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/lbF (4.2 KJ/kgC)
specific heat of ice = 0.5 Btu/lbF (2.1 KJ/kgC)
specific heat of steam = 0.5 Btu/lbF (2.1 KJ/kgC)
specific heat of air = 0.24 Btu/lbF (1 KJ/kgC)
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/lbF
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/lbF
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/lbF
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 KWhr = 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/KWhr, 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/KWhr = $0.07/KWhr
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 KWhr/day
c. Daily energy cost for heating in the winter season
Energy cost per day = 220 KWhr/day x $0.07/KWhr
Energy cost per day = $15
d. Monthly power consumption for heating
Power per month = 220 KWhr/day x 30 days/month
Power per month = 6600 KWhr/month
e. Monthly energy cost for heating in the winter season
Energy cost per month = 6600 KWhr/month x $0.07/KWhr
Energy cost per month = $462
AIR CONDITIONING
Airconditioning 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 airconditioner.
The living room and all common living spaces are cooled by a ductless split
airconditioner with a total cooling capacity of 2 tons. All airconditioning
equipment run continuously for 24 hours each day during the summer. The cost
of electricity is 7 cents/KWhr. 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/KWhr = $0.07/KWhr
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 KWhr/day
c. Daily energy cost of air conditioning in summer season
Energy cost per day = 288 KWhr/day x $0.07/KWhr
Energy cost per day = $20
d. Monthly power consumption for air conditioning
Power per month = 288 KWhr/day x 30 days/month
Power per month = 8640 KWhr/month
e. Monthly energy cost of air conditioning in summer season
Energy cost per month = 8640 KWhr/month x $0.07/KWhr
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 infrared 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 airconditioning 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 subcooled liquid with the removal of heat.
SUPERHEAT AND SUBCOOLING
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 subcooled 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.
Subcooling = 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 lowpressure side of the system and
raises pressure on the highpressure 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 AIRCONDITIONER 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
R12 (Freon12)
R22 (Freon22)
R500
R502
NEW/REPLACEMENT REFRIGERANTS
R134a
 replacement for R12
 medium and hightemperature refrigeration
 refrigerators, freezers, automotive airconditioning
R407C
 replacement for R22 in residential and commercial airconditioning
R410A
 replacement for R22 in residential and commercial airconditioning
 higher operating pressures than R22
R404A
 replacement for R502
 low and medium temperature refrigeration
R507
 replacement for R502
 low and mediumtemperature refrigeration applications
 higher pressures and capacity than R404A
R507A
 replacement for R502 and R22
 higher efficiency than R404A
 higher efficiency than lowtemperature R22
R407A
 medium and lowtemperature commercial refrigeration
 direct expansion residential and commercial airconditioning
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. Soapbubble solution
3. Ultrasonic leak detector
4. Electronic leak detector
5. Ultraviolet leak detection
REFRIGERANT CYLINDER COLOR CODES
R11 Orange
R12 White
R22 Green
R113 Purple
R114 Dark blue
R134a Light blue
R123 Light gray
R401A Coral red
R401B Mustard yellow
R401C Aqua
R402A Light brown
R402B Greenbrown
R404A Orange
R406A Light graygreen
R407A Bright green
R407B Cream
R407C Chocolate
R409A Tan
R410A Rose
R500 Yellow
R502 Orchid
R717 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
Pipethreading die
Awl
Measuring Tape (Rule)
Flashlight
Extension cord/lights
Drill bits
Hacksaw
Hole saw
Reciprocating saw
Jigsaw
Square
Levels
Fish tape
Utility knife
CClamp
Wrenches
 Socket wrench with ratchet handle
 Open end wrench
 Box end wrench
 Combination wrench
 Adjustable wrench
 Ratchet box wrench (refrigeration wrench)
 Airconditioning and refrigeration reversible ratchet hex wrench
 Airconditioning and refrigeration reversible ratchet box wrench
 Flare nut wrenches
 Pipe wrench
 THandle hex keys
 Combination crimping and stripping tool
 Automatic wire stripper
 Stainless inspection mirror
 Glass telescoping inspection mirror
 Stapling tackers
Pliers
 Generalpurpose pliers
 Needlenose pliers
 Side cutting pliers
 Slip joint (channel locks)
 Locking pliers (vise grip)
Hammers
 Ballpeen 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, slipon yoke, feed screw with flaring cone, handle)
 Tube deburring tool
 Swaging Tools (punchtype, levertype, dietype)
 Tube Benders (spring type, lever type, gear type)
 Plastic Tubing Shear (PVC and polyethylene tubing shear)
 Tubing Pinchoff tool
 Metalworker’s Hammer
 Tube brushes
Specialized Service and Installation Tools:
 Gauge manifold (high pressure, compound gauge (lowpressure 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

.
Engineering
Tuesday, January 3, 2017
Sunday, November 8, 2015
Signs/Causes/Reasons for Overload trip  Motor troubleshooting
 blown fuse
 improperly sized fuse
 tripped/open circuit breaker
 failed capacitor
 overheating
 poor ventilation/lack of cooling
 high ambient temperature
 broken motor fan
 low voltage
 overvoltage
 voltage fluctuation, transient voltages
 power imbalance to the phases (not same power to each phase)
 insufficient torque
 belts too tight
 one phase drawing excessive amps
 one phase is open (single phasing)
 phase imbalance (voltage on one phase lesser than others)
 one phase drawing no (zero) amps
 open contactor contacts
 open motor winding
 low insulation resistance
 low resistances causing high currents
 vibration
 damaged stator
 loose wiring
 improper wiring combinations
 loose contactor contacts
 grounded motor
 shorted motor
 defective overload heater
 malfunctioning overload block
 overload relay failure
 overload protection not properly sized/not properly set
 motor sitting for long period of time (e.g. not operated during winter)
 seized motor
 rotor not turning
 shaft not turning freely
 shaft misalignment
 shaft too loose
 rotating parts contact stationary parts
 soft footing/foundation
 pipe strain
 moisture
 rust, contamination, dirt, accumulated dust
 bearing failure
 overlubrication
 inadequate or incorrect lubrication
 jam/obstruction inside/outside motor
 inadequate wiring/undersize wires
 undersized motor
 tripping of thermal protection/overload trip
 system alteration (e.g load increase, power supply changes, etc)
 new component added
 flowrate increase (airflow, water flow, heat load increase)
 operating a motor at 10% to 15% above rated speed
 harmonic distortion causing high frequency voltages or high current
 wear, incorrect fit
 shaft voltages/currents while rotating
 variable frequency drive (VFD) has improper pulses generated
Saturday, October 10, 2015
How ddc works: HVAC Furnace direct digital control
HOW DDC WORKS
CENTRAL WORKSTATION (control, monitor, manage)

talks to global controllers/building
controllers

send, receive, process, store,
print the data

check component and system status

check trends and monitor
performance of equipment

change set points, schedules
and settings

ASCENT COMPASS  Alerton
GLOBAL CONTROLLERS (integrate devices)

talks to field
controllers/zone controllers

connects/provides gateway to
allow communication
with devices made by different manufacturers

integrates different devices

ACM, BCM, VLX  Alerton
FIELD CONTROLLERS (control terminal units)

operates actuators, fans,
blowers, pumps

field controller for VAV, AHU,
Fan coil unit, chiller
cooling tower, boiler, terminal unit

VLC, VAV  Alerton
SENSORS

provide input, send signal to
controllers

display room setpoints and fan
status

display room temperature, outside
air temperature, relative humidity

MICROSET  Alerton
Receive Input
Information From Sensors
Process Input
Information
Respond with an Output
Action
Receive Feedback when
set point is met
Respond with another
Output Action
Example: How
a Furnace DDC works?
1.
Thermostat (Sensor) calls for
heat
2.
Thermostat switch is closed, sends
voltage signal to control board
3.
Control board (“Brain”)
receives and processes input signal from thermostat
4.
Control board closes contactor,
inducer fan motor starts
5.
Air flow switch/Sail switch
closes to confirm air flow supplied by fan
6.
Control board will activate
the Spark Igniter
7.
Control board activates Gas
Valve
8.
Burner ignites
9.
Flame Sensor detects flame
10.
Spark Igniter deactivates
11.
Control board energizes Indoor Fan Blower
motor
12.
Furnace supplies heat to the structure (room, home,
building, etc.)
13.
Thermostat temperature set point is met
14.
Thermostat opens switch, sends feedback signal
voltage to control board
15.
Control board shuts off the
Gas Valve and Induced draft fan
HOW BUILDING AUTOMATION SYSTEM (BAS) WORKS
BAS uses an integrated network of Workstations,
Global Controllers, Field Controllers, Sensors, Gateways, Routers, Repeaters,
Cables and other associated components that are designed and programmed to provide
the following building services:
 Networking
 Control, Monitoring and Alarms
 Data collection
 HVAC – Heating, Ventilation, AirConditioning
 Realtime Adjustments of Setpoints, Percent valve openings, etc
 Scheduling, Remote Operation and Remote Access
 Changing Modes, e.g. Economizer mode (to save on energy costs)
 Load and Demand Control
 Fire Protection
 Security
 Lighting management
 Energy Management
Saturday, October 3, 2015
Refrigeration System Sequence of Operation
REFRIGERATION (COOLING) CYCLE:
Power is supplied to Timer Motor
Timer Motor controls time for Cooling and Defrost cycles
Room Thermostat Closes when the temperature rises above setpoint
Control Board will activate the Refrigeration/Cooling Mode
Timer Motor is on Refrigeration/Cooling Mode
Defrost Heater is Off
Defrost Limit Switch is Off
Defrost Termination Solenoid is Off
Defrost Termination Thermostat/Fan Delay Switch is On
Evaporator Fan Motor is On
Liquid Line Solenoid is energized
Liquid Line Solenoid Valve Opens to allow liquid refrigerant to Evaporator
Low Pressure Control/Switch is Closed
High Pressure Control/Switch is Closed
Overload Protection is Closed
Compressor Contactor Coil is Energized
Compressor Runs
Condenser Fan Motor Runs
When Desired Room Temperature is reached
Room Thermostat Opens
Liquid Line Solenoid is Deenergized
Liquid Line Solenoid Valve Closes
Compressor continues to run to allow Pump Down of refrigerant
Low Pressure Control/Switch will Open
Compressor Contactor Coil will Deenergized
Compressor will turn Off
Condenser Fan Motor will turn Off
When defrost time is reached by the Timer Motor
Defrost Mode/Cycle begins
===============
DEFROST CYCLE/MODE:
Compressor Pump Down process
Timer Motor will be on Defrost Mode/Cycle
Defrost Heater is On
Defrost Limit Switch is On
Defrost Termination Solenoid is On
Defrost Termination/Fan Delay Switch is Off
Evaporator Fan Motor is Off
When defrost time is completed and Evaporator warms up
Defrost Termination Thermostat Closes
Low Pressure Control Closes
Compressor starts
Condenser Fan Motor Starts
Refrigeration Cycle Starts
When Evaporator coil temperature cools down
Fan Delay Switch will Close
Evaporator Fan Motor will turn On
Refrigeration cycle continues until the next Defrost Cycle is set by Timer Motor
Saturday, September 19, 2015
Advantages Disadvantages of Pneumatic, Hydraulic, ElectricalElectronic systems
The following relates to general Pneumatic, Hydraulic, ElectricalElectronic systems
as well as Control Systems which uses the systems mentioned.
Pneumatic Systems advantages:
 cheap initial installation
 air availability
 ease of transfer through piping
 ease of power and speed transmission
 ease of use
 ease of maintenance
 safe, explosion proof
 clean
 works in wide temperature range
 reserve compressed air can be stored
Pneumatic Systems disadvantages:
 more energy cost compared to hydraulic
 easy to leak
 hard to find leaks
 noise
 condensation/moisture
 needs drying to avoid condensation
 control of speed needs additional devices
 control of position difficult
 does not work underwater
 does not work in extreme temperatures
 requires bigger cylinder to handle the same load as in hydraulic
Hydraulic Systems advantages:
 high horsepowertoweight ratio
 maintains constant torque and force
 hydraulic power can be transmitted in long distances
 motion reversal is fast
 handles strong, heavy loads, shock forces
 lesser overall wear because of oil lubrication
 does not generate sparks
 smooth operation/lifting/movement of loads
 costs less energy to operate
 leaks easier to find
 operates in hot environments
Hydraulic Systems disadvantages:
 more expensive initial installation than pneumatic system
 noisy
 risk of contamination
 requires more energy to operate
 requires more maintenance
 heavier components, parts
 hydraulic fluid dangerous to humans
 hydraulic fluid not environment friendly
Electronic/Electrical Systems advantages:
 high accuracy
 quiet
 longer life
 no moving parts
 lesser maintenance
 more reliable
 high efficiency
 higher energy savings due to better energy management
 less drift and recalibration problems
Electronic/Electrical Systems disadvantages:
 expensive initial cost
 complexity of algorithms/troubleshooting
 risk of radio frequency interference
 risk of fire hazards due to arcs, sparks
 risk of electricution, short circuits, grounds
as well as Control Systems which uses the systems mentioned.
Pneumatic Systems advantages:
 cheap initial installation
 air availability
 ease of transfer through piping
 ease of power and speed transmission
 ease of use
 ease of maintenance
 safe, explosion proof
 clean
 works in wide temperature range
 reserve compressed air can be stored
Pneumatic Systems disadvantages:
 more energy cost compared to hydraulic
 easy to leak
 hard to find leaks
 noise
 condensation/moisture
 needs drying to avoid condensation
 control of speed needs additional devices
 control of position difficult
 does not work underwater
 does not work in extreme temperatures
 requires bigger cylinder to handle the same load as in hydraulic
Hydraulic Systems advantages:
 high horsepowertoweight ratio
 maintains constant torque and force
 hydraulic power can be transmitted in long distances
 motion reversal is fast
 handles strong, heavy loads, shock forces
 lesser overall wear because of oil lubrication
 does not generate sparks
 smooth operation/lifting/movement of loads
 costs less energy to operate
 leaks easier to find
 operates in hot environments
Hydraulic Systems disadvantages:
 more expensive initial installation than pneumatic system
 noisy
 risk of contamination
 requires more energy to operate
 requires more maintenance
 heavier components, parts
 hydraulic fluid dangerous to humans
 hydraulic fluid not environment friendly
Electronic/Electrical Systems advantages:
 high accuracy
 quiet
 longer life
 no moving parts
 lesser maintenance
 more reliable
 high efficiency
 higher energy savings due to better energy management
 less drift and recalibration problems
Electronic/Electrical Systems disadvantages:
 expensive initial cost
 complexity of algorithms/troubleshooting
 risk of radio frequency interference
 risk of fire hazards due to arcs, sparks
 risk of electricution, short circuits, grounds
Sunday, June 21, 2015
8minute run time for 3/8 cordless drill 18 volts
Example:
18 volts Lithiumion
3/8 inch chuck
Max. Torque: 400 inchlbs/33 ftlbs/45 Nm
No Load RPM : 0400 / 01,300 rpm
Clutch Settings : 15 position
Length : 7.7 inch
Weight : 3 lbs
Battery Charger: 45Minute Fast Charger
Cordless drill run time depends on:
 Battery voltage: More volts, More run time
 Battery capacity: More amphours, More run time
 Drill Efficiency: More efficient, More run time
How long cordless drill run in a single charge?
?
Run time calculation
actual screws driven: 160 screws (3 inch size) per charge
assume: 3 seconds to drive per screw
Run time = 160 screws x 3 sec/screw
Run time = 480 sec
Run time = 480 sec/60 sec per minute
Run time = 8 minutes
Saturday, May 30, 2015
Pressurized Cylinder Projectile
OXYGEN, NITROGEN
Oxygen and Nitrogen cylinders typically have 2500 psig pressure
Oxygen is one of the three elements of the Fire Triangle (Fuel, Oxygen, Heat/Spark)
Oxygen cylinders are usually green in color
Nitrogen cylinders are usually black in color
ACETYLENE
Acetylene cylinders typically have 250 psig pressure
Acetylene is a highly explosive fuel gas used for brazing and welding
Acetylene burns at over 3,000 C/6,000 F (one of the hottest burning fuel gas)
Acetylene cylinders are usually maroon in color
Acetylene chemical formula is C2H2
Pressure relief valve set at 400 psig
ACETYLENE CYLINDER PROJECTILE FORCE
Example: 6 x 20 inch  small cylinder
Diameter = 6 inch
Radius = 3 inch
Height = 20 inch
Cylinder pressure = 250 psig
Weight = 22 lbs
Capacity = 40 cubic foot of acetylene gas (120 times compressed vs. empty)
Capacity of cylinder without gas = (pi/4) x D^2 x H
Capacity of cylinder without gas = (pi/4) x 6 in^2 x 20 in x 1 cubic foot/1728 cubic in
Capacity of cylinder without gas = 0.33 cubic feet (empty)
Full capacity comparison = 40 cu ft/0.33 cu ft = 120 times compressed
Cylinder Total Surface Area, TSA
TSA = 2 x pi x r x h + 2 x pi x r^2
TSA = pi x d x h + 2 x (pi/4) x d^2
TSA = pi x 6 x 20 + 2 x (pi/4) x 6^2
TSA = 377 + 57
TSA = 434 sq. in.
Total Cylinder Projectile Force = Pressure x Area + Cylinder Weight
Total Cylinder Projectile Force = 250 lb/sq in x TSA + 22 lbs
Total Cylinder Projectile Force = ( 250 x 434 ) + 22 lbs
Total Cylinder Projectile Force = 108,500 lbs + 22 lbs
Total Cylinder Projectile Force = 108,522 lbs
Total Cylinder Projectile Force = 54 tons approx.
*** Small pressurized cylinder with huge 50 ton force projectile !!!
PRESSURIZED CYLINDERS (PRESSURE VESSELS) SAFETY
WORKING
1. Before working with pressurized cylinders, check them first.
2. Check valves, regulators  broken, damage, leak.
3. Check hoses  wear, crack, leak, damage.
4. Pay attention to leaks, noises and unusual smell.
5. After checking, if you found any defects, don't use it.
6. Keep cylinders a distance away from the work area  don't braze/weld directly in front of the cylinder!
7. Always wear personal protective equipment when working with pressurized cylinders.
8. Open quarter turn only, this is enough for most jobs, faster to close in case of emergency.
9. Know location of fire extinguishers and nearest exits when working with pressure vessels.
TRANSPORT
1. Don't drop  damages valves, etc.
2. Transport in upright position.
3. When transporting heavy cylinders, they must have a safety cap securely fastened and be in an approved cart with chains to secure the cylinders.
STORAGE
1. Store in upright position and ensure safety cap in place  to prevent tripping, valve damage, etc.
2. They must have chains to prevent from falling over.
3. Oxygen and Fuel cylinders must be separated with a minimum wall height of 5 ft or 20 ft distance away from flammable substances such as paint.
CYLINDER SAFETY DEVICES
1. Pressure relief valves
2. Fusible plugs (low melting point)
3. Pressure regulators
4. Safety cap
EMERGENCY ACTIONS
If you find a leaking cylinder, use your discretion, safety first!!!
If the leak is small and it is safe to close the valve, then immediately close the valve without endangering yourself.
If the leak is big, evacuate the area, close the door, pull the nearest fire pull station to sound the alarm, call Fire dept...
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