Finding power consumed given a load

  • Thread starter eehelp150
  • Start date
  • Tags
    Load Power
In summary: How do calculate the capacitance required? I don't understand the delta/Y part. My notes have something like...The capacitance is calculated using the following equation:C = σ*f*LwhereC = Capacitance in faradsσ = Capacitive reactance in ohmsf = Frequency in hertzL = Length of the capacitor in meters
  • #1
eehelp150
237
0

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?
 
Physics news on Phys.org
  • #2
eehelp150 said:
Load 1: Q = 12kVA,
This is not Q. Check the unit.
eehelp150 said:
S = P/cos(θ) = 8kW/0.8 = 10000VA
Right.
 
  • #3
cnh1995 said:
This is not Q. Check the unit.

Right.
Is the complex power consumed by load 1 simply 12kVA?
 
  • #4
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.
 
  • Like
Likes cnh1995
  • #5
eehelp150 said:
Is the complex power consumed by load 1 simply 12kVA?
Yes. Unit of Q is VAR or kVAR.
 
  • #6
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?
 
  • #7
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.
 
  • #8
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.
 
  • #9
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).
 

1. How do I calculate power consumption for a specific load?

To calculate power consumption for a specific load, you will need to know the voltage and current of the load. Once you have this information, you can use the formula P = VI, where P is power, V is voltage, and I is current. This will give you the power consumed by the load in watts.

2. What is the difference between power consumption and energy consumption?

Power consumption refers to the amount of power used by a device or load at a specific moment in time. It is measured in watts. Energy consumption, on the other hand, refers to the total amount of energy used over a period of time. It is measured in watt-hours or kilowatt-hours.

3. How does the type of load affect power consumption?

The type of load can greatly affect power consumption. Some loads, such as resistive loads, have a constant power consumption regardless of the voltage and current. Other loads, such as inductive or capacitive loads, have a variable power consumption that depends on the voltage and current.

4. Can I use a multimeter to measure power consumption?

No, a multimeter is not designed to measure power consumption. It can only measure voltage, current, and resistance. To measure power consumption, you will need a wattmeter or a power analyzer.

5. What are some ways to reduce power consumption in a circuit?

There are several ways to reduce power consumption in a circuit. Some common methods include using energy-efficient components, optimizing the circuit design, and implementing power management techniques such as sleep modes or variable frequency operation. It is also important to regularly monitor and maintain the circuit to ensure it is running efficiently.

Similar threads

Three phase AC concepts and problems
    • Engineering and Comp Sci Homework Help
Replies
2
Views
5K
Engineering Source impedance using given voltage swing and power
    • Engineering and Comp Sci Homework Help
Replies
2
Views
765
Engineering How to Find the First Buckling Load from the Deflection Equation?
    • Engineering and Comp Sci Homework Help
Replies
3
Views
1K
Engineering Intro to Power Systems: Calculating Power
    • Engineering and Comp Sci Homework Help
Replies
7
Views
1K
Engineering How am I susposed to find the active power here?
    • Engineering and Comp Sci Homework Help
Replies
9
Views
1K
Supplied complex power from source vs consumed by load
    • Engineering and Comp Sci Homework Help
Replies
13
Views
2K
Parallel Capacitance - Power Factor Question
    • Engineering and Comp Sci Homework Help
Replies
6
Views
3K
Engineering Transformer -- find equivalent circuit from efficiency and V.R
    • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
Maximum pull-off load, axial load and radial load of two strapped pipes
    • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
How Do You Calculate Engine Power for Semi Trucks in Various Conditions?
    • Engineering and Comp Sci Homework Help
Replies
8
Views
1K

Hot Threads

Recent Insights

Back
Top