Load calcs when electrical specs cover a range of values?

[Jon456]

I'm trying to do load calcs and I've come across some equipment specifications that have me stumped. Specifically, what values to use when the specs cover a range of numbers.

Example 1:

208 - 240V, 1 Phase
9.5 - 10.3 A
1.8 - 2.3 KW​

Example 2:

200 - 240V, 1 Phase
21.3 - 25.5 A
4.2 - 6.1 KW
​

Example 3:

200 - 240, 3 Phase
10.0 A
2.7 HP motor​

Example 4:

115V, 1 Phase
6.9 A
3/4 HP compressor
​

Note that the motor in example 3 is powered by a VFD internal to the equipment; the compressor in example 4 is not. I'm not sure if or how the VFD affects the load calculations.

 

[Dennis Alwon]

In example 1 and 2 I would use the kw based on which voltage you have there-- 208 or 240V

Example 3- I would use the 10 amps

Example 4 I would use 6.9 amps

 

[Jon456]

In example 1 and 2 I would use the kw based on which voltage you have there-- 208 or 240V

Let's say the service is 208V.

So for Example 1, I would use 208V x 9.5A = 1,976 VA. Correct? But why would the wattage would be listed in the specs as 1,800W? This is a resistive load, so I thought VA = W.

And for Example 2, I interpolate the amperage, correct?

21.3A/200V x 208V = 22.2A
208V x 22.2A = 4,618 VA

But interpolating the wattage doesn't match.

 

[synchro]

Let's say the service is 208V.

So for Example 1, I would use 208V x 9.5A = 1,976 VA. Correct? But why would the wattage would be listed in the specs as 1,800W? This is a resistive load, so I thought VA = W

Motors are in general inductive loads, but their lagging power factor can be corrected using capacitors. A power factor of 1800/1976=0.91 is good for most applications.

And for Example 2, I interpolate the amperage, correct?

21.3A/200V x 208V = 22.2A
208V x 22.2A = 4,618 VA

But interpolating the wattage doesn't match.

First of all the wattage and KVA in Example 2 are nearly the same, and so the power factor is excellent and close to 1.0.
I'm assuming that you interpolated the wattage by the same 208/200 factor and the result wasn't close to the 4,618 VA you found above. That's because you have to interpolate by (208/200)² since both the voltage and current go up by the same factor of 208/200.

 

[Dennis Alwon]

Let's say the service is 208V.

So for Example 1, I would use 208V x 9.5A = 1,976 VA. Correct? But why would the wattage would be listed in the specs as 1,800W? This is a resistive load, so I thought VA = W.

And for Example 2, I interpolate the amperage, correct?

21.3A/200V x 208V = 22.2A
208V x 22.2A = 4,618 VA

But interpolating the wattage doesn't match.

*I don't know what to say about that. Obviously something is not correct on the nameplate. There seems to be a 10% difference. If you using it for calculating one branch circuit then it doesn't much matter.

 

[Dennis Alwon]

Motors are in general inductive loads, but their lagging power factor can be corrected using capacitors. A power factor of 1800/1976=0.91 is good for most applications.

OP said it is resistive load

 

[david luchini]

OP said it is resistive load

The numbers don't make sense for a resistive load, though.

 

[Dennis Alwon]

The numbers don't make sense for a resistive load, though.

I know-- I stated that something is not correct

 

[Jon456]

Example 1 is an oil heating tank with a resistive heating element, so it is a resistive load. The device does have a small pump for evacuating the hot oil out of the tank, but it must be manually activated by an NO switch and it would never operate for very long. So I didn't think it would be factored into the power rating for the device.

 

[Dennis Alwon]

Example 1 is an oil heating tank with a resistive heating element, so it is a resistive load. The device does have a small pump for evacuating the hot oil out of the tank, but it must be manually activated by an NO switch and it would never operate for very long. So I didn't think it would be factored into the power rating for the device.

Well yeah, the nameplate would have that factored into the numbers if the pump is part of the equipment. That may be the difference but you think there would some numbers for the pump on the nameplate

 

[gar]

190530-1449 EDT

Jon456:

Had you provided more detail in the first post it would have been easier.

Load 1 is resistive, thus current is proportional to voltage. 240*10.3 = 2472. Close enough. Use 10.3 A.

Load 2 is resistive. 240*25.5 = 6120. Very close, use 25.5 .

Load 3 is a VFD. Probably high harmonic (distorted current waveform). Use the 10 A. 240*10*1.732 = 4157 VA. 2.7*746 = 2014 W. Implies low power factor and inefficiency.

Load 4 is a plain motor. Use its current which looks reasonable.

.

 

[Jon456]

Had you provided more detail in the first post it would have been easier.

Sorry for that. What additional details should I have provided? I'm just going by what's listed on spec sheets.

Load 1 is resistive, thus current is proportional to voltage. 240*10.3 = 2472. Close enough. Use 10.3 A.

Load 2 is resistive. 240*25.5 = 6120. Very close, use 25.5 .

Load 3 is a VFD. Probably high harmonic (distorted current waveform). Use the 10 A. 240*10*1.732 = 4157 VA. 2.7*746 = 2014 W. Implies low power factor and inefficiency.

Load 4 is a plain motor. Use its current which looks reasonable.

Are you saying I should be basing all my load calculations on 240V, even though our electrical service is only 208V?

 

[gar]

190530-1937 EDT

Jon456:

In your first post you did not define your loads, except for 3 and 4 in an added paragraph. KW does not define resistive. Motor does not define VFD, and compressor does not imply a motor only. Within each example along with the given data you should have accurately defined what the load was. This was only partially done in the ending paragraph.

Nowhere in the first post did you define your source voltage. So now we know your source is 208. Thus, base your current on the 208.

Example 1 use the 9.5 A.

Example 2 use 25.5*208/240 = 22.1 A.

Example 3 I don't know. The specs are not clear (adequate).

Example 4 just use the 6.9 A. Possibly it is just slightly less at 120 V.

.

 

[Jon456]

Thanks to everyone for your instructive replies. I greatly appreciate your assistance!

 

[kwired]

I'm trying to do load calcs and I've come across some equipment specifications that have me stumped. Specifically, what values to use when the specs cover a range of numbers.

Example 1:

208 - 240V, 1 Phase
9.5 - 10.3 A
1.8 - 2.3 KW​

Example 2:

200 - 240V, 1 Phase
21.3 - 25.5 A
4.2 - 6.1 KW
​

Example 3:

200 - 240, 3 Phase
10.0 A
2.7 HP motor​

Example 4:

115V, 1 Phase
6.9 A
3/4 HP compressor
​

Note that the motor in example 3 is powered by a VFD internal to the equipment; the compressor in example 4 is not. I'm not sure if or how the VFD affects the load calculations.

Didn't check for accuracy, but example 1 to me means it draws 9.5 amps if you supply it with 208 volts or 10.3 amps if you supply it with 240 volts. Same concept with #2.

VFD, can have input current rating different from actual motor current rating. (current related to power factor of the motor is only flowing between the drive and the motor and not on the supply circuit)