Tuesday, January 3, 2017

BASICS OF REFRIGERATION & AIR CONDITIONING

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


----------------------------------------------------------------


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


 -----------------------------------------------------------------------------



.



No comments:

Post a Comment