## 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.

Thank you in advance.

## RE: CH water calculations

## RE: CH water calculations

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

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

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 T

_{air}Your algebra for determining GPM given Qt is correct.

## RE: CH water calculations

"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

## RE: CH water calculations

## RE: CH water calculations

To say Q

_{s}= 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*ΔT

_{water}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.

Q

_{s}= 1.08*CFM*ΔT_{air}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 Q_{t}= 4.5*CFM*Δh comes in. Where Q_{t}isTotalheat transferred.## RE: CH water calculations

## RE: CH water calculations

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

## RE: CH water calculations

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

## RE: CH water calculations

If you change one variable (ΔT

_{WATER}), one or more other variable(s) must change.## RE: CH water calculations

## RE: CH water calculations

Qs = 1.08xCFMx(ΔT)

Ql = 0.68xCFMx(ΔW)