Calculating Three Phase Power Load: Help & Homework Statement

In summary: Is that what is wrong?Thanks againIn summary, the conversation discusses a practice question involving a three-phase supply and a balanced star-connected load. The power delivered to the feeder-load combination is measured as 23.3kW at a lagging power factor of 0.9. The summary includes calculations for the magnitude of the supply line current, real and reactive power delivered to the load, and the line to line voltage at the load end of the feeder. It also discusses a delta-connected capacitor bank connected to improve the power factor of the feeder-load combination, and calculations for the required capacitor value and overall power factor if accidentally connected as a star instead of delta.
  • #1
Tomas_89
2
0

Homework Statement



I have an exam coming up and one of the practice questions is as follows
A 415V 50Hz three phase supply energises a balanced star-connected load through a three phase feeder with a per-phase series impedance of Zline = ( 0.165 + j0.413 ). A power meter at the source end of the feeder measures the power delivered to the feeder-load combination as 23.3kW at a lagging power factor of 0.9.

1. Calculate the magnitude of the supply line current in the feeder.

2. Calculate the real and reactive power delivered to the load, and hence determine the line to line voltage that is developed at the load end of the feeder. A delta-connected capacitor bank is connected across the supply at the source end of the feeder, so as to improve the power factor of the feeder-load combination.

3. Determine the capacitor value required to ensure that the power factor as seen by the

4. Calculate the overall power factor of the feeder-load combination if the capacitor bank supply is unity is accidentally connected in star rather than delta.[/B]

Homework Equations



The Attempt at a Solution


I have done the following

1. Power = 23.3kw
Reactive Power = 11.3 KVAR
Apparent Power = 25.9 KVA

Current = 36.03 amps

2. Load, by working out the power loss through the three-phase feeder
Ploss = 36.03^2 x 0.165 = 214.2 watts
Power = 23300 -214 = 23086 watts

Reactive Power loss = 36.03^2 x .413 = 536.14 VAR
Reactive Power = 11300 - 536.14 = 10764 VAR

Apparent Power at the load end of the feeder = 25472 VA

Voltage at the load end
V = 25472/sqrt(3)*36.03 = 412.49 volts

3. Capacitor value
Xc = 3*415^2/11300 = 45.72 ohms
C = 69.62 uF

4. Overall power factor if accidentally connected as a star instead of delta
Zy = 45.72/3 = 15.24 ohms

Q = 3*415^2/15.24 = 33902 VAR

pf = 23300/33902 = .68 power factor

Any help or guidance would be most appreciated. If any more working out is required let me know and I'll post the rest up.

Cheers
 
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  • #2
Tomas_89 said:

Homework Statement



I have an exam coming up and one of the practice questions is as follows
A 415V 50Hz three phase supply energises a balanced star-connected load through a three phase feeder with a per-phase series impedance of Zline = ( 0.165 + j0.413 ). A power meter at the source end of the feeder measures the power delivered to the feeder-load combination as 23.3kW at a lagging power factor of 0.9.

1. Calculate the magnitude of the supply line current in the feeder.

2. Calculate the real and reactive power delivered to the load, and hence determine the line to line voltage that is developed at the load end of the feeder.


A delta-connected capacitor bank is connected across the supply at the source end of the feeder, so as to improve the power factor of the feeder-load combination.

3. Determine the capacitor value required to ensure that the power factor as seen by the

4. Calculate the overall power factor of the feeder-load combination if the capacitor bank supply is unity is accidentally connected in star rather than delta.[/B]


Homework Equations






The Attempt at a Solution


I have done the following

1. Power = 23.3kw
Reactive Power = 11.3 KVAR
Apparent Power = 25.9 KVA

Current = 36.03 amps

2. Load, by working out the power loss through the three-phase feeder
Ploss = 36.03^2 x 0.165 = 214.2 watts
Power = 23300 -214 = 23086 watts

Reactive Power loss = 36.03^2 x .413 = 536.14 VAR
Reactive Power = 11300 - 536.14 = 10764 VAR

Apparent Power at the load end of the feeder = 25472 VA

Voltage at the load end
V = 25472/sqrt(3)*36.03 = 412.49 volts

3. Capacitor value
Xc = 3*415^2/11300 = 45.72 ohms
C = 69.62 uF

4. Overall power factor if accidentally connected as a star instead of delta
Zy = 45.72/3 = 15.24 ohms

Q = 3*415^2/15.24 = 33902 VAR

pf = 23300/33902 = .68 power factor

Any help or guidance would be most appreciated. If any more working out is required let me know and I'll post the rest up.

Cheers

Hi Tomas_89, Welcome to Physics Forums.

Where the question statement says: "A power meter at the source end of the feeder measures the power delivered to the feeder-load combination as 23.3kW at a lagging power factor of 0.9", does that refer to the per-phase power delivered, or is it to be interpreted as a total delivered by all phases to the feeder-load? I realize that the interpretation may depend upon conventions or assumptions made by the given text or course content.

Some of your values you provide without showing your calculation method, so it's hard to tell what might have gone wrong if the values look a bit "off" to someone checking (particularly if the question interpretation is not certain for the checker). For example, can you show the details of how you arrived at your current magnitude value of 36.03 amps?
 
  • #3
Hey thanks for that, it's been taught per phase in this course so that's the assumption.

Power is equal to 23.3KW as stated.

The reactive power was calculated as:
Q = ( 23.3x10^3/0.9 (power factor) ) x sin(inverse cos (0.9)
Q = 11.3 KVARS

Apparent power :
S = P + jQ = √23.3^2 + 11.3 ^2
S = 25.9 KVA
For the current I used the apparent power formula (S = VI) and rearranged as
I = 25900VA/√3x 415V
I = 36.03 A
 

FAQ: Calculating Three Phase Power Load: Help & Homework Statement

What is 3 phase power load?

3 phase power load refers to a type of electrical power distribution that uses three alternating currents with a 120 degree phase difference to power a system. This type of power distribution is commonly used in industrial and commercial settings.

How is 3 phase power load different from single phase power load?

The main difference between 3 phase and single phase power load is the number of alternating currents used. Single phase power load uses only one alternating current, while 3 phase power load uses three. This allows for a more balanced and efficient distribution of power in systems with high power demands.

How is 3 phase power load calculated?

3 phase power load is calculated by multiplying the voltage by the current and the square root of three. The result is then multiplied by the power factor, which represents the efficiency of the system. This calculation is used to determine the total power load in a 3 phase system.

What are the advantages of using 3 phase power load?

There are several advantages to using 3 phase power load, including a more balanced distribution of power, higher efficiency, and the ability to power larger and more complex systems. 3 phase power load also allows for smoother operation of motors and other equipment.

How can I determine the appropriate 3 phase power load for my system?

The appropriate 3 phase power load for a system can be determined by calculating the total power requirements of all the equipment and devices in the system. It is important to also consider the power factor and any potential future expansions or additions to the system. Consulting with an electrical engineer can also help determine the appropriate 3 phase power load for a specific system.

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