How to make a quick relation between CFM and Ton
How to make a quick relation between CFM and Ton
(OP)
Is it correct to use the assumption of 340CFM per ton?
Thankyou
Angel
MEP Engineer
Thankyou
Angel
MEP Engineer
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How to make a quick relation between CFM and Ton
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RE: How to make a quick relation between CFM and Ton
return air:
enthalpy = 31.43 Btu/lbm (dry air)
humidity ratio = 0.01067
specific volume = 13.89 ft3/lbm (dry air)
rh = 45.6 percent
supply air:
enthalpy = 24.59 Btu/lbm (dry air)
humidity ratio = 0.00890
(psychrometric data from PMTHERM)
Given we have 400 cfm of air, the mass flow of dry air becomes:
400 cfm / 13.89 ft3/lbm = 28.80 lbm/min (dry air)
Sensible load is then calculated by taking the difference in enthalpies, and multiplying by the mass flow of dry air:
28.80 * (31.43 - 24.59) = 197.0 Btu/min = 0.985 tons
But we also have a latent load, since our humidity ratio has decreased, i.e., water is condensing at the coil. Assuming the water is condensing at the final air temperature, 62°F, enthalpy to condense water is 35.11 Btu/lbm using a handy ASHRAE table. Latent load is then calculated:
28.80 * 35.11 * (0.01067 - 0.00890) = 1.790 Btu/min = 0.009 tons
Total cooling load is then: 0.985 + 0.009 = 0.994 tons.
Of course, if your state point are significantly different, simply follow this procedure to equate CFM to tons.
RE: How to make a quick relation between CFM and Ton
To calculate sensible load, we need to establish change in enthalpy at constant humidity ratio, therefore, at 62°F dry bulb and humidity ratio = 0.01067:
enthalpy = 26.52 Btu/lbm
Sensible load then becomes:
28.8 (31.43 - 26.52) = 141.4 Btu/min = 0.707 tons
Total load is solved using the conservation of energy equation:
ma * h1 = ma * h2 + mw * hw + q
where:
ma = mass flow of dry air
h1 = return air enthalpy
h2 = supply air enthalpy
mw = mass of water removed
hw = enthalpy to condense water
q = total load
And the conservation of mass equation:
ma * W1 = ma * W2 + mw
where:
ma = mass flow of dry air
W1 = return air humidity ratio
W2 = supply air humidity ratio
mw = mass of water removed
Combining these equations, we get:
q = ma * (h1 - h2) - ma * hw * (W1 - W2)
Using the above equation, and the numbers established in the previous post, we get:
Total cooling load: 0.985 - 0.009 = 0.976 tons
and not: 0.985 + 0.009 = 0.994 tons
The latent load becomes: 0.976 - 0.707 = 0.269 tons
Hopefully, I did not confuse the point I was trying to make...
RE: How to make a quick relation between CFM and Ton
That purely depends upon the type of load. Here are the two simplified equations.
Total Load (Btu/Hr) = 4.5 x CFM x (Entering air enthalpy - Leaving air Enthalpy)
Sensible Load (Btu/Hr) = 1.08 x CFM x (Entering air DB - Leaving air DB)
For example if I consider the air condition at entrance as 240C and 75% RH and that of leaving air as 200C and 50% RH across a coil then the CFM comes out to be around 300 for 1TR (1 TR = 12000 Btu/Hr)
If your load is purely sensible, the CFM can be quite high.
RE: How to make a quick relation between CFM and Ton
You must know that the value of cfm/TR varies with the application and the region, the figure you use in England for one application you can't use in Dubai for the same application.
This figure also depends on your refrigeration equipment,
ie. the rule of thumb for DX applications is 300 cfm/TR,
and the variations around this figure should be minimized for optimum performance.
Regards,
RE: How to make a quick relation between CFM and Ton
This doesn't work for all applications because if the central system is 100% outside air with a design condition of 91°F DB/74°F WB, each 244 cfm would be required to be capable of a ton of cooling to maintain 72°F/50% in the same space. So I don't think there's a definitive answer to your question as it depend's upon exterior design condition and percent of outside air at the AHU.
RE: How to make a quick relation between CFM and Ton
RE: How to make a quick relation between CFM and Ton
Heat pump operation generally uses 450 cfm per ton due to the need for a safety factor required to keep condensing pressures and temperatures lower when filters plug during the heat cycle.
Heat pumps are less affective at latent removal because of this.