Can't answer this one on my own.

[fletcher]

I am doing my homework again. How would you answer this question. It is the only one I couldn't do on my own!! Help!! :roll:

A 20,000 sq ft school building has the following estimated demand loads:

Lighting 20,000 W, 120 V one phase
Mechanical equipment: 100 hp, 480 V three phase

Convenience power: 1 W/sq ft, 120 V one phase

What power distribution system or systems would you choose? Note: 1 hp = 746 W, 0.75 kW or approximately 1 kVA@ 0.75 PF.

 

[fletcher]

found it

found it

I think I found the answer in the book. It says the distributed radial system is a typical application in shopping centers, apartment complexes and schools.

 

[Zifkwong]

Not sure I undertand your question almost all systems are radial especially in apartments and shopping centers.

 

[bob]

It says the distributed radial system is a typical application in shopping centers, apartment complexes and schools.

All Systems in these projectrs are radial systems.

What power distribution system or systems would you choose?

I think they are asking for the type of system you would need.
voltage = 480 volt.
Caculate the load to determine the service requirements.
install a 480 to 120/208 for the smaller loads.[/quote]

 

[charlie b]

The question gave you information that forces you into a specific type of distribution system. It told you the mechanical equipment was 480 volts. It also told you that the lights and convenience outlets were 120 volts, but that part would have been obvious anyway. So part of the answer is, as given in the question, that you need both 480 volts and 120 volts.

The question also gave you something important that is not obvious to the general public. They said that the mechanical equipment was three phase, but they didn't have to tell you that. That is because the question said that the mechanical equipment was 100 horsepower. They don't say how many machines there are, and how big the biggest would be. But it is a very safe bet that the biggest is bigger than 1 horsepower. That is significant because any motor that is bigger than 1 horsepower would almost certainly be a three-phase motor, rather than a single phase motor. So now you know that you need a 480 volt, three phase power system.

The other part of the question requires you to do something with the information that gave you concerning the loads. If you add up their list of loads, you will get 20,000 (lighting) plus 100,000 (motors) plus 20,000 (for 1 watt per square foot), for a total of 140,000 VA. I don't know if they taught you the following tidbit, but for a 480 volt, three phase system, you determine the amps by first dividing VA by 480, and then dividing the result by 1.732 (the square root of three). I get a result of about 168 amps. Therefore, you would need at least a 200 amp service.

Finally, they said the single phase load was 40,000 VA. The next higher size transformer (to go from 480 volts to 120 volts) is 45 KVA.

So my answer would be that you need a 200 amp, 480 volt, three phase service, and that you need at least a 45 KVA, 480-120/208V, three phase transformer.

One important note: The question gave you information that would be of interest to a buyer. However, that information is not what electricians and electrical engineers use as a basis for calculating the required service. We use similar information, and we have a calculation process that allows us to take credit for the high probability that not everything will be running at the same time.

Did you get the "book answer" for the homework assignment yet? I am curious as to what they consider the "right answer" for this question.

 

[ramsy]

Many thanks Charlie. Great excercise for me.

Made some changes here: 1? load
20,000 Sq.ft * 3 = 60,000 VA (Tbl.220-3a)
20,000 Sf * 0.75 = 15,000 VA (Table 220-34 Demand Factors for Schools)
--------- Net load = 75,000 / 240 = 312.5 A

Neutral Feeder per NEC 220-22
312.5 * 0.70 = 218.75
18.75 * 0.70 = 13
Neutral Feed = 213 A

3? Motors 100HP @ 480vac. 1HP = 1 kVA@ 0.75 PF
100,000 VA / 480 / 1.732 = 120.3 A

3? 480vac Xfmr Service Calc.
60k + 15k + 100k = 175k VA
175k / 480 / 1.732 = 210.5 or 225 A Service @ 480vac

Don't do this often. May have missed a few things, like Xfmrs sized for 1.25% load. That would be 219k VA or next size up.

 

[charlie b]

Made some changes here: 1? load
20,000 Sq.ft * 3 = 60,000 VA (Tbl.220-3a)
20,000 Sf * 0.75 = 15,000 VA (Table 220-34 Demand Factors for Schools)
--------- Net load = 75,000 / 240 = 312.5 A

You are using the NEC calculation process, as all of us would do. Please note two things, however. One is that the OP is not an electrician, but rather a home inspector. The other is that the question specifically told us the watts for lighting, and gave us the rule to use to get watts for the receptacles. So you would have been marked down on this test, for doing the problem the "right way." :wink: 8)

 

[bob]

Made some changes here: 1? load
20,000 Sq.ft * 3 = 60,000 VA (Tbl.220-3a)
20,000 Sf * 0.75 = 15,000 VA (Table 220-34 Demand Factors for Schools)
--------- Net load = 75,000 / 240 = 312.5 A

In the origional question, you posted the voltage at 120 volts. I would
assume that you would install a 120/208 volt 3 phase transformer for this load. The panels would be 120/208 3 phase. That said your caculations would be :
Net Load = 75000/(208 x 1.73) = 208 amps.

 

[ramsy]

..the question specifically told us the watts for lighting, and gave us the rule to use to get watts for the receptacles. So you would have been marked down on this test, for doing the problem the "right way." :wink: 8)

Yes, I see. They ignore all demand factors, including Tbl.220.13.

..you would install a 120/208 volt 3 phase transformer for this load. The panels would be 120/208 3 phase. That said your caculations would be : Net Load = 75000/(208 x 1.73) = 208 amps

Yes, I see. I like that better too.

 

[bob]

You are using the NEC calculation process, as all of us would do. Please note two things, however. One is that the OP is not an electrician, but rather a home inspector. The other is that the question specifically told us the watts for lighting, and gave us the rule to use to get watts for the receptacles. So you would have been marked down on this test, for doing the problem the "right way."

Charlie
It is my opinion that if ramsy had worked the problem using the 20000
watts of lighting, he would have been incorrect. He is required to use the larger of the 2 methods. Using the table 220.12(05NEC) & 220.3A(02NEC)
or the 20kw lts given you should use the method that gives you the larger figure.

 

[charlie b]

In a PM, I was asked to look into the methods used to calculate the load, as posted by ramsy and by bob. I have, and I conclude that both methods were wrong. I also believe that a proper load calculation cannot be performed, using the information given.

But first keep in mind that the original question did not ask about service load calculations, and it was not addressed to electricians or engineers.

Made some changes here: 1? load
20,000 Sq.ft * 3 = 60,000 VA (Tbl.220-3a)
20,000 Sf * 0.75 = 15,000 VA (Table 220-34 Demand Factors for Schools)
--------- Net load = 75,000 / 240 = 312.5 A

Don't do this often. May have missed a few things

I believe you have misinterpreted the intent of Table 220-34 (2002 NEC). That is part of the "Optional Method." You used a "VA/sq.ft." as though it were an amount of load. It is not. It is a threshold for applying a demand factor of 100%, versus 75%, versus 25%. But you have to add up the load separately, before applying the demand factors.

First, you add up all connected load (Bob missed this point as well). The question did not give us that information. It gave us the connected load for lights and a mysterious category called "mechanical equipment." But it used a rule of thumb, and not actual connected load, for receptacles. Plus it said nothing about a school cafeteria, or a laundry area for employee uniforms, or water heating, or any special equipment for classroom use, or any other possible load. Finally, it did not separate heating from cooling, so we do not get the luxury of disregarding the lesser of the two.

So we have to go with what they did give us. In my first post, I calculated a load (from the information given) of 140,000 VA. Let's call that the "total connected load," for the moment. Applying Table 220-34, we get

• First 60,000 VA at 100% = 60,000
• Next 80,000 at 75% = 60,000
• Total VA = 120,000.