Converting T/hr, BTU and kW

[CelsoIII]

I work at a high rise building and part of my job is to manage the energy intensity of the building.

In order to come up with the total energy use for the building we had to convert all energy types to a common energy unit, in this case kBTUs, and add the totals.

Chilled Water (CHW)
The building does not have chillers in house so we purchase CHW, measured in Ton/hr, from a district central plant. I know that 1 Ton = 12,000 BTU = 12 kBTU

Electrical
1 kW = 3412 BTU

The Issue:

Example “A” (Building purchasing CHW)
CHW use = 1,000 Ton/hrs = 12,000 kBTU
Electrical = 200 kWh = 682.4 kBTU
Total Energy Use = 12,682.4 kBTU

Example “B” (Building with chillers @ 0.7 kWh/Ton Eff.)
CHW use = 1,000 Ton/hrs = 700 kWh = 2,388.4 kBTU
Electrical (not including chiller load) = 200 kWh = 682.4 kBTU
Total Energy Use = 2,388.4 + 682.4 = 3,070.8 kBTU

Why are the two numbers (total energy use for each example) so different?

In school they teach us that 1 Ton = 12,000 BTU but it turns out that you can produce 1 Ton with less energy???

Chilled water production
Input .7 kWh, Output = 1 Ton/hr = 12,000 BTU
which means that 0.7 kWh = 12,000 BTU,
but per the book 3.5 kWh = 12,000 BTU (12,000 / 3412 = 3.5)

Can someone please explain?

 

[someguy79]

kWh is an energy unit.
Ton(refrigeration) is a power unit.

1 Ton = 12000 Btu/h = 3.5 kW
1 Ton * 1 h = 12000 Btu = 3.5 kWh

 

[CelsoIII]

Thanks, unfortunately, I am still not understanding.

If when producing CHW

I use .7 kWh which produces 1 Ton per hour (all this in 1 hr)

How do you explain that with the equivalencies that you mention above?

 

[someguy79]

So you really want to know why you're getting more energy output than you have energy input.

You say you're not including chiller load in example B. Why?

 

[tys90]

You are looking at power input versus cooling output. To be consistent, you would:

a) Look at the district chilled water efficiency rather than the amount of chw purchased in example A

If you have the COP of the district plant it would make things pretty easy. The efficiency would work too, just an extra step to convert to kBTUs. There are things like pumping losses for district plants too, I don't know if they would account for that in any COP or efficiency they have on record.

OR

b) Look at the chw produced in example B rather than the power input. But that doesn't help you in figuring energy intensity.

For example. If you purchased 1 ton of cooling, that's 12,000 BTUs. If you made 1 ton of cooling, that's 12,000 BTUs, regardless of how much it took to produce it.

Make sense?

 

[CelsoIII]

@tys90
Thanks for the replay. It does make sense if we knew the central plants' COP (coefficient of performance?)We could calculate the energy use that way.

Having some electrical experience, I am used to for example in motors: If you know the volts, amps, and motor efficiency, you can get a pretty good value for Hp. Granted that Hp is always going to be less than the power input due to inefficiencies.

Compared to CHW production it makes no sense. Maybe there is no real conversion from BTU or Ton/hrs to kWh and this value is always dependent upon individual equipment COPs. At any rate, this would mean that the conversion factor of 12,000 BTU/h = 1 Ton/h = 12,000 / 3412 = 3.5 kWh is at a minimum misleading. We know that typical CHW production can be achieved at a rate of .5 to .8 kWh/Ton with today’s equipment. I hope I am wrong and there is a better explanation for this.

@Someguy79
The electrical load is not included in example "B" because it is included in the CHW being produced. The chiller does not consume electricity; it converts it to thermal energy. So the net result should be cooling and energy loss in the form of heat. I think that if we included the electrical load of the chiller we would be doubling up on energy use.
For example in a motor: if you want to know what the energy use I would not include the electrical load and the Hp. We use either electrical load or the Hp plus the energy loss (Hp / Motor Eff.) to calculate energy use.

 

[tys90]

No, it's not misleading, 12,000 BTUs/h = 1 ton = ~3.5 kWh no matter what, that's just conversion. Like I said you need to differentiate between power input and power output.

 

[tys90]

Maybe I should explain a little better what I meant by differentiating.

What I am saying is you can only compare power output to power output or power input to power input. A conversion is a conversion. What you are showing in example B is the relationship between power input and power output, then applying the conversion to the power input to get your "energy usage". Example A you only have power output, which you convert and use as the "energy usage". That's why there is a discrepancy.

 

[CelsoIII]

@tys90
Thanks again. I agree, for this application I need to keep them separated

 

[CelsoIII]

My own conclusion,
In refrigeration, the cooling effect is not the direct transformation of the energy input. The operation of cooling equipment has many energy outputs, manly in heat and the mechanical forces driving pumps, compressors, fans, etc. One of the effects of all these activities is the transfer of heat from one place to the other, which we call cooling… so in other words, in chillers, electrical energy is used to transfer (move) heat energy from one place to another…or something like that.

 

[katmar]

Yes, that is exactly the right conclusion. In refrigeration you are moving heat (energy) from one body to another. You cannot create "coolth" in the way that you can create heat. All you can do is move the energy from one place to another and the COP is the ratio between the amount of energy you move and the amount of energy required to accomplish that move.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[mauricestoker]

The numbers don't work because you reach a conclusion not related to the examples. You are comparing apples to oranges.

You include a conversion factor of 0.7 kWhr/TN, which I would assime is a chiller plant overall full load or NPLV, for example B. That would include weather variation, pumping, cooling towers and chillers. That is left out of A, then you proceed to a total energy comparison. The 0.7 kWhr overall efficeincy could very well change between loading, chiller efficiencies, pumping, etc.

I think better points would be how the refrigeration tonnage is metered, if any limitation exists on delta T, and whether the refrigeration can be delivered at lower cost/energy in-house. You can't change the supplier's chiller plant or pumping, you can (perhaps) use less tonnage or aim for a higher delta T, depending on how you are getting billed.

 

[ChrisConley]

The whole point of a refrigeration cycle is that we aren't producing 'cooling' but moving energy from a low temperature source to a high temperature source using a compressor (or absorption).

A chiller that required 1kW of energy to produce 1kW of cooling would be a very bad chiller.