# Solve Delta Power System Homework - P=I^2 R

• camino
In summary, the problem involves calculating the power and efficiency in a power system with various loops and resistors. The power in the resistors must be calculated using the magnitude of the current and the angle, and the efficiency can be determined by taking into account the power loss in the transmission lines.
camino

## Homework Statement

http://img42.imageshack.us/img42/3537/powersyst.jpg

P = I^2 R

## The Attempt at a Solution

Here's what I have so far:

I1 = 49.68A<-68.6 degrees (top loop)
I2 = 49.68A<-128.6 degrees (bottom loop)
I3 = 28.68<-98.6 degrees (triangle loop)

a)
PA12 = (49.68A)^2(12) = 29,617.23 W
PB12 = (49.68A-49.68A)^2(12) = 0 W
PC12 = (49.68A)^2(12) = 29,617.23 W
Ptotal12 = 29,617.23W + 0W + 29,617.23W = [59,234 W] answer

b)
Pab400 = (49.68A-28.68A)^2(400) = 176,400 W
Pbc400 = (49.68A-28.68A)^2(400) = 176,400 W
Pca400 = (28.68A)^2(400) = 329,017 W
Ptotal400 = 176,400W + 176,400W + 329,017W = [681,817 W] answer

Now c) and d) I am unsure of. I was thinking power factor would be 59,234 W/681,817W which would give me a p.f. of 0.087 but that doesn't seem right, and efficiency I have no clue how to calculate.

I'm not 100% sure I even did parts a) and b) right. Any help/explanations would be greatly appreciated!

Last edited by a moderator:
I believe the power lost in the line is the loss hence providing you efficiency of less than 100% (power consumed by the delta load)

I don't check whether the numerical values are true or not, but there are errors in calculating the real power in the resistors.

On assumption that the currents values are true.

PA12 is OK.

PB12 = I1 - I2 = (|(49.68A<-68.6 - 49.68A<-128.6)|^2) * 12 = something
By "|" I mean the magnitude.

Here we are dealing with phasors, so first obtain the current through the resistor (don't forget the angle of the current) and then calculate the power.

The same should be done for Pab400 and Pbc400.

By considering lines "Aa", "Bb", and "Cc" as transmssion lines, the impedance (12+j16) is the line impedance and there are power loss in the line. The loss is the power dissipated in the 12 Ohm resistors. So, the efficiency can be calculated.

## 1. What is the purpose of solving for the Delta Power System homework?

The purpose of solving for the Delta Power System homework is to understand the relationship between power, current, and resistance in an electrical circuit. This knowledge is essential for designing and troubleshooting electrical systems.

## 2. How do you calculate power in a Delta Power System?

The formula for calculating power in a Delta Power System is P = I^2R, where P is power in watts, I is current in amps, and R is resistance in ohms.

## 3. What is the significance of the Delta Power System in electrical engineering?

The Delta Power System is important in electrical engineering because it is a commonly used configuration for three-phase power systems. It allows for efficient distribution of power and is used in many industrial and commercial applications.

## 4. Can you explain the concept of power factor in a Delta Power System?

Power factor is a measure of how efficiently a Delta Power System is using the power it receives. A power factor of 1 means all the power is being used effectively, while a power factor less than 1 indicates some power is being lost due to reactive elements in the circuit.

## 5. How does solving for the Delta Power System homework relate to real-world applications?

Solving for the Delta Power System homework is directly applicable to real-world electrical systems. Understanding the relationship between power, current, and resistance is crucial for designing, maintaining, and troubleshooting electrical systems in industries such as manufacturing, transportation, and energy production.

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