Leaving conditions of DX HELP REQUIRED!!!

[abbasi]

Hi

One of enquiries came to me with data as under

Entering DB/WB(F) 104/83
CFM 1200
Capacity 53 MBH

Now I want to design the DX coil Please tell me what will be leaving dry bulb/wet bulb (capacity=4.5xCFMx enthalpy difference) what will be RH of leaving air and how we can draw it on psychometric chartMoreover what will be evaporating temperature?

[Peter_1]

This can only be determined with a practical test.
You have at first the bypass factor, the capacity will also vary if ripple fine tubes are used, flat fins of preformed fins to increase turbulence between the fins, if tubes are positioned in line or in an X-pattern, injection pattern,...

What I do is measure an excisting coil and look in the catalogue what capacity the manufacturer gives.
This gives a very accurate estimate of it.

Or run a simulation program of a manufacturer and look what size of coil it calculates.

[wambat]

To find the enthalpy difference is 4.5 x 1200 cfm x (h1-h2) = 53,000 btu/hr or (h1-h2) = 53,000/5400 = 9.815 btu/lb air
So if 104*f and 83*f WB = Enthalpy 46.764 btu/lb – 9.815 btu/lb = Enthalpy 36.949 btu/lb final condition.
Now working this out on my trusty abacus I get the following final conditions
DB = 74*F
WB 73.279*F
RH = 96.7*F
DP = 73.01*F This is air off ADP ~ 72*f
122.78 = moisture content
13.712 = SV
36.949 = enthalpy
The sensible heat = lbs air x 0.24 x TD = 87.51lbs x 0.24 x 30*F x 60 min = 37,806 btu/hr.
The latent heat = 138.12 gr/lb-122.78gr/lb = 15.34gr/lb x 87.51lb/air = 1342btu/min x 60min = 80,545gr/hr devided by 7000gr/lb = 11.5lbs/hr x 970btu/lb = 11,161btu of latent heat
Total heat = 37,806+11,161 = 48,967 btu. So the SHF is 11,161/48,967 = ~ 77%
I estimated a by pass factor ~ 10%
To prove maybe you should run a test

[Lc_shi]

Hi wambat,
What formula your calculation based on ?It depends on the DX coil paremeters. I don't understand how you estimate. Pls give some explanation. thx!

rgds
LC

[wambat]

Ic-Shi the parameters were given as 1200 CFM, 53000 BTUS, 104*F, and 83*F WB so if you use the formula Q = 4.5 x CFM x (h1-h2) you get 4.5 x CFM x (H1-H2) = 53000 Btu's. So I need to find (h1-h2) and that’s = to 53,000 btu/4.5x1200 = 53,000 btu/5400 = 9.815 btu/lb of air. So if I determine what h1 is by the parameters given and subtract 9.815 btu from it I then know what h2 should be, once I know this I know what the final enthalpy is, when I know what the final enthalpy is I can find the final air conditions

[Lc_shi]

Oh yeah. I see. But you supposed the coil is properly designed. Actually the Q is changeable according to the actual heat transfer performance of the coil. However it could be used by estimation-

thx.
LC

[Peter_1]

I agree your psychometrics Wambat but what are now the actual dimensions of the coil given that we need to evaporate at 0°C/32F?
What velocity will you use?
And how to determine how many rows, tube diameter, injection pattern...
These can't be calculated.

I think that was the underlaying question of Abbasi.

[abbasi]

I will second with PETER


WAMBAT

we all know Sensible heat= 1.08 x cfm x Temp difference

Here u calculated 38000BTU/hr

addin bypass factor in above and subtractin from total gives latent heat. thats basic thermodynamics.

But mate question is how to select evaporating temperature without it no DX coil in world may be designed

also how can u guess RH of leaving air?

we can all calculate wet bulb what about dry bulb

again i wud repeat my question and it is

"what will be evaporating temperature?
and Dry bulb of leaving can u calculate these by drawing on psychometric chart?
what is criteria for choosing evaporating temperature in relation to leaving DB/WB?"

Its a question that i have been trying to find out for somewhile now. I will appreciate if someone comes up with some clue about it

[Lc_shi]

abb,
Based on your data,the leaving air DB/WB can be decided by capacity=air volume*dh formula. As for evaporation temp ,there are more than one solutions which depend on the coil paremeters(rows/FPI/face area/refrigerant type etc). The equation Q=K*LMt is not a linear equation which need trial calcualtion to get the optimal solution.
I don't think the leaving point can be directly draw by pschrometric chart because the out air decided by coil characteristics.

rgds
LC