What's the load on this heater panel?

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ron

ron

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New York, 40.7514,-73.9925
I was having a discussion with an electrician recently regarding an electric heater installation.
Here?s the installation:
Six-208V, single phase, 2.5kW heaters circuited to six-two pole breakers in a 208V, three phase panelboard installed solely for this project arranged to balance the load across the three phases.
The discussion resulted in a disagreement regarding the total connected load on this panelboard.
I believe that it will be 17.3kW of load, where he felt it would be 15kW. Who's right?
I thought this brain teaser would be an interesting discussion, before I tell you how I got my answer.

[ June 30, 2004, 05:54 PM: Message edited by: ron ]
 
Re: What's the load on this heater panel?

I?ll bite. A watt is a watt. 6 times 2.5 is 15. I can't guess why you are adding 15% to the total.

Now are you trying to calculate the "service or feeder load" per NEC methods? If so, then it would be incorrect to be asking for ?total connected load.?
 
Re: What's the load on this heater panel?

Charlie,
Not calculating service or feeder loads. A very innocent question about the electrical load on the panel.
I will admit that my first instinct was to multiply 2.5kW x 6 too.
(I surely hope I'm right, or I'll be eating portions of my foot).
BTW, the reason for the higher load calc on my part is related to how these single phase loads are distributed on the panelboard.

[ June 30, 2004, 08:20 PM: Message edited by: ron ]
 
Re: What's the load on this heater panel?

I will go with the 15kw also, even if you look at each phase at 1250kw per connection, everything is balanced with four points. :confused:

Roger
 
Re: What's the load on this heater panel?

?Power,? as measured in watts or kW, is the rate at which energy, as measured in Joules, is being expended. If device #1 is expending energy at the rate of 2,500 Joules every second, and device #2 is expending energy at the rate of 2,500 Joules every second, and heaters #3 though 6 are each expending energy at the rate of 2,500 Joules every second, then the total energy being expended every second is 15,000 Joules.

It does not matter how the heaters are connected. It doesn?t even matter whether they are all the same type of device. Device #1 can be a heater expending energy at 2,500 Joules per second, and device #2 could be a motor running under a load that causes it to expend 2,500 Joules per second, and device #3 could be a dog running down the road at a speed that causes it to expend 2,500 Joules per second, and devices #4, 5, and 6 could be a car, a bicycle, and a baseball player. If each is expending 2,500 Joules every second, then the total energy being expended every second is 15,000 Joules.

I say again, "A watt is a watt."
 
Re: What's the load on this heater panel?

As you have described it, it's 15kW. Other configurations could change the current per phase, but if each heater is fed at 208 the total power utilization is 15kW.

[ June 30, 2004, 10:37 PM: Message edited by: rbalex ]
 
Re: What's the load on this heater panel?

Let's layout this installation in the panelboard.
There will be four heaters drawing approx 12A from each of the three phases from the panel. So will you agree that each of the three phases of the panelboard will draw approximately 48.07692308A.
The three phase panelboard will see a total of:
208V * 48.07692308A * sqrt(3)= 17320.50808W

or 17.3kW of load.
 
Re: What's the load on this heater panel?

Ron:

Your error is in your second sentence:

So will you agree that each of the three phases of the panelboard will draw approximately 48.07692308A.
Click to expand...
No, you will not have 48 amps. Look at two heaters connected to line A. Each will have about 12 A peak. But one heater will be connected across lines AB, and one across lines AC. Thus, the currents don't simply add.

If you are going to do a 3 phase calc. by considering each phase at a time, it is best to avoid adding currents. Too easy to make a mistake like yours. It is easier to add the powers like Charlie did, and then go back to find the current.

For example, 2.5 KW * 6 = 15KW.
Then: 15KW/(208 * sqrt(3)) = 41.6 amps per phase.

Steve
 
Re: What's the load on this heater panel?

My apologies to all for the exercise. I have realized the error of my ways. Now I will have to follow the SOP.
Step 1: Open mouth
Step 2: Insert foot
Step 3: Learn for next time

My error was to consider the loads device by device entered on a panel schedule, without considering the vectorial combination of loads on the phases.

[ July 01, 2004, 09:41 AM: Message edited by: ron ]
 
Re: What's the load on this heater panel?

Ron:

I wanted to prove that you can calculate the power the way you were doing it if you keep track of all the angles and use vector math.

It took me a while to figure out why I kept getting the wrong answer. If we draw out the circuit (its a delta of course), and give all the currents reference directions (assume all the currents in the delta flow counterclockwise, also assume all the line currents flow into the delta) then it becomes apparent that:

Line current a = Load current ab minus Load current ac

Where the above subtraction is done with vectors. So for 2 heaters we have (use a ^ to show angle) 12^0 - 12^120 = 20.8. The easy way to find the 20.8 is 12* sqrt(3).

So now that we know the line current is 20.8 amps for two heaters, we have to multiply that by two since you have two heaters between each phase. That gives 41.6 amps.

total power is 41.6 * 208 * sqrt(3) = 14990 Watts which is basically 15KW (but I rounded off the 12 amps a little).

So you were on the right track, there were only a few math errors. This does demonstrate that it is a lot easier to just add the power of each heater.

Steve
 
Re: What's the load on this heater panel?

Why wouldn't the total load be multiplied by 125% as states in 424.22(E) Conductors for Subdivided Loads thus the load being 14.99 times 125% which equals 18.74kW? Still learning from you guys. :confused:
 
Re: What's the load on this heater panel?

The original question was basically "What is the actual load?". Your statement is more like "What load should the circuit conductors be capable of supplying?"

Steve
 
Re: What's the load on this heater panel?

I agree with Steve. I made the same comment on my first post.
 
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