Calculation Help

[awc]

Ok, I want to make sure i'm thinking correctly on this one. I had this question on a test and it asked if you have 208V 3 phase with 111amp per phase how much power will it supply in watts. It would be 23,088Watts. Is this correct?

 

[charlie b]

Sorry, but that is not correct. In fact, the question does not give you enough information to get the answer.

You need to multiply your answer by the square root of three (about 1.732) to get the amount of volt-amps associated with that voltage and that current. That is the ?apparent power,? and it is measured in units of VA (or KVA). But the question is asking about ?real power,? a term that is measured in units of watts (or KW). To get KW from KVA, you need to multiply KVA times the power factor. But the question did not give you a value for power factor, so you can?t get an answer.

My guess is that they are looking for the answer of 39,988. That is the 208 times the 111 that you have, but multiplied again by 1.732.

 

[chris kennedy]

208?333?1.73=119,826.72W

Thats with no power factor multiplier.

 

[charlie b]

208?333?1.73=119,826.72W

Chris, you fell into the "amps per phase" trap. That is a bad combination of words, and it should be forever banned from use by our industry. If Phase A has 111 amps, and Phase B has 111 amps, and Phase C has 111 amps, then the "total amps" is 111, not 333. Actually, "total amps" is another phrase that should be banned.

 

[chris kennedy]

Oops, I added the phase amps.

 

[chris kennedy]

Chris, you fell into the "amps per phase" trap. That is a bad combination of words, and it should be forever banned from use by our industry.

Part of test taking skill is to read the question carefully twice. I'm slipping.

 

[awc]

Charlie & Chris, thank you both for your input! I see what you're saying I got sloppy and missed the 1.732 multiplier for the 3 phase...I'll be curious to see if that is in fact the correct method we were suppose to use after it's graded????

Charlie & Chris, I'm also wanting to straighten something else out.... I'm a newbie into this field so i'm quite inquizzitive...........I tried to get my instructor to help me on this but i'm afraid he didn't help much. With True Power and Apparent power i'm so confused please help:-? :-?

 

[Dennis Alwon]

With True Power and Apparent power i'm so confused please help:-? :-?

See if this helps True, Reactive & Apparent Power

 

[mivey]

Chris, you fell into the "amps per phase" trap. That is a bad combination of words, and it should be forever banned from use by our industry..

Don't know why I've never thought about it, but I use that phrase as well. Now that you point it out, it is kind of silly. Consider me reformed (or at least on the road until I quit doing that).

 

[gar]

080428-1952 EST

awc:

Instantaneous power is p = v*i
If v = Vp*sin b, and i = Ip*sin b -- in other words current and voltage are in phase,
then p = Vp*Ip*sin^2 b using a trig identity this is
p = Vp*Ip*(1-cos 2b)/2
in other words the instantaneous power varies as the cos of the double rate frequency.
If you average this over 1/2 cycle the result is
Pave = Vp*Ip/2 converting to RMS values
Pave = Vrms*Irms and this is real power.

If one or the other is shifted by 90 deg, then
p = Vp*Ip*(sin b)*(cos b)
now the trig identity is (sin 2b)/2. This has no DC component and when averaged over a multiple of 1/2 cycles it is 0.
Thus, no real power.
But the instantaneous power is
p = Vrms*Irms*sin 2b

.

 

[chris kennedy]

080428-1952 EST

awc:

Instantaneous power is p = v*i
If v = Vp*sin b, and i = Ip*sin b -- in other words current and voltage are in phase,
then p = Vp*Ip*sin^2 b using a trig identity this is
p = Vp*Ip*(1-cos 2b)/2
in other words the instantaneous power varies as the cos of the double rate frequency.
If you average this over 1/2 cycle the result is
Pave = Vp*Ip/2 converting to RMS values
Pave = Vrms*Irms and this is real power.

If one or the other is shifted by 90 deg, then
p = Vp*Ip*(sin b)*(cos b)
now the trig identity is (sin 2b)/2. This has no DC component and when averaged over a multiple of 1/2 cycles it is 0.
Thus, no real power.
But the instantaneous power is
p = Vrms*Irms*sin 2b

.

So is there a number you like gar or should we do the math?

 

[mivey]

With True Power and Apparent power i'm so confused please help:-? :-?

How about
(Apparent Power)^2 = (Real Power)^2 + (Reactive Power)^2
or
kVA^2 = kW^2 + kVAR^2
or
VA^2 = W^2 + VAR^2

Apparent Power is the vector sum of Real Power and Reactive Power. Plot Real Power on the x-axis and Reactive Power on the y-axis.

Real power does the useful work. Reactive power is needed to create the magnetic fields in motors and such but does no real work. The apparent power is the total power delivered to the system, real plus reactive.

 

[Dennis Alwon]

How about
(Apparent Power)^2 = (Real Power)^2 + (Reactive Power)^2
or
kVA^2 = kW^2 + kVAR^2
or
VA^2 = W^2 + VAR^2

Apparent Power is the vector sum of Real Power and Reactive Power. Plot Real Power on the x-axis and Reactive Power on the y-axis.

Real power does the useful work. Reactive power is needed to create the magnetic fields in motors and such but does no real work. The apparent power is the total power delivered to the system, real plus reactive.

Yeah that is equivalent to trying to decipher this
A geriatric human female proceeded to a storage compartment for the purpose of procuring a fragment of asseous tissue from an unidentified deceased specimen to transfer to an indigent carnivorous domesticated mammal. Upon arrival at her destination, she found the storage compartment in a denuded condition, with the consequence that the indegent carnivore was deprived of the intended donation.

 

[mivey]

Yeah that is equivalent to trying to decipher this
A geriatric human female proceeded to a storage compartment for the purpose of procuring a fragment of asseous tissue from an unidentified deceased specimen to transfer to an indigent carnivorous domesticated mammal. Upon arrival at her destination, she found the storage compartment in a denuded condition, with the consequence that the indegent carnivore was deprived of the intended donation.

Can you say delivery? Mine was the Reader's Digest Condensed version of the link you provided.

 

[gar]

080428-2044 EST

My previous post was in response to awc's ending question in post #7.

My answer to post #1 would be to consider the load as a Y and then a leg's current is in phase with the load current.

For a balanced source and load the load voltage is leg to neutral and this is leg to leg voltage divided by the sq-root of three. Since the load is balanced each element of the load receives 1/3 of the total power.

Power to each element is Irms * Vll * PF / 1.732.

Total power is 3*111*208*PF/sq-rt 3 = 1.732*111*208*PF = 39,988*PF.

As an old professor told me when I ask about a question that seemed to be in error on a final exam --- present your analysis of what is wrong.

So in this case since the characteristics of the load were not defined one fills in the missing information. Either state the assumption that the load is resistive, or add the PF element to the answer.

.

 

[KentAT]

Yeah that is equivalent to trying to decipher this
A geriatric human female proceeded to a storage compartment for the purpose of procuring a fragment of asseous tissue from an unidentified deceased specimen to transfer to an indigent carnivorous domesticated mammal. Upon arrival at her destination, she found the storage compartment in a denuded condition, with the consequence that the indegent carnivore was deprived of the intended donation.

No, yours was much easier...the poor dog's still hungry...

 

[awc]

Thanks again to all of you....I did some research and through that along with all of your answers I have myself on the right page :wink: . mivey, your answer helped me alot at first, breaking it down to the basics help me build from there. All of you guys input and answers were greatly appreciated, Thank you!