HEAT LOAD CALCULATION

HEAT LOAD CALCULATION

Heat load calculation is very important part of design .

Well heat load calculation is nothing but finding out the amount of heat present in an area which is to be air conditioned.

Basically if we want to repair any staff it is necessary to identify and convert it into excel to plan accordingly.

In the same way we calculate the amount of heat present in an space which is to be designed.

Basically heat which is present in the room are of two types

Sensible heat
Latent heat

Basically we need to calculate this both and we are done with it.

Lets find out What are heat giving body ……

Heat giving object:

Well as we know what are source through we get heat now lets classify it by sensible and latent heat.

HEAT GIVING OBJECTS CATEGORISED

Well as we classified it, now we need to find out how much amount of heat does it emits so lets find the steps to calculate the heat.

Overview of heat load calculation

1. Geographical condition

2. Ambient Condition

3. Transmission coefficient (U value)

4. Temp diff (∆T)

5. Complete building information.

6. Internal sensible heat (ISH)

7. External sensible heat (ESH)

8. Total sensible heat

9. Latent Heat

10. Total latent heat

11. Effective room sensible heat (ERSH)

12. Effective room latent heat (ERLH)

13. Effective room total heat (ERTH)

14. Load at cooling coil

15. Grand total heat (GTH)

16. Effective sensible heat factor (ESHF)

17. Apparatus dew point temp. (ADP)

18. Dehumidified rise (DR)

19. Dehumidified (CFM)

Friends there are basically 19 steps we need to follow to obtain the amount of heat present in the an space to be conditioned.

well this a manual method through which which we can find how much Ton of air conditioner required to overcome the heat load.

STEP 1 GEOGRAPHICAL CONDITION:

1.Location: Mumbai

2.Building Application: Residence

3.Floor to floor height:12 feet

4. False ceiling height:9 feet

5.Longitude :19.0760°North

6. Latitude:18.54°North

7. Altitude:11 Metre

8. Orientation: N,S,E,W

well it is necessary to find the details of geographical condition and the data of area which is to be conditioned.

PG-68

PG-69

This are the outside Design Data information for India which is refereed from ISHRAE handbook.

STEP 2 AMBIENT CONDITION:

	DBT°F	WBT°F	RH %	HR	Daily range
Outside	95	83	60	154	12
Inside	75	62.8	50	66
	∆T=20°F			∆HR=88Grains/pound

Outside DBT, WBT, RH, Daily range is taken from PG-68 & 69.

Inside DBT & RH % is as per comfort DBT-75°F & RH-50%, WBT and HR of inside as well as of outside is taken from the psychometric chart.

Correction factor- PG 83

Correction factor for Mumbai location with daily range of 12 and ∆T which is 20°F = 9°F.

PG-83

STEP 3 TRANSMISSION FACTOR (U-VALUE) :

The amount of heat transfer through 1 sq.ft area at 1°F.

Temperature difference is known as transmission co-efficient.

UNIT- BTU/hr.sq.ft°F

Q= U X Area of the wall X ∆T.

Where Q= Heat = BTU/Hr

U= Resistivity factor = BTU/Hr.sq.ft° F

A= area of the wall in sq.ft.

∆T= Temp diff in °F

Transmission Co-efficient (U-value)

Paint POP Cement Wall Cement POP Paint

U- Value = 1/R1+R2+R3+R4+R5+R6+R7

To find transmission for

1. Wall

2. Partition

3. Roof/Ceiling

4. Floor

5. Glass

Material	Specification	U-value (BTU/HR.SQ.FT.°F)
WALL (PG-92)	Solid brick (commonly only) of thickness 12’’ with 3/8 of plaster On sand aggregate (weight 120).	0.30
PARTITION (PG-95)	4’’ Face brick venear, ½’’ insulating board, No interior finish.	0.38
ROOF/ CEILING (PG-97)	Concrete sand and gravel aggregates with 8’’ thickness suspended plaster of ½’’ (weight 93 lb/sq.ft)	0.20
FLOOR (PG-99)	Floor tile sand aggregates of 8’’ thickness with ½’’ light weight plaster (weight 82 lb/sq.ft)	0.25
GLASS (PG-100) TRANSMISSION	DOUBLE PLANE, VERTICAL GLASS, AIR SPACE THICKNESS ½’’ WITHOUT STORM WINDOWS	0.55
GLASS (PG-90) RADIATION (SOLAR HEAT GAIN FACTOR)	ORDINARY GLASS INSIDE VENETIAL BLIND MEDIUM COLOUR	0.65

depending on the building construction material data used in walls , floors , roof .ceilings, glass etc u value is been choosen with the following data from Ishrae.

PG-92

PG-95

PG-97

PG-99

PG-100

PG-90

STEP 4 Temperature Difference (∆T):

Find temperature difference for,

1. Partition

2. Floor

3. Ceiling

4. Wall

5. Roof

6. Glass (Transmission)

7. Glass (Radiation)

Temperature Difference (∆T)

MATERIAL	Non AC & AC	∆T
Partition (partition are room which is connected to room which is to be conditioned) If the room beside is having AC than ∆T will be 0.	1. Non AC =((outside temperature-5)-75°F) =((95-5)-75) 2. AC =75-75	15°F 0°F
FLOOR	1. Non AC =((outside temperature-5)-75°F) =((95-5)-75) 2. AC =75-75	15°F 0°F
CEILING	1. Non AC =((outside temperature-5)-75°F) =((95-5)-75) 2. AC =75-75	15°F 0°F

MATERIAL	DIRECTIONS (A)	CORRECTION FACTOR (B)	∆T (A+B)
Wall (PG 82), weight of wall = 120 lb/sq.ft	N-3 S-13 E-18 W-11	9 9 9 9	12 22 27 20

MATERIAL	DESCRIPTION (A)	CORRECTION FACTOR (B)	∆T (A+B)
ROOF (PG-83)	Exposed to sun, 80 weight at 4 pm (32)	9	41
GLASS (TRANSMISSION)	Outside-Inside 95°F-75°F		20°F
GLASS (RADIATION) (PG 77-81)	Solar heat gain factor for each direction		N-23 S-12 E-12 W-163

PG-82

PG-83

PG-77

PG-78

PG-79

PG-80

PG-81

Step-05 Complete Building Information:

FLOOR	FLAT NO	ROOM	LENGTH	WIDTH	HEIGHT	AREA	VOLUME	PEOPLE	TR	CFM

Step-06 INTERNAL SENSIBLE HEAT (ISH):

This is the heat which is present in the equipment

A. Light (Qlighting)

For residential- 1 to 1.25 watt/sq.ft.

For commercial- 1.25 to 1.50 watt/sq.ft.

For industrial -1.5 to 4 watt/sq.ft.

Q light= (Watt /sq.ft) X (room area in sq.ft)

But 1 watt =3.415 BTU/hr

B. Equipments (Q Equipments)

For Reidential :- 0.5 watt/sq.ft

For commercial- 0.5 to 0.73 watt/sq.ft.

For industrial -From clients (watt/sq.ft)

Q equipments= (watt/sq.ft) X (room area in sq.ft)

But 1 watt =3.415 BTU/hr

C. People (Q people) (PG-84)

Q people = Number of people X (Sensible heat/persons)

From pg 84 Sensible heat per person = 245.

D. Floor (Q floor)

CASE 1. If below floor is Non-AC.

Q floor = U X A X ∆T

CASE 2. If below floor is AC.

∆T = 0

Q floor = U X A X ∆T

Q floor = 0

E. Ceiling (Q ceiling)

CASE 1. If above floor is Non-AC.

Q ceiling = U X A X ∆T

CASE 2. If above floor is AC.

Q ceiling = U X A X ∆T, than ∆T=0.

F. Partition

Case 1.If other side of the partition is non Ac .

Q Partition = U X A X ∆T

Case 1.If other side of the partition is Ac .

Q Partition = U X A X ∆T, than ∆T=0.

Step-07 EXTERNAL SENSIBLE HEAT (ESH):

A. Wall

Q = U X A X ∆T (Area -E,W,N,S),( ∆T - E,W,N,S)

If there is glass in wall than A wall (Total area of wall-Area of window).

B. Roof

Q = U X A X ∆T

C. Glass

Q = U X A X ∆T

Transmission: Q = U X A X ∆T(NOTE: Take U from step 3 and for transmission consider the window as ∆T ,will not vary i.e Q (glass transmission E,W,S,N)= U X Sum of Area of window E,W,S,N x ∆T

Radiation : Q = U X A X ∆T (NOTE: Take U from step 3 and for Radiation consider the window ∆T will vary for each direction, i.e Q (glass radiation E,W,S,N)= U X Sum of Area of window E,W,S,N x ∆T of window E,W,S,N.

Total External sensible heat of glass

Q Glass total= Q glass transmission += Q glass radiation

D. Outside air (PG :73)

PG 73.

1. According to people

30CFM per person

=No. of person x CFM per person( PG :73)

2. According to Area

0.33CFM per person

=CFM per sq.ft. ( PG :73) x Floor area

3. According to Volume

Room volume x NACPH

Number of air change per hour (NACPH)

1. Residential:1 to 3

2. Commercial:3 to 7

3. Industrail:7 and above

Bypass factor(Page 84)

BPF=0.30( For residential)

Sensible Heat Constant

The sensible heat gain through 1 CFM of air with 1°F ∆T .

SHC =1.08BTU/hr CFM °F

Q outside =SHC X Bypass factor x outside air CFM x ∆T

E. Infiltration air

Q Infiltration Air=Sensible heat constant x CFM /feet x Length ( consider 1)x ∆T(outside -inside temp)

Infiltartion std (In CFM)

Residential -1 to 3

Commercial-3 to 7

Industrial-7 and above

Step-08 TOTAL SENSIBLE HEAT (TSH):

TSH =Internal sensible heat + external sensible heat

Step-09 LATENT HEAT (LH)

1.PEOPLE (PG 84)

Latent heat generated by per person =205 btu/hr

Q People =LH/. Person X NO. Person

2.Outside

Latent heat constant

The heat generated by 1 CFM OF Air when 1 GR per pound difference

LHC =0.68 BTU/hr CFM Grains per pound

Q Outside =LHC x bypass factor x max. CFM x ∆HR

3.Infiltration

Q infiltration = LHC X (CFM /FT. x length) x ∆HR

Step-10 TOTAL LATENT HEAT (TLH)

Q total latent heat =Q PEOPLE +Q OUTSIDE +Q INFILTRATION

Step-11 EFFECTIVE ROOM SENSIBLE HEAT (ERSH)

ERSH=Total sensible heat + factor of safety x total sensible heat

FOS =10 TO 15 %

Step-12 EFFECTIVE ROOM LATENT HEAT (ERLH)

ERLH=Total Latent heat + factor of safety x total latent heat

FOS = 2.5 TO 5 %

Step-13 EFFECTIVE ROOM TOTAL HEAT (ERTH)

EFFECTIVE ROOM TOTAL HEAT = EFFECTIVE ROOM SENSIBLE HEAT + EFFECTIVE ROOM LATENT HEAT

ERTH = ERSH + ERLH

Step-14 LOAD AT COOLING COIL

Contact factor = 1 – Bypass Factor

= 1 – 0.30

= 0.70

A. Q coil sensible = Sensible heat constant X contact factor X outside air x ∆T.

B. Q coil latent = Latent heat constant X contact factor X outside air x ∆HR.

TOTAL LOAD AT COOLING COIL

Q Total coil = Q coil sensible + Q coil latent

Step 15: Grand total heat:

Q grand total heat = Q effective room total heat + Q total load at cooling coil.

1 TR = 12000 Btu/hr

Step 16: Effective sensible heat factor (ESHF)

ESHF = Effective room sensible heat / Effective room total heat

Step 17: Apparatus dew point temperature (PG 87 to 89)

Dry Bulb temperature = 75°F

Relative humidity = 50%

Effective sensible heat factor = 0.67

ADPT = 44°F

NOTE : To maintain comfort level.

PG-87

PG-88

PG-89

Step 18: Dehumidified rise

Dehumidified rise = Contact Factor X (Room temperature – ADP)

Note- Increase in temperature will decrease in moisture level.

Step 19: Dehumidified CFM

Dehumidified CFM = Effective room sensible heat / sensible heat constant X dehumidified rise

NOTE- TO maintain 75°F how much CFM need to be added to get rid of dehumidified rise.

Cross check

CFM/TR < 400 CFM

well this are the steps which will help you out to determine the heat load of the space .

This is quite lengthy but it is foundation through which we can determine what we want to.

Heat load calculation can be also determined by different softwares but before that it is necessary to know all the steps inorder to be a good designer .

hope you liked this article ..............

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