) humidity

(3) air flow

(4) air quality

COOLING COST IN SUMMER SEASON

Example:

A house has 3 bedrooms each with a 6000 Btu/hr window air-conditioner.

The living room and all common living spaces are cooled by a ductless split

air-conditioner with a total cooling capacity of 2 tons. All air-conditioning

equipment run continuously for 24 hours each day during the summer. The cost

of electricity is 7 cents/KW-hr. How much is the daily and monthly power

consumption and electric bill spent for cooling and air conditioning the house.

Solution:

Total Power = (3 x 6,000 Btu/hr) + (2 tons x 12,000 Btu/hr per ton)

Total Power = 18,000 + 24,000

Total Power = 42,000 Btu/hr ==> 3.5 tons

Energy cost = 7 cents/KW-hr = $0.07/KW-hr

Air Conditioner running time = 24 hr/day

a. Total Power in KW

Power = 42,000 Btu/hr x 1 KW/3412 Btu/hr

Power = 42,000/3412

Power = 12 KW

b. Daily power consumption for air conditioning

Power per day = 12 KW x 24 hr/day

Power per day = 288 KW-hr/day

c. Daily energy cost of air conditioning in summer season

Energy cost per day = 288 KW-hr/day x $0.07/KW-hr

Energy cost per day = $20

d. Monthly power consumption for air conditioning

Power per month = 288 KW-hr/day x 30 days/month

Power per month = 8640 KW-hr/month

e. Monthly energy cost of air conditioning in summer season

Energy cost per month = 8640 KW-hr/month x $0.07/KW-hr

Energy cost per month = $600

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CONDUCTION

Conduction is a method of heat transfer through direct contact. An example

of conduction is a cooking pot directly over a burner hot plate. Heat directly

transfers from the stove hot plate to the cooking pot.

Metals such as copper, aluminum, brass, tin and silver are good conductors of

heat and electricity because of free electrons that travel to another substance.

Good conductors of heat take a short amount of time for heat transfer.

Copper is one of the best conductors and is widely used in the refrigeration

industry for refrigerant piping as well as wires for power and controls.

Poor conductors of heat are called insulators. Glass, fiberglass, cork, styrofoam

and vermiculite are examples of heat insulating materials. Glass is one of the

poorest conductors of heat and is typically used in many applications.

CONVECTION

Convection is a method of heat transfer by moving currents in gases or liquids.

There are two forms of convection: natural and forced. Hot air currents rising

above a furnace and spreading around the room is an example of natural

convection. The hot air is replaced by the colder denser air and thereby provides

continuous air current movement. Another type of convection is forced convection,

whereby a fan or blower creates a pressure difference (pressure flows from a

region of higher pressure to a region of lower pressure). This ensures air is

circulated throughout the room.

Air and water are the most common mediums used in the heating and air

conditioning industry. Common examples are centralized hot water heating

systems in most buildings. Other applications are the use of hot air and steam

for heating.

RADIATION

Radiation is a method of heat transfer whereby the heat travels in the form of

high speed infra-red radiation waves. Radiation is the process in which heat is

transferred without affecting the space between the heat source and the

absorbing material. An example of this is the heat from the sun to the earth.

Heat is absorbed by the earth without heating the outer space.

The amount of heat radiated by a hot body depends on the:

a. area of radiating surface

b. texture of radiating surface

c. colour of radiating surface

d. temperature difference between the object and its surroundings

The best radiators and absorbers of radiant heat are those materials that have

rough dull black surfaces. The best reflectors of heat are those materials

with shiny or white surfaces.

MAIN USES OF REFRIGERATION

Food preservation

As the temperature of food products is lowered, the growth of bacteria which

causes food spoilage will be diminished.

Comfort cooling

Comfort cooling, also known as air-conditioning is used to provide comfortable

living conditions for homes and offices.

REFRIGERATION TEMPERATURE RANGES

The temperature ranges in the refrigeration industry are:

1. High temperature: 45°F to 70°F (7 C to 21 C)

2. Medium temperature: 35°F to 45°F (2 C to 7 C)

3. Low temperature: 0°F to -10°F (-18 C to -23 C)

AIR CONDITIONING EQUIPMENT DESIGN GUIDELINES

40°F (4 C): evaporator cooling coil design temperature

55°F (13 C): evaporator fan air temperature (supply to room)

75°F (24 C): inside room design temperature (return air from room)

95°F (35 C): outside ambient design temperature (condenser air supply)

125°F (52 C): condenser outside coil design temperature

AIR PRESSURE, ELEVATION & BOILING POINT

The atmospheric pressure of air at sea level is 29.92 in Hg or 14.7 psia. The

boiling point of water at these conditions is 212 F.

The atmospheric pressure of air at higher elevations is reduced by approx

1 in Hg per 1000 ft (0.5 psi/1000 ft). Mount Everest has a height of 29,029 ft

(8,848 m). The actual measured air pressure at the top of Mount Everest is

about 10 in Hg (5 psi). The air pressure is 1/3 that of sea level, resulting in the

reduction of oxygen which is the reason why it is very difficult to breathe at

higher elevations. Also, the boiling point of water is reduced to 71 °C (160 °F).

These conditions will make cooking difficult and will require more time to cook.

REFRIGERANT SATURATION PRESSURE & TEMPERATURE

A refrigerant is a substance used in refrigeration process that can quickly turn

into vapor by the addition of heat and into a liquid by the removal of heat.

Saturation temperature of a refrigerant is its boiling point. In this state, the

refrigerant is a mixture of liquid and vapour. The saturation temperature

increases with an increase in saturation pressure. A saturated vapor is a vapor

in which it will readily become superheated vapor with the addition of heat and

will readily condense to a sub-cooled liquid with the removal of heat.

SUPERHEAT AND SUB-COOLING

Superheat is the temperature above saturation. A superheated vapor is a vapor

with a temperature higher than the saturation temperature. Subcooling is the

temperature below saturation. A sub-cooled liquid is a liquid with a temperature

lower than the saturation temperature.

Example:

Steam at 222 F is superheated by 10 F because the boiling point (saturation

temperature of water is 212 F).

Superheat = 222 F - 212 F = 10 F

Water at 200 F is subcooled by 12 F because the saturation temperature of

water is 212 F.

Sub-cooling = 212 F - 200 F = 12 F

BASIC COMPONENTS OF REFRIGERATION & AIR CONDITIONING

1. Compressor

2. Condenser

3. Metering device

4. Evaporator

COMPRESSOR

The function of the compressor is to circulate the refrigerant throughout the

system. It lowers the pressure on the low-pressure side of the system and

raises pressure on the high-pressure side of the system. This difference in

pressure (pressure differential) is what causes the refrigerant to flow. The

compressor does this by drawing the refrigerant vapor from the evaporator,

compressing it (reducing its volume, while increasing its pressure) and then

discharges it to the condenser.

CONDENSER

The condenser removes heat from the refrigeration system. The condenser

performs three basics purposes. First, it desuperheats the refrigerant from

superheated vapor to saturated vapor. Second, it condenses the saturated vapor

and convert it to saturated liquid. Third, it subcools the saturated liquid to a

temperature below the boiling point.

The Liquid Line is the tubing from the condenser outlet to the metering device.

It contains subcooled liquid and its temperature is higher than the suction line.

METERING DEVICE

The metering device or expansion device regulates the amount and flow of

liquid refrigerant into the evaporator. It lowers the pressure of the refrigerant,

causing it to turn into a mixture of flash gas and liquid. This will cause the

refrigerant to have low pressure and low temperature so that it can evaporate

(boil) at a low temperature, absorb heat from the load and produce a cooling

effect in the evaporator.

EVAPORATOR

The evaporator provides cooling by absorbing heat. It performs three main

functions: absorb heat, turn liquid refrigerant into saturated vapor and superheat

the refrigerant before being sent to the compressor.

The Suction Line is the tubing used to connect the evaporator outlet to the

compressor suction. The suction line is usually made of copper and is insulated

to prevent heat from the surroundings to be absorbed by the system. Another

reason why suction lines are insulated is to prevent sweating (condensation).

A suction line with insulation provides higher efficiency than a suction line with

no insulation. Higher efficiency will translate into savings on energy costs.

18 apr 2015

HOME (DOMESTIC) ROOM AIR-CONDITIONER TEMPERATURES & PRESSURES

COMPRESSOR

60 F = vapor refrigerant entering temperature

200 F = vapor refrigerant exiting temperature

delta T = 200 - 60 = 140 F

70 psig = vapor refrigerant entering pressure

280 psig = vapor refrigerant exiting pressure

delta P = 280 - 70 = 210 psig

compression ratio = 280/70 = 4

CONDENSER

280 psig = saturation pressure (head pressure, condensing pressure)

280 psig = refrigerant inlet pressure

280 psig = refrigerant outlet pressure (constant pressure condenser)

95 F = outside ambient air temperature

30 F = temperature difference between ambient and condensing temperature

200 F = vapor refrigerant entering temperature

125 F = saturation temperature at condenser (condensing temperature)

105 F = subcooled liquid temperature leaving condenser

subcooling = 125 - 105 = 20 F

METERING DEVICE

105 F = subcooled liquid refrigerant temperature entering metering device

40 F = refrigerant temperature exiting metering device

delta T = 105 - 40 = 65 F

280 psig = subcooled liquid refrigerant pressure entering metering device

70 psig = refrigerant pressure exiting metering device

delta P = 280 - 70 = 210 psig

75% liquid, 25% vapor = state of refrigerant exiting metering device

EVAPORATOR

40 F = refrigerant inlet temperature

50 F = refrigerant outlet temperature

evaporator superheat = 50 - 40 = 10 F

70 psig = refrigerant inlet pressure

70 psig = refrigerant outlet pressure (constant pressure evaporator)

60 F = compressor suction line temperature

system superheat (compressor superheat) = 60 - 40 = 20 F

50% = relative humidity of air entering evaporator

75 F = temperature of air entering evaporator (room air return temperature)

55 F = temperature of air leaving evaporator (room air supply temperature)

room air delta T = 75 - 55 = 20 F

20 F = outside ambient air & inside room air temperature differential

20 F = 95 F - 75 F

BANNED/FOR PHASEOUT REFRIGERANTS

R-12 (Freon-12)

R-22 (Freon-22)

R-500

R-502

NEW/REPLACEMENT REFRIGERANTS

R-134a

- replacement for R-12

- medium and high-temperature refrigeration

- refrigerators, freezers, automotive air-conditioning

R-407C

- replacement for R-22 in residential and commercial air-conditioning

R-410A

- replacement for R-22 in residential and commercial air-conditioning

- higher operating pressures than R-22

R-404A

- replacement for R-502

- low and medium temperature refrigeration

R-507

- replacement for R-502

- low and medium-temperature refrigeration applications

- higher pressures and capacity than R-404A

R-507A

- replacement for R-502 and R-22

- higher efficiency than R-404A

- higher efficiency than low-temperature R-22

R-407A

- medium and low-temperature commercial refrigeration

- direct expansion residential and commercial air-conditioning

REFRIGERANT SELECTION CRITERIA

1. Safe

2. Detectable

3. Boiling point above atmospheric pressure

4. Economical compressor to handle vapor volume

Note:

Water is not a practical refrigerant because it will require a large compressor.

The vapor volume of water is 2445 cubit feet per 1 pound at 40°F.

REFRIGERANT SAFETY PRECAUTIONS

1. Refrigerants displace oxygen because refrigerants are heavier than air.

2. When working with refrigerants, ensure proper ventilation.

3. Ensure there are no sources of arcs or open flames.

4. When refrigerants react with flame, they will emit toxic and corrosive gas.

5. Don't mix and match refrigerants.

METHODS OF REFRIGERANT LEAK DETECTION

1. Hearing/Visual

2. Soap-bubble solution

3. Ultrasonic leak detector

4. Electronic leak detector

5. Ultraviolet leak detection

REFRIGERANT CYLINDER COLOR CODES

R-11 Orange

R-12 White

R-22 Green

R-113 Purple

R-114 Dark blue

R-134a Light blue

R-123 Light gray

R-401A Coral red

R-401B Mustard yellow

R-401C Aqua

R-402A Light brown

R-402B Green-brown

R-404A Orange

R-406A Light gray-green

R-407A Bright green

R-407B Cream

R-407C Chocolate

R-409A Tan

R-410A Rose

R-500 Yellow

R-502 Orchid

R-717 Silver

LAW

It is illegal to intentionally vent refrigerant into the atmosphere.

It is mandatory for technicians to recover and sometimes recycle refrigerants

during installation and servicing.

TOOLS OF THE TRADE

The following are the tools generally used by refrigeration and air conditioning

service technicians:

Hand Tools:

Screwdrivers, bits & tips

Nut drivers

Files

Cold chisel

Straight metal snips

Aviation metal snips

Taps (internal thread cutter)

Die (external thread cutter

Pipe-threading die

Awl

Measuring Tape (Rule)

Flashlight

Extension cord/lights

Drill bits

Hacksaw

Hole saw

Reciprocating saw

Jigsaw

Square

Levels

Fish tape

Utility knife

C-Clamp

Wrenches

- Socket wrench with ratchet handle

- Open end wrench

- Box end wrench

- Combination wrench

- Adjustable wrench

- Ratchet box wrench (refrigeration wrench)

- Air-conditioning and refrigeration reversible ratchet hex wrench

- Air-conditioning and refrigeration reversible ratchet box wrench

- Flare nut wrenches

- Pipe wrench

- T-Handle hex keys

- Combination crimping and stripping tool

- Automatic wire stripper

- Stainless inspection mirror

- Glass telescoping inspection mirror

- Stapling tackers

Pliers

- General-purpose pliers

- Needle-nose pliers

- Side cutting pliers

- Slip joint (channel locks)

- Locking pliers (vise grip)

Hammers

- Ball-peen hammer

- Soft head (rubber mallet)

- Carpenter’s claw hammer

Power Tools:

- Portable electric drill, cord type

- Portable electric drill, cordless

Tubing Tools:

- Tube Cutter

- Inner–outer reamers

- Flaring Tools (flaring bar, slip-on yoke, feed screw with flaring cone, handle)

- Tube deburring tool

- Swaging Tools (punch-type, lever-type, die-type)

- Tube Benders (spring type, lever type, gear type)

- Plastic Tubing Shear (PVC and polyethylene tubing shear)

- Tubing Pinch-off tool

- Metalworker’s Hammer

- Tube brushes

Specialized Service and Installation Tools:

- Gauge manifold (high pressure, compound gauge (low-pressure and vacuum)

- Electronic gauge manifold (superheat and subcooling, vacuum, thermometer)

- Programmable charging meter or scale

- Digital vacuum gauge

- Electronic thermistor vacuum gauge (up to 50 microns or 0.050 mm Hg)

- Vacuum Pump

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