Trying to give a schematic level estimate for a new project I have a good handle on the lighting w/per sq. ft, and power w/per sq. ft., my question is there a rule of thumb (conservative) for kva per ton for heating and cooling loads? For this project in particular (office building) it will utilize packaged roof top units with dx cooling, any suggestions, also if anyone would be able to provide a conservative number for tons per sq.ft. for an office application. Just curious with everyone?s experience was...
kva per ton???
- Thread starter Mike01
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- Chapel Hill, NC
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- Retired Electrical Contractor
1 ton= 12000 BTU
and 1 kw = 3413 BTU
and 1 kw = 3413 BTU
- Location
- Massachusetts
Dennis Alwon said:
How do you come up with that?
The efficiency of each unit will be different, I would not count on anything until I got real numbers from the Mech. contractor.
sceepe
Senior Member
1 kwh = 3412 btu. But if you use this you assume that the hvac equipment is 100% efficient in converting electrical to thermal energy. It is far from it.
This question has been asked many times check out the search function of this forum.
Alternatly, find out who the ME's basis of design is (Trane, Carrier, etc). Then you can go to their site and get ideas of MCA for various tonnages. How you gonna guess at the tonnage the ME is gonna use?
Here's what I do,
From tranes info for RTU's at 208/1 phase:
2 ton, MCA = 64.8 amps
2.5 ton, MCA = 86.7 amps
3 ton , MCA = 102 amps
4 ton, MCA = 145 amps
5 ton, MCA = 155 amps
should be enough to ballpark it if you guess the tons/sqr ft
This question has been asked many times check out the search function of this forum.
Alternatly, find out who the ME's basis of design is (Trane, Carrier, etc). Then you can go to their site and get ideas of MCA for various tonnages. How you gonna guess at the tonnage the ME is gonna use?
Here's what I do,
From tranes info for RTU's at 208/1 phase:
2 ton, MCA = 64.8 amps
2.5 ton, MCA = 86.7 amps
3 ton , MCA = 102 amps
4 ton, MCA = 145 amps
5 ton, MCA = 155 amps
should be enough to ballpark it if you guess the tons/sqr ft
Effiency/efficacy:
Effiency/efficacy:
The correct term is "efficacy".
Your numbers from Trane seem high to me. My old 3-ton unit drew 19A at 240V single phase. I now have a new 4-ton unit, but have not measured the current yet.
Effiency/efficacy:
Fact is that AC units pump heat instead of transforming electrical energy into heat energy. Strictly speaking, it is incorrect to assign an efficiency rating to a cooling unit. If we do compute efficiency though, modern AC units would have an efficency of 1000% or more.sceepe said:
The correct term is "efficacy".
Your numbers from Trane seem high to me. My old 3-ton unit drew 19A at 240V single phase. I now have a new 4-ton unit, but have not measured the current yet.
Sizing
Sizing
I am only asking because the mechanical engineer has roughy indicated they need 25 tons of cooling / heating for the application, I want to convert this to kva I am not trying to size my cb's or overcurrent protection but to just indicate my distribution board ampacities so the contractor can get a estimated cost durring the sd phase of the project for budgetting reasons as we progress into DD and finally into CD's if the distribution board gets smaller than it can be a savings a decrease in the project cost, instead of having to explain why I had to increase my panelboard size at the last minnute. What is the best way to do this if you could convert the tons to kva I could provide a rough sizing for my distribution the mechanical already indictes that the size is larger than will be required but it's a starting point and I do not want to fall behind the 8-ball and have the electrical portion under estimated. Thanks for all you opinions and help.
Sizing
I am only asking because the mechanical engineer has roughy indicated they need 25 tons of cooling / heating for the application, I want to convert this to kva I am not trying to size my cb's or overcurrent protection but to just indicate my distribution board ampacities so the contractor can get a estimated cost durring the sd phase of the project for budgetting reasons as we progress into DD and finally into CD's if the distribution board gets smaller than it can be a savings a decrease in the project cost, instead of having to explain why I had to increase my panelboard size at the last minnute. What is the best way to do this if you could convert the tons to kva I could provide a rough sizing for my distribution the mechanical already indictes that the size is larger than will be required but it's a starting point and I do not want to fall behind the 8-ball and have the electrical portion under estimated. Thanks for all you opinions and help.
dbuckley
Senior Member
- Location
- Canterbury, New Zealand
I thought it was just me 
My rule of thumb is that most A/C plant moves about three times as much heat as it uses power, so a 2 ton A/C is about a 7KW unit, so I would expect it to use about 2.5KW.
But the real problem the OP has to solve is how much A/C he needs to start with? As a minimum its the lighting and power loads already calculated, and then people, which you can guess (1 body / 100 sq foot * 120W from memory). But the real kicker is solar load though the windows, which could be much much greater.
UPDATE: Just seen it: 25 tons, so... 25 * 12000 / 3413 is about 90KW, so about 30KW electrical load.
My rule of thumb is that most A/C plant moves about three times as much heat as it uses power, so a 2 ton A/C is about a 7KW unit, so I would expect it to use about 2.5KW.
But the real problem the OP has to solve is how much A/C he needs to start with? As a minimum its the lighting and power loads already calculated, and then people, which you can guess (1 body / 100 sq foot * 120W from memory). But the real kicker is solar load though the windows, which could be much much greater.
UPDATE: Just seen it: 25 tons, so... 25 * 12000 / 3413 is about 90KW, so about 30KW electrical load.
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What part of the country are you in?
I generally use 1.7 kW/ton for dx equipment, so 25 tons could be about 42 kW, but it also depends on the type of equipment your using and the location of the project. If you have 5 RTUs and each one has a 10kW heater, then your loads jumps up to 53kW. If you're in Central Florida like me, we don't use that much heat, so "I" would figure something less.
If your ME assummed 25 Tons of AC, ask him what kind of equipment he is planning....(5) splits, (2) 12.5 Ton RTU's, (4) 6.5, etc. That might help you guess better.
I generally use 1.7 kW/ton for dx equipment, so 25 tons could be about 42 kW, but it also depends on the type of equipment your using and the location of the project. If you have 5 RTUs and each one has a 10kW heater, then your loads jumps up to 53kW. If you're in Central Florida like me, we don't use that much heat, so "I" would figure something less.
If your ME assummed 25 Tons of AC, ask him what kind of equipment he is planning....(5) splits, (2) 12.5 Ton RTU's, (4) 6.5, etc. That might help you guess better.
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mdshunk
Senior Member
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- Right here.
The amount of energy consumed per ton of refrigeration is absolutely dependant on the condensing temperature and the refrigerant used. These effect the head pressure, which is the controlling factor in how heavily the compressor is loaded. This will necessarily vary with the outdoor ambient. For instance, for R-22:
S.e.e.r.
S.e.e.r.
The SEER number is defined to be,
SEER = cooling rate (BTU/hr)/input power (W),
e.g., consider a 3-ton unit drawing 3KW.
SEER = (36,000BTU/hr)/3,000W = 12
The input power is easily computed from the SEER.
Convert BTUs to watts to obtain,
EFFICACY = 3.52 --> 352% which is close to the number posted by dbuckley.
PS: Don't forget to factor in PF when computing the current requirement.
My earlier post claiming 1000% is WRONG:
S.e.e.r.
The SEER number is defined to be,
SEER = cooling rate (BTU/hr)/input power (W),
e.g., consider a 3-ton unit drawing 3KW.
SEER = (36,000BTU/hr)/3,000W = 12
The input power is easily computed from the SEER.
Convert BTUs to watts to obtain,
EFFICACY = 3.52 --> 352% which is close to the number posted by dbuckley.
PS: Don't forget to factor in PF when computing the current requirement.
My earlier post claiming 1000% is WRONG:
would this work???
would this work???
:-?
Thanks everyone rattus I understand that power factor will have impact on the final decision and have been thinking about this too. What I was going to do was to estimate how dbuckley indicated above, and then add a percentage I guess you could call ?safety factor? for this portion of the project and then for the next issue when it goes out I will have preliminary unit selections and can refine my sizes from that, my only question is if you get 90kw why only 30kw electrical load?
would this work???
:-?
Thanks everyone rattus I understand that power factor will have impact on the final decision and have been thinking about this too. What I was going to do was to estimate how dbuckley indicated above, and then add a percentage I guess you could call ?safety factor? for this portion of the project and then for the next issue when it goes out I will have preliminary unit selections and can refine my sizes from that, my only question is if you get 90kw why only 30kw electrical load?
Mike01 said:
Because you are not transforming 30KW into 90KW.You are pumping heat from the inside to the outside.
You boil liquid Freon in the evaporator by lowering the pressure. The Freon absorbs heat from the inside air.
You then compress the Freon vapor which raises its temperature.
You then liquefy the vapor by blowing outside air through the condenser. This removes heat from the vapor to complete the pumping cycle.
The liquid is then pumped back to the evaporator.
With resistive heating, the 3413BTU/KWH applies.
I've used 1.5 KVA per ton for AC or heat pumps. But sometimes that number is too low. For a safety factor, I would probably use more like 2 or 2.5 KVA per ton.
Then, after you have sized your service, they will add a kitchen with a 125KVA make up air unit, and you will be back at square one.
Then, after you have sized your service, they will add a kitchen with a 125KVA make up air unit, and you will be back at square one.
LarryFine
Master Electrician Electric Contractor Richmond VA
- Location
- Henrico County, VA
- Occupation
- Electrical Contractor
Holey moley! According to that chart, my house should have about 9 tons of cooling!ed downey said:
Robert T. Nethken
New member
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- 250 Rue Jonathan, Slidell, La 70461
I usually use a very simple relationship for KVA per tons of A/C , 1.8Kva /ton. If I have better information I will use it but so often I can't get the data from the M.E. before I have to complete my calculations. This will always give me an answer I can live with.
Robert T. Nethken, P.E.
Robert T. Nethken, P.E.
tallgirl
Senior Member
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- Glendale, WI
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- Controls Systems firmware engineer
ed downey said:
I didn't see any correction for climate. My father used to size AC units for houses, back when he was building them all the time. When I told him what I have on this house he wanted to know why I had so much. Then I explained what it's like when the summer time high reaches 110F from time to time ...
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