A-Z HVAC Practical DesginA-Z HVAC Practical Desgin

Bank ground floorBank ground floor

Heat load calculationsHeat load calculations:: I intended to calculate it with two I intended to calculate it with two

ways and compare:ways and compare: 11stst method is more complicated and method is more complicated and

more real.more real. 22ndnd method is less complicated and method is less complicated and

also real.also real. Beginning is with the 1Beginning is with the 1stst method: method:

Calculations of internal Calculations of internal loadsloads::

I) Heat transmission load: (sensible)I) Heat transmission load: (sensible)It’s heat load transmitted from outside media It’s heat load transmitted from outside media

to the inside media through walls.to the inside media through walls.Simply : QSimply : Qtt==ƩU*A*(TƩU*A*(Too-T-Tii))U: overall heat transfer coefficient ;assuming walls are formed from U: overall heat transfer coefficient ;assuming walls are formed from

concentric bricks so U=0.72 w/m2 .Celsius from table(3-4).concentric bricks so U=0.72 w/m2 .Celsius from table(3-4).A: Area=175 m2 from AutoCAD file.A: Area=175 m2 from AutoCAD file.TToo:outside dry bulb heat (without any moisture)=40 C.:outside dry bulb heat (without any moisture)=40 C.TTi:i:inside wet bulb heat(100% moisture)=24.inside wet bulb heat(100% moisture)=24.This values are standards for Alexandria and depends on location.This values are standards for Alexandria and depends on location.Then Then QQt=t=2016 W.2016 W.

II)Q solar heat gain:(sensible)II)Q solar heat gain:(sensible)QQss=Q=Qwallswalls+Q+Qwindowswindows+Q+QroofroofGenerally Q=U*A*(TGenerally Q=U*A*(Toutsideoutside-T-Tinsideinside)) as U as defined and as U as defined and

differs from glass to bricksdiffers from glass to bricks T Too=40,T=40,Tii=24=24..1.Area of walls = area of walls –area of glass. 1.Area of walls = area of walls –area of glass. area of walls from CAD =(29.12*3)-area of walls from CAD =(29.12*3)-

[2*1+2*1]=83.36m squared.[2*1+2*1]=83.36m squared.2.As bank is 2 floors so roof heat gain will be 2.As bank is 2 floors so roof heat gain will be

discarded.discarded.3.Q3.Qglassglass=SHGC*A*SHC=SHGC*A*SHC as SHC is the shading coefficient =0.6,SHGC is solar as SHC is the shading coefficient =0.6,SHGC is solar

heat gain coefficient=400 W/m2 in alexandria. heat gain coefficient=400 W/m2 in alexandria.QQss==960.3+960960.3+960=1920.3 W=1920.3 WNote: Alexandria’s location make direct sun rays only Note: Alexandria’s location make direct sun rays only

from south so I used this rule in design.from south so I used this rule in design.

III) III) Personnel’s heat load:Personnel’s heat load:(sensible and latent):(sensible and latent):

Latent heat loadLatent heat load related to the relative humidityrelated to the relative humidity in air.in air.

Sensible heat loadSensible heat load is the sensed heat from 1 is the sensed heat from 1 person.person.

Qp=Qp Qp=Qp sensiblesensible+Qp +Qp latentlatent Qp Qp sensible sensible =number of persons*Q =number of persons*Q per person sensible per person sensible Qp Qp latent latent =number of persons*Q =number of persons*Q per person latent per person latent

Values of Q Values of Q per person per person is 120 W & Q latent is 160 W.is 120 W & Q latent is 160 W. Average personnel except stairs and bathrooms Average personnel except stairs and bathrooms

and supervisor room is 20.and supervisor room is 20. QQpp=5600 W=5600 W

IV)Light load:(sensible)IV)Light load:(sensible) QQlightlight=N*(watt per =N*(watt per

lamp)lamp)==1111(4*14)+19(2*18)+3*7=1321w(4*14)+19(2*18)+3*7=1321w V)Equipment load:(sensible)V)Equipment load:(sensible)Q=Q=Ʃ wattage for all equipments.Ʃ wattage for all equipments.Assuming 8 computers ,4 printers & 6 phonesAssuming 8 computers ,4 printers & 6 phonesFor pc average wattage is 200 ,70 for printers and 5 For pc average wattage is 200 ,70 for printers and 5

for phones.for phones.Q=1910 WQ=1910 WTotal load =2016+1920.3+5600+1910=11.45 KW.Total load =2016+1920.3+5600+1910=11.45 KW.

22ndnd method method Area=175 meter squared.Area=175 meter squared. So 12 TR is needed then use 20 TR Chiller.So 12 TR is needed then use 20 TR Chiller. Circulation rate 12*195=2340 L/sCirculation rate 12*195=2340 L/s Required fresh air =30* 7=210 L/sRequired fresh air =30* 7=210 L/s AHU circulated air=0.2*2340=468 L/sAHU circulated air=0.2*2340=468 L/s FCU circulated air=1872 L/sFCU circulated air=1872 L/s AHU’s power =468*3=1.4KWAHU’s power =468*3=1.4KW FCU’s power=1872*0.8=1.5KWFCU’s power=1872*0.8=1.5KW Chiller’s power=20*2=40KWChiller’s power=20*2=40KW Due to cost we use 2 AHU 1*1KW &1 *0.5KW.Due to cost we use 2 AHU 1*1KW &1 *0.5KW.

FCU is distributed as following:FCU is distributed as following: 2*0.2KW FCUs at reciption,2*0.2KW at 2*0.2KW FCUs at reciption,2*0.2KW at

Customer services ,2*0.2Customer services ,2*0.2 KW at KW at Customer waiting list Customer waiting list & 1*0.3KW at the & 1*0.3KW at the stairs.stairs.

FFor the supervisor room 1TR is required or the supervisor room 1TR is required so we use 1*0.2KW FCU and for the so we use 1*0.2KW FCU and for the other stairs onther 0.3KW FCU is other stairs onther 0.3KW FCU is installed.installed.

So the total chillerSo the total chiller’s power =44KW’s power =44KW Finally we choose 1 chiller of 50KWFinally we choose 1 chiller of 50KW