Quick calculations check

[Smart $]

I missed yours when I went to get my calculator. 28.4205255521 uF for #2

So what assumptions are you two making?

I know you are making at least one, and that is the size of the existing capacitor. Without the value of the existing capacitor being given, you cannot calculate (or even guess with any certainty) a capacitor value to double it.

Maybe throwing you two a curve will be enough for you to see it. Let's explore hypothetical conditions to the problem.

1. Assume for a moment that the existing is 10µF...

2. Assume for a different moment the existing capacitor corrects the power factor to .95...

 

[GoldDigger]

1. Assume for a moment that the existing is 10µF...

That assumption is not consistent with the stated measurement of the capacitor current (not the motor current) at 6A.

It is fair to ask why it is desired to double the capacitance. Evaluating that would require knowledge of the uncorrected (and corrected) power factor of the motor and the motor current.
But the OP did not ask for that. He asked, essentially, how to find the value of the existing capacitor given measurements of the voltage and current at the capacitor terminals.

 

[Smart $]

That assumption is not consistent with the stated measurement of the capacitor current (not the motor current) at 6A.
...

Aha! I was reading the 6A as machine current, i.e. all inclusive. The problem could be better stated... :happyyes:

 

[GoldDigger]

Aha! I was reading the 6A as machine current, i.e. all inclusive. The problem could be better stated... :happyyes:

"The capacitor is connected to a 560V line and [the capacitor] has a current draw of 6A."

Makes it totally unambiguous, but does not really seem necessary to me.

 

[Smart $]

"The capacitor is connected to a 560V line and [the capacitor] has a current draw of 6A."

Makes it totally unambiguous, but does not really seem necessary to me.

The "line" is what threw me off. If omitted, I probably would have seen as cap' current on first go round. I admit I should have seen through it... but it is what it is.

"The capacitor is connected to 560V and has a current draw of 6A."

 

[Besoeker]

Good afternoon gents,

Can anyone check my calculations on these (2) questions:

1. Residential Heat Pump compressor has current draw of 34A when operating. Compressor has a PF of 70%. Back up heat strip is rated at 10 KW. What is the total current draw when heat strip comes on while the compressor is operating?

My answer: 69.8A (Still new to the postings so don't have the cool symbols yet so I will not show my work)

2. Working in an industrial plant. I have been instructed to double the capacitance connected to a machine. The markings on the capacitor are not visible. The capacitor is connected to a 560V line and has a current draw of 6A. What size capacitor is needed to be connected in parallel with the existing capacitor to double the capacitance?

My answer: 28.42 micro farads (Still new to the postings so don't have the cool symbols yet so I will not show my work)


Thanks to all who view! Happy Thanksgiving!

FWIW, my calcs yeilded the same results.

 

[NewtonLaw]

#1) 34 amps at 0.7pf = 34 amps @-45.57° and 10kw/240 = 41.67 amps

Adding currents you get Compressor -> 23.80 -j 24.28 amps
............................................Heater -> 41.67 +j 50.0
.............................................Total -> 57.80 -j 24.28 = 62.69 amps @ -22.79°

#2) Assuming single phase, 60Hz so 560V*6Amps = 3,360 VARs

C = VARs / (377*V^2) = 3360 / (377*560^2) = 28.4 micro-Farads

Hope this helps

 

[Dennis Alwon]

10000/240= 41 amps + 34 amps = 75 amps... what is the issue here with all this vector stuff

 

[Smart $]

10000/240= 41 amps + 34 amps = 75 amps... what is the issue here with all this vector stuff

I asked the same thing earlier. Typical load calculations do not use vector math.

 

[Chamuit]

10000/240= 41 amps + 34 amps = 75 amps... what is the issue here with all this vector stuff

KISS is not a normal SOP here in the forum...:happyno:

 

[NewtonLaw]

Load Current Calculation

Load Current Calculation

10000/240= 41 amps + 34 amps = 75 amps... what is the issue here with all this vector stuff

The heater load is resistive in nature, i.e., phase angle displacement of current would be zero. The pump motor is given as 34 amps at power factor of 70% thus it is not in phase with the resistive current from the heater. The currents do not add as a simple algebraic value from what I can see. Am I missing something? These loads are in parallel correct?

 

[Julius Right]

In my opinion, if Fishn sparky said the result is 53.8 A that means the heat strip is three-phase 240 V supplied too.
So, if we take the compressor motor current as 34*(0.7-0.714i) [ cos(fi)=0.7 sin(fi)=sqrt(1-0.7^2)=0.714], and strip heater current 10*1000/sqrt(3)/240=24.06 then Itotal=sqrt((34*.7+24.06)^2+(0.71414*34)^2)=53.67.Close!

 

[Smart $]

The heater load is resistive in nature, i.e., phase angle displacement of current would be zero. The pump motor is given as 34 amps at power factor of 70% thus it is not in phase with the resistive current from the heater. The currents do not add as a simple algebraic value from what I can see. Am I missing something? These loads are in parallel correct?

Load calculations under the NEC use volt-ampere values with no consideration for power factor. Volt-ampere values add algebraically just fine.

 

[Julius Right]

Load calculations under the NEC use volt-ampere values with no consideration for power factor. Volt-ampere values add algebraically just fine.

Theoretically you cannot take Stotal=S1+S2 since S1=P1+/-iQ1 and S2=P2+/-iQ2.
So absolute value will be Stot=sqrt((P1+P2)^2+(Q1+Q2)^2).
In our case the error it is only 8%.
Since the NEC load appreciation cannot be so accurate and for pf of 0.7-0.8 the error is less than 1% we would live with that. However for pf1=0[pure reactive] and pf2=1[pure active] the error could be 30%.:weeping:

 

[Smart $]

Theoretically you cannot take Stotal=S1+S2 since S1=P1+/-iQ1 and S2=P2+/-iQ2.
So absolute value will be Stot=sqrt((P1+P2)^2+(Q1+Q2)^2).
In our case the error it is only 8%.
Since the NEC load appreciation cannot be so accurate and for pf of 0.7-0.8 the error is less than 1% we would live with that. However for pf1=0[pure reactive] and pf2=1[pure active] the error could be 30%.:weeping:

I am aware of all this. However, it does not change the NEC approach to load calculations, as error prone as it is. Be that as it may, this is in part the reason why NEC load calculations are much more often than not on the conservative side.

 

[Julius Right]

I am aware of all this. However, it does not change the NEC approach to load calculations, as error prone as it is. Be that as it may, this is in part the reason why NEC load calculations are much more often than not on the conservative side.

Agreed!:ashamed1: