How to make a quick relation between CFM and Ton 1

[puertorico]

Is it correct to use the assumption of 340CFM per ton?

Thankyou
Angel
MEP Engineer

 

[ProfSporlan]

Actually, 400 cfm per ton has been considered the rule of thumb. One can see where this value comes from if we pick a couple of appropriate state points, say: 82°F DB/67°F WB return air, 62°F DB/75 RH supply air:

return air:
enthalpy = 31.43 Btu/lb_m (dry air)
humidity ratio = 0.01067
specific volume = 13.89 ft³/lb_m (dry air)
rh = 45.6 percent

supply air:
enthalpy = 24.59 Btu/lb_m (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/lb_m = 28.80 lb_m/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/lb_m 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.

 

[ProfSporlan]

Mmmmm... I noticed I've buggered up this solution :-(

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/lb_m

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:
m_a * h₁ = m_a * h₂ + m_w * h_w + q

where:
m_a = mass flow of dry air
h₁ = return air enthalpy
h₂ = supply air enthalpy
m_w = mass of water removed
h_w = enthalpy to condense water
q = total load

And the conservation of mass equation:
m_a * W₁ = m_a * W₂ + m_w

where:
m_a = mass flow of dry air
W₁ = return air humidity ratio
W₂ = supply air humidity ratio
m_w = mass of water removed

Combining these equations, we get:
q = m_a * (h₁ - h₂) - m_a * h_w * (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...

 

[quark]

Peurtico!

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.

 

[arabei]

Hi puertorico,

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,

 

[ChasBean1]

I agree with the last two posts - the 400 cfm per ton is a sizing rule of thumb for a typical office type space with given design conditions. For example it's the flow rate and refrigeration I would need to cool (sensibly and latently) 82°F DB/69°F WB entering air to keep a room 72°F/50% RH.

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.

 

[mt31]

350 to 400 cfm per ton of cooling capacity is the proper figures to use

 

[r22]

For small a/c systems (direct expansion evaporators) 400 cfm per ton is the accepted nominal value for general applications that must keep the coil temperature above 32 degrees F to avoid freeze ups. These small systems cannot unload as fan speed goes down and load decreases. Cfm can be reduced for more latent removal, in fact some systems will sense outdoor humidity and slow the fan.

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.