RTU Ton to KW multiplication factor

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Designer69

Senior Member
hey guys if you have a rooftop unit with 15 ton cooling capacity, you get the KW consumption by multiplying by 2000W right?

so that would be 15 x 2000 = 30kW

does this sound familiar? I think this is the conversion I used to apply just want to get confirmation


thanks!!
 

topgone

Senior Member
hey guys if you have a rooftop unit with 15 ton cooling capacity, you get the KW consumption by multiplying by 2000W right?

so that would be 15 x 2000 = 30kW

does this sound familiar? I think this is the conversion I used to apply just want to get confirmation


thanks!!

Please look for the energy efficiency rating of your heat machine (EER). EER is expressed in BTU/Watts. Assuming an ERR of 10, and a ton of cooling is equal to 12,000 BTU; your 15 tons of cooling translates into 15 tons x 12000 BTU per ton = 180,000 BTU; Your electrical input requirement will be 180,000 BTU/ 10 BTU per Watt = 18,000 Watts or 18 kW (a rule of thumb of tons X 1,200).

However, if your heat machine is not very efficient, say EER=8, then your electrical requirement becomes 22.5 kW (15 X 12000/8 = 22,500 W, or a rule of thumb of tons X 1,500). IMO, tons X 2,000 Watts is an over-design.

I found other people use a 1kW per ton of cooling too. It is so because it seldom happens users operate a cooling unit to its rating. HVAC designers usually do provide extra cooling capacity for every space considered. Hope that helps.
 

Npstewart

Senior Member
There really is absolutely no real world conversion you could apply here.

Technically speaking however, there is 12,000 btu per 1 ton of cooling, and there is 3.41 btus in 1 watt. So in a 15 ton rtu that would be (15*12,000/3.41)/1000 = total KW. That equals about 52 kw total, seems
A little high.
 

topgone

Senior Member
There really is absolutely no real world conversion you could apply here.

Technically speaking however, there is 12,000 btu per 1 ton of cooling, and there is 3.41 btus in 1 watt. So in a 15 ton rtu that would be (15*12,000/3.41)/1000 = total KW. That equals about 52 kw total, seems
A little high.

Converting a heat quantity into equivalent electrical unit is wrong. We need to know how efficient your "heat pump" is that's why we need to know the EER. For a 15-ton cooling, we will be "pumping" that amount of "heat" in the refrigerant medium out of the controlled space and throw those heat outside. The "real world conversion" will then be the machine electrical input requirement ratio to the amount of heat the machine pumps out, in terms of BTU handled per watts input, simply called EER.
 

kingpb

Senior Member
Location
SE USA as far as you can go
Occupation
Engineer, Registered
Rules of thumb used on the engineering side, barring actual nameplate information, are as:

1 KW = 3412 BTU/HR
1 Ton = 12,000 BTU/HR
1 Hp = 2,545 BTU/HR

Doing the math;

1 Ton = 3.517kW (approx)

Also,

1 Ton = 4.72 Hp (approx)

These rules of thumb came from a person that I would consider an expert in chilled water and CHP systems. They may be on the conservative side, but they planning electrical systems without nameplate or cut sheet data is tricky, and they haven't failed me yet.
 

Designer69

Senior Member
wow very informative. 2kw/ton does not appear to be conservative enough although I am sure the 3.5kw/ton includes units that heat by an electric heater instead of heat exchanger.

Is this correct?
 

skeshesh

Senior Member
Location
Los Angeles, Ca
Why would you think electric heat is included?

If he's aiming for worst case scenario that's what I'd think you need to assume.

I sometimes use methods similar to kingpb's. Other times I look for a bit of detail first: does the unit have electrical heat? Next I approximate the square footage of the area and multiply by 1 - 1.5 cfm depending on the location of the project. 1 ton (approximately)= 400 cfm, so I divide the cfms by 400 and select a tonnage. Then jump on carrier or trane and check out some units that fit the tonnage.
I've found both methods to work decently well. At the end I sometimes add a 10-15% contingency to the load just in case. But bottom line is trying to approximate the unit is a crude approach and the methods suggested here don't give you an exact conversion, rather a real conservative approach that aims only not to undersize the electrical source.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
To me, the bottom line is that you cannot properly size without seeing the equipment labeling. If that can't be supplied in a timely fashion, you must plan for worst-case.

You have to figure out the cost difference between the as-planned installation and the over-engineering, and make it known that the customer must cover the difference.
 
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