How to size the chilled water pipe for AHU and FCU?

First Step -  We need to know the total cooling capacity of AHU or FCU. 

Let's say , Total Cooling Capacity of FCU is 22kW and Total Cooling Capacity of AHU is 250kW.

Second Step - We need to calculate the chilled water flow rate. 

Based on Sensible Heat Equation  : q = m c ΔT  

q  = total cooling capacity of AHU or FCU or Chiller  (kW)
c   = specific heat capacity of water  ( 4.19 kJ/kgK)
ΔT = Chilled water supply temp and chilled water return temp (K) . (Normally Ts- 6.7C and Tr- 12.2C)

m = Chilled water flow (kg/s) For Water (kg/s = l/s) Since density of water is 1000 kg/m3. 

For FCU - 22 kW , 

Chilled Water Flow Rate = q / ( c ΔT) = 22 / (4.19 x 5.5) 

So chilled water flow rate for FCU (22-kW) is 0.955 l/s (kg/s).

For Chilled Water Piping ,As per ASHRAE Guideline; Less than or equal of 50mm Pipe size ,

Water Velocity shall be less than 1.2 m/s . 

Normally we design 1 m/s for less than or equal 50mm pipe size . 

Based on Continuity Equation          Q = A V  

Q = Volumetric Flow Rate (m3/s)
A = Area ( m2 ) 
V = Average Water Velocity  (m/s)

A = Q / V = 0.955 x 10^-3 / 1 

A = 955 mm2 

r = 17.4 mm ( A = π r2 ) 

D = 34.8 mm  ,  Nominal Diameter is 32mm and Actual inside diameter of 32mm is 35mm. 

So that chilled water pipe size for FCU (22kW) is 32mm.

For AHU (250 kW) , 

q = mc ΔT 

m = Q = q/ c ΔT   = 250 / (4.19 x 5.5 ) = 10.85 l/s ( kg/s) 

Q = 10.85 l/s . 

For AHU Chilled Water Pipe size ,  As per ASHRAE Guideline ,

Water Pressure Drop Shall be not more than 400 Pa/meter run. 

Most of the project design are using 250 Pa/m  

Water Velocity shall be less than 2.4 m/s 


Based on Flow Rate - 10.85 l/s and not more than 250 Pa/m , Pipe size shall be 100mm. 

If we choose pipe size 100mm, Water Pressure drop shall be 175 Pa/m and

Water Velocity shall be 1.32 m/s . 

Thanks a lot. 

Zaw Moe Khine

RSED Engineering.

Design Supply Air Temp for Peak Load

When we design the air con, we need to specifies how much the supply air volume is. If we need to calculate the supply air volume, we need to specifies supply air temp. So, this post I'm going to discuss about design supply air temp.
Most common design supply air temp is 13 C to 15C dependent on cooling source and cooling coil configuration.

My personal default Design Supply Air Temp is 14C because most common coil default leaving air temp is 12.8C (55 F). So, If I choose design supply temp is 14C, leaving air temp from coil will be around 12.7C to 13C.  


By right,before we specifies the supply air temp,we need to specifies operation temp and RH %. So, typical common set point is 23+-1 & 50% RH.

For Singapore Design Parameter,Based on SS-554, maximum RH % is less than 65 on both Peak & Part Load Condition. So, our Peak Load condition should be 50 % ~ 55% coz if peak load is about 60%, part load may be more than 65 % and may not comply.

Just check with sample.
Space Area is 100m2 and Ceiling High is 3m.
Sensible Heat Load is 10 kW and Latent Load is 1.76 kW.
Meaning that the space have 85% sensible heat load & latent heat load is 15%.
So,SHR is 0.85 (SH/TH).
Space Operation Temp is 24C & RH-50% (Less than 55%) at Peak Load condition.
What's the design supply air temp and air volume?

First step, draw the line on Psychrometric chart by Standard Point and SHR.

So,  As per Psychrometric Chart, If the space have 0.85 SHR and we need to get exactly 50% RH, cooling coil ADP should be 11.3C. And then Space dew point is about 13C, so that cooling coil temp should be less than or equal 13C in order to get not more than 55% RH.
As I said above, normal cooling coil leaving air temp is about 12.8C (55F) and cooling coil bypass factor is 0.1 (10%). The room operation temp is 24C and then mixing temp with fresh air is about 25C. So back work with formula, Coil ADP = {Ts-(TmxBF) }/ 1-BF.
Supply air temp will be 14C. So, normally I did choose Design Supply Air temp is 14C.
Let check again with Design Supply Air Temp-14C, how much the RH in the space.

Just draw the line from point Coil ADP-12.8 to parallel with the line.
(The line from SHR to standard point).

Intersection point with 14C DB (Supply Air Temp) is leaving air temp 
and the room operation temp is 24C & 54% RH at Peak Load.
So, the leaving air enthalpy is about 37 kJ/kg.
Most common leaving air enthalpy is 35 ~38 kJ/kg coz supply air temp is 13C~15C.

So, The room Operation temp 24C & RH-54% is acceptable,
Make the supply air temp is 14C.

If supply air temp is 14C and the space sensible load is 10kw, we can calculate the design supply air volume. 
Q =  m x c x ∆T 
Q = 10 kw (kJ/s)
 c  = 1.004 kJ/kg
∆T = the room operation temp - design supply air temp (24C - 14C)

So, m = Q/( c x ∆T)............. (kg/s)
m = 10 / 1.004 x 10 = 0.996 kg/s.
V = m x v 
v = Humid Volume (m3/kg)
From Psychrometric Chart, Supply Air v is 0.825 m3/kg.
V = m x v = 0.996 x 0.825 (m3/s)
V = 0.822 m/s = 2960 m3/hr (CMH).

So, The space sensible load is 10 kw ,SHR is 0.85 & Supply air temp is 14C,
 Supply air volume should be 2960 cmh. After calculate the supply air volume,
we need to check it that it's sufficient air circulation to meet minimum air circulation.
It's checked by minimun MV requirement. For office building type, minimum air circulation should be 6 ACH. That's mean Area x Ceiling High x mini ACH= 100 x 3 x 6 =1800 cmh.
So,our supply air volume 2960 cmh is more than enough compare with minimum circulation.

Let check the cooling coil load for the room if mixing air temp is 25C and RH is 60% by fresh air make up.

Draw the line 14C db for supply air temp and then Mixing Air Temp-25C & RH-60%.
The cooling coil rate is 
Q = m x ∆h
m = 0.996 kg/s
∆h = (h2-h1) (kJ/kg)
h2 = mixing air temp enthalpy -55 kJ/kg
h1= Leaving Air temp enthalpy- 37 kJ/kg.
Q = m x ∆h = 0.996 x (55-37) = 17.93 kW.

Summary will be-
Total Cooling Coil is 18 kW.
Design Supply Air Temp is 14C.
Design Supply Air Volume is 2960 cmh.

Thanks.
Zaw Moe Khine.

Sample Heat Load Calculation for General Office Meeting Room

Sample Heat Load Calculation for General Office Meeting Room


Heat Load Calculation -
-Cooling Capacity (kW) ,
-Supply Air Volume (CMH),
-Supply Fresh Air Volume (CMH),
-Chilled Water Flow Rate (l/s) if CHW System is used. 


When we calculate the heat load,
We need some information for space.
1-Location of space
(Ambient Temp for out door air )
(Singapore -Summer Ambient Temp - 32.8C/26.1C (DB/WB)
2- Room/ Space Datas & Orientation (Facade Orientation)
(Area-50 m2  & Room is located at North Side of Building)
3-Room Construction Materials
(Such as Window Glass (U-value & Sc-Value) ,Wall, Partition & Floor.)
4-No of Occupancy & Room Usage
(20-Persons & Meeting Room)

I-Internal Heat Gain
i-Heat 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-Electrical Equipments Load ( General Office 25 W/m2)
Laptop-55 W x 20 & Projector-300 W.
Electrical 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 brick 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 
 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

iv- 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 load- 
Sensible 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 = enthalpy differential of entering air and leaving  air (kJ/kg)

For  ∆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 ℓ c ΔT

Q = Cooling Capacity (kJ/s) (kW)

V = Water Flow (m3/s)

ℓ = 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 / ℓ 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 








 

HEAT LOAD CALCULATION

When we calculate the heat load, we have to know the location of space (Country,City)
And then building information such as building type,material,orientation & authority standard (Which standard we have to follow? local or international ).
Main Heat Source are Internal Heat Gain , External Heat Gain and Heat Gain from OA (FA)
See following Heat Load Chart.

1-Internal Heat Gain

• Heat Gain from Lighting 
• Heat Gain from Human
• Heat Gain from Electrical Equipments

2-External Heat Gain

•  Heat Gain from Wall 
•  Heat Gain from Window
•  Heat Gain from Roof
•  Heat Gain from Floor
•  Heat Gain from Partition

3-Heat Gain from Outdoor Air /Fresh Air 

•   Fresh Air Requirement for Comfort Air
•   Fresh Air by infiltration  

http://www.scribd.com/doc/111569258/Load-Calculation-Spreadsheets-Quick-Answers-Without-Relying-on-Rules-of-Thumb?secret_password=1lkzrbj8jnpsqznoyf54

PRECOOLED AHU CALCULATION

Just start with precooled AHU Calculation!!

When you want to size the precooled AHU,

you need to know fresh air volume and temperature set point.

Once you know this two info, you can size the AHU.

Just start calculate for 3000 l/s fresh air volume and set  point temp is 16 C.

We all know that, there are two kinds of heat for fresh air,Sensible and Latent Heat.

Cooling process for Sensible and De-humidification for  removing of moisture.

For Sensible Heat Load,(Cooling Process).

Q = m c ΔT 

Q = V ℓ c ΔT

Q = Cooling Capacity (kJ/s) (kW)

V = Air Volume (m3/s)

ℓ = Density of air (kg/m3) (1.204 kg/m3 for standard air )

c = Specific Heat Capacity of air (kJ/kgK) (1.004 kJ/kgK for standard air)

ΔT = Temperature differential.

m= mass of air (kg/s)

V= 3000 l/s = 3 m3/s. (1000 l/s = 1 m3/s)
ℓ = 1.204 kg/m3 for standard air @Sea level.

c = 1.004 kJ/kgK for standard air.
ΔT = Ambient temperature - Set  point temp. 

ΔT =(33 - 16 ) (Singapore Summer Ambient Temp - Set point temp)

Q  = 3 x 1.204 x 1.004 x ( 33-16)

Q = (m3/s) x ( kg/m3) x (kJ/kgK) x (K)

Q = 61.65 kW.

For Latent Heat Load, (De-humidification Process/Removing of Moisture )

Q = m hfg Δg

Q = V ℓ hfg Δg

Q = Cooling Capacity (kJ/s) (kW)

V = Air Volume (m3/s)

ℓ = Density of air (kg/m3) (1.204 kg/m3 for standard air )

hfg = heat of vaporization of water (kJ/kg) (2454 kJ/kg for standard air )
hfg= 2500 kJ/kg (Estimated for 16C)

Δg = Humidity Ratio (Moisture Content of dry air) (kg/kg)

m= mass of air (kg/s)

Q = V ℓ hfg Δg

Q = (m3/s) x ( kg/m3) x (kJ/kg) x (kg/kg)

Q = 3 x 1.204 x 2500 x  Δg

So that we need to see Psychrometric Chart for Humidity Ratio.
First Point is Ambient Temperature. (Singapore Summer Temp - (33 DB / 26.1 WB )
From Psychrometric Chart, g1 is 0.0186 kg/kg (18.6 g/kg) Based on 33DB/26.1 WB.
For Second Point, Supply Air Temp and Coil ADP or Coil Curve.
Coil ADP is based on Coil bypass Factor.
Normal Standard coil bypass factor is 0.1.
If bypass factor is 0.1 , Coil ADP is {16 - (33x0.1) }/(1 - 0.1) = 14.11 C.
Based on Supply Air temp 16C and Coil ADP (14.11 C),
From Psychrometric Chart , g2 is 0.0108 kg/kg (10.8 g/kg).
So that Δg = g1-g2

Δg = 0.0186 - 0.0108 = 0.0078 kg/kg.

Q = 3 x 1.204 x 2500 x  Δg

Q = 3 x 1.204 x 2500 x  0.0078

Q = 70.434 kW

Total Cooling Capacity for AHU is Sensible Load + Latent Load.

Total Load = 61.65 + 70.434 = 132.084 kW

NEXT OPTION CALCULATION WITH ENERGY METHOD

Q = m Δh
Q = V  ℓ  Δh

Q = Cooling Capacity (kJ/s) (kW)

V = Air Volume (m3/s)

ℓ = Density of air (kg/m3) (1.204 kg/m3 for standard air )

Δh = Enthalpy Differential of entering and leaving air to the coil (kJ/kg)

m= mass of air (kg/s)

Q = V  ℓ  Δh

V = 3 m3/s
ℓ = 1.204 m3/kg
Δh = Enthalpy Differential of entering and leaving air to the coil (kJ/kg)
Δh = Enthalpy of entering air (Off coil air) - Enthalpy of leaving air (On coil air)
From Psychrometric Chart,
First Point, h1 , Enthalpy of Entering Air is based on Ambient Air temp
(33 C DB/ 26.1 C WB), h1 = 81 kJ/kg.
Second Point, h2 , Enthalpy of Leaving air is
Based on Supply Air Temp  16C ,Coil ADP & Entering Point.

h2= 44 kJ/kg.
So that, Δh = h1-h2 = 81-44 
            Δh = 37 kJ/kg.

Q = V  ℓ Δh 

Q = 3 x 1.204 x 37

Q = 133.644 kW.