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# CH water calculations

## CH water calculations

(OP)

Can I used this equation to determine Chilled water GPM: Qt= GPM x 500x Delta T
Qt= Total heat @ BTU/hr. Latent and sensible heat
Delta T= Temp. difference between entering and leaving cooling coil.

My point:Since the chilled water can handle sensible and latent load (Qt which is equal to: CFM x 4.5(Delta H)
Delta H= enthalpy difference, in and out for Air.
Based on the above equations, CFM x 4.5 (H2-H1)= GPM x500 x Delta T. Is this correct assumption. something is telling me that this cant be Qt, it must be the Qs, sensible load so, Qs= GPM x 500 x delta T. I am sure this works for heating as this is only sensible load but for cooling? need some assistance Brothers. What is the better equations to use.

### RE: CH water calculations

Tell "something" to go away. The equations are correct. To confirm, check ASHRAE Handbook of Fundamentals.

### RE: CH water calculations

(OP)
Thank you Bro Trashcan man.

I know this will be redundant but I'll confirm it anyway. If I was given the total load (Qt=Btu/hr) of a cooling coil then I can use
the equation Qt=500 x GPM x delta T in determining the GPM for the coil.
Thank you in advance for those inputs.

### RE: CH water calculations

Yes correct, but the Qt of the coil was sized by entering and leaving air and water conditions. If your conditions are not the same as the coil selection, you will not get the same performance out of it. In other words, let's say your air side load has a Qt of 100,000 BTUH. A water side equation to reach 100,000 BTUH would be (500)(20 GPM) (10 deg). However, (500)(40 GPM)(5 deg) also yields 100,000 BTHU. So more information is needed to determine the appropriate flow for your application. You need to have a flow and delta-T appropriate for the coil.

But to directly answer your question, ignoring small elevation and density differences:

Water (no glycol): Qt=(500)(GPM)(Delta-T)
Air: Qt=(4.5)(CFM)(Delta-Enthalpy)

### RE: CH water calculations

Important distinction:
In Qt= GPM x 500 x Delta T
Delta T is the Delta T of the water.

In Qt = CFM x 4.5(Delta H)
Delta H is the enthalpy change of the air.

Your Qs equation is incorrect; correct equation is
Qs = CFM x 1.1 x Delta Tair

Your algebra for determining GPM given Qt is correct.

### RE: CH water calculations

(OP)
@ Bro dbill74
"Your Qs equation is incorrect; correct equation is
Qs = CFM x 1.1 x Delta T air"
So based on your comment I cant use the equation: Qs= 500 x GPM x Delta T even for heating? I qualified this in my OP
@BronYrAur
Now, can I used the energy balance equation Qs (air side)= Qs (water side). CFM x 1.08 x Delta Tair =
GPM x500 x Delta Twater. will this energy balance work, even for cooling? I am quite positive that the air side and water side of the equation applies to both heating and cooling but I maybe missing something.
Thanks a lot for the help.

### RE: CH water calculations

I am not in a place where I can type very well but no. you're air equation in the last post is sensible only. You have forgotten about the latent. you must use the equation with the change in enthalpy for cooling. if your application is strictly Heating then there will not be a latent difference in the air so you can get away with CFM x 1.08 x delta T

### RE: CH water calculations

(OP)
@ Brony, Thank you and to the others. Till the next follow up question later.

### RE: CH water calculations

To say Qs = 500*GPM*ΔT is somewhat erroneous.

This equation is only applicable to the water side of the coil. In these water filled coils, the water does not undergo a phase transition like a refrigerant does. ALL heat goes into or out of the water. Therefore the correct form of the equation is:

Q = 500*GPM*ΔTwater

Although the "water" label of the Temperature variable is often left off as being implied.

Now, the air side of the coil gets a bit more complicated because we are dealing with air. And not just air, but moist air. The water vapor in the air has to be taken into consideration and is the reason we have multiple equations for this side of cooling coils.

Qs = 1.08*CFM*ΔTair assumes water vapor in the air does not change phase. This is fine for heating air, but only partially works for cooling. As air cools, it has less ability to hold water vapor; defining the shape of your psychometric chart. Cool air enough and water vapor is forced to undergo a phase change (a.k.a. condensate). To account for the water that is condensing out of the air we need another equation, which is where Qt = 4.5*CFM*Δh comes in. Where Qt is Total heat transferred.

### RE: CH water calculations

To dovetail on dbill74, the latent side of the air equation is Ql=(CFM)*(0.7)*(Delta Humidity ratio), where humidity ratios are in grain per pound of dry air. So on the air side, you can use the Qs and Ql separately to calculate sensible and latent loads. Or you can use the delta-enthalpy equation to calculate the total load in one shot.

### RE: CH water calculations

for energy calculation in real estate offices to understand and impose the energy bill to tenants, using this same formula 500*gpm*delt T. virtually it is same.
example for 1-ton machine

waterside: 2.4*500*10f=12,000 btu.

for air side: 4.5*400(cfm)*7(delta enthalpy)=12,600btu.

the concept is different but the result is approximately same.

### RE: CH water calculations

(OP)
Thank you Bros. I really appreciate the inputs. Hope to return the favor later to other engineers.

### RE: CH water calculations

Dear Engineers .
If the chilled water DT is 12F ( 42(in) 54(out).
is the Air flow will be the same if the DT reduced to 10f ( 44/54 F ).
and what the air flow temp. (out off coil) will be the same ?

### RE: CH water calculations

the question is a little confusing, but this is where the specific construction of the coil comes into play. also the wet-bulb temperatures. there is not enough information given. with a lower water temperature differential, you should expect a lower air temperature differential at the same flow rate, but it can't be Quantified without more information.

### RE: CH water calculations

In the equations provide above, except where there is a constant (500, 4.5, 1.08), EVERY variable is ... variable.

If you change one variable (ΔTWATER), one or more other variable(s) must change.

### RE: CH water calculations

You can not define the required enthalpy change to meet the load without separately calculate sensible and latent loads to determine to total load.

### RE: CH water calculations

(Qt= GPM x 500x Delta T )& (Qs = 4.5xCFMx(ΔH) )is right for Total Load Qt and not Qs Sensible Heat, Sensible Heat or Latent heat in a System depends on the room/area to be cooled/heated which can be found by cooling Load/ Heating Load Calculations, Sensible Heat and Latent Heat is used in coil sizing but the overall flow requirement for the coil is really as per your given equation

Qs = 1.08xCFMx(ΔT)

Ql = 0.68xCFMx(ΔW)

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