Cooling Coil question

[hvacn00b]

Hello everyone

I'm a newcomer in the industry and I was hoping that I could get some advice regarding the problem that my boss experienced at one of our job sites.

We have an existing chilled water cooling coil (1000 mm wide x 900 mm high x 200 mm deep). Number of rows is unknown but we think it's 6. Carried out a performance test last week and here are the results:

Air Side
Flowrate: 1500 l/s
Entering temperature: 16.3 C
Leaving temperature: 8.8 C

Water Side
Flowrate: 3.3 l/s
Entering temperature: 7.6 C
Leaving temperature: 8.5 C

Pressure drop across the coil: 68 kPa

What he wanted to know is why we are getting very low TD of water.

Any advice would be greatly appreciated. Thanks

 

[ChasBean1]

Your boss is challenging your engineering skills. The energy extracted from the air should be the same as the energy gained in the water. Shake off all the extraneous givens...

Assuming the air has no latent change (no condensation on the coil; meaning the entering air dew point is below the coil surface temperature), each side can be represented by:

Q = m Cp dT
changing mass flow rate to volume flow rate:
Q = rho V Cp dT

• Q is energy in watts or BTU/hr
• rho is water or air density
• V is the volume flow rate of water or air
• dT is the coil differential temperature

The math works out to 13.6 KW on the air side and 12.4 KW on the water side. Pretty darn good considering potential metering errors.

-CB

 

[hvacn00b]

Hi ChasBean1,

Thanks so much for your reply.

Anyway, no my boss is not testing me this time.. I've actually worked out the same calculations and had the same figures as you but according to him, the coil should give more than 12-13 kW.

I've also gone through my textbook and found a formula for calculating the maximum possible heat transfer given the measured parameters:

Q max = Cmin x (EAT - EWT)

Cmin = the lowest value of mass flow rate x specific heat
EAT = entering air temperature
EWT = entering water temperature

Now, theoreticaly we should be able to get around 29 kW..

Also, after speaking to a sales consultant giving him the coil size, and based on 2.5 l/s water flow, 6 rows of coil, 12 fins per inch - we should be getting around 53 kW.

 

[Yorkman]

I noticed that none of the calculations dealt with dehumidification at 7.6[°]C I would bet your seeing some latent heat removal. I wonder is the chilled water a %glycol solution and is this affecting your calculations. Also could it be that the water side was not piped in counter flow arrangement, that could be affecting your coil performance.

I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[IRstuff]

Your airflow doesn't appear to be fast enough to hit the HT values calculated.

I'm getting something like 3.7 mph of air flow, resulting in a htc of about 9 W/m²ºC, which results in a total power transfer of about 12 kW, assuming a total fin area of 170 m².

In order to hit 29 kW, you'd need a higher htc, viz. more air flow, or a higher delta temp.


TTFN

 

[hvacn00b]

Thanks for your reply Yorkman..

I will have someone to check on the humidity, glycol and also the piping arrangement.

 

[hvacn00b]

IRstuff,

I've looked around on the net and found that the recommended airflow velocity for chilled water is:
Sensible cooling: 1-6 m/s
Latent cooling: 1-2.8 m/s

Say if I use the value 2.5 m/s, for coil of this size I can increase the airflow to 2250 l/s. But now the maximum possible heat transfer becomes 43.76 kW. Hmm... I'll see if we can increase the airflow and will post the results...

 

[Yorkman]

IRstuff is right, look at your approach of the coil
L.A.T. - E.A.W. your doing 1.2[°]C usually I see 2.3 - 3.6[°]C Your leaving air temperature can't get any colder than your entering water temperature. The only way to get more heat removal from the coiil is to increase air flow, this will increase the [Δ]T on the water side.

I'm not a real engineer, but I play one on T.V.
A.J. Gest, York Int./JCI

 

[AbbyNormal]

Add glycol, run the coil colder. That air is cool to begin with, not much temperature differential between the air and the water.

As I mentioned elsewhere, if that air was say 24 to 27C with a dewpoint of say 13C, that coil could be doing 3 to 3.5 times as much cooling as it is doing now with some dehumidification.

That airflow is low for that coil, almost like you were trying to remove as much moisture as possible except, the entering air dewpoint has to be low so nothing is coming out.

Based on that face area, typically you could be moving 2260 l/s if there was dehumidification, or maybe even 2700 l/s since it seems to be all sensible cooling.

I still think you could be piped for parallel flow