air conditioning (ch2)

7

Chapter Two

Base case

8

(2-1)Air Conditioning Analysis

Air conditioning system is designed to control the environment of occupied

spaces, so the thermal comfort approached by occupants. So, according to seasonal

variation of weather, air conditioning system has to cope with heating and cooling

loads to achieve comfort ability.

By conduction, convection and radiation sensible heat transmitted out of the

building in winter, and transmitted to the building in summer, and by the same way;

moisture can be transmitted in and out of the building, sensible and latent heat can be

produced within the building according to occupant’s activity, lighting and machines

also.

The first step before any design process for any air conditioning system;

engineers must carefully determine the amount of heat removal needed in summer

season and the amount of heat to be produced in winter season . Before an air

conditioning system can be designed, all these loads must be analyzed and summed

up together with great care in order to select the most suitable equipment for the

building systems.

9

(2-2) Overall Heat Transmission Factor

RU

1 (2-1)

External wall heat transmission factor(U)[R]

Figure (2-1):” A schematic for external wall construction”.

Table (2-1):-”external wall construction materials& thickness of material and

thermal resistance”.

Material x(m) k(W/m.K) R(m2.K/W)

cement plaster 0.025 1.2 0.02

hollow brick 0.1 0.9 0.11

air gap 0.1 0.28 0.36

insulation 0.02 0.045 0.44

concrete 0.15 1.75 0.09

face stone 0.1 1.7 0.06

outside air - - 0.029

1nside air - - 0.12

Floor heat transmission factor(U) UExternal = 0.815 (W/m

2.K)

10

Figure (2-2):” A schematic for Floor construction”.

Table (2-2):” Floor construction materials & thickness of material and thermal

resistance’’

material x(m) k(W/m.K) R(m2.K/W)

ceramic 0.02 1.05 0.02

sand and gravel 0.1 0.7 0.14

concrete baking 0.15 1.75 0.09

hollow brick 0.18 0.9 0.2


1nside air - - 0.12

1nside air - - 0.12

UFloor= 1.42(W/m2.K)

11

Roof heat transmission factor(U)

Figure (2-3):” A schematic for roof construction”. [1]

Table (2-3): “Roof construction materials& thickness of material and thermal

resistance”

Material x(m) k(W/m.K) R(m2.K/W)

asphalt water proving 0.02 0.7 0.029

concrete baking 0.15 1.75 0.086

Hollow brick 0.18 1.1 0.164


outside air - - 0.029

1nside air - - 0.12

URoof=2.26 (W/m2.K)

12

(2-3)Design Conditions and Desired Indoor Conditions

Inside and Outside Design Condition:

Inside design factors such as the following must be considered:

Type and use of building

Length or duration of occupancy

Degree of activity of occupants

Radiant heat source such as large lighting load or glass exposure.

Outside design conditions and daily peak temperatures.

Outside design conditions vary markedly with the locality. They are

determined by averaging conditions, which occur over a number of years.

Recommendations for Inside/Outside Design Conditions:

For preparing any load calculation or in evaluating the equipment

requirements, it is recommended to take into consideration the following

points:-

1- The desired inside design conditions are largely dependent on the load

components and on the use and occupancy of the building.

2- The most comfort applications, exact maintenance of space conditions is not

required if the system can limit the variation to acceptable tolerances.

3- Seldom do loads peak simultaneously in building.

4- Off peak or partial load conditions may and frequently do produce more

problems than the peak conditions. A successful system cannot be designed

solely for a single set of load conditions.

5- Economic operation of the system is to a large extent dependent upon the

analysis of the load variation, kind of building, and building use.

13

(2-3-1)Design Conditions

I. Outdoor Design Conditions:

The building is located at Ramallah so the design conditions related to

(Palestinian Guidelines for energy efficient building design) are as follows:

a. winter (heating):

Tdb = 2 oC

Ф=50%

No humidity adjustment is needed.

b. Summer (cooling):

Tdb = 35 oC

Twb = 24 oC

And also no humidity adjustment is needed

Ф=60%

Where:

Tdb: the dry bulb temperature at Ramallah

Twb: the wet bulb temperature at Ramallah

Ф: the relative humidity

II. Indoor Design Conditions:

The recommended design conditions for Office Buildings refer to (Palestinian

Guidelines for energy efficient building design) were:-

a. In winter (heating):

Tdb= 23

0C. (For patient room)

Tdb= 21

0C. (For CCU room)

Φ = 50%.

b. In summer (cooling)

Tdb= 23


Tdb= 21

0C. (For Corridors)

Φ = 50%.

14

(2-4) Cooling Load

Cooling load calculations deal with two types of heat gain:-

a. Sensible Heat: heat that flows into or produced in a space will tend to

cause a temperature rise in the space.

b. Latent Heat: in the form of moisture which although it does not cause a

temperature rise does change the condition of the air in the space resulting in a

higher relative humidity.

a)Sensible heat gains to the space include:

a1. Heat transmission through the building structure as a result of

conduction, convection and radiation.

a2. Sensible heat brought in with the outside air, which is introduced, for

ventilation.

a3. Sensible heat produced by occupants.

a4. Sensible heat produced in the space by lights, appliances, motors and

the like.

a5. Sensible heat to be extracted from materials or products brought into the

space, (industrial air taming).

b)Latent heat (moisture) gain may be classified as follows:

b1. Latent heat from outside air ;( both that introduced for ventilation and

that which infiltrates into the space).

b2. Latent heat from occupants.

b3. Latent heat from cooling, hot baths or other vaporization processes in the

space.

b4. Latent heat from products or materials brought into the space

b5. Latent heat from equipment.

15

(2-4-1)Cooling Load Calculations

The following features concern the Office building that we are going to

calculate the cooling load for:

Design Conditions:

Outdoors design conditions:

Tdb=35 0C.

Twb = 240C.

WO= 0.014

Indoors design conditions:

Tdb= 23


Tdb= 21

0C. (For CCU room)

WI=0.0085

Figure (2-4):-“Cooling design Conditions on the psychometric chart”.

16

(2-4-2) Transmission Load

A-External Walls:

By using the general heat transfer equation which is:

Q = U×A × (CLTD) (2 - 2) [2]

Where:

Q =Heat transmitted. (Watts).

U =Overall heat transfer coefficient (W/m2.K).

A=Area of exposed surface (m2).

T=Temperature difference (To-Ti).

But since all external walls in this building were exposed to the sun, so no

need to this equation and instead using solar load equation which is described in the

next section.

B-Internal Walls:

By using the general heat transfer equation (2-2)as the external walls, by

noticing that T was equal to one or two degrees Celsius because of using the

different temperatures for the offices, corridor and WC’s.

C-Windows:

There are two type of heat gain through the windows which are by conduction

and by solar; all windows were assumed to be made of single clear glass.

1) Heat gain by conduction:

From equation (2-2)

2) Heat gain by solar

Q solar heat gain = 𝐴 𝑆𝐶 𝐶𝐿𝐹 + 𝑈𝐴∆𝑇 (2-3)

Where:

SC: Shading coefficient

CLF: Cooling load factor for glass

SHGF: Solar heat gain factor for sunlit glass (W/m2)

U: overall heat transfer coefficient (W/m2.K)

17

D- Doors:

The external door was assumed to be made of single glass, so it was treated as

windows. The Internal doors were made from wood.

E- Roofs:

Transmission load through the roof will be calculated by the solar load for the

third (last) floor, while the roof for the other internal floors will not transmit heat

since the floors are cooled, so there is no temperature difference just CCU room.

F- Floors:

Since the Office Building structure contains three floors, the first floor will

transmit heat from the ground because it was assumed to be 12 0C.

(2-4-3) Solar Load

(a) Solar load through external walls and roof were calculated by using the following

equation[R]:-

adjCLTDUAq )( (2 – 4)[2]

Where:

U= Overall heat transfer coefficient (w/m2.oC).

A= Area of the roof.

CLTD=cooling load temperature difference. (From Appendix A).

(CLTD) adj =CLTD + (25-Ti) + (Tav-29) (2 –5)[2]

Where:

Ti = inside design dry-bulb temperature (oC).

Tav = Average outdoor dry-bulb temperature for design day, (oC).

(From appendix A)

CLTD = 12 (Roof with suspended ceiling at 1 Pm).

(CLTD) adj = 12 + (25-23) + (30-30)

CLTD=14

(b) Solar load through windows and external doors were calculated by using the Heat

gain by solar by equation (2-3)

Qsolar heat gain= 𝐴 𝑆𝐶 𝐶𝐿𝐹 𝑆𝐻𝐺𝐹 + 𝑈𝐴∆𝑇

18

(2-4-4) Internal Load

Internal Sensible Load:

a- Lights:

For lamps and using the following equation[2]:

CLFffgLampsRatinq bu (2 – 6)

Where:

Lamps rating = 60 Watt.

Fu = utilization factor (it expresses fraction of installed lamps in use) =1

Fb = ballast factor = 0.9.

CLF = cooling load factor.

From table (Appendix A), for 8 hours so, CLF = 1

b- Occupants:

For persons where people are doing office works, waking and standing; and

utilizing (Appendix A)

𝑞 = 𝑆𝐺𝐻𝑥𝐶𝐿𝐹𝑥𝑁𝑜. 𝑜𝑓 𝑝𝑒𝑟𝑠𝑜𝑛𝑠 (2-7)[2]

Where:-

SGH: sensible gain heat

CLF: cooling load factor

Internal Latent Load:

a- Occupants:

From (Appendix A), latent load for standing persons [2].

QL= 𝐿𝐻𝐺𝑥𝑁𝑜. 𝑜𝑓 𝑝𝑒𝑟𝑠𝑜𝑛𝑠 (2-8)

Where:

QL : Latent heat load.

LHG: latent heat gain

19

(2-4-5)Infiltration and ventilation Load

Sensible infiltration was calculated for the building by the following equation

qS= 1.23𝑉 ∆𝑇 (2-9) [2]

Where

QS=sensible infiltration load

𝑉 =

Latent infiltration load:

qL=3000𝑥𝑉𝑥∆𝜔 (2-10)[2]

Where

QL: Latent heat load (kW)

𝑉 : Infiltration volume flow rate (l/s)

∆ω :(ω out – ω in)

ω: Humidity ratio, (kg moisture /kg dry air )from psychometric chart.

*Ventilation load is considered as coil load not as room load because this load will be

covered on coil.

*The previous equations are used in cooling load calculation and the final results are

tabulated below with sample calculation for (ground floor-doctor exam room 1) and

all detailed calculation results are in Appendix (A)

Table (2- 4): cooling Load Calculations for the Ground Floor “doctor exam

room1”

Transmission

Element Length

(m)

Height

(m)

U

(W/m.K)

A

(m )

CLTD

( K)

Q

(W)

NW External Wall - - 0.815 - 8.0 0.0

NE External Wall - - 0.815 - 16.0 0.0

SW External Wall - - 0.815 - 9.0 0.0

SE External Wall - - 0.815 - 21.0 0.0

Western External Wall 3.7 4.0 0.815 13.0 8.0 96

20

Load from external

wall, ground and

roof

table(2-4),continued

Southern External

Wall

- - 0.815 - 12.0 -

Eastern External Wall - - 0.815 - 22.0 -

Northern External

Wall

- - 0.815 - 8.0 -

External Roof - - 2.26 - - -

Floor 6.0 3.0 1.42 18.0 12.0 306.7

Total 402 W

Glass Load by

Solar

Element Length

(m)

Height

(m)

A

(m )

SC SHGF

(W/m2 )

CLF Q

( W )

Door - - - 0.8 - - -

Table(2-4),continued

Window 1 1.5 1.2 1.8 0.95 685.0 0.17 199

Window 2 - - - 0.8 - - -

Total 199W

Glass/Door Load

by

Transmission

Element U

(W/m.K)

A

(m )

CLTD

( K)

Q

(W)

Window 1 6.2 1.8 12.0 133.9

Window 2 6.2 - 12.0 -

Door 3.0 - 12.0 -

Total 133.9W

People load #of person HG(w) CLF Q(W)

3 60 0.8 144.4

3 40 1 120

Total 264.4W

Light Load

Light Rating (W/m2 ) # of

lights

Fu Fb CLF Q ( W )

60 6.0 0.9 0.9 1 324

V (m3 ) Air Change Q ( W )

Infiltration Load 72.0 0.8 211

Total 221 W

21

Table(2-4),continued

ventilation Load

#of person

Air change

Q(W)

3 0.68 560

Total 501.8

# of machine machine (W) Q ( W ) computer

&lamp machine load

1 100 100 sensible

1 20 20 latent

Total 120W

Total latent load for this space is equal to

281W

Total sensible load for this space is equal to

1374W

Total Cooling Load Required for this Space is equal to about

1655W

*Load calculation for all room in GROUND FLOOR with same values of ∆T.

* Load calculation for all room in FIRST FLOOR with different values of ∆T in

some rooms that is because heat transfers from first to second floor with ∆T=2 oC.

*The load calculation for SECOND FLOOR of the last roof with temperature

difference for roof =14 oC and inside temperature in CCU rooms=21

oC which caused

the difference in load calculations:

Cooling load =90kW=25.7TR



22

(2-5)Heating Load

In winter, the factors affecting the heating load depends only on transmission

load and ventilation and infiltration load .since solar and lights considered as load to

the room which helps in heating of the room.

Indoor and out door conditions

a. Outdoor design conditions

Tdb = 2 0C.

b. Indoor design Condition:

Tdb= 23


Tdb=21(For CCU room)

Figure (2-5):-“Heating Design Conditions on psychometric Chart”

Transmission Load:

By using the general heat-transfer equation

RU

1

23

For calculating the heat loss through:

A- External walls.

B- Internal walls.

C- External doors.

D- Roof.

E- Floor.

Infiltration and ventilation load

The same procedure followed in cooling case.

Table (2- 5): “Ground floor-doctor exam room (1) heating load calculations”

width external wall A ext. ΔT U ext q ext(W)

3.7 12.55 21 0.815 140

width internal wall A int. ΔT U int. q int(W )

- - - 1.98 -

width window A window ΔT U window qwind(W )

1.5 1.8 21 6.2 200.88

width Door A door ΔT U door q door

- - - 3 -

#of windows A floor ΔT U ceiling Q ceiling(W)

1 18 21 1.42 460.08

A room #of persons Vroom Δω Q Tran.(W)

18 3 72 0.002 851.8

air change/person.min 𝑽 inft.(l/S) q inf.(W)

0.75 15 421

Q tot(kcal/hr) Q tot(W)

1093 1273

24

*Final results are in appendix (A)

* Load calculation for all room of GROUND FLOOR with same values

Where:- 1kW=859.58kcal/hr.

* Load calculation for FIRST FLOOR, ∆T=2 o

C in some rooms which causes heat

transfer from first floor to second floor.

* Load calculation for SECOND FLOOR of the last roof with differences which

caused the difference in load calculations:

∆T for roof = 18 oC.

And in (CCU room) inside temperature=21 oC

Table (2-6):”cooling and heating load for base case”.

Floor

Case

Ground floor First floor Second floor

Load(kW) Load(kW) Load(kW)

Heating 61 36 65

Cooling 90 65 88

Heating load =61kW=53900kcal/hr.



25

(2-6)Coil Capacity

(2-6-1) Cooling Consideration and design conditions

Cooling coil capacities for the selected area in the building are shown below:

Sensible cooling load (Qs) =213.96kW

Latent cooling load (QL) = 32.64kW

SHR = 𝑞𝑠

𝑞𝑠+𝑞𝐿 =

213.96

213.96+32.64= 0.867

Where:

SHR: is the sensible heat gain ratio.

Figure (2- 6): “Supply and Return Air for cooling” [3]

Outdoor Design Conditions:

Tdb = 35 oC

Twb = 24 oC

Where:

Tdb: is the dry bulb temperature.

Twb: is the wet bulb temperature.

Indoor Design Conditions:

Tdb = 23 oC

Φ = 50%

Special Considerations:

For hospital it’s accepted to use 50% mass as fresh air, and 50% mass recalculated.

26

For central cooling:

Coil Temperature is 13 oC

Fan Effect +1 oC

Duct Effect +2 oC

Supply Temperature = 16 oC

Return Temperature = 23 oC

Figure (2-7): “Supply and Return Air on psychometric chart for cooling”

At Point O (which represents outside conditions):

Tdb = 35 oC

Twb = 24oC

From Psychometric Chart

ho = 71.78 kJ/kg

wo = 0.0142 kg/kg

At Point R (represents Room conditions):

Tdb = 23 oC

Φ = 50%


hR = 45.37 kJ/kg

wR = 0.0..8747 kg/kg

27

At Point W (represents coil conditions):

Tw = 13 oC

h w = 35.9 kJ/kg

At Point B (represents Fan effect on temperature rise

TB = TW +1

TB = 13 + 1 = 14 0C

h b = 37 kJ/kg

By Locating W, B. we can find Point M (which represents mixing point conditions) at

the Psychometric Chart.

Mixing state can be obtained by mean conservation of mass:

am

x hm = ( rm

x hr) + ( om

x ho) (2 – 11)

But:

rm

= 0.5 am

om

= 0.5 am

Where:

rm

= mass flow rate of air recalculated [kg/s].

om

= mass flow rate of fresh air [kg/s].

hm = (0.5 am

x hr + 0.5 am

x ho) / am

By eliminating am

:

hm = 0.5 hr + 0.5 ho = 58.26 kJ/kg

Tm = 28.88 oC

At point C

hC = 47.5 KJ/Kg

TC = 24 0C

At Point S (this represents Supply air state):

TS = 16 0C

wS = 0.009 kg/kg

hS = 40 KJ/Kg

28

(2-6-2)Cooling Coil Capacity Calculations

Table (2-7):-“Cooling and ventilation load for base case “

Ground floor First floor Second floor Total load

Cooling load 90kW

25.6TR

65kW

18.4TR

88 kW

25 TR

243kW

69TR

Ventilation 19.2 kW 15.07 kW 16.2 kW 50.5kW

14.4 TR

Cooling coil capacity = total cooling load + ventilation load

Cooling coil capacity for base case =243+50.5=293.5kW=83.4TR

Cooling coil capacity (base case)=83.4TR

29

(2-6-3)Heating Coil Calculations

As we are using central heating by a boiler we must find out the heating coil

capacity to ensure that the machine will meet the load we want, and by following a

similar steps as in cooling coil calculation we are going to find the heating coil

capacity:

As there is no humidity adjustment in heating load calculations, our calculation

will be based on sensible heating load.

Sensible cooling load (Qs) = 72594 W.

Figure (2- 8): “Supply and Return Air for heating “. [3]

Outdoor Design Conditions:

Tdb = 2 oC

Φ = 60%

Where Tdb: is the dry bulb temperature.

Indoor Design Conditions:

Tdb = 23 oC

Φ = 30%

Special Consideration:

For office buildings it’s accepted to use 20% mass as fresh air, and 80% mass

recalculated.

30

Since cooling load is always larger than heating load, duct design is calculated based

on the total air flow of the cooling season; this requires that the flow rate in the

heating season should be unchanged.

The supply temperature is calculated based on the sensible load obtained previously

Ts = Tr +𝑅𝑆𝐻

𝑄 (2 –12)

Where:

RSH: Room’s sensible heat [kW]

Q: Volumetric air flow rate in [m3/s]

The same mixing conditions are used (0.5 re-circulated and 0.5 fresh air supplied)

Figure (2- 9):”heating conditions on psychometric chart”.[3]

At Point O (which represents outside conditions):

Tdb = 2 oC

Φ = 60%


ho = 8.55 kJ/kg

At Point R (represents Room conditions):

Tdb = 23 oC

Φ = 30%


hR = 36.4 kJ/kg

31

Mixing state can be obtained by mean conservation of mass by equation

am

x hm = ( rm

x hr) + ( om

x ho)

But:

rm

= 0.5 am

om

= 0.5 am

where:

rm

= mass flow rate of air recalculated [kg/s].

om

= mass flow rate of fresh air [kg/s].

hm = (0.75 am

x hr + 0.25 am

x ho) / am

By eliminating am

:

hm = 0.5 hr + 0.5 ho = 30.3kJ/kg

Tm = 21.9 oC

At Point S (this represents Supply air state):

TS = 29 0C

hS = 40 kJ/kg

Table (2-8):-“Heating and ventilation load for base case “

Ground floor First floor Second floor Total load

Heating load 61kW

53936kcal/hr

36 kW

31853kcal/hr

65kW

56495.2kcal/hr

162 kW

142284.2kcal/hr

Ventilation 26 kW 20 kW 21kW 67kW

Heating coil capacity for base case =163+67=229kW=196711kcal/h

H.Cbasecase =229kW

32

(2-7)Equipment Selection

(2-7-1)Chiller Selection:

We selected an AIR-COOLED SCREW COMPRESSOR LIQUID

CHILLER Module and the selection was based on the Carrier Products Catalogue,

the following data was obtained.

1. The chiller used was selected to be an Air-Cooled because it can handle the

large load we had.

2. 30 GX series

3. The load required for the cooling coil was found to be 297kW

4. This series can cover 282-1203 kW

(Model 30GX 092) has the following properties:

Net nominal capacity =305kW = 88 TR

Operating weight of 3097 kg

Semi hermetic compressors, twin SCREW POWER , Pro-dialog plus control

HFC-134a used as refrigerant

Copper tubes and aluminum fins condensers

Water connections inlet & outlet = 4 in.

Nominal Unit Power input of 108 kW

Note: More details in Appendix(C)

33

(2-7-2)Air Handling Unit selection (AHU):

The air-handling unit is equipment with a fan, cooling coil, air filter, motor

assembly, comprising shaft, bearings, pulley... etc.

The basic function of the AHU is to suck air from the rooms, let it pass through

chilled water cooling coils (or hot water heating coils) and then discharging the

cooled air (or heating air) back to the rooms. Normally, letting it pass through panel

or bag filters also filters the air. A certain amount of fresh air may be introduced at the

suction duct so that air in the rooms may be gradually replaced.

AHU's come in many sizes and shapes. Usually, the air conditioning designer will

choose a particular AHU based on the air flow requirements and the cooling

capacity.

We selected a 39CD/CX/CH Central Station Air Handling Units Module

and the selection was based on the Carrier Products Catalogue, the following data

was obtained.

So we selected from Carrier Catalogue three devices as following:

1. (Model 39CD-450) for load of (137.7 kW).



(Model 39CD) has the following properties:

23 different basic sizes with an air flow range 280 to 26400l/s.

Static pressure up to 3000 pa.

Complete range of matching equipment modules and accessories.

Customized and site assembly versions available.

Heating and cooling coils mounted on slide tracks for easy removal.

Note: More details in Appendix(C)

34

(2-7-3)Boiler selection:

1. The boiler required to cover the load is selected from the CHAPPEE NXR 3,

this type belong to the new range of cast iron sectional boilers. It has been

designed to operate on oil or gas fuel.

2. The boiler was designed to operate in winter for heating.

3. The selections based on total Heating Coils Load with a safety factor of 5 %,

so from the boiler catalog MODEL, NXR 3-39 with a range capacity (250-

290) kW was selected, so it covered our load which are 283 kW

4. Technical data :

Net output= 290 kW = 990402 Btu/hr = 249582 kcal/hr

Number of sections = 9

Water content = 2,088 L

The boiler contains a burner already.

Weight = 1000 Kg

Max operating pressure (primary) = 6 bar

Efficiency = 93 %

Note: More details in Appendix C

35

(2-7-4)Chimney Selection:

𝐶ℎ𝑖𝑚𝑛𝑒𝑦 = 𝐵𝑜𝑖𝑙𝑒𝑟 𝐾𝑤 × 12000 ÷ 3.5 =𝐵𝑡𝑢

ℎ𝑟.

= 290×12000

3.5= 994285

𝐵𝑡𝑢

ℎ𝑟.

Selection of chimney depended on:

Vent Height=100 ft.

Vent type ‘‘V’’, which are used.

994285 Btu/hr.

From catalog 8" common steel chimney, 3 mm thickness, insulated by 2'',

Rockwool protected by sheet metal.


(2-7-5)Exhaust Fans selection

Exhaust fans will be used as a mechanical ventilation device for the WC’s in

the building. The return duct of them is shown in the duct design section. We should

notice here that there were some important considerations used to select the suitable

fans for this duty.

The exhaust fans have a minimum static pressure of 800 Pa; (according to the

exhaust fan catalogue) they were from the WC return ducts in each floor and the

main return ducts which are in the utility shaft.

The total air flow of the fan in the right side =1500 CFM = 0.7 m3/s

The total air flow of the fan in the left side =9000 CFM = 4.2 m3/s

The exhausted flow rate is more than the calculated to maintain a negative pressure

in the WC’s, this to avoid any bad air quality exfiltrated to the corridor.

So we selected two exhaust fans which are Roof Crub Mounted (shown in

detailed drawings)


36

(2-7-6) Pumps Selection

Pump Efficiency:

Pump efficiency refers to the ability of a pump to convert electrical power to

pumping power. Pump efficiency commonly vary from 50% to 80% depending on the

application. Individual pump physical characteristics determine if pump is best suited

for a specific application. Great care should be given to match a pump with its

application. Refer to manufacture’s published data to determine which pump is best

for the desired application.

Pump Construction:

Construction refers to the material of construction in of the pump impeller and

casing. In order to accommodate differing system and mediums, pump construction

must be altered. Most hydraulic systems employ cast iron construction. Domestic

water should have all bronze construction. Bronze fitted may also be used on

domestic systems to save cost. Applications where extreme duty is expected may

employ stainless steel construction.

In order to select any pump, two main properties should be taken into

consideration:

1. Total Head loss which is the sum of all pressure drops across equipments and the

head loss due to friction.

2. The flow rate, which calculated previously in the coil load calculations.

Chilled Water pump

Air cooled Chiller is located at the roof of the building

Pump must re-circulated chilled water for all AHUs in different elevations

Pipe work has a considerable losses

A pipe line connect chilled water to the main collector in boiler room, then

pump circulate water

hp = ΔZ + hf +hm (2-13)

Where:

hp = pump head required in m.

37

ΔZ = Vertical length in m. (ΔZ =0 here according to Bernoulli equation)

hf = head friction losses in m.

hm = head minor losses in m.

hf = ƒ L V2/ 2 D g (2-14)

Where:

ƒ = friction factor

L = pipe length

V2 = square of water velocity m

2/s

2

ρ = is the density of water (kg/ m3)

D = is the inner diameter of the pipe (m)

Here we assume that: V =2 m/s, Є for steel pipe = 0.000046, ρ water = 1000 kg/ m3

Table (2-9): “Chilled Water Pump hf Calculations”

Name of

Dia.

Pipe

Dia

(in)

Є/D Total

length

[m]

Rynolds

no.

ƒ hf(m)

Main 1 4 0.0005 36 144226 0.0194 1.51

AHU 1 3 0.0007 10 98805 0.0215 0.68

AHU 2 2½ 0.0008 10 86074 0.0218 0.79

AHU 3 2 0.0012 10 60249 0.024 1.24

hf = 1.51+ 0.68+ 0.79+ 1.24= 4.21m

hm = K × (V2/2g) (2-15)

Where:

hm = minor head losses.

K = resistance coefficient.

V2 = squared velocity.

g = ground acceleration m/s2

38

Table (2-10): “Chilled Water Pump hm Calculations”

Name of

Dia.

Pipe

Dia

(in)

Check

Valves

no.

K Gate

Valves

no.

K Elbow

no.

K Tees K hm(m)

Main 1 4 4 2.1 4 0.16 5 0.95 4 0.9 3.46

AHU 1 3 1 2.1 2 0.16 0 0.95 1 0.9 0.66

AHU 2 2½ 1 2.1 2 0.16 0 0.95 1 0.9 0.66

AHU 3 2 1 2.1 2 0.16 0 0.95 1 0.9 0.66

hm = 3.46+ 0.66+ 0.66+ 0.66 = 5.43m

hp = 0+ 5.43 + 4.21 = 10m + 3 = 13m

We added 3m to the head of the pump to compensate any losses in the

AHU or any other losses.

To evaluate chilled water flow rate, we need to do the following.

From chiller data sheet

Tsupply = 7 oC

Treturn = 12 oC

Where:

Tw1 = supply chilled water temperature.

Tw2 = return water temperature.

wm

= chilled water flow rate.

= (chiller capacity/ Δhw)

= 293.4 / (50.83-29.4) = 13.69 kg/sec= 0.01369 m3/s

wm

= 49.3 m3 / hr

From Salmons general catalogue, required pump is,

Model , 50-125-H1 (2poles - 50Hz, DN 65-50 )

39

Hot Water Pump

The same procedure followed in chilled water pump we found the following:

Table (2-11): “Hot Water Pump hf Calculations”

Name of

Dia.

Pipe

Diameter

(in)

Є/D

Total

length

[m]

Reynolds

no. ƒ hf(m)

Main 1 2 0.0010 16 70573 0.0235 1.66

AHU 1 1¼ 0.0016 24 43140 0.024 4.16

AHU 2 1 0.0019 18 36943 0.027 4.09

AHU 3 ¾ 0.0025 12 28242 0.028 3.70

DHW 1 0.0023 40 30937 0.026 10.46

hf = 1.66+ 4.16+ 4.09+ 3.70= 13.62 m

For Domestic Hot Water hf = 10.46 m

Table (2-12): “Hot Water Pump hm Calculations”

Name of

Dia.

Pipe

Dia

(in)

Check

Valves

no.

K Gate

Valves

no.

K Elbow

no.

K Tees K hm(m)

Main 1 2 2 2.1 2 0.16 3 0.95 2 0.9 1.83

AHU 1 1¼ 1 2.1 2 0.16 0 0.95 1 0.9 0.66

AHU 2 1 1 2.1 2 0.16 0 0.95 1 0.9 0.66

AHU 3 ¾ 1 2.1 2 0.16 0 0.95 1 0.9 0.66

DHW 1 2 2.1 5 0.16 5 0.95 4 0.9 3.50

hm = 1.83+ 0.66+ 0.66+ 0.66= 3.81m

For Domestic Hot Water hm = 3.5 m

hp = 0 + 3.81 + 13.62 = 18 m + 3 = 21m

For Domestic Hot Water hp = 12 + 3.5 + 10.46 = 26 m

For Domestic hot water [ ΔZ =12 m ]

From boiler data sheet

Tsupply = 80 oC

40

Treturn = 60 oC

Where:

Tw1 = supply hot water temperature.

Tw2 = return water temperature.

mw = hot water flow rate.

mw = (boiler capacity/ Δhw)

mw = 283 / (334.92-251.09) = 3.38 Kg/sec

Vw = 0.00338 m3/s = 12.17 m

3 / hr

From Salmons general catalogue, required pump is,


For Domestic Hot Water:


(2-7-7) Boiler expansion tank selection

1. Calculate the volume of water in the piping, coils, boiler….etc:

Boiler = 9 × 232 = 2088 liter

Heating coils 260 liter

Pipes fittings and valves = 1260 liter 3608 liter

According to SALMSON catalogue:

1. Expansion Volume:

Vexp = Vt × (Cm – Cr) (2-16)

Where:

Vt = total installation Volume.

Cm = Expansion Coefficient at filling temperature,

41

(To Boiler outlet + T

o return)/2

Cr = Expansion coefficient at filling temperature (10o to 12

o)

2. Total Vessel Volume:

V = Vt ×(Cm – Cr)/1-(P1/P2)

And

P1: Effective inflating precision corresponding to static head + 1bar

P2: Valve Cracking pressure + 1bar.

P1 = 442.4 kPa

P2 = 550 kPa

V = 3608 × (0.0225 – 0.0018)/ 1 – (442.4/550) = 382 liter

Vessel type is: SALMSON, 382 L capacity, order 020847.

Note: More details in (Appendix C)

(2-7-8) steam boiler selection

Steam is used in hospitals for the purpose of sterilization of equipments, for

such hospital that contains 100 people and wants steam for laundries and

sterilization we need a steam boiler with capacity 63kW.

Specifications of the steam boiler:

Tuttnauer Model GEN350-63kW

Volume 70 liter

Serial No. 2804040

400V,3ph,50Hz,95A

42

(2-8)Solar heating system

(2-8-1)Hot water consumption

We saw that, for hospital, the average daily consumption of hot water is,

136.4 l/Person. Day

from (1999 ASHRAE hand book, HVAC Applications, Chapter 48)

(Table – 7; hot water demands and use for various types of buildings):

The daily hot water demand is given by

Average personal demand × Number of persons

For 100 occupants,

136.4 × 100 = 13640 liter/ day

Heat load calculations

The building heat load varies during the year, depending on the temperature outside

and the building will be under service all over the year seasons,

The calculation must be based on the following equation

Q =

m × Cp × ΔT (2-17)

Where:

m = hot water flow rate (Kg/s)

ΔT = Temperature difference between required Temp. Hot water supply & inlet water

Temperature of the collector (o C)

Cp = specific heat of water (kJ/kg. K) = 4.197

Q = heat load of the system (kW)

The mass flow rate of water is calculated by the following equation:

m = ρ × V

Where:

ρ = water density (kg/m3)

V = Volume flow rate of hot water (m3/s)

m = hot water flow rate (kg/s)

43

For 16 hour operation per day

m = (1000×13.64)/ (16 × 3600) = 0.236 kg/sec.

We noticed that May has the lowest ambient temperature which will be our design

temperature for cold water supply.

Tin = 22.6 o C

Tout = 60 o C

Where our design temperature for hot water usage = 60 o C for hospitals (1999

ASHRAE hand book, HVAC Applications, Chapter 48),

Q = 0.236 × 4.197 × (60 – 22.6)

51.9 kW without heat losses.

Heat losses estimation

Piping system (steel pipes 40) estimated to have:

A- 3/4" main supply with total length = 60 m.

B- 1/2" supply branches of total length = 10 m.

Pipes system insulated using VEEDO FLEX insulator with low thermal conductivity;

K = 0.027 W/m.K

Total pipe length = 60+ 10 = 70m

From (1999 ASHRAE Handbook, HVAC Applications, and Chapter 48):

Heat losses in insulated Piping System = 30 W/ m.

Pipe losses = (70 × 30)/1000 = 2.1 kW

Q tot = heat load + piping losses

Q tot = 51.9+ 2.1= 54 kW

44

(2-9)Solar System Design

(2-9-1)Solar collectors’ selection

To find the useful heat from the collectors based on the efficiency equation of

"AMCOR COMPANY" data:

The effective absorber area for (AM80- single glazing collector) = 1.32 m2.

η col = 0.8 – (30 ×[ T m – T a] / I (2-18)

Where:

T m = is the mean water temperature of the collector (Tm=47.5 C).

T a = is the ambient temperature of the collector (Ta =Tavg).

I = is the total solar insulation on the tilted collector (kcal/ m2 day).

For various ambient and insulation; efficiency will vary depending on these

variables as seen from the listed table:

Table (2-13):”Solar energy”

Tavg.

(ºC)

Hot water

(Qneeded)

Energy/day

(kJ/day)

Solar

Energy(kJ/m2)

Solar Energy

(kcal/m^2.day)

ηcol(%) Ecol.use

kJ/m2 .day

22.6 54.00 3110502.6 24840 5932.9 67.4 16744

25.8 50.83 2927935.1 24120 5761.0 68.7 16570

27.4 49.25 2836651.3 24120 5761.0 69.5 16771

28.3 48.36 2785304.2 25920 6190.9 70.7 18324

Note: Ramallah is located at 32 o latitude Northern hemispheres;

Collectors must be oriented to the southern face.

From the table above we see that collector efficiency at May:

η col = 67.4 %

The useful heat get from the collector given by the following equation:

E col .use = η col * I

Where:

45

η col : collector efficiency.

I: is the total solar insolation on the tilted collector (kJ/ m2 day).

E col .use: useful heat (KJ/m2 day).

In May:

E col .use = 0.674× 24840 = 16744(kJ/m2 day).

Collectors area could be found by dividing the total heat by the useful from

the collector;

A = Q tot / E col .use

At May:

A = (54×3600×16)/16744= 185.8 m2

Number of collectors needed = A/1.32

(2-9-2)Optimization of flat plate solar collector

Collector area must be large enough to cover the load without the aid of

auxiliary system in the sunny days, but the optimum collector's area must meet the

economical requirements.

Knowing the collector area and the unit price, the collector cost could be found.

The total cost of the collector could be calculated as follows:

C col,(tot) = fixed cost + operating cost

C col,(tot) = FCR x Investment + maintenance. (2-19)

Where:

FCR = i + [i /{1 + i )n – 1}] + t + j

Where:

i = interest rate in Palestine = 10%

Number of collectors = 141

46

n = is the number of expected serving years of the system = 20 years.

t = annual taxes = 0

j = annual insurance = 0.3 %

FCR = 0.1 + [0.1 / {1 + 0.1)20

– 1}] + 0 + 0.003 = 11.6 %

This cost is calculated for different values of collector's area as shown below,

Then these values plotted against collector's area.

The break even point could be obtained from the figures and it shows the optimum

collectors area and its cost. Above this point the addition of solar collectors will

not be justified economically. As the collectors solar energy is more expensive

than fuel cost.

From the figure below break even point gives us

130 collectors.(LPG as auxiliary heater).

With annual heating cost = 2324$/year

And the total collectors produced energy = 2024945KJ / day

Figure (2-10):”Cost optimization for flat plate solar collector”

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

10000

0 20 40 60 80 100 120 140 160

cost

(do

llar/

year

)

number of collector

(Cost) Vs.(# of collector)

deisel

solar

Gas

47

(2-9-3)Storage tank selection

The daily hot water demand is given by

Average personal demand x Number of persons

For 100 occupants,

136.4 × 100 = 13640 l/ day

It is recommended by the designer to design the volume of storage tank assuming

that 70% of hot water in the tank is usable;

storage tank capacity =

(13640)/0.7 = 19486 L

Storage tank capacity = 19.5 m3

To satisfy this volume choose four tank at the same volume

Vone tank=19.5/4= 5 m3

Assume that L = 3D = 6r

Where:

L = the length of storage tank. (m)

D = storage tank base diameter (m)

r = storage tank base diameter (m)

Storage tank volume can be estimated using the following equation:

V = r2 × π ×L (2-20)

V=6πr3

Where:

V = storage tank Volume (m3)

r = (V/6π)1/3

r = 0.64 m

D = 1.28m

L = 3.82m

Storage tank surfaces area:

Storage tank bottom and top area = 2× r2 × π = 2.55 m

2

Storage tank side surface area = 2 r × π× L = 12 π r2 = 15.3m2

Storage tank total surface area = 2.55 + 15.3 = 17.85m2

Storage tank insulation:

48

From (1996 ASHRAE Handbook, HVAC Systems and Equipment,

Chapter 33).

We can determine the vertical tank insulation factor (w/m2) (ƒQ/AӨ)

)(

11

aavg ttA

fQ

R

(2-21)

Where:-

R: thermal resistivity of insulation, m2· K / W.

f : specified fraction of stored energy that can be lost in time θ

Q: stored energy, J

A: exposed surface area of storage unit, m2

θ : given time period, s

tavg: average temperature in storage unit, °C

ta : ambient temperature surrounding storage unit during season when it will be

heated, °C

(ƒQ/AӨ) = 15.36

t avg = 60 °C

t a = 22.6 °C

1/R = 0.293 (W/ m2· K)

R = 3.41 (m2· K/W)

But R = L/K

Where:

L = insulation thickness (m).

K = thermal conductivity (W/m.k)

(From Palestinian energy efficient building code, page 112)

For extruded polystyrene as insulator;

K = 0.028 (w/m.K)

L = K x R = 0.028 × 3.41= 0.0955 m = 9.5Cm.

U = thermal transmittance (w/ m2· K) = 1/R

U(insulation) = 0.293 (W/ m2· K)

49

(2-9-4)Solar Boiler selection

Boiler Capacity(kJ /day) Capacity(kW)

Gas 2516051.3 54

In the previous section the number of collectors was found and so the quantity of

energy produced by the collectors.

Now assuming that there is no solar energy system under service, and the load of

coldest month must be covered by the conventional system then the load on the boiler

will be as follows

As the maximum load reach to = 54 kW, since there are 13 hours /day for solar

radiation.

Assuming that gas boiler efficiency = 0.9

Boiler capacity = 54/ η boiler = 65.8kW

Boiler needed is “Heat master- model 60N”

Gas fired boiler with capacity reach to 69.9kW

Determination of auxiliary load and fuel consumption is as follows:

Q auxiliary = Q load avg – Q useful (2-22)

Where:

Q auxiliary = is the heat supplied by the auxiliary heater (Boiler) (kJ / day)

Q load AVG = is the yearly average of the total heat load.

Q useful = is the useful heat from the collector.

The average useful heat in (KJ/day) from the collector was calculated as follows:

Q useful = A × η avg × Ecol.use avg (2-23)

Where:

A = area of collectors

η avg = yearly average efficiency of the collectors

Ecol.use avg =Yearly average useful heat from the collector

Q useful = 130× 1.32 × 0.69 ×17102 = 2024945.2 kJ / day

Q auxiliary = 2915098 – 2024945.2= 890152.8 kJ / day

50

m fuel = Q auxiliary /( H.Vgas × ŋburning) (2-24)

Where:

m fuel = fuel mass flow rate ( kg/ day)

H.V of gas = 50000 kJ/kg K

Burning efficiency in selected boiler = 0.9

m fuel = 890152.8 / (50000×0.9) = 19.8 kg/day

(2-9-5)Solar Pump Selection:

Selecting a specific pump needs two basic parameters, the flow rate that the pump

should operate at, and the head of the pump.

The selected flow rate should overcome heat load and heat losses flow rate,

Q total =

m × Cp × ΔT (2-25)

Q total = total heat load + heat losses (kJ/sec)

Cp = specific heat for water kJ/kg.K = (4.197)

ΔT = Temperature difference between required Temp hot water supply & inlet water

Temperature of the collector (o C)

m = water mass flow rate in kg / sec

m = Q total / Cp × ΔT

m = 54 / (4.197 ×(75- 22,6)) = 0.244 kg / sec

The second step is to determine the head required by the pump to over come.

At the beginning we have to calculate the velocity in all sizes of the pipes which are:

3/4" and 1/2"

It can be calculated from the following equation:

V =

m / (ρ × A) (2-26)

Where:

m = is water mass flow rate (kg / sec)

ρ = is the density of water (kg / m3).

A = is the pipe cross sectional area (m2)

51

Then we have to find the friction factor from moody chart for commercial steel pipes.

Roughness constant for commercial steel "Є" = 0.046

D = 3/4" = 0.01905 m

D = 1/2" = 0.0127 m

Reynolds number = 4 m / (π D μ)

Where:

m = water mass flow rate in (kg/s)

D = pipe diameter (m).

μ = the viscosity of water (N.s/m2) = 489 × 10

-6


ƒ; obtained from moody chart if flow is turbulent.

If the flow is laminar:

ƒ = 64 / Re

Table (2-14): “Friction factor ƒ for solar pump”.

Pipe

type

D [m] Velocity

( m/sec)

m

kg /sec

Є / D Re Flow

state

ƒ

3/4 " 0.01905 0.85826074 0.244 0.0023 33435 Turbulent 0.036

1/2" 0.0127 0.437712977 0.05542 0.00362 11368 Turbulent 0.034

We can get pressure drop in piping system using above data, and pressure

drop can be evaluated using the following equation.

ΔP = ƒ L V2 ρ / 2 D (2-27)

Or, directly we can calculate head friction losses from following equation:

hf = ƒ L V2/ 2 D g

Where:

ΔP = is the pressure drop (Pa/m)


L = pipe length

V2 = square of water velocity m

2 / sec

2

52

ρ = is the density of water (kg/ m3)

D = is the inner diameter of the pipe (m)

hf = head friction losses (m)

Table (2-15): “head friction losses for solar pump”

Pipe type ƒ V2

(m2/s)

L(m) 2D

(mm)

hf

(m)

3/4 " 0.036 0.736 60 0.0381 4.11

1/2" 0.034 0.191 10 0.0254 0.303

hf (total) =4.11+ 0.26= 4.37m

Then we have to calculate pressure drop in fittings, because it is important to note the

pressure drop in these fittings.

Table (2-16):” pressure drop in fittings for solar pump”

fitting 3/4" 1/2"

Elbows 8 6

Tees 6 4

Valves 4 6

hm = (V2

K/2g)

Where:

hm = minor head losses

K = resistance coefficient.

V2 = squared velocity.

g = ground acceleration m/s2

For 3/4" pipe: ((8×0.8) + (6×0.9) + (4×0.27))×(0.736)/ (2×9.81) = 0.63 m

For 1/2" pipe: ((6×0.3) + (4×0.9) + (6×0.9))×(0.191)/ (2×9.81) =0.105 m

hm total = 0.63 + 0.105 = 0.735m

htotal = ΔZ + hf + hm = 15 +4.37 +0.735 = 20.1m

Pump Specifications:

Circulating pump must circulate 1.08 (m3/ hr.)

Head to overcome = 20.1m

“SALMSON-JRL204-15/3”

53

(2-10)Air Distribution & Duct work

The supply air should in the right temperature, humidity and in the right quantity

so that when it is mixed with the room air, the resultant room air condition falls within

the comfort condition.

In the hospitals proper temperature and humidity control depends on

delivering the proper amount of supply air at a carefully controlled temperature and

humidity. The delivery and distribution of conditioned air must be accomplished

without appreciable draft or noise.

Important consideration:

The correct amount of air (m3/s or CFM) that passes in each section or branch

of duct must be known and can be calculated from room sensible heat gain

Noise level.

Pressure drop in dampers, outlets, coils…etc. are obtainable from

manufacturer catalogues.

Pressure drop in straight ducts and fittings is calculated or selected based on

recommended velocities.

Components of a duct system:

1) Straight sections.

2) Fittings (Bends, Branches … etc).

3) Fire fighting dampers.

4) Terminal units and air outlets as grills, diffusers, register … etc).

5) Heating coil.

6) Cooling coil.

7) Filters.

Methods of duct sizing:

(1) Velocity method (simple, not accurate)

Volume flow rate in each branch and main duct has been given.

Velocity of air in each branch and main duct has been selected from recommended

velocities.

(2) Equal pressure drop (or equal friction method, accurate enough, widely used)

54

This method gives better results. It reduces size of duct and cost and is suitable for

completer system.

(3) Static region method (for balancing of branches).

(4) T-method (for computer simulation)

The method used is:-

Equal friction method

Main duct recommendation velocity is from 7m/s.

Branches recommendation velocity is from 6 m/s

Steps of using this method:-

1- Determine the air quantity required for each zone to cover the cooling load (since

the cooling load is more critical than the heating load. So the duct design must depend

on the cooling load).

2- Locating the supply diffusers on the plane.

3-take the main duct velocity 7 m/sec (From velocity recommended for the buildings).

And by using the ductulator;

We can determine the remaining unknown values:

Pressure Friction (Pa/m)

Rectangular Duct Dimensions (Width X Depth)

In each Floor there is 3 Air Handling Units (numbered from 1 to 3), they cover the

whole floor load, the table below show the floors CFM needed and each room will be

covered by the proper AHU

Sample calculations of duct sizing:

A section of the duct connected to air handing unit # (1) (which will meet the

load of second floor)

55

Figure (2-11) : “Section of duct line and grills distribution”

Line [A-B]

Assuming main duct velocity at A = 7m/s

Qtot=4928(l/s)

From ductulator (catalogue)

Qtot vs. V→ (ΔP/m)=0.494 (pa/m)

Duct dimensions = 87.5 (cm) × 85(cm)

Line [B-C]

ΔP= 0.494 pa/m Qtot =2763 L/s

From ductulator

Duct dimensions = 70(cm) × 70(cm

Line [C-D]

ΔP= 0.494 Pa/m Qtot =2486L/s

From ductulator

Duct dimensions = 65(cm) × 70(cm)

Line [D-E]

ΔP= 0.494 pa/m Qtot =2301 L/s

From ductulator

Duct dimensions =60(cm) × 70(cm)

4-Way Ceiling Supply

Diffuser

56

Line [E-F]

ΔP= 0.494pa/m Qtot =2077 L/s

From ductulator

Duct dimensions =55 (cm) × 70(cm)

Exhaust duct sizing

The same procedure followed by sizing the supply duct, by taking the ventilation load

only.

Return air :-

All return grills are located in the corridors and we added door grills for each room

in order to allow return air to pass through these door grills to return grills exists in

corridors.

50% of the supplied air is to be returned to the AHU by false ceiling the remaining

100% of fresh air for CCU rooms (AHU3)

50% of supplied air is exhausted by exhaust grills in bath rooms through door grills.

Duct area calculation

Area of one section = (2xdepth+2xwidth) × length of section

= (2x0.85+2x0.875) ×3.5

A=12.075 (m2)

The total area of the duct = ∑ (section area) +∑ (fitting area)

=136.7+4.3=141 (m2)

And the same procedure for all A.H.U

The total area of duct for all = 941.8 (m2)

The A.H.U should be covered all friction loss in the duct and in the system,

and for it was noticed that the maximum pressure loss was 1.115

Maximum pressure loss in the duct = (Total length x pressure drop/meter) + (pressure

drop in diffuser)

= (95×1.2) + (0.03 in water × 350 pa/in water)

= 125 Pa.

Assume pressure loss in the air handling unit to be 375 Pascal (losses in the air intake,

filters cooling coil and other fittings).

57

So the fan pressure = 125 + 375 = 500 Pascal

And fan flow rate = 32204 CFM = 15.2m3/sec.

*all detailed duct sizing results are in Appendix (A)

Grills selection

Supply grills

To determine the position of grill we shoud calculate the throw that grill can coverd it

Calculation of throw

V0.25 → T0.25 /L

Throw = (T0.25 /L) × L

All supply grills are selected as supply ceiling diffuser –four way

From catalogues (throw vs.CFM of grill)

L: is measured by dividing each room and taking the farthest point from the middle or

where the grill is to be located.

Return Grills

All return grills are located in the corridors and we added door grills for each room in

order to allow return air to pass through these door grills to return grills exists in

corridors.

Exhaust grills the same as return grills taking the ventilation load, bath room, and

kitchen.

air conditioning (ch2)

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