Air Cooled Condenser Design

[Samarjit Sen]

When the ambient temperature is 42 Deg. C and the designed condensing temperature of the compressor is 50 Deg. C, what should the air temperature on the condenser be considered. The Air Cooled Condenser shall be located outside in the open. The refrigerant is R 22

[Andy]

When the ambient temperature is 42 Deg. C and the designed condensing temperature of the compressor is 50 Deg. C, what should the air temperature on the condenser be considered. The Air Cooled Condenser shall be located outside in the open. The refrigerant is R 22

Hi Samarjit Sen

Ambient temperature in the warmest expected day is your air on. You should not hav eto allow for any air recirculation. If this happens baffles and air straightners on the fans will be required.

Your design has an 8K TD I usually deisn on 10 but then air never gets above 34 deg c in Ireland and seldom above 27 deg c

Kind Regards Andy

[US Iceman]

The ambient dry bulb temperature is considered the entering air temperature. This can only change if you have some air re-circulation back into the inlet side of the condenser, or if there is a warmer exhaust air stream entering the condenser.

If you want to know the exhaust air temperature, you need to know the exact entering air temperature, the exact heat rejection load going to the condenser, and the total volume of air flowing through the condenser. You have to do a heat balance to find the leaving air temp.

[Samarjit Sen]

Thank you Andy, I have designed my previous coils at 8 K and they are working efficiently.

[Lc_shi]

my previous coils at 8 K and they are working effi ciently

.
8K is no problem but you need more heat exchanger surface. I don't know how about the RH of your place. Maybe evaporative consenser is more efficient

regards
LC

[mannym]

Hi guys!
I've just registered in this site. I have a huge problem here trying to design an airconditioning system for a computer equipment container.
I worked out a plant capacity of 6.5kW, the ambient temperature being 40 deg C, room tempereature being 12 deg C with 75RH.

I selected a danfoss/maneurop compressor MT028-4 (R22 refr)with cooling capacity of 7.3kW, Power = 2.55kW, To=7.2deg C (satureted suction), Tc=54.4deg C (saturated discharge).

According to my calculations, the THR at the condenser is: 6.5 + 2.55 = 9.05kW.

I have calculated the TD condenser to be: 54.4 - 40 = 14.4 K. I plan to use a fan/fans with 2m/s air velocity.

For the evaporator TD is: 12 - 7.2 = 4.8K. I plan to use a fan/fans with 2m/s air velocity.

Now i want to calculate the area of the evap & condensor coils including the depth (number of rows).

I was thinking that the this formula would help:
Q = U . A. dT.

Lets assume that both coils will be made from copper fins & tubes. I think the U factor can be looked up in the material specificifications. Change in temperature is already determined in TD. and Q is already calculated. I can obviously find the Area by making A the subject of the formula, but, now how do i know how deep the coils should be?

Please help guys.

[NoNickName]

The number of rows for the two coils depend on a number of factors.
In the evaporating coil, for such a low evaporating temperature, a low number of rows would maximise sensible heat, while a higher number of rows would increase the latent heat. This is a design factor which is related to how much dehumidification you might require.
In a tropical environment 12°C 75% may not be possible. 12°C 95% is more likely. If you want to dehumidify more, than you need to cool more and then reheat, just like in a air handling unit.

For the condensing coil, heat is all sensible despite the number of rows. Here, the determining factor is the fan static pressure. Generally speaking, the higher the number of rows, the more pressure drop. You need to design the coil increasing its depth until the pressure drop reaches the fan static pressure at the desired air flow.

Keep an eye on the geometry, because the turns of the tubes must be even to ensure the inlet and the outlet are on the same side of the coil.
Multiply the number of tubes in height times the number of circuits, and divide by the number of rows. The result must be even.

[lonthia]

What is Power = 2.55 kW ? Power required to drive the compressor ? And 6.55 kW is the heat load of the room, right ?