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CALCULATION SHEET

Space used for Office : 27 x 17 x 4 m3

= 1836 m3

PSYCHROMETRIC PROPERTIES

CONDITION DBT WBT % RH DPT h W [kg/kg]

OUTDOOR 43 27 29 21.3 85 0.016

ROOM 25 18 50 15.7 50.85 0.010

DIFFERENCE 18 34.15 0.006

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EFFECTIVE ROOM SENSIBLE HEAT

SL.

NO

TYPE OF LOAD LOAD IN

WATT

1 SENSIBLE HEAT SOLAR HEAT GAIN THROUGH WINDOW

GLASS

W

2 SOLAR TRANSMISSION GAIN –WALLS & ROOF

W

3 TRANSMISSION GAIN –OTHERS

W

4 INTERNAL HEAT GAIN

W

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A. ROOM SENSIBLE HEAT LOAD CALCULATION

1. SENSIBLE HEAT SOLAR HEAT GAIN THROUGH WINDOW GLASS

ITEM AREA (m2) /

QUANTITY

SOLAR HEAT GAIN

(W/m2)

CONVERSION /

MULTIPLICATION

FACTOR

LOAD IN

WATT (W)

EAST GLASS ---------- ------------------------ -------------- ----------

WEST GLASS m2 W/m2 --------------

W

NORTH

GLASS

m2 W/m2 --------------

W

SOUTH GLASS m2 W/m2 --------------

W

SKY LIGHT ------------ ----------- --------------

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

TOTAL W

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2. SOLAR TRANSMISSION GAIN –WALLS & ROOF

ITEM AREA (m2) /

QUANTITY

EQUIVALENT

TEMP.DIFFERENCE

(Δ0C)

CONVERSION /

MULTIPLICATION

FACTOR (U)

W/m2/K

LOAD IN

WATT (W)

UA ΔT

EAST WALL ---------- ------------------------ -------------- ----------

WEST WALL m2 0C W/m2/K W

NORTH

WALL

m2 0C

W/m2/K W

SOUTH

WALL

m2 0C

W/m2/K W

ROOF SUN m2

0C

W/m2/K W

TOTAL W

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3. TRANSMISSION GAIN –OTHERS

ITEM AREA (m2) /

QUANTITY

TEMP.DIFFERENCE

(Δ0C)

CONVERSION /

MULTIPLICATION

FACTOR (U) W/m2/K

LOAD IN WATT (W)

UA ΔT

DOORS m2 0C W

ALL GLASS m2 0C W

PARTITION m2 0C

W

FLOOR m2 0C

W

INFILTRATED

LOAD

VENTILATED

LOAD

CMM

CMM

0C

0C

W

20.4x(cmm) x ΔT x B.F

= W

TOTAL W

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4. INTERNAL HEAT GAIN

ITEM PERSON/ QUANTITY LOAD/PERSON

OR

QUANTITY

LOAD IN WATT (W)

PEOPLE W

POWER W

LIGHTS W

APPLAINS W

ADDITIONAL W

TOTAL W

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SOLAR HEAT GAIN THROUGH WALL/ROOF

Solar heat gain through wall can be calculated using the

equation

Q = U A Δ T

Where ΔT = Equivalent temperature difference obtained

from Table

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From Table 18.9 following values OF EQUIVALENT TEMPERTURE

DIFFERENCE in 0C are tabulated for walls & Roof

WALL 2 PM

WEST WALL

16.5

NORTH WALL

11.3

SOUTH WALL

17.4

ROOF

29.7

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HEAT LOAD CALCULATIONS

To calculate heat load Q through Wall, roof & floor, We require U , the

overall heat transfer coefficient as

Q = U x A x Δ T

A. To calculate U for Outside wall:-

The thermal conductivity K is available in Table 18.1

Uoutside wall = 3.5 W/m2/K

oi h

1

h

1

1

plasterbrickconcrete K

x

K

x

K

x

U

23

1

65.8

01251.0

73.1

2.0

32.1

1.0

7

1

1

U

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B. Partition wall:

U = 1.86 W/m2/K

ii h

1 2

h

1

1

plasterbrick K

xx

K

x

U

7

1

65.8

0125.02

32.1

33.0

7

1

1 x

U

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C. Roof:

URoof = 2.13 W/m2/K

o i h

1

h

1

1

asbestosplasterroofconcrete K

x

K

x

K

x

U

23

1

154.0

04.0

65.8

0125.0

9

2.0

7

1

1

U

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D. floor

UFloor = 6.05 W/m2/K

h

1

1

i concretefloorK

x

U

9

2.0

7

1

1

U

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Rates of solar heat gain through glass on June 21st in W/m2

TABLE 17.9D

DIRECTION 2PM

WEST GLASS 492

NORTH GLASS 91

SOUTH GLASS 32

The heat gain includes the direct + diffused solar radiation

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A. ROOM SENSIBLE HEAT LOAD CALCULATION

1. SENSIBLE HEAT SOLAR HEAT GAIN THROUGH WINDOW GLASS

ITEM AREA (m2) /

QUANTITY

SOLAR HEAT GAIN

(W/m2)

CONVERSION /

MULTIPLICATION

FACTOR

LOAD IN

WATT (W)

EAST GLASS ---------- ------------------------ -------------- ----------

WEST GLASS 12 m2 492 W/m2 --------------

5900 W

NORTH

GLASS

3 m2 91 W/m2 --------------

270 W

SOUTH GLASS 6 m2 32 W/m2 --------------

190 W

SKY LIGHT ------------ ----------- --------------

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

TOTAL 6360 W

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2. SOLAR TRANSMISSION GAIN –WALLS & ROOF

ITEM AREA (m2) /

QUANTITY

TEMP.DIFFERENCE

(Δ0C)

CONVERSION /

MULTIPLICATION

FACTOR (U)

W/m2/K

LOAD IN

WATT (W)

UA ΔT

EAST WALL ---------- ------------------------ -------------- ----------

WEST WALL 96 m2 16.5 0C 3.5 W/m2/K 5540 W

NORTH

WALL

34 m2 11.3 0C

3.5 W/m2/K 1345 W

SOUTH

WALL

34 m2 17.4 0C

3.5 W/m2/K 3590 W

ROOF SUN 459 m2

29.7 0C

2.13 W/m2/K 29035 W

TOTAL 39510 W

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Two types of air circulated:

1. Ventilated Air

2. Infiltrated Air

Load due to Infiltrated air is

a. Through open door

b. Exhaust fan

c. Crack through Windows.

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INFILTRATION RATE FOR WINDOWS

The flow of air due to wind over a building creates regions in which the static

pressure is higher or lower than the static pressure in the undisturbed area.

The pressure is positive on the wind side resulting in the infiltration of air

There are two methods of estimating the infiltration of air into conditioned space

due to wind action. They are

i) Crack method

ii) Air change method

In crack method, the estimate is based on measured leakage characteristics

and width and length of cracks( perimeter) around windows or doors.

The air change method assumes a certain number of air changes per hour for

each space depending on its usage.

The crack method is generally regarded as more accurate and is used in the

case of windows.

The air change method is more convenient to use for doors.

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The leakage of air in this case is a function of the wind pressure ΔP which can

be determined by knowing the wind velocity C using the equation

ΔP = 0.00047 C2

Where ΔP is in cm of water and c is the wind velocity in km/hr

Here assuming wind velocity 15 km/hr,

ΔP = 0.00047 x 152 = 0.11cm of water

Using Table 18. 11 for Infiltration through double Huge windows in m2/h/m

For weather-stripped, loose fit type Window, and for minimum ΔP = 0.25 ,

Infiltration = 2.5 m3/h/m crack

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Length of the crack:

Length of the crack = No. of windows x perimeter

= 7 x [ 2 ( 2+1.5) ]

= 49 m

Infiltration load in cmm:

Infiltration load in cmm = Infiltration in m3/h/ length of crack x Length

of crack

= 2.5 x 49 /60

= 2.04 cmm

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INFILTRATION DUE TO DOOR OPENING:

Infiltration through door openings depends on the type of door, as well

as usage.

Use Table 18.13- 18.15 for this.

Here Door is on adjacent wall, wooden door for average use

From the Table 18.13,

Infiltration in cmm /m2 = 1.98

Here No.of doors = 3

Area of the door = 1.5 x 2 = 3m2

Total infiltration rate = infiltration rate in cmm /m2 x area x No.of doors

= 1.98 x 3 x 3

= 17.8 cmm

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Total load due to Infiltration:

= Infiltration due to windows + Infiltration due to door openings

= 2.04 cmm + 17.8 cmm

TOTAL INFILTRATION LOAD = 19.84 cmm

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VENTILATED AIR

Ventilated is the fresh air coming from outside conditions.

Total ventilation in cmm = No. of occupancy x ventilation rate

Ventilation rate for different application can be obtained from Table 16.2

Here application is for Office use

From table 16.2 recommended cmm/person = 0.28-0.6

VENTILATION LOAD = No.of occupancy x 0.28

= 100 x0.28

= 28 cmm

VENTILATION LOAD = 28 cmm

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OCCUPANCY LOAD

The occupancy load ( both Sensible heat load (SHL) and Latent heat

load (LHL) is obtained form the Table 19.1

Corresponds to the activity as office work and the DBT 250C, take the

average value between 240Cand 260C

SHL = (80 + 70)/2 = 75 W

LHL = (60 + 70)/2 = 65 W

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3. TRANSMISSION GAIN –OTHERS

ITEM AREA (m2) /

QUANTITY

SOLAR HEAT GAIN

(W/m2) /

TEMP.DIFFERENCE

(Δ0C)

HUMIDITY DIFFERENCE

CONVERSION /

MULTIPLICATION

FACTOR (U) W/m2/K

LOAD IN WATT (W)

UA ΔT

DOORS 9 m2 180C 0.63 100 W

ALL GLASS 12+3+6 =

21 m2

18 0C 5.9 W/m2/K 2230 W

PARTITION 108+28 =131

m2

*15.5 0C

1.86 W/m2/K 3930 W

FLOOR 459 m2 ** 2.5 0C

6.05 W/m2/K 36940 W

INFILTRATED

LOAD

VENTILATED

LOAD

19.8 CMM

28CMM

18 0C

18 0C

20.4 ***

20.4

7270 W

20.4x(cmm) x ΔT x B.F

= 20.4x 28x18x0.15

=1542 W

TOTAL 20470 W

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* Assume a temperature difference of 2.50C across the floor,

ΔT for Partition = 18-2.5 = 15.5 0C

** Assume a temperature difference of 2.50C across the floor

*** 20.4 is the conversion factor from cmm into W for Infiltration

W60

C x x cmm

pTQonInfiltrati s

W (cmm) 20.4

60

1.0216 x 1.2 x cmm

T

TQonInfiltrati s

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4. INTERNAL HEAT GAIN

ITEM AREA (m2) /

QUANTITY

SOLAR HEAT GAIN

(W/m2) /

TEMP.DIFFERENCE

(Δ0C)

HUMIDITY DIFFERENCE

CONVERSION /

MULTIPLICATION

FACTOR (U)

W/m2/K

LOAD IN

WATT (W)

UA ΔT

PEOPLE 100 -------------- 75W/PERSON 7500 W

POWER --------- -------------- -------------- ---------

LIGHTS 15,000 W -------------- 1.25* 18750 W

APPLAINS -------------- -------------- -------------- --------------

ADDITIONAL -------------- -------------- -------------- ------------

TOTAL 26250 W

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*Lighting load :Electric lights generates sensible heat w]equal to the

amount of the electric power consumed.

Most of the energy is liberated as heat, and the rest as light which also

eventually becomes heat after multiple reflections.

As rough calculation, one may use the lighting load equal to 33.5 W/m2

to produce a lighting standard of 540 lumens/m2 in an office space;

20W/m2 is minimum.

After wattage is known, the calculation of the heat gain is done as

follows

Fluorescent : Q = Total watts x 1.25

Incandescent : Q = Total watts

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EFFECTIVE ROOM SENSIBLE HEAT

SL.

NO

TYPE OF LOAD LOAD IN WATT

1 SENSIBLE HEAT SOLAR HEAT GAIN THROUGH WINDOW

GLASS

6360 W

2 SOLAR TRANSMISSION GAIN –WALLS & ROOF 39510 W

3 TRANSMISSION GAIN –OTHERS 20470 W

4 INTERNAL HEAT GAIN 26250 W

SUB TOTAL 1 92690 W

5 STORAGE LOAD( HERE NEGLECTED) ---------------

6 SAFTEY FACTOR - 5% OF THE SUB TOTAL 4635 W

SUB TOTAL 2 97325 W

7 FAN POWER FOR SUPPLY DUCT(5%)+ LEAKAGE LOSS(0.5%) OF

SUB TOTAL- 2

5352 W

EFFECTIVE ROOM SENSIBLE HEAT 10422 W

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B.LATENT HEAT LOAD CALCULATION

ITEM AREA (m2) /

QUANTITY

SOLAR HEAT GAIN

(W/m2) /

TEMP.DIFFERENCE

(Δ0C)

HUMIDITY

DIFFERENCE

CONVERSION /

MULTIPLICATION

FACTOR (U) W/m2/K

LOAD IN WATT (W)

INFILTRATION

VENTILATED

LOAD

19.8cmm

28 CMM

0.006 kg/kg

0.006 kg/kg

50,000*

50,000*

75940 W

50, 000 x(cmm) x

Δw x B.F

=1260 W

PEOPLE 100 -------------- 65 W 6500 W

STEAM

APPLAINCES

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

ADDITIONAL

VAPOUR

TRANS.

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

SUB TOTAL 11,440 W

W (cmm) 50,000

60

2500 x 1.2 x cmm *

LQLoadHeatLatent

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EFFECTIVE ROOM LATENT HEAT

ITEM

AREA (m2) /

QUANTITY

SOLAR HEAT GAIN

(W/m2) /

TEMP.DIFFERENCE

(Δ0C)

HUMIDITY DIFFERENCE

CONVERSION /

MULTIPLICATION

FACTOR (U) W/m2/K

LOAD IN WATT

(W)

SUB TOTAL 1

11,440 W

SAFTETY FACTOR – 5 % OF THE SUB TOTAL

570 W

SUB TOTAL 2

12010 W

SUPPLY DUCT LEAKAGE LOSS(0.5%) OF SUB TOTAL- 2

60 W

EFFECTIVE ROOM LATENT HEAT LOAD

13,330 W

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EFFECTIVE ROOM TOTAL HEAT LOAD

= ROOM SHL + ROOM LHL

= 104,425 + 13,330

= 117, 755 W

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C.OUTDOOR AIR TOTAL HEAT LOAD

(ON EQUIPMENT)

ITEM

AREA (m2) /

QUANTITY

SOLAR HEAT GAIN

(W/m2) /

TEMP.DIFFERENCE

(Δ0C)

HUMIDITY DIFFERENCE

CONVERSION /

MULTIPLICATION

FACTOR (U)

W/m2/K

LOAD IN WATT

(W)

SENSIBLE

HEAT

28 cmm 180C 20x (1-0.15) 8740 W

LATENT

HEAT

28cmm 0.006 kg/kg 50,000x(1-0.15) 7140 W

RETURN --------- -------------- ----------- ---------

EFFECTIVE OUTDOOR HEAT 15880 W

GRANT TOTAL

HEAT

EFFECTIVE ROOM TOTAL HEAT

+ EFFECTIVE OUTDOOR HEAT

133635 W

OR

38 T.R

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