Finding power consumed given a load

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Discussion Overview

The discussion revolves around calculating the power consumed by two 3-phase parallel loads, determining their complex power, and exploring power factor correction through capacitor banks. It includes theoretical and practical aspects of electrical engineering, particularly focusing on complex power, reactive power compensation, and the implications of different capacitor connections.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Participants discuss the calculation of complex power for Load 1 and Load 2, with some asserting that the 12 kVA for Load 1 represents apparent power, not reactive power.
  • There is a debate regarding the correct interpretation of units for reactive power (Q) and apparent power (S), with some participants suggesting that the values need to be revisited.
  • Participants propose methods for calculating the combined load power factor and the total complex power consumed by both loads.
  • Discussion includes the need to find the capacitance of a capacitor bank to improve the total power factor to 0.95 lagging, considering different configurations (delta vs. Y) and the implications of line-to-line voltage.
  • Some participants express uncertainty about how to perform power factor correction without a specified frequency, while others suggest using standard frequencies (50Hz or 60Hz).
  • There are questions about the calculations for reactive power changes and the relationship between delta and star connections in terms of impedance and capacitance.

Areas of Agreement / Disagreement

Participants generally do not reach a consensus on the interpretation of complex power for Load 1, with multiple competing views on the correct approach to calculating reactive power and the implications of capacitor connections. The discussion remains unresolved on several technical points, particularly regarding the calculations and assumptions made.

Contextual Notes

Limitations include the absence of a specified frequency for calculations, which affects the determination of capacitance values. Additionally, there are unresolved mathematical steps regarding the calculations of reactive power and the implications of different capacitor configurations.

Who May Find This Useful

This discussion may be useful for students and professionals in electrical engineering, particularly those interested in power systems, reactive power compensation, and complex power calculations.

eehelp150
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Homework Statement


The power consumed by 2 3-phase parallel loads are:
Load 1: 12 kVA with power factor 0.7 lagging
Load 2: 8 kW with power factor 0.8 lagging

A)Find the complex power consumed by Load 1 and Load 2.
B)Find the total complex power consumed by these two loads
C)Determine the combined load power factor

Homework Equations


S = V * I*
S = sqrt(P^2 + Q^2)
P = V * I

The Attempt at a Solution


A)
Load 1: Q = 12kVA, cos(θ) = 0.7, sin(θ) = 0.71414284272
S = Q/sin(θ) = 16803.36VA

Load 2: P = 8kW, cos(θ) = 0.8
S = P/cos(θ) = 8kW/0.8 = 10000VA

B)
Total Complex = S1 + S2 = 26803.36VA

Am I doing this correctly?
 
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eehelp150 said:
Load 1: Q = 12kVA,
This is not Q. Check the unit.
eehelp150 said:
S = P/cos(θ) = 8kW/0.8 = 10000VA
Right.
 
cnh1995 said:
This is not Q. Check the unit.

Right.
Is the complex power consumed by load 1 simply 12kVA?
 
Complex power is expressed as a complex number, with both real and imaginary parts. In the power triangle, the components of that number are represented by the real and reactive power, while the apparent power is the hypotenuse (and thus the magnitude of the complex power). Just like vectors, you can't add them by adding their magnitudes. You have to add the components separately.

In this problem I would argue that the 12 kVA of first load is not the reactive power but rather the apparent power (i.e. the magnitude of the complex power). The second load has units of watts, so that can only be the real power and you've treated it as such.

So I would revisit the first load and assume that the given value is the apparent power. Then I would sum the real and reactive components of the loads to find the combined power.
 
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eehelp150 said:
Is the complex power consumed by load 1 simply 12kVA?
Yes. Unit of Q is VAR or kVAR.
 
cnh1995 said:
Yes. Unit of Q is VAR or kVAR.
So complex power consumed by Load1 = 12kVA
Complex power consumed by Load2 = 10kVA

P/S = cos(θ)
Load 1:
P/12k = 0.7
P = 8400W
S = sqrt(P^2 + Q^2)
12000 = sqrt(8400^2 + Q^2)
144000000 = 8400^2 + Q^2
Q = 8569.7
Load 1 = 8400W + 8569.7VAR

Load2:
10000VA = sqrt(8000^2 + Q^2)
100000000 = 64000000 + Q^2
Q = 6000
Load 2 = 8000W + 6000VAR

Combined = 16400W + 14569.7VAR

Stotal = 21937.1VA

P/S = cos(θ)
16400/21937.1 = cos(θ)
PF = cos(θ) = 0.7476

correct?
 
eehelp150 said:
So complex power consumed by Load1 = 12kVA
Complex power consumed by Load2 = 10kVA

P/S = cos(θ)
Load 1:
P/12k = 0.7
P = 8400W
S = sqrt(P^2 + Q^2)
12000 = sqrt(8400^2 + Q^2)
144000000 = 8400^2 + Q^2
Q = 8569.7
Load 1 = 8400W + 8569.7VAR

Load2:
10000VA = sqrt(8000^2 + Q^2)
100000000 = 64000000 + Q^2
Q = 6000
Load 2 = 8000W + 6000VAR

Combined = 16400W + 14569.7VAR

Stotal = 21937.1VA

P/S = cos(θ)
16400/21937.1 = cos(θ)
PF = cos(θ) = 0.7476

correct?
Looks good.
 
cnh1995 said:
Looks good.
I forgot the last part:
Find the capacitance of a 3-phase capacitor to be connected in parallel with the load to raise the total power factor to 0.95 lagging.
Consider
a) the capacitor is connected in delta
b) The capacitor is connected in Y.
Load line-line voltage is 207.85 V.

What role does delta vs Y connection play? The only example in my notes did not mention how the capacitor would be connected. The example in my notes also gave the circuit frequency, which is not given for this problem.
 
eehelp150 said:
I forgot the last part:
Find the capacitance of a 3-phase capacitor to be connected in parallel with the load to raise the total power factor to 0.95 lagging.
Consider
a) the capacitor is connected in delta
b) The capacitor is connected in Y.
Load line-line voltage is 207.85 V.

What role does delta vs Y connection play? The only example in my notes did not mention how the capacitor would be connected. The example in my notes also gave the circuit frequency, which is not given for this problem.
The capacitor bank supplies reactive power to the load and improves the power factor. This is called reactive power compensation.
Current through the capacitors will depend on their connection (star or delta). Also, frequency of the supply is used to calculate the capacitance value from the capacitive reactance obatined from reactive power calculations.
 
  • #10
cnh1995 said:
The capacitor bank supplies reactive power to the load and improves the power factor. This is called reactive power compensation.
Current through the capacitors will depend on their connection (star or delta). Also, frequency of the supply is used to calculate the capacitance value from the capacitive reactance obatined from reactive power calculations.
How would I do power factor correction without a given frequency?
 
  • #11
eehelp150 said:
How would I do power factor correction without a given frequency?
You can take the standard power frequency. It's either 50Hz or 60Hz, depending on your country.
 
  • #12
cnh1995 said:
You can take the standard power frequency. It's either 50Hz or 60Hz, depending on your country.
I have:
current power factor = 0.74
desired power factor = 0.95 lagging
Load line-line voltage: 207.85

How do calculate the capacitance required? I don't understand the delta/Y part. My notes have something like this:

V = 120V<60, I = 10<25
S = 1200<35
P1 = 982.98
Q1 = 688.3
desired power factor: 0.95
current power factor: 0.82
arccos(0.95) = 18.19 degrees
Q2 = tan(18.19) * P1 = 323
Qchange = Q1-Q2 = 365.3
Zcapacitor = (120*120)/365.3 = 39.42 ohm
C = 1/(w(Zcapacitor)) =67.27uF
 
  • #13
eehelp150 said:
current power factor = 0.74
desired power factor = 0.95 lagging
Load line-line voltage: 207.85
So how much reactive power should be supplied by the capacitor bank? It will be the difference between current reactive power (with pf=0.74) and reactive power with the desired power factor (0.95). Star or delta connection determines the capacitor current. For the line voltage of 207.85V (you sure this is line-line and not line to neutral?), capacitor current will be 207.85/Zc for delta and 207.85/√3Zc for star.
 
  • #14
cnh1995 said:
So how much reactive power should be supplied by the capacitor bank? It will be the difference between current reactive power (with pf=0.74) and reactive power with the desired power factor (0.95). Star or delta connection determines the capacitor current. For the line voltage of 207.85V (you sure this is line-line and not line to neutral?), capacitor current will be 207.85/Zc for delta and 207.85/√3Zc for star.
Prompt says load line-to-line voltage
 
  • #15
eehelp150 said:
Prompt says load line-to-line voltage
Ok..That's 120V line to neutral, USA's standard distribution voltage.
Well, how much is the difference between the reactive powers?
 
  • #16
cnh1995 said:
Ok..That's 120V line to neutral, USA's standard distribution voltage.
Well, how much is the difference between the reactive powers?
Q1 = 14569.7
new pf = 0.95
new Q = Q2 = 5390.46
Q change = 9179.24
Zcapacitor = 120^2/9179.24
Zcapacitor = 1.57ohm

For same capacitance, Delta = Star/3, correct?

Are these correct:
Zcapacitor = 120^2/(3*9179.24)
and for Y/star:
Zcapacitor = 120^2/(9179.24)
?
 
Last edited:
  • #17
eehelp150 said:
Q1 = 14569.7
new pf = 0.95
new Q = Q2 = 5390.46
Q change = 9179.24
Zcapacitor = 120^2/9179.24
Zcapacitor = 1.57ohm

For same capacitance, Delta = Star/3, correct?

Are these correct:
Zcapacitor = 120^2/(3*9179.24)
and for Y/star:
Zcapacitor = 120^2/(9179.24)
I can't double check your calculations right now. You have the difference in reactive powers as Qc. This is the power supplied by the capacitor bank.
Qc=√3VLIL
From this, you can find the line current IL. For star connection, IL=Iph=I, hence, you can use Qc=3I2Xc and calculate Xc per phase.
For delta connection, you have VL=Vph=V, hence, Qc=3V2/Xc. You can get Xc from this and then C=1/(ωXc).
 
  • #18
cnh1995 said:
I can't double check your calculations right now. You have the difference in reactive powers as Qc. This is the power supplied by the capacitor bank.
Qc=√3VLIL
From this, you can find the line current IL. For star connection, IL=Iph=I, hence, you can use Qc=3I2Xc and calculate Xc per phase.
For delta connection, you have VL=Vph=V, hence, Qc=3V2/Xc. You can get Xc from this and then C=1/(ωXc).
I don't have I, how do I find Xc for star?
 
  • #19
eehelp150 said:
I don't have I, how do I find Xc for star?
I=Qc/(√3VL).
Then,
cnh1995 said:
Qc=3I2Xc
∴Xc=Qc/3I2.
And
C=1/(ω*Xc).
 

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