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Sample Heat Load Calculation for General Office Meeting Room
January (2) Sample Heat Load Calculation for General Office Me...
Sample Heat Load Calculation for General Office Meeting Room
Zaw Moe Khine !"##"$
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HEAT LOAD CALCULATION
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Popular Posts !
2012 (3) Sample Heat Load General Office Mee Sample Heat Load General Office Mee Calculation - -Cooli upply Air V olume olume ( PRECOOLED AHU Just start with prec Calculation!! When precooled AHU, yo air volume and tem
Heat Load Calculation -Cooling Capacity (kW) ,
Design Supply Air When we design design th specifies how much volume is. If we we nee supply air volume,
-Supply Air Volume (CMH), -Supply Fresh Air Volume (CMH), -Chilled Water Flow Rate (l/s) if CHW System is used.
HEAT LOAD CALC When we calculate calculate t to know the locatio (Country,City) And information such as
W hen hen we calculate the heat load, W e need some information for space. 1-Location 1-Location of space (A mbient mbient Temp for out door air ) (Singapore (S ingapore -Summer Ambient Temp - 32.8C/26.1C (DB/WB) 2- Room/ 2- Room/ Space Datas & Orientation (Facade Orientation) (A rea-50 rea-50 m2 & Room is located at North Side of Building) 3-Room 3-Room Construction Materials (Such (S uch as Window Glass (U-value & Sc-Value) ,Wall, Partition & Floor.) Floor.) 4-No 4-No of Occupancy & Room Usage (20-Persons (2 0-Persons & Meeting Room)
ACMV DESIGN DESIGN COURSE ACMV DESIG DESIGN COURSE ACMV E-20(Hourly Analysis Program-H Program-H ACMV DESIG DESIGN COURSE (SATU YOU WA... (no title) http://www.scribd.com/doc/1115 Calculation-Spreadsheets-QuickRelying-on-Rules-of-Thumb? secret_password=1lkzrbj8jn...
I-Internal IInternal Heat Gain i-Heat iHeat Gain from Lighting (Based on SS-530-Office Lighting Power - 15 W/m2) Lighting Load = 50 x 15 = 750 W ii-Heat Gain from Human (20-person) (Based on ASHRAE Office Human Sensible -75 W/person & Latent 55 W/person) Human Load (S) 75 x 20 = 1500 W ,(L) 55 x 20 = 1100 W iii-Electricall Equipments Load ( General Office 25 W/m2) iii-Electrica Laptop-55 W x 20 & Projector-30 Projector-300 0 W. Electrical Equipments Equipments Load= 30 x 50 = 1500 W II-External Heat Gain i-Heat Gain from Wall (Conduction Heat Gain) (Q= A x U x "T) Wall Area = 10 m x 1m = 10m2 Wall U-value(200mm U-value(200mm brick wall wall = 2.254 W/m2) "T for North side facing 200mm brick wall =
14
So, Q= 10 x 2.254 x 14 = 315.56 W ii- Heat Gain from Glass Window. ( Conduction + Radiation Heat Gain) Glass Area = 10 m x 3 m = 30m2 Glass U-value = 1.7 W/m2 Glass Sc-Value = 0.3 0.3
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CF = 0.8 (Q= (3.4 x A x U ) + (211 x A x Sc x CF) Q = (3.4 x 30 x 1.7 ) + (211 x 30 x 0.3 x 0.8 ) = 1692.6 W iii- Heat Gain from Partition Wall ( Conduction Heat Gain) (Q= A x U x "T) Partition Area = 20 x 4 = 80 m2 Partition Wall U-Value = 0.573 W/m2 "T (30-24) = 6
Q = 80 x 0.573 x 6 = 275 .04 W i v - Heat Gain from Floor ( Conduction Heat Gain) (Q= A x U x "T) Floor Area = 50 m2 Floor U-Value = 2.452 W/m2 "T (30-24) = 6
Q = 50 x 2.452 x 6 = 735.6 W After adding out the total heat loadSensible Heat Load is 6746.8 W (10% SF) 7421.53 W = 7.421 kW. Latent Heat Load is 1100 W (10% SF)-1210 W =1.21 kW SHR = SH/(SH+LH) = 7.421/(7.421+1.21) = 0.86 That's mean 86% is Sensible and 14 % is Latent heat. Once we get the total sensible heat load, we can get the supply air volume. Qs = m c
"T
Qs = 7.421 kW (kJ/s) c = 1.004 kJ/kg If supply air temp is 14 C, "T = 24-14 = 10. m = Qs / (c x "T)
- (kg/s)
m = 7.421 / (1.004 x 10 ) = 0.73919 kg/s V,Air Volume (m3/s) = m x v (Humid Volume-m3/kg) If supply air temp is 14 C, v is about 0.825 m3/kg. So, Air Volume , V = 0.73919 (kg/s) x 0.825 (m3/kg) V = 0.6098 m3/s = 2195 CMH. So, Supply Air Volume is 2195 CMH. Next step is Fresh Air Volume Supply, If the space type is under office, Minimum FA Requirement is based on 62.1 ASHRAE (2.5 l/s/person + 0.3 l/s/m2)= (2.5x20) + (0.3x50)=65 l/s Minimum FA Requirement based on local code (SS-553). 5.5 l/s/person or 0.6 l/s/m2 (Choose higher volume) 5.5 x 20 = 110 l/s. So, We have to choose 110 l/s for this meeting room. Fresh Air Supply will be 110 l/s (396 CMH).
Air Con Sketch
So,supply air volume 2195 cmh &
fresh air volume-396 cmh for this meeting room. Layout will be as per air con sketch. Just need to calculate how much capacity we need. For Cooling Load Capacity , Q = m !h m = mass of air (kg/s) (we already calculated) !h
For
= enthalpy differential of entering air and leaving air (kJ/kg) !h = h1-h2
h1= enthalpy of entering air into coil (maxing air) h2= enthalpy of leaving air from coil (supply air) So that we need to calculate maxing air temp. Mixing air volume is same as supply air volume. See Air con Sketch dwg. In our design condition, Supply Air Temp is 14 C & Room Operation Temp is 24C(maxi) & RH is 50~55 %. Fresh air temp is Singapore Summer Temp , 32.8 C (DB) & 26.1 C (WB). So that we can calculate maxing air temp (entering air temp/on coil air temp). Supply air volume (maxing air volume) is addition of return air volume and fresh air volume. So, maxing air temp is depended on ratio of return and fresh air volume. Equation will be Tm x Mixing Air Volume = (Tr x Return Air Volume) + (Tf x Fresh Air Volume) Other way is Tm x 100% = (Tr x Return Air % of Supply Air) + (Tf x Fresh Air % of Supply) So we just calculate fresh air volume % of supply air volume = 396 / 2195 x 100 = 18% Fresh air volume is 18%, return air volume will be 100-18= 82%. Tm x 100 = ( Tr x 82) + (Tf x 18) Tr = Room Operation Temp (24 C) (It should add 1 or 2 C for duct heat gain and plenum heat gain) Tf = Fresh Air Temp (32.8 C). Tm x 100 = (24x82) + (32.8 x 18)
Tm = 25.58 C Mixing Air Temp (DB) is 25.58 C. Next step is maxing air temp (WB ) or Humidity Ratio ( "g) Same as maxing air temp (DB)equation will be gm x Mixing Air Volume = (gr x Return Air Volume) + (gf x Fresh Air Volume) gm x 100 = ( gr x 82) + (gf x 18) gr = Return Air Humidity Ratio Read from Psychrometric Chart by Room Operation Temp ( 24 C & RH-54%) gr=9.8 g/kg = 0.0098 kg/kg gf = Fresh Air Humidity Ratio. Read from Psychrometric Chart by Singapore Summer Temp ( 32.8 C & 26.1 C) gf = 18.6 g/kg = 0.0186 kg/kg. So gmx100 = (0.0098 x 82)+(0.0186x18)
gm = 11.83 g/kg = 0.01183 kg/kg By Maxing Air Temp (DB)
Tm,25.58C & Mixing Air Humidity Ratio (gm) 11.83 g/kg,
h1=55 kJ/kg (From Psychrometric Chart) For h2, Read from Psychrometric Chart Based on supply air temp,coil ADP,Space SHR & intersection line. Supply Air temp is 14 C. Space SHR = 0.86 Coil ADP = 12.71 C (By coil bypass factor is 0.1)
h2= 37.2 kJ/kg. So, Cooling Capacity is
Q = m !h m = 7.421 / (1.004 x 10 ) = 0.73919 kg/s !h = h1-h2 = 55- 37.2 = 17.8 kJ/kg
Q = 0.73919 x 17.8 Q = 13.15 kJ/s (kW)
Fan Coil Unit minimum cooling capacity is 13.15 kW. Cooling Rate is 13.15/50x100 = 263 W/m2. Supply Air Volume is 2195 CMH Fresh Air Volume is 396 CMH.
If FCU is chilled water system and CHW !T is 5.5, Chilled Water Flow rate ?
Q = m c !T Q = V " c #T Q = Cooling Capacity (kJ/s) (kW) V = Water Flow (m3/s) " =
Density of water (kg/m3) (1000 kg/m3 for water)
c = Specific Heat Capacity of Water (4.19 kJ/kg) m= mass of water (kg/s) !T = 5.5
V = Q / " c !T = 13.15 / (1000x4.19 x 5.5) V = 0.00057 m3/s (1m3/s = 1000 l/s) So, V = 0.57 l/s. So Chilled Water Flow Rate is 0.57 l/s. So that for CHW System, m = V = .... (Unit is l/s) Q = m c !T m = Q / (c x !T) ... (l/s)
Thanks, zaw moe khine
Posted by Zaw Moe Khine at 01:18 !#5 %&'())&*+ ,-./ (* 0((12&
Labels: Sample Heat Load Calculation
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