HVAC PPT.pptx

HVAC
ENRUN INDIA ELECTRICAL
CONSULTANTS AND CONTRACTORS
PREPARED BY ER MAJEED ALI
B.E (MECHANICAL) OSMANIA-
UNIVERSITY

BUILDING SERVICES (OR) FACILITY
MANAGEMENT
MEP
MECHANICAL(HVAC) ELECTRICAL PLUMBING &FIREFIGHTING

MEP
• MEP IS NOTHING BUT PUTTING ALL TOGETHER THE VARIOUS SERVICES SUCH AS
(HVAC, LIGHTING AND POWER LOAD, PLUMBING & FIREFIGHTING, BMS) AND
USING THEM IN ENERGY EFFICIENT WAY TO LIVE A LUXURIOUS LIFE.

APPLICATION OF HVAC(HEATING, VENTILATON AND AIR
CONDITIONING.
• RESIDENTIAL BUILDINDS
• HOSPITALS
• MALLS & HYPER MALLS
• SUERMARKETS
• AUDITORIUMS
• THEATRES
• SCHOOLS
• PHARMACIES
• HOTELS
• TRANSPORTATION ( AIRPLANE, TRAINS, AUTOMOBILES) etc.

HVAC
AIR CONDITIONING WAS FIRST INTRODUCED BY AN AMERICAN SCIENTIST KNOWN
AS DR HENRY WILLIS CARRIER SO HE IS CALLED AS FATHER OF HVAC.
THE CARRIER COMPANY HAS NAMED UPON HIM WHO MANUFACTURES,
RESEARCH AND DEVELOPS THE AIR CONDITIONING MACHINES AND ITS
FUNCTIONALITY.

HVAC
• H HEATING ( IT IS THE RAISING OF TEMPERATURE AND USUALLY DONE USED
FOR COLD CLIMATE CONDITIONS.
• VVENTILATION (IT IS THE SYSTEM TO INTAKE OUTSIDE FRESH AIR INSIDE AND
BAD ODOURS TO OUTSIDE(EXHAUST)

HVAC
• AC AIR CONDITIONING ( IT IS THE CONDITIONING OF AIR ACCORDING TO THE
HUMAN COMFORT CONDITIONS.DESIGNING THE SYSTEM FOR DRY,HOT AND
HUMID CLIMATE.
• HEAT: ACCORDING TO FIRST LAW OF THERMODYNAMICS
HEAT IS A FORM OF ENERGY THAT TRANSFER FROM HIGH TEMPERAURE
BODY TO LOW TEMPERATURE BODY.

HVAC
• IN SUMMER HEAT TRANSFER FROM OUTSIDE TO INSIDE

HVAC
• IN WINTER HEAT TRANSFER FROM INSIDE TO OUTSIDE

HVAC
• HEAT TRANSFER TAKES PLACE IN THREE WAYS
1. CONDUCTION (HEAT TRANSFER THROUGH SOLIDS)
2. CONVECTION (HEAT TRANSFER THROUGH FLUID(AIR OR LIQUID)
3. RADIATION (HEAT TRANSFER THROUGH LIGHT IN FORM OF RAYS)

HVAC
HVAC CONTROLING BODIES
ASHRAEAMERICAN SOCIETY OF HEATING REFRIGERATION AND AIR(US)
CONDITIONING ENGINEERS.
SMACNASHEET METAL AND AIR CONDITIONING NATIONAL ASSCIATION
CIBSECHARTERED INSTITUTE OF BUILDING SERVICES ENGINEER(UK)
ISHRAEINDIAN SOCIETY OF HEATING REFRIGERATION AND AIR CONDITIONING
ENGINEERS.
ISINTERNATIONAL STANDARD
ARIAIR CONDITIONING AND REFRIGERATION ENGINEERS
OSHAOCUPATIONAL SAFETY & HEALTH ADMINISTRATION.
IAQINDOOR AIR QUALITY

HVAC
• AIR: AIR IS A MIXURE OF VARIOUS ODOURLESS GASES SUCH AS NITROGEN 78%,
21% OXYGEN, 1% ARGON AND OTHER GASES WITH VARIOUS AMOUNTS OF
WATER VAPOUR.
1. SATURATED AIR=100% MOISTURE
2. UNSATURATED AIR= SOME AMOUNT OF MOISTURE

HVAC
• PSYCHROMERIC PROPERTIES:THE PROPERTIES WHICH GOVERNS THE PHYSICAL
LAWS OF AIR-WATER MIXTURE IS CALLED PSYCHROMETRY.
1. HUMIDITY RATIO OR ABSOLUTE HUMIDITY
2. RELATIVE HUMIDITY (%)
3. SENSIBLE HEAT
4. LATENT HEAT (OR) HIDDEN HEAT
5. WET BULB TEMPERATURE
6. DRY BULB TEMPERATURE
7. DEW POINT TEMPERATURE
8. ENTHALPY
9. DENSITY(MASS/SPACE)(LB/FT3)

HVAC
• HUMIDITY: THE AMOUNT OF MOISTURE PRESENT IN THE AIR IS CALLED
HUMIDITY.
• TWO TYPES OF HUMIDITY.
1. RELATIVE HUMIDITY: IT IS DEFINED AS THE RATIO OF ACTUAL AMOUNT OF MOISTURE
PRESENT IN THE AIR TO THE AMOUNT OF MOISTURE IT CAN HOLD.
2. SPECIFIC HUMIDITY OR HUMIDITY RATIO: IT IS EXPRESSED AS THE GRAINS OF
MOISTURE PER POUND OF DRY AIR (OR) HOW MUCH MOISTURE IS PRESENT IN
1 POUND(0.454 KG) OF DRY AIR.
NOTE: HOT AIR HAS LESS WATER THAN COLD AIR.
ENTHALPY: THE AMOUNT OF HEAT CONTENT IS CALLED AS ENTHALPY.
H=U + PV

HVAC
• TWO TYPES OF HEAT:
1. SENSIBLE HEAT: THE AMOUNT OF HEAT REALEASED OR ABSORBED BY A
CHEMICAL SUBSTANCE DURING CHANGE IN TEMPERATURE IS CALLED SENSIBLE
HEAT.(NO CHANGE IN PHASE ONLY CHANGE IN TEMPERATURE).
2. LATENT HEAT: THE AMOUNT OF HEAT RELEASED OR ABSORBED DURING
CHANGE OF PHASE i.e.; solid to liquid or gas to liquid etc. is CALLED AS LATENT
HEAT(CHANGE IN PHASE & NO CHANGE IN TEMPERATURE).

HVAC
• TEMPERATURE:THE DEGREE OF HOTNESS OR COLDNESS OF A BOBY OR A
SUBSTANCE MEASURED BY ORDINARY DRY BULB THERMOMETER IS CALLED
TEMPERATURE. CELCIUS(C) AND FARENHEIT(F).
• F=32F-212F(180 DIVISIONS)
• C=0C-115C
F=1.8C+32
C=(F-32)/1.8
EG: CONVERT 78F TO CELCIUS = (78-32)/1.8 = 25.5C ==26C.
CONVERT 24C TO FARENHEIT = (1.8X24)+32 = 75.5F ==76F.

HVAC
• TYPES OF TEMPERATURES
1. DRY BULB TEMPERATURE: IT IS THE PEAK TEMPERATURE OF AMBIENT AIR
MEASURED BY THERMOMETER.
2. WET BULB TEMPERATURE: IT IS THE TEMPERATURE OF AIR MEASURED BY A
THERMOMETER WHOSE BULB IS COVERED WITH A WET CLOTH.THE WET BULB
TEMPERATURE OF THE AIR IS RECORDED AFTER RAPIDLY MOVING THE
THERMOMETER IN THE AIR AND ALLOWING THE TEMPERATURE TO
STABILIES.IT IS CALLED WET BULB BECAUSE THE BULB IS WET WHEN
TEMPERATURE READING IS TAKEN.
3. DEW POINT TEMPERATURE:IT IS THE
TEMP OF AIR TAKEN WHEN IT IS FULLY
SATURATED AND MOISTURE CONDENSES
ON SURFACE.

PSYCHROMETRIC SOFTWARE
• PSYCHROMETRIC SOFTWARE :
STEPS:
1. OPEN PSYCHROMETRIC SOFTWARE BY DOUBLE CLICK ON ICON
2. GO TO RIBBON BAR AND SELECT ANALYSIS
3. AFTER SELECTING ANALYSIS CLICK ON STATE POINT AND PROCESSES or press F2
4. THEN CLICK ON ADD POINT THEN WINDOW POPS-UP NAME CURRENT POINT
5. THEN IN OPEN SPACES WE SHOULD ENTER ONLY 2 VALUES AND IT WILL PLOT
ALL THE PSYCHROMERTIC PROPERTIES.

HVAC
• AIR CONDITIONING IS NOTHING BUT CONDITIONING OF AIR TO PROVIDE
HUMAN COMFORT BY MAINTAINING ARTIFICIAL TEMPERATURE , RELATIVE
HUMIDITY, PURIFYING THE AIR THROUGH AIR FILTERS, NOISE LEVELS &
PROVIDING FRESH AIR TO IMPROVE (IAQ)INDOOR AIR QUALITY.
• TEMPERATURES =22-24C or 72-76F & RH=50%-60% ACCORDING TO ASHRAE
HUMAN FEEL COMFORT AT THIS CRITERIA.
• CONDENSATION:IT IS THE PROCESS WHERE CHANGE OF PHASE TAKES PLACE
FROM VAPOUR TO LIQUID BY COOLING.
• EVAPORATION:IT IS THE PROCESS WHERE CHANGE OF PHASE TAKES PLACE FROM
LIQUID TO VAPOUR BY HEATING.
4

• For Hyderabad comfort conditions.

HVAC
• UNIT OF HEAT IN HVAC IS TERMED AS BTU(BRITISH THERMAL UNIT).
BTU: BTU IS THE AMOUNT OF HEAT REQUIRED TO CHANGE THE TEMPERATURE OF
1POUND(0.454 KG) OF LIQUID WATER BY 1DEGREE F
WHAT IS TON OF REFRIGERATION ?
TR: WHEN 1LB(POUND)(0.45kg) OF ICE MELTS, IT ABSORBS 144 BTU i.e.; 1lb of ice =144BTU
THEREFORE 1TON=2000LB(POUNDS)
144*2000=288,000 BTU
WITHIN 24 HOURS OF TIME THROUGH DAY SO 288,000/24=12,000 BTU/HR.
TR IS NOTHING BUT AMOUNT OF COLLING EFFECT WE WANT IN COOLING SPACE
BTU IS NOTHING BUT AMOUNT OF HEAT
CFM(CUBIC FEET PER MINUTE):IT IS VOLUME OF AIR REQUIRED IN SPACE

HVAC
• UNITS UNITS ARE THE REPRESENTATIONS OF PHYSICAL QUANTITIES LIKE
DISTANCE, AREA, HEAT, TEMPERATUTE ETC.
UNITS ARE OF TWO TYPES :
1. SI (or) METRIC (or) INTERNATIONAL STANDARD eg: Millimeter, Centimeter, Meter.
2. US-IMPERIAL (or) IPs(Inch Pound second)
Units of Heat: IN SI (KCAL,KJ) AND IN US IMPERIAL (BTU-BRITISH THERMAL UNIT).
Units of Distance: IN SI (MM,CM,M) AMD IN US IMPERIAL (FEETS,INCHS).
Units of volume: IN SI (m3,cm3,mm3) and In US IMPERIAL (CFM-CUBIC FEET PER
MINUTE, CUBIC METER PER HOUR).
Units of velocity: IN SI (M/SEC) & IN US IMPERIAL(FPM-FEET PER MINUTE)
Units of temperature: IN SI CENTIGRADE(C) & IN US IMPERIAL (F) FARENHEIT.
FARENHEIT SCALE TOTAL 180 DIVISIONS(32F-212F).

HVAC
UNIT CONVERSIONS
1) LENGTH (OR) DISTANCE:
1METER=3.28 FEETS (or) 3.28’ ( 12 YARD=1 GUNTA, 40 GUNTA= 1ACRE, 4840SQYD=1 ACRE)
1 feet= 12 inches (or) 12” (1 ACRE = 43560 SQFT, 1 ACRE=4046 SQMETER)
1 inch = 2.54 cm (or) 25.4 mm
1 feet = 305 mm (304.8 mm)
1meter = 39 inch
1 yard (gaz) = 3 feet
2) AREA :
1M2 = 10.76 SQFT (or) ft2
1ft2 = 0.093 m2
1 ft2 = 144 in2
1 SQYARD = 9 ft2
36 FEET= 1 GUNTA

HVAC
3) TONNAGE AND CFM:
1TR=210 KJ/MIN
1TR=3.5 KW cooling kw
1 tr = 0.6-1.4 kw of power consumsion
1TR=12000 BTU/HR
1TR=400CFM (CUBIC FEET PER MINUTE)
1TR=2.4 GPM (GALLONS PER MINUTE)
1 TR= 3 GPM FOR OPEN PIPING (COOLING TOWER)
1TR=3000KCAL/HR
1 US GALLON=3.786 LITERS
1 CMH=1.7*CFM
1 CFM=2 LITERS/SEC (2.119) AIR FLOW
1 CFM=2118 M3/SEC
1 M/S = 197 FPM

• 4. GPM ( GALLONS PER MINUTE ):
1 GPM = 2.4 X TR(TONNAGE) FOR AIR COOLED CHILLERS
1 GPM = 3 X TR FOR WATER COOLED CHILERS FOR COOLING TOWER.
1 GPM = O.OO3785 m3/min.
1gpm (UK) = 4.54 liters.
1 GPM (US) = 3.78 LITERS.
1 m3= 1000 liters

HVAC
• INDOOR AIR CFM (or) FRESH AIR CFM:VOLUME OF AIR REQUIRED.
IAQ=(CFM/PERSON)*NO OF PEOPLE+CFM/AREA IN SQFT*AREA IN SQFT
SEE ASHRAE TABLE 6.1 VENTILATION IN BREATHING ZONE.

INDOOR AIR QUALITY PROBLEM
• CALCULATE THE AMOUNT OF FRESH AIR IN THE CLASS ROOM OF 25’ X 20’ FEET
FA CFM = (CFM/PERSON X NO OF PEOPLE) + (CFM/SQFT X AREA IN SQFT)
AREA = 25’ X 20’ = 500 ft2
from table 1000 ft2 = 35 for class room
500 ft2 = 35/2 = 17.5 ==18 people
from table cfm/person = 10
from table cfm/SQFT = 0.12
PUT ALL VALUES IN FORMULA= (10 X 18) + (0.12 X 500) =60 + 180=240 CFM
FA CFM = 240 CFM.

HVAC
• HVAC HEAT LOAD OR COOLING LOAD CALCULATION: THE MAIN AIM OF HEAT
LOAD OR COOLING LOAD CALCULATION IS TO DETERMINE HOW MUCH HEAT IS
ENTERING INTO THE COOLING SPACE(IN SUMMER) OR HOW MUCH HEAT IS
TRANSFERING FROM THE CONDITIONING SPACE(IN WINTER) & TO DETERMINE
THE CAPACITY OF THE MACHINE IN TR(TONNAGE) & HEAT GAIN IN (BTU/HR) &
VOLUME OF CHILLED AIR REQUIRED IN CFM(CUBIC FEET PER MINUTE).
• THERE ARE THREE METHODS TO DO HEAT LOAD CALCULATION:
1. THUMB RULE
2. MANUAL CALCULATIONS = OGENERAL METHOD(JAPAN).
= E-20 FORM (CARRIER US).
3. SOFTWARE (HAP SOFTWARE BY CARRIER,US).

HVAC
• THUMB RULE METHOD: THUMB RULE METHOD BY ASHRAE ITSELF
1. FOR LOW LOAD STRUCTURE TR=0.07*AREA IN SQUAREMETER(M2)
EG: corridors, lounges, receptions, balconies, walking bays etc.
2. FOR MEDIUM LOAD STRUCTURESTR=0.08*AREA IN SQUAREMETER(M2)
EG: offices, living rooms, shop etc.
3. FOR HIGH LOAD STRUCTURE TR=0.09*AREA IN SQUAREMETER(M2)
EG: malls, hyper malls, super markets, airports, auditoriums etc.
1TR=400CFM
1M=3.28 FT & 1M2=10.76FT2
1 FT2= 0.093 M2

CONDITIONS THAT PRODUCE HEAT IN SPACE
• PEOPLES
• LIGHTS
• GLASS (DOORS AND WINDOWS)
• ROOF
• FLOOR
• WALLS
• INFILTRATION: THE UNWANTED AIR WHICH IS COMING INTO THE ROOM
THROUGH CRACKS,DOOR VENTS or undercuts, MINOR OPENINGS etc. is called as
INFILTRATION
• EQUIPMENTS
• PRODUCTS etc.

HVAC heat load : HEAT TRANSFER

RCP(REFLECTED CEILING PLAN) DRAWINGS

• REFLECTED CEILING PLAN SHOWS THE COMPLETE DETAILS OF THE FALSE CELING
• PLENUM SPACE OR RCP HEIGHT IS THE SPACE IN WHICH ALL THE MEP SERVICES
RUN
• ROOF:THE EXTERNAL COVERING AT THE TOP OF A BUILDING
• CEILING: THE OVERHEAD CLOSURE OF A ROOM
• PARTITION: THE WALL BETWEEN THE CONDITIONED AND UNCONDITIONED
SPACE IS CALLED AS PARTITION
• FLOOR IS THE INTERIOR BOTTOM SURFACE OF THE ROOM
• Types of RCP (REFLECTED CEILING PLAN)
1. PLAIN FALSE CEILING
2. STEP FALSE CEILING
3. TILE FALSE CEILING

HVAC
OGENERAL
HEAT LOAD
CALCULATION
FORM
(APPROXIMATE
METHOD)

HVAC OGENERAL HEAT LOAD CALCULATION
• LOCATION : HYDERABAD = DBT 106 F, WBT 78 F & HEIGHT OF WALLS = 10FT

ACCORDING TO ASHRAE TABLE 62.1 VENTILATION STANDARD
NO OF PEOPLES IN OFFICE = 5 PEOPLE PER 1000 SQFT
5P=1000 SQFT (AREA= 20 X 30 = 600 SQFT)
?=600 SQFT 3 PEOPLE
1. PEOPLE= 3 PEOPLE (NUMBER) X 600 = 1800 BTU/HR.
2. WINDOWS EXPOSED TO SUN= [(5X4)+(5X4)] {40}TOTAL SQFT X(A) SELECT VALUE FROM
TABLE (A) {5}=200 BTU/HR.
3. LIGHTS AND ELECTRICAL APPLIANCES =
a) LIGHTS : FOR ALL CASES 0.5 TO 1.5 WATTS/SQUARE FEET & 2 IN FEW CASES
1 WATT = 1 SQFT
? WATT = 600 SQFT600WATT
b) ELECTRICAL APPLIACES :
1. COMPUTERS = quantity x watts x hours used /day
4 x 500 w x 8/24 =666.66 watts
2. printer/scanner= 1 x 300 w x 3/24 = 37.5 watts
3. water cooler = 1 x 200 w x 10/24 = 83.33 watts
4. others = 2000x 1/24 = 83.33 watts 1+2+3+4 = 870.87 watts
ADD (a+b) = 1471 watts x 3.4 (BTU conversion factor) = 5001.4 BTU/HR.

4. SUB-TOTAL (SUM OF ALL STEPS 1+2+3) = 1800 + 200 + 5001.4 = 7001.41 BTU/HR 
1EQUATION.
5. WINDOWS (TOTAL OF ALL WINDOWS) = NORTH WINDOW + WEST WINDOW
= (5X4)+(5X4)+(10X3) X (B) FACTOR TABLE
= 70 SQFT X 23 (HYD DRY BULB).
= 1610 BTU/HR.
6. WALLS a.) 1.)Outside wall north exposure = SQFT x (C) FACTOR = BTU/HR
= (30 X 10) – [(5X4)+(5X4)] X©
= 300-40 X (C)
= 260 X 4 (heavy construction BCZ outside wall)= 1040BTU/HR.
2.) Other than north exposure = (20x10)-(10x3) x © factor
= 170 x 7 = 1190 BTU/HR.
b.) Inside walls = (10x 10) SQFT x c factor
= 100 x 10 =1000 BTU/HR
NOTE: OUTSIDE WALL IS THICK THAN INSIDE WALL BECAUSE OF LAYERS OF PAINT,
PLASTERS etc.

7.) FLOOR(DISREGARD IF FLOOR IS ON THE GROUND OR BASEMENT)= SQFT X D FACTOR
 IF ON GROUND = 0 SQFT X D FACTOR = 0 BTU/HR
 IF NOT ON GROUND = 600 SQFT X 5 = 3000 BTU/HR(not considered in
this project).
8.) ROOF/ CEILING: THREE TYPES OF CEILING
1. PLANE FALSE CEILING : NORMAL
2. TILE FALSE CEILING : FOR MAINTENANCE PURPOSE
3. STEP FALSE CEILING : FASCINATING LOOK
===> ROOF= AREA IN SQFT X (E) FACTOR
= 600 SQFT X 8 = 4800 BTU/HR.
9.) VENTILATION/INFILTRATION = F X G
a.) VENTILATION (F) = L X B X H/60 = 20 X 30 X 10/60 = 100 CFM
b.) G = 37 FOM FORM BY WBT
FXG=100X37=3700 BTU/HR.
10.) ADD STEP 5+6+7+8+9= 13,340 BTU/HR  2 EQUATION

• ADD 1 EQUATION AND 2 EQUATION =7001.41+13,340 = 20341.41 BTU/HR IS
TOTAL HEAT GAIN IN SPACE , (CONVERT BTU/HR TO TR)
SINCE WE KNOW 1TR = 12000 BTU/HR
? = 20341.41 (CROSS MULTIPLY)
= 1.69 TR == 2 TR IS COOLING CAPICITY
SINCE 1TR = 400 CFM THEN 2TR = 800 CFM IS VOLUME
OF CHILLED AIR REQUIRED IN SPACE.
 IF DONE BY THUMB RULE
TR=0.08 X AREA IN SQMT
= 0.08 X 600 X 0.093
= 4.46 TR == 5 TR.
(O GENERAL METHOD).

E – 20 FORM
PROJECT DETAILS:
LOCATION : HYDERABAD
DRY BULB TEMPERATURE: 106 F
WET BULB TEMPERATURE: 78 F
ALTITUDE :505 METERS
LATTITUDE:17.86 N0RTH
DAILY RANGE : 14 F
WEIGHT OF WALL = 100 LB/SQFT
WEIGHT OF ROOF = 60/80
U-VALUES:
WALL =0.49
ROOF = 0.08, PARTITION= 0.4, GLASS= 0.56

E – 20 FORM
• JOB NAME : ABC
• ADDRESS : TOLICHOWKI
• SPACE USED FOR : HOSPITAL
• SIZE : 7.5’ X 7.5’ = 56.25 SQFT
AREA = 56.25 SQFT
VOLUME= 56.25 X 10 = 562.5 CUBIC FEET (ft3)

HEAT LOAD CALCULATION SPACE SHEET:

CARRIER US (E-20 FORM) FOR HEAT LOAD ANALYSIS
• EXAMPLE HYDERABAD :
DESIGN
CONDITIONS
DRY
BULB
TEMP
WET BULB
TEMP
RELATIVE
HUMIDITY
DEW POINT
TEMPERATU
RE
HUMIDITY
RATIO
GRAINS/PO
UND(GR/LB)
OUTSIDE TEMP 96 84 60 81 158
ROOM TEMP 72 62 57 56 66
TEMPERATURE
DIFFERENCE
24 XXXXXXXX XXXXXXXX XXXXXXX
92

REFRIGERENTS USED IN HVAC SYSTEM
• REFRIGENT: REFRIGENT IS A HEAT CARRYING MEDIUM AND ACT AS A WORKING
MEDIUM IN THE REFRIGERATION CYCLE WHICH PRODUCES CHILLNESS OR
COOLNESS EFFECT AS BY PRODUCT.
REFRIGERANT EVAPORATES AND GIVES SENSATION OF COOLING
• PROPERTIES OF REFRIGENT:
1. IT MUST HAVE LOW BOILING POINT
2. NON-CORROSIVE ( DOESN’T REACT WITH METAL)
3. NON-FLAMMABLE & NON EXPLOSIVE AS WELL
4. NON TOXIC OR NON POISONOUS
5. HIGH LATENT HEAT OF VAPOURISATION

CLASSIFICATION OF REFRIGERATION:
1. CFC ( CHLORO FLURO CARBON): IT CONTAINS CHLORINE WHICH RESULTS IN THE
DEPLETION OF OZONE LAYER.
EX: R-12, R-502,FREON.
2. HCFC ( HYDRO CHLORO FLURO CARBON): IT HAS LESS CHOLRINE THAN CFC
EX: R-22, R-409A
3. HFC ( HYDRO FLURO CARBON): NO CHLORINE CONTENT(OZONE FRIENDLY)
EX: R134-A, R-404, R-410A, R-507,R-32 etc.
R-134A IS KNOWN AS TETRA FLURO ETHANE FROM FAMILY OF HFC
BOILING POINT= -149F OR -26.1C
NOTE: PRIMARY REFRIGERANT= CHANGE OF PHASE GAS TO LIQUID REFRIGERANT USED
IN DX SYSTEM eg: R-134A,R 54, R32,R22 ETC
SECONDARY REFRIGERANT= CHANGE OF TEMPERATURE EG: CHILLED
WATER,ETHYLENE GLYCOL,BRINE

WORKING OF AIR CONDITIONING SYSTEM
• AIR CONDITIONG MACHINES WORK ON THE PRINCIPLE OF VCC( VAPOUR
COMPRESSION CYCLE)

PARTS OF VCC
• COMPRESSOR: IT IS A DEVICE WHICH COMPRESSES THE LOW TEMPERATURE AND LOW PRESSURE GAS
REFRIGERANT(Freon) TO HIGH TEMPERATURE AND PRESSURE GAS.
• CONDENSER: IT IS DEVICE WITH FAN BEHIND ITS COILS WHICH HELPS IN CONVERTING THE HOT GAS INTO
LIQUID IN CONDENSER COILS WITH MEDIUM PRESSURE AND TEMPERATURE
• EXPANSION VALVE: IT IS A DEVICE WHICH REDUCES THE PRESSURE OF THE LIQUID REFRIGERANT FROM
HIGH PRESSURE TO LOW PRESSURE BY PASSING THROUGH A NARROW SPACE, THEREBY REDUCES THE
TEMPERATURE OF THE REFRIGERANT SO, REFRIGERANT CONDITION AT THE OUTLET OF THE THROTTLE
VALVE IS LOW PRESSURE AND LOW TEMPERATURE AND SEND TO THE EVAPORATOR COILS.
• EVAPORATOR: IT IS A DEVICE WHICH IS PLACED INSIDE THE SPACE AND IT IS ALSO FITTED WITH FAN BEHIND
THE COILS WHICH ABSORBS THE HEAT OF THE REFRIGERANT BY EVAPORATION AND THUS WE GET
SENSATION OF COOLING, REFRIGERANT CONDITION IS NORMAL TEMPERATURE AND PRESSURE.
CYCLE :1.) COMPRESSOR IS THE WORK FORCE OF THE SYSTEM WHICH RECEIVES COOL AND LOW
TEMPERATURE & LOW PRESSURE GAS WHICH IS COMPRESSED TO HIGH PRESSURE AND HIGH TEMPERATURE
GAS .
2.) THE HIGH TEMPERATURE & HIGH PRESSURE GAS AFTER ENTERING INTO THE CONDENSER THE COOL AIR
THROUGH FAN IS BLOWN ACROSS THE COILS OF THE CONDENSER AND THEN IT BECOMES HIGH PRESSURE
LIQUID WITH SOME AMOUNT OF HEAT & THEN PASSED THROUGH FILTER DRYER TO REMOVE MOISTURE
WHICH CONTAMINATE THE REFRIGENT AND THEN TO EXPANSION VALVE.

3.) EXPANSION VALVE: REDUCES THE HIGH PRESSURE OF THE LIQUID
REFRIGERANT TO ALLOW EXPANSION OR CHANGE OF PHASE FROM LIQUID TO GAS
IN THE EVAPORATOR.
4.) EVAPORATOR: THE LOW PRESSURE LIQUID FROM THE EXPANSION VALVE TO
EVAPORATOR IT BEGINS TO BOIL AND TURN BACK TO LOW PRESSURE GAS
ABSORBING HEAT BY FAN BEHIND ITS COILS AND GIVES SENSATION COOLING.
 EXPANSION VALVE OR THROTTLE DEVICE

TYPES OF COMPRESSORS IN HVAC
• There are basically 5 types of air conditioner compressor that are commonly
used in the HVAC industry:
• Reciprocating.
• Scroll.
• Screw.
• Rotary.
• Centrifugal.
TYPES OF CONDENSORS USED IN HVAC:
• AIR COOLED CONDENSOR
• WATER COOLED CONDENSORS

TYPES OF AIR CONDITIONERS
• AIR CONDITIONING
1) NON CENTRAL AC
2) CENTRAL AC
A. NON CENTAL AC
WINDOW AC (0.5 TR- 5 TR)
SPLIT AC
FLOOR STAND or floor mounted (1-5TR)
CEILING MOUNTED (cassette) = EXPOSED TYPE, CONCEALED, SEMI CONCEALED (1-5TR)
DUCTABLE SPLIT AC(1-8.5TR)
B. CENTAL AC
i. PACKAGE AC (1TR-150TR)
ii. CENTAL PLANTS
 AIR COOLED CHILLERS ( 2000 TR FOR AREA OF 4000 SQFT AND ABOVE)
 WATER COOLED CHILLERS ( 2000 TR FOR AREA OF 4000 SQFT AND ABOVE)
VRV ( ONE OUT DOOR UNIT AND MULTIPLE INDOORS UNITS ) [6.5TR- 24 TR]

• EXAMPLE : CASETTE AC CATALOG

PACKAGE AC (1-125TR) & VRV (VARIABLE REFRIGENT
VOLUME (6.5 - 150 TR)

CHILLERS ( AIR COOLED AND WATER COOLED)

• CEILING SUSPENDED DISTRIBUTION UNIT (UPTO 25TR).

AIR HANDLING AND DISTRIBUTION DEVICES
• A fan coil unit (FCU) is a simple device consisting of a heating and/or cooling heat
exchanger or 'coil' and fan. It is part of an HVAC system found in residential,
commercial, and industrial buildings.( 2000 CFM) 5TR
• An air handler, or air handling unit (AHU), is a device used to regulate and
circulate air as part of a heating, ventilating, and air-conditioning (HVAC) system
(100,000 CFM), 10TR-250TR AND UP 500 TR ALSO.

TYPES OF FILTERS USED IN HVAC
• Several types of air filters are common in commercial HVAC systems
• Different Types Of HVAC Filters Explained
• Fiberglass Air Filters. Disposable, most affordable yet with the lowest straining
ability of all the types are fiberglass air filters. ...
• Pleated filters. ...
• Washable Air Filters. ...
• Electrostatic Air Filters. ...
• HEPA Filters. ...
• UV Filters

TYPES OF FILTERS USED IN HVAC
• Several types of air filters are common in commercial HVAC systems
• Different Types Of HVAC Filters Explained
• Fiberglass Air Filters. Disposable, most affordable yet with the lowest straining
ability of all the types are fiberglass air filters. ...

AHU
• ELETROSTATIC AIR FILTER

AHU SCHEMATIC AUTOCAD
• (AHU) AIR HANDLING UNIT

AHU SCHEMATIC AUTOCAD
• (AHU) AIR HANDLING UNIT AND FITTINGS CONNECTED TO IT

FCU SCHEMATIC AUTOCAD
• FCU FAN COIL UNI 1-8.5 TR

DIFFERENCE BETWEEN DX-SYSTEM AND CHILLER SYSTEM
• (DX) or DIRECT EXPANSION SYSTEM: IN DIRECT EXPANSION SYSTEM THE
REFRIGERANT ITSELF IS RESPONSIBLE FOR THE COOLING OF THE SPACE TO BE AIR
CONDITIONED
IN DX SYSTEM WE USE JET AXIAL FANS OR BLOWERS AS A MODE OF COOLING
THE AIR AND THIS NON CENTALISED SYSTEM WHOSE CAPITAL COST IS LOW AND
HAS LOW LIFE SPAN LIKE 5-8 YEARS ONLY.
CHILLER: CHILLER IS A MACHINE WHICH IS USED TO REDUCE THE TEMPERATURE
OF (MEDIUM) WATER OR TO COOL THE WATER BY VAPOUR COMPRESSION CYCLE
• CHILLER SYSTEM (or) CHILLED WATER SYSTEM: IN CHILLED WATER WATER
SYSTEM FIRST THE WATER IS CHILLED or COOLED BY THE REFRIGENT AND THEN
THIS CHILLED WATER IS RESPONSIBLE FOR COOLING THE SPACE TO BE
CONDITIONED.
IN CHILLER SYSTEM WE USE COOLING TOWERS & ALSO JET BLOWER FANS AS A
MODE OF COOLING THE AIR.THIS IS CENTRALISED COOLING SYSTEM AND HAS
GOOD LIFE SPAN

TYPES OF COMPRESSORS IN HVAC
• There are basically 5 types of air conditioner compressor that are commonly
used in the HVAC industry:
• Reciprocating.
• Scroll.
• Screw.
• Rotary.
• Centrifugal.
TYPES OF CONDENSORS USED IN HVAC:
• AIR COOLED CONDENSOR
• WATER COOLED CONDENSORS
• BYPASS: Commonly termed DECOUPLER is an open bypass line before secondary
variable pump suction to bypass back to chiller on the NO DEMAND for cooling.

AIR COOLED CHILLER
• AIR COOLED CHILLERS WORKING TERMINOLOGY:
1. AIR COOLED CHILLER DO NOT USE COOLING TOWER FOR PROCESS OF COOLING
INSTEAD THEY DUMP THE HEAT OUTSIDE INTO THE AMBIENT AIR (or)
ATMOSPHERIC AIR HENCE THEY NEED TO BE INSTALLED IN OPEN SPACE eg: roof of
the building, ground floor, car parking etc.
2. THEY TAKE LESS SPACE AS COMPARED TO THE WATER COOLED CHILLER & DON’T
NEED WATER TO DISSIPATE HEAT
3. AIR COOLED CHILLERS ARE LESS EFFICIENT AS COMPARED TO WATER COOLED
CHILLERS BECAUSE AIR COOLED CHILLERS USE LATENT HEAT OF VAPORIZATION (
MEANS CHANGE OF PHASE OF THE REFRIGERANT FROM GAS TO LIQUID AFTER
CONDENSATION OR SUB-COOLING).
working: CONDENSER COOLING FANS WHEN THEY ROTATE THEY SUCK THE AIR
FROM THE ATMOSPHERE & PASS FROM BETWEEN THE GROOVES OF THE COOLING
COIL PIPES THEREBY CARRY HEAT AND PUSH TO THE ATMOSPHERE

• COMPRESSOR IS A DRIVING FORCE OF THE REFRIGERANT. WE CAN USE ROTATING,
SCROLL,SCREW,CENTRIFUGAL,RECIPROCATING TYPE.
• THE REFRIGERANT FROM THE COMPRESSOR IS AT HIGH TEMPERATURE AND HIGH
PRESSURE IT BECOMES SUPERHEATED VAPOUR AND THEN IT PASSES THROUGH THE
CONDENSER AND WHEN IT LEAVES IT BECOMES HIGH PRESSURE MEDIUM
TEMPERATURE LIQUID
• AFTER BEING COLLECTED IN THE TUBES OF THE CONDENSER IT PASSES THROUGH THE
FILTER DRIER TO REMOVE MOISTURE FROM REFRIGERANT AND THEN PASSES
THROUGH THE EXPANSION VALVE AND BECOMES LOW PRESSURE LOW TEMP LIQUID-
VAPOUR MIXTURE AND GIVES COOLING EFFECT TO THE TUBES OF WATER IN
EVAPORATOR AND THROUGH SECONDARY PUMP THIS CHILLED WATER IS CIRCULATED
TO THE AHU,FCU THEN IT CHILLS THE SPACE AND GO BACK TO PRIMARY PUMP THEN
CYCLE REPEATS.
• REFRIGERANT NEVER LEAVES THE SYSTEM IT IS A CLOSED SYSTEM
• PRIMARY PUMP SUCKS CHILLED WATER FROM AHU, FAHU ETC AND GIVE BACK TO
CHILLER EVAPORATOR.
• SECONDARY PUMP SUCKS CHILLED WATER FROM CHILLER EVAPORATOR AND GIVES TO
AHU, FCU ETC.
• BYPASS PIPE IS CONNECTED FROM SUCTION TO SUCTION OF BOTH PRIMARY AND
SECONDARY.

WATER COOLED CHILLER
• Water cooled chiller system :
IT CONSISTS OF THREE CYCLES:
1. REFRIGERNT CYCLE: IT CONSISTS THE COMPONENTS LIKE CONDENSER,
COMPRESSOR,EVAPORATOR,FILTER DRIER AND EXPANSION VALVE.
2. CHILLED WATER CYCLE : IT CONSISTS THE COMPONENTS LIKE AIR HANDLING
UNIT AND WATER PUMP (PRIMARY AND SECONDARY PUMP).
3. CONDENSER WATER CYCLE: THE COMPONENTS NEEDED ARE COOLING TOWER
AND CONDENSER WATER PUMP.
A. REFRIGERANT CYCLE  IN THE REFRIGERANT CYCLE THE REFRIGERANT IS
COMPRESSED TO HIGH PRESSURE AND TEMPERATURE COMPRESSOR (SCREW
OR SCROLL OR RECIPROCATING) AND THEN PASSED TO CONDENSING UNIT
 THE REFRIGERANT IS FLOODED IN SHELL AND WATER IS FLOWING
THROUGH TUBES OF SHELL AND TUBE TYPE CONDENSOR

IN CONDENSOR THE HEAT WILL BE ELIMINATED TO THE CONDENSOR WATER
FROM COOLING TOWER BY THE PROCESS CALLED CONDUCTION AND CONVECTION
THEN IT CONVERTS TO HIGH PRESSURE LIQUID AND THEN IT PASSES THROUGH
FILTER DRIER TO REMOVE MOISTURE FROM THE REFRIGERANT AND ASSURE THAT
THE REFRIGERANT IS CLEAN & DRY BEFORE ENTERING THE EXPANSION VALVE.
WHEN IT PASSES THROUGH EXPANSION VALVE IT PRESSURE GETS REDUCED AND
ALSO EXPANSION VALVE CONTROLS THE FLOW OF THE REFRIGERANT GOING INTO
THE EVAPORATOR AND WHEN ITS PRESSURE DROPS THEN ITS TEMPERATURE ALSO
DROPS WHICH IS LOWER THAN CHILLED WATER TEMPERATURE.
THE REFRIGERANT AT LOW PRESSURE LIQUID FORM COMING OUT OF THE
EXPANSION VALVE WILL FLOW INTO THE EVAPORATOR AT LOW TEMPERATURE
IN THE EVAPORATOR THE HEAT FROM THE CHILLED WATER WHICH COMES OF
THE AHU WILL BE ABSORBED BY THE REFRIGERANT DUE TO TEMPERATURE
DIFFERENCE BETWEEN THEM & WHEN THE REFRIGERANT LIQUID WITH LOW
PRESSURE AND TEMPERATURE ABSORBS HEAT OF THE CHILLED WATER
CONVERTS TO LOW PRESSURE GAS

THE EVAPORATOR USED HERE IS SHELL AND TUBE TYPE WHERE REFRIGERANT IS
FLODDED IN THE SHELL AND WATER IS FLOWING INTO THE TUBES & IT IS
INSULATED WITH INSULATION TO REDUCE HEAT TRANSFER & CYCLE CONTINUES.
2. CHILLED WATER CYCLE:
AHU: AIR HANDLING IS THE PLACE WHERE THE COLD AIR IS PRODUCED BEFORE IT
IS DISTRIBUTED.
PARTS OF AHU: FAN, FAN MOTOR, COOLING COIL, FILTER(HEPA), CASING
 CHILLED WATER WHICH IS PRODUCED IN THE EVAPORATOR IS FLOWS IN AHU
THROUGH A PIPE LINE CALLED AS CHILLED WATER SUPPLY THROUGH THE
SECONDARY PUMP AND THEN IT RETURN BACK TO THE EVAPORATOR OF THE
CHILLER THROUGH PRIMARY PUMP TO COOL THE RETURN HOT CHILLED WATER
WHERE HEAT EXCHANGE TAKES PLACE AND KNOWN AS CHILLED WATER RETURN.
PRIMARY AND SECONDARY PUMPS ARE CENTRIFUGAL PUMPS.
3. CONDENSER WATER CYCLE: COOLING TOWER=COOLING TOWER SERVES TO
COOL DOWN THE TEMPERATURE OF THE CONDENSER WATER THAT WILL BE
USEFUL TO ABSORB THE HEAT IN THE REFRIGERANT SET IN THE CONDENSER.

• COOLING TOWER ELIMINATES THE HEAT OF THE REFRIGERANT IN SHELL AND
TUBE CONDENSER.THE CONDENSER WATER AFTER COOLING IN THE COOLING
TOWER IS SUPPLID TO THE TUBES OF THE SHELL AND TUBE CONDENSER WHERE
IT COOLS THE REFRIGRANT WHICH IS IN THE SHELL SO THAT IT CONDENSES
DOWN TO LIQUID AND SUPPLID TO CHILLER BY CONDENSER WATER PUMP AND
CYCLE CONTINUES. REFRIGERANT NEVER LEAVES THE CYCLE, IT IS CLOSED CYCLE.

GPM CALCULATION
• HOW TO CALCULATE GPM (gallon per minute) IT IS A FLOW OF WATER REQUIRED
IN A PIPE and FPS (FOOT PER SECOND) IS VELOCITY FOR FCU,AHU etc.
GPM=TR*12000/500*DELTA (T)
OR
GPM=TR*24/DELTA(T) OR GPM=TR*2.4 { DELTA (T) = SUPPLY-RETURN}
GPM=TR*3 FOR WATER COOLED CHILLER (OPEN PIPING)
{DELTA (T)=55-45=10}
TR=TONNAGE OF FCU OR AHU
DELTA (T)=CHILLED WATER RETURN(CHWR) FROM AHU OR FCU - CHILLED
WATER SUPPLY(CHWS) TO AHU OR FCU

DESIGN OF COOLING TOWER
• DESIGNING OF COOLING TOWER:
ASSUME FLOW (m)=30 GPM
WE KNOW, HOT WATER SUPPLY (HWS) = 95F (35C)
HOT WATER RETURN (HWR) = 85F (30C)
COOLING CAPACITY (HEAT REJECTION) Q=McpDELTAT
Q= DISCHARGE or FLOW RATE IN
GPM or BTU/min
m= MASS FLOW RATE IN LB/min
Cp= specific heat of water at const pressure =1 BTU/LB DEGREE F
delta T = TEMPERATURE DIFFERENCE = T2-T1 IN F (95-85)=10F
A cooling tower is a heat rejection device, which extracts waste heat to the atmosphere
though the cooling of a water stream to a lower temperature. ...
Common applications for cooling towers are providing cooled water for air-
conditioning, manufacturing and electric power generation.

1. IN FPS=FOOT POUND SECOND ( US IMPERIAL SYSTEM)
Q=M*Cp*deltaT
M=30 GPM  GPM TO POUNDS(LB)= 30 X 8.33 = 250 LB/MIN.
M=250 LB/MIN, Cp=1BTU/LB F, delta T= 95-85=10F
Q=250 X 1 X 10 = 2500 BTU/MIN.
2. IN MKS
M=30 GPM  GPM TO KG 0.003785 X 1000 or liters
M=30 X 0.003785 X 1000 = 113.4 KG/MIN
Cp= 1 kcal/kg c
Delta T =35-30=5 celcius
Q=113.4 X 1 X 5 = 567 KCAL/min

3. IN SI (STANDARD INTERNATIONAL) SYSTEM :
M=113.4 kg/min
Cp= 4.187 kJ/kg k
Delta T= (308 k – 303 k)= 5 k
Q=113.4 X 4.187 X 5 =2374 KJ/MIN.
TYPES OF COOLING TOWERS:
1. NATURAL DRAFT TOWER
2. MECHANICAL DRAFT TOWER
a. INDUCED DRAFT TOWER
b. FORCED DRAFT TOWER
c. CENTER or COUNTER FLOW
d. CROSS FLOW

MAKE UP WATER TANK CAPACITY DESIGNING OF COOLING
TOWER
• THERE ARE WATER LOSSES DURING COOLING TOWER PROCESS SO INORDER TO
HAVE CONTINUOUS WATER COOLING OF WATER THE LOSS WATER SHOULD BE
ADDED INTO COOLING TOWER.
• THERE ARE THREE LOSSES
1. EVAPORATIVE LOSS: THIS LOSS IS DUE TO EVAPORATION OF WATER FROM
COOLING TOWER IN PEAK SUMMER CONDITIONS AS IT IS EXPOSED TO OPEN
ATMOSPHERE.THE WATER OF THE COOLING TOWER GETS EVAPORAED AND
BECOMES LESS
2. DRIFT LOSS: UNTREATED WATER DROPLETS ARE CARRIED AWAY BY THE
COOLING TOWER OR BLOWN OFF DUE TO AIR (USUALLY IT IS NEGLIGIBLE DUE TO
ADVANCEMENT OF TECHNOLOGY)= 0 VALUE. Because we fit orifice
(0.0006-0.0012 GPM/TR)
3. BLEED OFF LOSS: WATER INTENSIONALLY DISCHAGES OUT OF THE COOLING
TOWER INORDER TO MAINTAIN WATER QUALITY IS CALLED BLEED OFF.

MAKE UP TANK SIZING
1.EVAPORATIVE LOSS: VOLUME OF WATER DECRESE IN COOLING TOWER IS TAKEN IN m3/hr.
WATER EVAPORATIVE LOSS=0.00085 X 1.8 X Wcr x delta T (m3/hr)
Wcr= water circulative rate in M3/HR & it is taken with respect to gpm of mass flow rate
1 gpm = 0.003785 m3/min
delta T IN C(CENTIGRADE)
Wcr=0.00378 X 30gpm X 60 FOR HOUR = 6.813 m3/hr
EVAPORATIVE LOSS = 0.00085 X 1.8 X 6.813 X 5 celcius =0.052 m3/hr.
2. DRIFT LOSS:
0 m3/hr because we get fit orifice
3. BLEED OFF: BLEED OFF = EVAPORATIVE LOSS/COC-1
COC= CYCLE OF CONCENTRATION = 2 TO 3.
0.052/3-1=0.052/2=0.026 m3/hr
ADD 1 2 & 3 = 0.052 +0+ 0.026= 0.078 m3/hr
0.078 X 1000= 78 lit/hr x 12=936 lit/day  sump capacity for 3 days= 936*3=2808 lit/day
is sump capacity  2808 lits or 2.8 m3 =2.8 x 1 x 1 m3 (l x b x h)

PUMP HEAD CALCULATION
• CHILLED WATER PUMP HEAD CALCULATION IS DONE INORDER TO IDENTIFY THE
HORSE POWER (HP) HOW MUCH HORSE POWER IS REQUIRED TO SUPPLY THE
CHILLED WATER TO THE SYSTEM TILL THE END.
1. PRIMARY PUMP HEAD CALCULATION same (ahu to chiller)
2. SECONDARY PUMP HEAD CALCULATION same (chiller to ahu)

• WE DO THE CALCULATION FROM PUMP DISCHARGE POINT TO AHU OR FCU
RECEIVING COIL END AND THE BRINGING BACK THE WATER TO CHILLER.
2. SECONDARY PUMP HEAD:

• WE HAVE TO CALCULATE THESE THREE PARAMETERS FOR HEAD LOSS OF BOTH
PRIMARY PUMP AND SECONDARY PUMP
1. CHILLED WATER PIPE PRESSURE DROP.
2. PIPE FITTINGS PRESSURE DROP.
3. EQUIPMENT PRESSURE DROP.
FOR PUMP HEAD CALCULATION WE NEED TO CALCULATE HORSE POWER OF THE
PUMP DEPENDING UPON THE FLOW RATE GPM AND HEAD LOSS IN FEET
FORMULA : HEAD LOSS IN FORMULA IS NOTHING BUT PUMP HEAD

• FOR 4”(INCH) PIPE; HEAD LOSSES FOR PIPE LENGTH, ELBOW TEE, DRV IS IN FEET.

• PIPE FITTING LOSSES ( FROM ASHRAE BOOK TABLE 11)

• EQUIPMENT PRESSURE DROP: THE PRESSURE DROP DUE TO NUMBER OR
FITTINGS IN BETWEEN THE SUCTION AND DISCHAGE PIPES OF PUMP AND
FCU,AHU IS CALLED EQUIPMENT PRESSURE DROP.

192 2 FC
90 ELBOW
CHK VALVE
DRV
1
1
1
1
20
10
40
120
190 192 2.14 4.10
576 5 90 ELBOW
TEE
1
2
16
30
76 81 2.19 1.77
576 57.4 90 ELBOW
TEE
2
3
16
30
122 179 2.19 3.92
192 2.82 FC
90 ELBOW
MBV
GLOBE VALVE
1
1
1
1
20
10
4.5
120
154.5 157 2.14 3.35
192 2.82 FC
90 ELBOW
DRV
1
1
1
20
10
120
150 153 2.14 3.27
576 59.6 90 ELBOW
TEE
2
3
16
30
122 182 2.19 3.98
576 32 90 ELBOW
TEE
2
1
16
30
62 94 2.19 2.05

• ENTER PUMP HEAD VALUE FEETS IN HP FORMULA
 HP=GPM*HEADLOSS*SPECIFIC GRAVITY OF WATER/3960* % EFFICIENCY
 HP=3648*38.28*1/ 3960*0.70
 HP= 50.3 HP==51HP (PUMP)

CHILLED WATER PIPE SIZING
• PIPE IS CONDUIT THROUGH WHICH ANY LIQUID OR GASSES PASS
IN AIR CONDITIONING SYSTEM THE PIPE WHICH CONNECTS,COMPRESSOR,
CONDENSOR, EXPANSION VALVE AND EVAPORATOR THROUGH WHICH
REFRIGERANT FLOW IS CALLED PRIMARY PIPING Or REFRIGERANT PIPING.
IT IS MADE WITH SOFT COPPER UPTO 2 TR AND CAPACITY HARD COPPER ABOVE
2TR.
IN CHILLERS THE PIPING WHICH JOINS CHILLERS, PUMPS, FCUS, AHUS through
WHICH CHILLED WATER IS SUPPLIED IS CALLED AS CHILLED WATER PIPING OR
SECONDARY PIPING.
IN WATER COOLED CHILLERS THE PIPING EHICH JOINS COOLING TOWER,
CONDENSER PUMP AND SHEEL AND TUBE OF THE CONDENSER IS CALLED AS HOT
WATER PIPING OR OPEN PIPING.
MILD STEEL AND LOW CARBON STEEL MATERIAL PIPES FOR CHILLED WATER
PIPING IS USED (SCHEDULE-40)

• CHILLED WATER PIPE SIZING CAN BE DONE BY
1. MANUAL CALCULATION
2. MCQUARY PIPE SIZER SOFTWARE
 BY MANUAL CALCULATION FOR EXAMPLE:
IF WE HAVE AHU OF 24TR THEN
GPM = TR*2.4= 24*2.4 = 57.6 GPM FROM CHART @ 57.6 LIES IN BETWEEN 40-
70GPM SO PIPE SIZE IS 2 ½” INCHES IMPERIAL AND 65 MM IN SI.

Pipe Size Pipe Size
Min Max (inch) Min Max (mm)
0 2.8 ½" 0.0 0.2 15
2.9 5 ¾" 0.2 0.3 20
5.1 8 1" 0.3 0.5 25
8.1 14 1¼" 0.5 0.9 32
14.1 20 1½" 0.9 1.3 40
20.1 40 2" 1.3 2.5 50
40.1 70 2½" 2.5 4.4 65
70.1 120 3" 4.4 7.6 80
120.1 250 4" 7.6 15.8 100
250.1 450 5" 15.8 28.4 125
450.1 700 6" 28.4 44.1 150
700.1 1300 8" 44.1 81.9 200
1300.1 2500 10" 81.9 157.5 250
2500.1 4000 12" 157.5 252.0 300
4000.1 5000 14" 252.0 315.0 350
5000.1 7000 16" 315.0 441.0 400
7000.1 9000 18" 441.0 567.0 450
9000.1 12000 20" 567.0 756.0 500
CHILLED WATER PIPE SIZING AS PER ASHRAE STANDARD
If Φ < 2" - 4 FPS AND IF Φ > 2" - 4 FT/ 100 FT OF PIPE LENGTH
IMPERIAL SI
Remark
Water flow rate in GPM Water flow rate in LPS

CHILLED WATER PIPE SIZING INSULATION
• CHILLED WATER PIPE INSULATION FOR HARD RUN COPPER PIPE NITRILE RUBBER
IS USED FOR INSULATION
• THERMAL INSULATION IS DONE FOR TWO PURPOSE TO SAVE HEAT TRANSFER
AND CONDENSATION.

INSULATION MATERIALS(FOR PIPES AND DUCT)
• WOOL
• FIBRE GLASS
• CORK RUBBER
• NITRILE RUBBER
• THERMOCOL
• PUF(POLY URETHANE foam)
• ASBESTORS CEMENT
• COALTAR
• BIRDS NET
ADDITIONAL MATERIALS
ALUMINIUM SHEET OF 26 TO 18 GAUGE
ALUMINIUM TAPE ACT AS A VAPOUR BARRIER

INSULATION THICKNESS AND PIPE FITTINGS
• INSULATION THICKNESS = NORMALLY 2”-3” FOR ROOF PIPING BECAUSE EXPOSED TO
SUNLIGHT AND 1”-2” FOR RISER AND FLOOR PIPING.
• PIPE FITTINGS:
ELBOWS(90 DEGEE, 45 DEGREE etc)
TEES
REDUCERS TEES AND ELBOWS
COUPLINGS
UNIONS
CROSS ELBOW
• MACHINE FITTINGS:
VALVES ( BUTTERFLY VALVE, GATE VALVE OR ISOLATING VALVE, GLOBE VALVE(REGULATING
VALVE OR DOUBLE REGULATING VALVE), 2WAY AND 3WAY VALVE, STRAINER.
 BUTTERFLY VALVE -- IT IS MOSTLY USED DUE TO LESS COST AND MAINTENANCE AND CAN
OPERATED AUTOMATICALLY
 CHECK VALVE -- SUPPLY ONLY IN ONE DIRECTION(NON-RETURN VALVE).
 STRAINER -- CLEANING THE DEBRIS AND IS CONNECTED BEFORE THE VALVES .

VALVES USED IN CHILLED WATER SYSTEM

VALVES IN HVAC
• A gate valve is the most common type of valve that used in any process plant. It
is a linear motion valve used to start or stop fluid flow. In service, these
valves are either in fully open or fully closed position. When the gate valve is
fully open, the disk of a gate valve is completely removed from the flow.
• A globe valve is a linear motion valve used to stop, start, and regulate the fluid
flow. The globe valve disk can be removed entirely from the flow path, or it can
completely close the flow path. During opening and closing of globe valve, disc
moves perpendicularly to the seat.

HVAC ZAMIL SOFTWARE:
• ZAMIL EQUIPMENT SELECTION SOFTWARE:

HVAC DUCTING
• DUCT : A DUCT IS A CONDUIT THROUGH WHICH AIR FLOWS
THE FUNCTION OF THE DUCT IS TO TRANSMIT THE AIR FROM
THE AIR DISTRIBUTION DEVICES (AHU,FCU,CSAHU) TO THE SPACE
WHICH IS TO BE AIR CONDITIONED.
------> TO FIT A DUCT FITTING IN A PRACTICAL MANNER THE DUCTING SYSTEM
MUST BE DESIGNED WITHIN THE AVALIABLE PLENUM SPACE HEIGHT, FRICTION
LOSS, VELOCITY, SOUND LEVEL, LEAKAGE LOSSES etc.
HT IN INCHES & FEETS, FRICTION LOSS IN INCHES OF WATER GUAGE, Velocity in
FPM (FOOT PER MINUTE), SOUND LEVEL IN HZ (HERTZ) FREQUENCY.
NOTE: WHEN VELOCITY(SPEED) OF AIR IN THE DUCT INCREASES ITS AREA WILL
BECOME LESS FOR VOLUME OF AIR PASSING THROUGH IT SO FRICTION
LOSSES(FITTINGS) INCREASES.

SHAPES OF THE DUCT
• SHAPES OF THE DUCT:
1. RECTANGULAR DUCT (MOST COMMONLY USED DUE TO EASY MANUFACTURE
AND SPACING IN PLENUM HEIGHT) MORE LOSSES
2. SQUARE DUCT (PROBLEM OF HEIGHT IN PLENUM SPACE)
3. CIRCULAR DUCT (COSTLIER, HIGH IN COST TO MANUFACTURE AND TIME
CONSUMING) LESS LOSSES DUE TO NO SHARP ENDS
4. SPIRAL DUCT, OVAL DUCT( FOR FASCINATING LOOK)
5. FLEXIBLE DUCT (THESE ARE CONNECTED IN AIR TERMINAL DEVICES LIKE
DIFFUSER & PLENUM BOX AND CAN BE EASILY ADJUSTABLE DUE TO ITS
FLEXIBILITY)
PLENUM BOX: PLENUM BOX ACTS LIKE A HUB(JUNCTION) IN WHICH WE CAN
INSTALL VARIOUS FLAXIBLE DUCT CONNECTIONS NEAR MACHINES AND DIFFUSER

HVAC DUCTING
• RECTANGULAR DUCT

PASSAGE OF AIR THROUGH DUCTS REPRESENTATION
• SUPPLY AIR DUCT (SAD):USED TO SUPPLY CHILLED AIR.
• RETURN AIR DUCT (RAD):IT CONSISTS OF USED CHILLED AIR AFTER COOLING
CALLED AS RETURN AIR.
• FRESH AIR DUCT (FAD):USED TO SUPPLY FRESH AIR IN VENTILATION AND AHU.
• EXHAUST AIR DUCT (EAD) or CLEAN ROOM DUCT : USED TO EXHAUST THE AIR
FROM KITCHEN, CAR PARKINGS, TOILET etc.

• ACCORDING TO THE VELOCITY THE DUCTS ARE OF
1. LOW AND MEDIUM VELOCITY DUCT RUN: IF THE VELOCITY IS LESS THAN = 600 MPM(METER PER MINUTE) OR
IF VELOCITY IS IN BETWEEN 900-2500 FPM(FEET PER MINTE IS UNDER THIS CRITERIA
2. HIGH VELOCITY DUCT RUN: IF VELOCITY IS GREATER THAN MPM AND 2500-4000 FPM THEN IT COMES UNDER
THIS CRITERIA (1MPS=197 FPM).
BASICALLY WE USE LOW AND MEDIUM DUCT RUNS 900-2500 FPM.
• ACCORDING TO THE PRESSURE INSIDE THE DUCT:
1) LOW PRESSURE DUCT WORK: USED FOR RESIDENTIAL AND COMMERCIAL BUILDINGS
VELOCITY LESS THAN < 600 MPM AND STATIC PRESSURE <= 5CM OF H20 GUAGE
2) MEDIUM PRESSURE DUCT WORK: USED FOR CLEAN ROOM APPLICATIONS
VELOCITY < = 600 MPM AND STATIC PRESSURE 5-15 CM OF H20 GUAGE (ashrae)
3) HIGH PRESSURE DUCT WORK: if it is greater than 600 > mpm and static pressure is > 15cm of water guageUSED
FOR INDUSTRIAL BUILDINGS
LEAKAGE TEST IS DONE BY
 SMOKE TEST OR PRESSURE TEST
LIGHT TEST
GOOSE NECK TO PROTECT THE OUTDOOR UNITS, EXHAUST AIR AND FRESH AIR FAS FROM RAINFALL , SUNHEAT
etc.

DUCT DESIGN SMACNA
MATERIAL OF THE DUCT
GALVANIZED IRON DUCT (GI) {80% DUCTING MATERIAL}-low cost and easy fabrication
STAINLESS STEEL DUCT :marine applications (no sadaa formation)
ALUMINIUM DUCT :used due to light weight and resistance to moisture application
BLACK STEEL DUCT OR CARBON STEEL : used for exhaust ducting (ventilation)
PUF DUCT: DO not require insulation as it is only an insulation material and it is costly
and catches fire easily and generate toxic gases
COPPER : IN HIGH HUMID AREAS AND VERY HIGH COST
FIBER GLASS DUCT: low velocity duct and it generates vibration
PIR (POLYISOSINORATE DUCTING) DUBAI STANDARD
INSULATION OF DUCT: MATERIAL USED IS FIBRE GLASS OR GLASS WOOL
THICKNESS 40MM FOR INTERNAL DUCTING 24 KG/M3 DENSITY
THICKNESS 50MM FOR EXPOSED DUCTING 48 KG/M3 DENSITY
ACOUSTIC LINING:FIBRE GLASS 35MM THICKNESS INSIDE DUCT FROM THE MACHINE
UPTO 10 FEET OR 3 METER

DUCT SIZING (GUAGE SELECTION)-SMACNA
• SELECTION OF A GUAGE OF DUCT MATERIAL
• DUCT JOINING METHOD IS DONE BY “C” CLIP OR “S” CLIP
• DUCT FLANGE BY RUBBER GASKET
LOW PRESSURE MEDIUM
PRESSURE
HIGH PRESSURE GI METAL AND
SHEET
GUAGE
UP TO 30 CM - - 26 0.5MM
30-75 CM UP TO 45 - 24 0.6MM
75-135 45-120 UP TO 120 22 0.7MM
135-210 120-180 120-180 20 0.9MM
ABOV3 120 ABOVE 180 180-240 18 1.2MM
- - ABOVE 240 16

DUCTING (GI SHEET DUCTING GUAGE)-SMACNA
DUCT SIZE (INCHES) GUAGE THICKNESS(mm) WEIGHT(KG/FT2)
UPTO 12 INCHES 26 0.551 0.4119
13-24 24 0.701 0.5255
25-30 22 0.851 0.6391
31-48 20 1.006 0.7528
49-60 18 1.30 0.98
61-90 16 1.613 1.207

DUCT DESIGNING METHODS
• BY CONTINUITY EQUATION ( Q=A*V)
• BY EQUAL FRICTION METHOD (MOST IMPORTANT METHOD)
• BY VELOCITY REDUCTION METHOD
• BY STATIC REGAIN METHOD
• BY CONSTANT VELOCITY METHOD
• BY MCQUARY DUCT SIZER (SOFTWARE)
METHOD RESIDENTIAL
AREA COMMERCIAL
AREA
PUBLIC FACILITIES
EQUAL FRICTION 0.05-0.07 INCHES
OF H20 GUAGE
0.08-0.1 INCHES OF
H20 GUAGE
0.1-0.15 INCHES OF
H20 GUAGE
VELOCITY 700-900 FPM 1000-1300 FPM 1200-1800 FPM

EQUATION OF CONTINUITY
DESIGN DUCT IF FLOW THROUGH IS 3000 CFM AND SPEED OR VELOCITY IS 700
FPM, CALCULATE AREA AND ASPECT RATIO.
GIVEN DESIGN DETAILS: Q = 3000 CFM , V = 700 FPM
SINCE WE KNOW Q = A X V
A= Q/V = (3000 FT3/MIN)/ 700 FT/MIN = 4.286 FT2
A = 4.286 FT2 ( SINCE 1 SQFT = 0.093 M2 )
A = 4.286 X 0.093 = 0.3986 M2
A = 0.3986 X 39 X 39 ( SINCE 1 INCH = 39.4 METER)
A = 606.27 in2.
Assume rcp height = 1meter
duct ht= ½ x rcp ht = ½ x 1meter = ½ or 0.5 meter
W x H = AREA
w x 0.5 meter = 0.3986
W= 0.7972 METER

• ASPECT RATIO : ASPECT RATIO IS THE RATIO OF LONGEST SIDE OF THE DUCT TO
THE SHORTEST SIDE OF THE DUCT .
ITS VALUE LIES IN BETWEEN 1- 4 ( safe design )
A.R= LONGEST SIDE / SHORTEST SIDE = 0.7952/0.5= 1.3 (LIES IN 1-4 ) DESIGN IS
SAFE.

BY CONTINUITY EQUATION
• EQUATION OF CONTINUITY Q=A X V
WHERE Q = FLOW RATE (OR) DISCHARGE (OR) VOLUME OF AIR IN CFM (OR)
m3/sec.
A = CROSS-SECTIONAL AREA OF DUCT (SQFT)
V = VELOCITY OF AIR (FPM,M/SEC).
1 M/S = 197 FPM
• DESIGN THE DUCT AREA AND ASPECT RATIO IF VOLUME OF AIR IS 3000 CFM
AND VELOCITY OR SPEED OF AIR IS 1700 FPM.
GIVEN DESIGN DETAILS: Q=3000 CFM AND V=1700 FPM
SINCE WE KNOW Q=A X V
A= Q/V=3000/1700=1.769 FT2
=1.769 X 0.093 = 0.1642 m2.
RCP HEIGHT = 14ft X 0.3048 = 4.2672 meter.
duct height = ½ x rcp height = ½ x 4.2672 = 2.1336 meter.

Duct design by continuity equation
• A = W X H = 0.1642 M2
0.1642= W X 2.1336
W = 0.1642/2.1336 = 0.077 M
W=0.077 METER
ASPECT RATIO : ASPECT RATIO IS RATIO OF LONGEST SIDE OF THE DUCT TO THE
SHORTEST SIDE
HERE,WIDTH / HEIGHT = 0.077/2.1336 = 0.361 (DESIGN IS not SAFE)
IN SUCH CASE LARGEST VALUE BY SMALLEST VALUE
A.R= 2.1336/0.77= 3 (SAFE DESIGN)
NOTE: WHEN WIDTH IS LESS THAN HEIGHT THEN DECREASE THE HEIGHT AND
INCREASE THE WIDTH TO REDUCE THE ASPECT RATIO FOR SAFE DESIGN.

CONTINUITY EQUATION
• IF CFM=2000 V=1200 FPM LENGTH = 12 METER FIND ALL SHAPES OF DUCT
SIZING AND LATERAL SURFACE AREA OF DUCTS
GIVEN DESIGN DETAILS: Q=2000 CFM
V = 1200 FPM AND AREA = ?
SINCE WE KNOW Q=A X V A = Q/V = 2000/1200=1.667 FT2
A=1.667 FT2
1. FOR CIRCULAR OR ROUND DUCT AREA (A)= 𝑃𝐼𝐸/4 X d2 1.667 = 3.14/4 x d2
d2= 2.1225  d= 1.456 ft
d= 1.456 x 12 =17.5” inch
d= 444 mm
2. For square duct AREA (A) = side square (s2)
s= under root (A)= UNDER ROOT 1.667= 1.288 FT
S = 15.49” == 15.5”

CONTINUITY EQUATION
• FOR RECTANGULAR DUCT
A = WIDTH X HEIGTH
1.667= W X H
ASSUME RCP HT = 1 METER
THEREFORE HT OF THE DUCT = ½ X RCP HT = 0.5METER X 39 = 19.5” OR
1.625’.
1.667 = W X 1.625 W= 1.025FT X 12 (FOR INCH)= 12.31”
A.R= WIDTH OF THE DUCT/ HEIGHT OF THE DUCT= 19.5/12.31= 1.584(SAFE)
NOW, LATERAL SURFACE AREA :
1. CIRCULAR DUCT LATERAL SURFACE AREA: 2 X 3.14 X r x L
 2 X 3.14 X 17.5/2 X 12m X (39) for inch conversion from meter = 25716.6 in2
25716.6/144= 178.58 FT2 178.58/10.76=16.597 m2.

2. LATERAL SURFACE AREA OF SQUARE :
L.S.A OF SQUARE = 4SL
= 4 X 15.5 X 12 X 39 = 29016 in2
=29016 /144=201.5 ft2
=201.5/10.76=18.72 m2.
3.LATERAL SURFACE AREA OF RECTANGLE:
L.S.A OF RECTANGLE = 2(W+H) X L
= 2 X (19.5+12.31) X 12METER X 39(FOR INCH CONVERSION)
=29774.16 in2
=29774.16 in2/144 for ft2=2066.765 ft2
=2066.765/10.76 for m2=19.21 m2

EQUAL FRICTION METHOD/CONSTANT PRESSURE METHOD
• IN EQUAL FRICTION METHOD SAME PRESSURE LOSS OR FRICTIONAL LOSSES ARE
CONSIDERED THROUGHOUT THE DUCT LENGHTH AND SIZING OF THE DUCT IS
DONE ACCORDINGLY
• MOST UNIVERSALLY USED FLEXIBLE METHOD FOR DESIGNING THE DUCTS
• IF HIGH FRICTION LOSS IS CONSIDERED THERE WILL BE SHORT RUN DUCTS
WHICH RESULTS IN VIBRATIONS IN THE DUCT DUE TO HIGH FAN CAPACITY SO
DAMPERS MUST BE USED TO REGULATE THE FLOW OF AIR THROUGH DUCT.
• FOR LOW AND MEDIUM VELOCITY DUCTS 0.08”– 0.15” INCHES OF WATER
GUAGE PER 100’ OF DUCT LENGTH/RUN
AND VELOCITY RANGES FROM 900-2500 FPM.
• FOR HIGH VELOCITY DUCTS SYSTEM 0.16” – 0.6” INCHES OF WATER GUAGEPER
100’ OF DUCT LENGTH/RUN
AND VELOCITY RANGES FROM 2600 – 4000 FPM.

• DESIGN THE FOLLOWING THROUGH EQUAL FRICTION DUCT DESIGNING METHOD

• EQUAL FRICTION METHOD
DUCT
NAME
FLOW
CFM
VELOCITY
FPM
FRICTION
IN INCHES
OF WATER
GUAGE
OR HEAD
LOSS
DUCT SIZE
ROUND
DUCT
EQUIVALE
NT
RECTANG
ULAR
DUCT
ASPECT
RATIO
A 3200 1262 0.1 22” 29” X 14” 2.28
B 2200 1152 0.1 19” 25” X 12” 1.91
C 1500 1049 0.1 16” 23” X 10” 2
D 600 836 0.1 11.5” 14” X 8” 1.62
E 1000 950 0.1 14” 17” X 9.5” 1.89

DUCT DESIGN BY DUCT SIZING SOFTWARE
• WE HAVE TO JUST ASSUME THE HEIGHT BECAUSE RCP HT IS MAX 1.5 METER

DUCT DESIGN BY CONSTANT VELOCITY METHOD
S NO DUCT FLOW (CFM) VELOCITY (FPM) FRICTION IN
INCHES OF
WATER GUAGE
DUCT SIZE
CIRCULAR/RECT
ANGULAR
ASPECT RATIO
1 A 5000 1200
2 B 4000 1200
3 C 3000 1200
4 D 2000 1200
5 E 1000 1200
6 F 500 1200

STATIC REGAIN METHOD
• STATIC PRESSURE IS THE PRESSURE LOSS DUE TO THE DUCT FITTINGS SUCH AS
ELBOWS,TEES,Y-JOINTS ETC WHICH EXERTS BACK PRESSURE ON THE FAN IS
CALLED STATIC PRESSURE
STATIC REGAIN METHOD:
STATIC PRESSURE(SP) = R {(V1/4005)2 – (V2/4005)2}
WHERE R = RECOVERY FACTOR THAT DEPENDS UPON THE SHAPE OF FITTING THAT
CHANGES VELOCITY SPEED
R= 0.7 – 0.9
DESIGN BY STATIC REGAIN METHOD:

Static regain method
SECTION FLOW
RATE
(CFM)
VELOCITY
(FPM)
FRICTION(
FT/100 FT)
LENGTH
OF THE
DUCT (FT)
FRICTION
IN INCHES
OF H20
GUAGE
STATIC
PRESSURE
(IN OF
WG)
EQUIVALE
NT DUCT
DIA
RECTANG
ULAR
DUCT
SELECTED
OR NOT
A-B 8000 3000 0.511 50 FEET 0.255 NOT
CONSIDER
ED
22.1 16” X 26” SELECTED
B-C 6000 2900 0.559 40 FEET 0.223 0.025 19.8 16” X 21” XXXXX
B-C 6000 2500 0.383 40 FEET 0.153 0.154 21 16”X 23” B-C
C-D
C-D
4000
4000
2350
2100
0.419
0.315
30 FEET
30 FEET
0.12
0.09
0.03
0.10 18.7 16 X 18
XXXXXXX
C-D

REDUCER SIZING
• DUCT REDUCERS ARE OF TWO TYPES
1. CONCENTRIC REDUCER
2. ECCENTRIC REDUCER
FOR LOW VELOCITY DUCT SYSTEM WE TAKE RATIO AS 4:1 AND REDUCER LENGTH
AS 400-850 MM
FOR HIGH VELOCITY DUCT SYSTEM WE TAKE RATIO AS 7:1 AND REDUCER LENGTH
AS 850-1500 MM
NOTE: IF W1-W2>=100MM
THEN DO REDUCER CALCUL
-ATION AND IF W1-W2<100MM
THEN PUT A STRAIGHT DUCT.

DUCT DESIGN BY REDUCERS
• DESIGN THE DUCT BY REDUCING
FOR SECTION A-B:
W1=700 , W2=550
THEREFORE W1-W2=700-550=150MM
W1-W2>100MM SO REDUCER IS MUST
FOR LOW VELOCITY DUCT SYSTEM REDUCER LENGTH = DIFFERENCE X 4=150 X 4
L1=600 MM

DUCT DESIGN OF REDUCERS
FOR SECTION B-C:
W2=550, W3=450
W2-W3= 550-450=100MM
W2-W3>=100MM SO GO FOR REDUCER
FOR LOW VELOCITY DUCT SYSTEM REDUCER LENGTH=DIFFERENCE X 4
= 100 X 4 =400MM
FOR SECTION D-E:
W4=400 , W5=250
W4-W5=400-250=150MM
FOR LOW VELOCITY DUCT SYSTEM REDUCER LENGTH=DIFFERENCE X 4
= 150 X 4= 600MM

DUCT DESIGN BY REDUCERS
• AFTER DOING REDUCER CALCULATION AUTOCAD DRAWING
• FIRST DRAW WIDTH 1 OF DUCT THEN TAKE REDUCER LENGTH AS OFF SET AND
THEN DRAW WIDTH 2 THEN JOIN THE REDUCING LENGTH.

AIR TERMINAL DEVICES
TYPES OF AIR TERMINAL DEVICES:
• CEILING DIFFUSER
• JET DIFFUSER
• LINEAR SLOT DIFFUSER
• ROUND DIFFUSER
• GRILLS AND REGISTERS
• LOUVERS
• DISC VALVES
PLENUM BOX: PLENUM BOX ACTS LIKE A HUB IN WHICH MULTIPLE DUCTS CAN BE
CONNECTED IT IS CONNECTED TO DIFFUSERS.

• Air TERMINAL Devices: Air terminal devices control the direction, height, and
amount of airflow delivered to a space. These devices, are often called registers,
grilles or diffusers, louvers & can be fixed or adjustable.
• DIFFERENCE BETWEEN GRILLS AND REGISTERS
THE GRILLS WHICH IS FITTED WITH THE DAMPERS BEHIND IS CALLED AS REGISTER.
HENCE GRILLS ARE WITHOUT DAMPERS AND REGISTER ARE WITH DAMPER.
• DISC VALVE: Disc valves offer a simple yet effective means of providing supply
and extract air form the bad odors places like toilet, bathrooms etc.

• DAMPERS IN HVAC DUCTING
VCD (VOLUME CONTROL DAMPER): VOLUME CONTROL DAMPER USED TO
REGULATE THE FLOW OF AIR THROUGH THE DUCT.
1. MANUAL VCD
2. MOTORISED VCD

• VOLUME CONTROL DAMPERS ARE CONNECTED IN THE DUCTS TO REGULATE THE
FLOW OF AIR IN DUCTS OR IN BETWEEN THE TWO DUCTS.
• FLEXIBLE VOLUME CONTROL DAMPER ( VCD ) IS USED IN FLEXIBLE DUCTS
WHICH ARE ROUND IN SHAPE TO REGULATE THE FLOW THROUGH FLEXIBLE
DUCT. LOUVERS

Air terminal device sizing
• AIR TERMINAL DEVICE MANUAL SIZE CALCULATION:

AIR TERMINAL DEVICE SIZE CALCULATION
• EXAMPLE: ASSUME, AIR FLOW(Q) = 600 CFM
VELOCITY(V) = 200-500 FPM (FOOT PER MINUTE) (STANDARD FOR AIR
TERMINAL)
WE KNOW CONTINUTY EQUATION Q=A X V
A=Q/V = 600/500 = 1.2 FT2 (FREE AREA)
A= FREE AREA X 1.25(FREE SPACE)
A = 1.2 X 1.25 = 1.5 FT2
A=1.5 FT2 X144 = 216 INCH2
A=216 INCH2 = UNDER ROOT 216= 15” X 15”
A= 15” X 15”.

ROUND DUCT SIZES AND CFM FLOW THROUGH IT

HVAC-VENTILATION
• VENTILATION IS THE SUPPLY OF AIR MOTION IN A SPACE BY CIRCULATION OR BY
MOVING THE AIR THROUGH THE SPACE (or) VENTILATION IS SUPPLY OF OUTSIDE FRESH
AIR IN THE SPACE TO IMPROVE ITS INDOOR AIR QUALITY.
• VENTILATION AIR AS PER ASHRAE STANDARD IS THE AIR USED FOR PROVIDING
ACCEPTABLE INDOOR AIR QUALITY.
TWO TYPES OF VENTILATION:
1. NATURAL VENTILATION: NATURAL VENTILATION OCCURS WHEN AIR IN A SPACE IS
CHANGED WITH OUTDOOR AIR WITHOUT USE OF MECHANICAL EQUIPMENTS LIKE
FAN BUT THROUGH WINDOWS etc.
2. FORCED VENTILATION (OR) MECHANICAL VENTILATION: THE EXCHANGE OF INDOOR
BAD ODOUR AIR WITH THE OUTSIDE FREASH AIR THROUGH MECHANICAL EXHAUST
FANS, fan UNITS etc IS CALLED AS MECHANICAL VENTILATION.
WE HAVE TO DO VENTILATION FOR THE FOLLOWING AREAS LIKE
A. TOILET VENTILATION .
B. KITCHEN VENTILATION .
C. CAR PARKING VENTILATION.

HVAC TOILET EXHAUST SYSTEM (VENTILATION)

TOILET VENTILATION SYSTEM
PASSAGE AREA OF TOILET = 10800 mm X 2500 mm = 27000000 mm2 = 27 m2
PASSAGE AREA IN SQFT = 290 SQFT
SINCE WE KNOW 1 CFM /1SQFT
THERE FORE PASSAGE AREA CFM = 290 CFM
DISC VALVE SELECTION : SINCE DISC VALVE MAXIMUM CFM IS 100 CFM
SO 290/100= 2.9 == 3 DISC VALVES { DISC VALVE STANDARD SIZES
15-100 = 4”
100-180= 6”
180-240= 8”
> 240 = 10”}
EACH DISC CFM IS 97 CFM
NOW, TOILET AREA SINGLE = 2000 mm X 2000 mm = 4000000 mm2= 4 m2=43ft2
43 x 10 toilets = 430 sqft = 430 cfm for toilets
430 /10= 43 cfm disc valve in each valve

TOILET DISC VALVES = 4” FOR EACH TOILET ( 43 CFM IN EACH TOILET ).
NOW , TOTAL TOILET AREA CFM+TOTAL PASSAGE AREA CFM = 430+290=720 CFM.
THEREFORE, EXHAUST AIR CFM = 720 CFM. (FAN CAPACITY = 720 CFM)
FRESH AIR CFM IS 80% OF EXHAUST AIR CFM BY ASHRAE 720 X 0.8= 576 CFM
THEREFORE FRESH AIR CFM=576 CFM.
DISC VALVE IS AN AIR TERMINAL DEVICE FOR TOILET EXHAUST SYSTEMS.
450 T0 500 FPM SPEED/VELOCITY IS MAINTAINED IN AIR TERMINAL DEVICE.
METHOD 2:
CFM = (L X W X H) m3 X ACH/1.7 ( FOR m3) ASHRAE VENTILATION STANDARD
CFM = (L X W X H) ft3 X ACH/60 ( FOR ft3) (ACH = 6-8 for BATHROOMS)
CMH = 1.7 X CFM
CFM=(3X4X3)X8/1.7=169 CFM
CMH = 1.7 X 169 = 287 ==290 CMH
DOOR UNDERCUT SIZE:
BY CONTINUITY EQUATION Q=A X V

TOILET VENTILATION
Q= FLOW = 169, A = AREA =?, V = VELOCITY = (200-500)FPM (STANDARD)
Q = A X V
169=A X 500 width (0.75 – 1.25 meter const)
A = 0.33 ft2
A = 0.33 x 0.093 = 0.03 m2
AREA = WIDTH X DEPTH
0.03 = 0.8 ( CONST) X DEPTH
DEPTH (d) = 0.03/0.8 = 0.037 meter
d=37 mm, W=80 mm IS DOOR UNDERCUT SIZE.
METHOD 3: 10 TOILETS OF AREA 2M X 2M (FOR COMMERCIAL)
25/50 RESIDENTIAL (CONTINOUS AND NON-CONTINUOUS)
50/70 COMMERCIAL (CONTINOUS AND NON-CONTINUOUS).
10 TOILETS X 50 =500 CFM.

• AIR CHANGES PER HOUR (ACH) TABLE.

VENTILATION (FAN SELECTION)
• FAN SELECTION PROCESS
1. AIR FLOW (720 CFM)
2. STATIC PRESSURE
FOR 1 METER STRAIGHT DUCT = 0.004” OF H2O GUAGE
90 DEGREE ELBOW = 0.2” OF H2O GUAGE
TOTAL LENGTH OF DUCT = 14 METER(ASSUME) X 0.004 = 0.056 “ OF H2O GUAGE
ELBOW = 2 NUMBERS X 0.2 = 0.4” OF WATER GUAGE
TOTAL STATIC PRESSURE = 0.056 + 0.4 = 0.456” OF H20 GUAGE = 12mm of hg.

PERFORMANCE CURVES FOR FAN SELECTION
---------- = 50 HZ
= 60 HZ

KITCHEN VENTILATION SYSTEM
• IN KITCHEN VENTILATION SYSTEM WE HAVE TO DO THE DESIGNING OF KITCHEN
HOOD FOR THE EXHAUST OF HEAT SMOKE AND HUMIDITY.
KITCHEN HOOD TYPES
• SINGLE ISLAND HOOD
• DOUBLE ISLAND HOOD
• WALL MOUNTED CANOPY HOOD
• BACK SHELF OR PROXIMITY OR PASS OVER HOOD
FRESH AIR VENTILATION CAN BE SUPPLIED THROUGH DUCTLESS AND UNDUCTED
SYSTEM

KITCHEN VENTILATION SYSTEM
• DESIGN CRITERIA FOR KITCHEN VENTILATION
1. AN INTERNAL TEMPERATURE OF 28C-30C MAXIMUM
2. A MAXIMUM HUMIDITY LEVEL OF 70%
3. INTERNAL NOISE LEVEL SHOULD BE :
NOISE RATING (40 TO 50) & NOISE CRITERIA ( 40 TO 50) FOR KITCHENS
NOISE RATING (35 TO 45) & NOISE CRITERIA (35 TO 45) FOR RESTAURANTS,BARS AND
CAFETARIAS.
4. EXHAUST/MAKEUP AIR 1500-2200 FPM VELOCITY
5. HUMIDITY IS HARD TO CONTROL SO, VENTILATION SHOULD BE GOOD ENOUGH TO PROVIDE
MAXIMUM COMFORT.
6. FRESH AIR CFM IS TAKEN AS 80% OF THE EXHAUST AIR CFM.
7. BAFFLE FILTERS “V” TYPE ARE USED AS COMPARED TO MESH TYPE
8. STOVE TO HOOD HEIGHT IS 600-900 mm (2 feet to 3 feet)
9. BLACK STEEL DUCTS and STAINLESS STEEL DUCTS ARE USED FOR ALL KITCHEN DUCTS.

KITCHEN HOOD DESIGINING
• DESIGN THE HOOD SIZE, EXHAUST AIR CFM AND EXHAUST AIR DUCT SIZE, FRESH
AIR CFM AND FRESH AIR CFM DUCT SIZE FOR WALL TYPE & MEDIUM COOKING.
ASSUME STOVE SIZE : LENGTH (l)= 750 mm, WIDTH(w) = 500 mm, HEIGHT(h)=
800 mm
DESIGN OF HOOD :
HOOD SIZE
LENGTH OF HOOD (L)=l+0.8(h)
= 750 + (0.8 x 800)
= 1390 mm
= 1.39 meters
WIDTH OF HOOD(W)= w + 0.8(h)
= 500 + (0.8 x 800)
= 1140 mm
= 1.14 meter

Kitchen ventilation
convert meters to feet L=1.39 METERS X 3.28 FEET FACTOR =4.55 FEET
THEREFORE L = 4.55’
W = 1.14 X 3.28 = 3.75 FEET
W = 3.75’
NOW,
EXHAUST AIR CFM: EXHAUST:
CFM = P X 150 (FOR MEDIUM) Px100= (LOW)
P=2L+W PX150 = (MEDIUM)
P= 2 x 4.55 + 3.75 PX200 = (HIGH)
p= 12.85 P=2L+2W FOR ISLAND TYPE
CFM = 12.85 X 150 = 1928 CFM

KITCHEN VENTILATION
EXHAUST AIR DUCT SIZE:
BY CONTINUITY EQUATION Q = A x V
Q= FLOW IN CFM
V = VELOCITY IN FPM (1500 – 2200)
A = AREA IN ft2
A= Q/V= 1928/1800= 1.07 ft2
Area (A)= 1.07 x 0.093 = 0.09 m2
a) ROUND DUCT = 3.14/4 X d2 = 0.33m=330mm==350mm
b) Rectangular duct
AREA = W x H
0.09= W x 0.25
W= 0.36 = 360mm==400mm
c) Square duct = side2= UNDEROOT 0.09 = 0.3meter

KITCHEN VENTILATION
• FRESH AIR CFM
FRESH AIR CFM IS TAKEN AS 80% OF EXHAUST AIR CFM
FRESH AIR CFM (1928) x 0.8 = 1542 CFM
Q= A x V
1542 = A x 1800 ( v = 1500 -2200 for kitchen)
A = 0.85 ft2= 0.85 x 0.093 = 0.07 m2
FRESH AIR DUCT:
a) Round duct = area = 3.14/4 x d2 d= 0.3m=300mm
b) Rectangular duct = area = w x h  W = 300mm
c) Square duct = UNDEROOT AREA = 0.25==0.3m =300mm.

CAR-PARKING VENTILATION
• CAR PARKING IS OF TWO TYPES
1. DUCTLESS SYSTEM (IF HEIGHT OF THE SLAB IS LESS THAN 3 METERS FROM
FINISHED FLOOR LEVEL)
2. DUTED SYSTEM ( IF HEIGHT OF THE SLAB IS MORE THAN 3 METERS FROM FFL)
NATURAL VENTILATION(wall openings) AND MECHANICAL VENTILATION(FANS)
FRESH AIR CAN BE SUPPLIED THROUGH DUCTLED AND UNDUCTED SYSTEM.

CAR PARKING VENTILATION (IMPLULSE FAN)
• DIAMETER 315MM – 32 NEWTON
THRUST – 400 SQMT = 20M X 20M
• DIAMETER 400 MM – 50 NEWTON

CAR PARKING VENTILATION (INDUCTION FAN)
• DIAMETER 315MM – 32 NEWTON
• DIAMETER 400 MM – 50 NEWTON

• TRADITIONAL CAR PARKING VENTILATION SYSTEM USE EXHAUST OR EXTRACT
FANS IN CONJUGATION WITH DUCT WORK TO DISTRIBUTE THE AIR AROUND THE
CAR PARKING, PROVIDING A COMMON SYSTEM FOR BOTH POLLUTION AND
SMOKE REMOVAL
• THERE ARE TWO FUNCTIONS OF CAR PARKING VENTILATION
A) PROVIDE NORMAL VENTILATION TO PREVENT THE BUILD UP OF CARBON
MONOXIDE DURING THE DAY-TO-DAY OF THE CAR PARK.
B) PROVIDE SMOKE CLEARANCE IN THE EVENT OF FIRE.
INDIAN STANDARD: NORMAL MODE = 12-30 ACH
FIRE MODE = 30 ACH
INTERNATIONAL
STANDARD: NORMAL MODE = 6-10 ACH
FIRE MODE = 10- 20 ACH(AIR CHANGES PER HOUR)

• CALCULATE THE EXHAUST & FRESH AIR FLOW FOR CAR PARK volume OF
25x15x2.8 meter AS PER INTERNATIONAL STANDARD
CFM=m3 x ACH/1.7
INDIA= 12-30 ACH
EXHAUST FLOW RATE CFM = (25 x 15 x 2.8)*15/1.7 = 11117 CFM
11117/NO OF FANS (LETS TAKE 2 FANS)
11117/2= 5558 CFM IS EACH FAN CFM
FRESH FLOW RATE CFM = FRESH AIR CFM IS 70-80% OF EXHAUST AIR CFM
FRESH FLOW CFM = 11117 x 0.8 = 8893 CFM
8893/NO OF FANS (LETS TAKE 2 FANS)
8893/2=4446 CFM OF EACH FAN.

ENERGY EFFICIENT RATIO
Energy Efficiency of Room Air Conditioners. A room air conditioner's efficiency is measured by theenergy efficiency
ratio (EER). The EER is the ratioof the cooling capacity (in British thermal units [Btu] per hour) to the power input (in
watts). The higher the EER rating, the more efficient the air conditioner.

EXPANSION TANK
WHAT IS EXPANSION TANK ?
• An HVAC system essentially contains a fluid to transfer heat between spaces.
When this is a liquid, like in chilled water system, its volume is bound to change
across temperature range. But the containment, which are vessels, coils or pipes,
are fixed in volume. Hence it becomes necessary to provide a free space for
accommodating this change in volume. Otherwise pressure of the system will go
abnormally high. Expansion tank provides this free space for expansion or
contraction.
• It can be a closed vessel with liquid and a column of cushioned air or an open
vessel with space for liquid expansion.

EXPANSION TANK SIZING
• FORMULA FOR EXPANSION TANK SIZING
Vt = Vs{(V2/V1)-1}- 3(ALPHA) DELTA (T)/1-(P1/P2)
Vs=GALLONS(TOTAL SYSTEM VOLUME AS PER ATTACH FILE)
V1=0.01602 FT3/LB (ASHRAE HAND BOOK)
V2=0.01615 FT3/LB (ASHRAE HAND BOOK)
ALPHA= 6.5 X 10POWER (-6) INCH/INCH FAREHEIT
ALPHA VALUE CHANCES ACCORDING TO THE PIPE MATERIAL
(ASHRAE 2000 HAND BOOK CHAPTER-12)
DELTA (T)=T2-T1
T1=CHILLED WATER SUPPLY TEMPERATURE (45F)
T2=AMBIENT TEMPERATURE(106)
DELTA(T)=106-45=61F
(ACCORDING TO ASHRAE 10 PSIG) SO ATMOSPHERIC PRESSURE 14.7 AT SEA LEVEL
P1=24.7 PSIA
P2=(HEAD PP OR SP X 0.4335)+14.7}=(44.44(ASSUME) X 0.4335)+14.7=33.9 PSI

HVAC
ENRUN INDIA ELECTRICAL
CONSULTANTS AND CONTRACTORS
PREPARED BY ER MAJEED ALI
B.E (MECHANICAL) OSMANIA-
UNIVERSITY, ph no:9676989280

HVAC PPT.pptx

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