Room Heating 1

[RJwall]

Im trying to verify correct sizing of a blower system. A 12" blower w/ 1/2HP blows approx. 1800 cfm of 100 deg F air into a 6000 cu ft shack, that ranges to lows of 0 deg F ambient. How do I calculate supplied BTU's and also required BTU's to raise the temp to 45 deg F?
Thanks for the help.

 

[CESSNA1]

RJWALL: For the heat requir4ed just calculate the heat lost through the walls/doors/windows/etc for the heat required. ASHRAE has a form fo doing this.

For the heat supplied it is simply q = m*cp*Dt

m = mass flow of air
cp = specific heat of air at constant pressur
Dt = temperature difference of air

Regards
Dave

 

[tombmech]

To be more specific, you can actually use the relationship:

Q, dry air = 1.08 X CFM X DeltaT (deg. F).

However, if I interpret your problem correctly, it seems somewhat superfluous. If you are really supplying that much air into the 6000 cu. ft. shack, all of the air will be displaced in 6000/1800 = 3.33 minutes.

Unless you are pressurizing the space, you must ensure that 1800 cfm is exhausted at the same time you are supplying it. A small point considering a "shack", but that's how things work: if the resistance to exhausting or pushing that much air out of the shack is great enough, then the flow from the fan will drop, and it will not really be 1800 cfm.

However, assuming that works, then the balance can be drawn this way:

Qtot = Qin - Qout
Qtot = 1.08 X 1800 x (100-45) = 106,920 Btuh (Btu/hr).

As noted, the air itself will be replaced in 3.33 minutes. However, the actual heat loss must be balanced with the steady-state condition that will exist at the desired final temperature, hence 45 deg. F.

Therefore, the heat loss from the shack should be calculated using the Temperature Difference of 45 deg. - 0 deg. Transmission can then be calculated using resistance values from the shack. Given the scenario, other factors will be trivial - given the high flow rate compared to volume.

However, all of this is superfluous - the real key is the energy required to provide the 1800 cfm @ 100 deg. F., and you haven't provided enough info to determine that.

 

[RJwall]

I feel ashamed being an ME, but I havent had to deal with energy balances and HVAC in over 20 years.

The air is being drawn from the top level of a Kiln, which is always at 100-110 deg F. The manufacturer spec'd the blower @ 1800 CFM with 1/2hp motor.

The "shack" is a sheet metal building that isnt insulated nor fully enclosed (gaps around doors and entrance points). The real problem I am facing is the heat loss of the building. Outside conditions may have temps to 0 deg F, and wind speeds of 30 mph.

I had taken my energy balance @ 100-45 and had found similiar numbers as you stated, but I really dont know how to find my heat loss through the building.

Thanks for the help!!

 

[tombmech]

In that case, transmission through the steel will be 100%. However, if there were no other effects, then you might as well dump the fan into the ambient air - but we know that's not the same. The fact is, there will be a heat transfer (resistance) that you can calculate.

Check your ASHRAE documentation for determining loads. Typically, they list an organized set of equations for every heat transfer component of both heating and cooling modes. Heating is the more limited of the two, but there are several that should be calculated in your case, using 45-0 deg.F. as the DeltaT basis:

1. Skin convective heat transfer. Yours will be high, due to the 30 mph wind, 15 is usually assumed. Note that this has an inside and outside component. You typically add these resistances along with the steel skin (zero), to determine an overall wall-section thermal resistance, and calculate the heat transfer through the wall and roof.
2. Slab/ground transfer. Compared to the steel with zero insulation (perfect conductor), there will be an insulating effect from the ground. This depends on more details of your shack construction.
3. Radiation - this will add heat, and will have both roof and wall components, depending on the time of year/day, location, and orientation.
4. Exfiltration - this is still unclear in your case - whether you have an exhaust, or force out the flow through cracks and doors.

I'm doing this from memory at home - some of these components may still be trivial, and there may be some I haven't remembered. Regardless, your answer will be some finite value that will allow you a reasonable comparison with the heat flow input from the fan. I still believe it's trivial, but this will let you prove it.

 

[cme]

I would measure the crack length and size .... then knowing velocity you can get cfm .....

Just make sure that you take one side in the calcs and have the opposing side measuring the same