Question
A three-phase, balanced load is connected to a 525V, 50Hz supply. The impedance of each phase is (7+j15)ohms. If the power drawn by this load is measured using two-wattmeter method, determine the readings on the two wattmeter.
Answers
GPT-4o mini
To determine the readings on the two wattmeters in the two-wattmeter method connected to a balanced three-phase load, we need to go through the following steps:
1. **Calculate the phase current (I)**:
The phase voltage \( V_{ph} \) for a star-connected load is given by:
\[
V_{ph} = \frac{V_{line}}{\sqrt{3}} = \frac{525V}{\sqrt{3}} \approx 303.12V
\]
The phase current \( I \) can be calculated using Ohm's Law:
\[
I = \frac{V_{ph}}{Z}
\]
where \( Z = 7 + j15 \, \Omega \) and its magnitude \( |Z| \) is calculated as:
\[
|Z| = \sqrt{7^2 + 15^2} = \sqrt{49 + 225} = \sqrt{274} \approx 16.52 \, \Omega
\]
Now, using the phase voltage:
\[
I = \frac{303.12}{16.52} \approx 18.35 \, A
\]
2. **Calculate the total active power (P)**:
The active power in a three-phase system is calculated as:
\[
P = 3 \cdot V_{ph} \cdot I \cdot \cos(\phi)
\]
where \( \phi \) is the phase angle. The phase angle \( \phi \) can be determined from the impedance:
\[
\tan(\phi) = \frac{15}{7} \quad \Rightarrow \quad \phi \approx \tan^{-1}(2.14) \approx 64.62^\circ
\]
Hence, the cosine of the phase angle is:
\[
\cos(\phi) \approx \cos(64.62^\circ) \approx 0.444
\]
Now substituting the values:
\[
P = 3 \cdot 303.12V \cdot 18.35A \cdot 0.444 \approx 3 \cdot 303.12 \cdot 18.35 \cdot 0.444 \approx 3 \cdot 303.12 \cdot 18.35 \cdot 0.444 \approx 4,072.34 W \quad (approximately)
\]
3. **Determine the wattmeter readings**:
Using the two-wattmeter method:
\[
W_1 = V_{line} \cdot I \cdot \cos(\phi) = V_{line} \cdot I_A \cdot \cos(\phi)
\]
\[
W_2 = V_{line} \cdot I \cdot \cos(\phi - 60^\circ)
\]
Now we can compute \( W_1 \) and \( W_2 \):
Using \( V_{line} = 525V \):
\[
W_1 = 525V \cdot 18.35A \cdot \cos(64.62^\circ)
\]
\[
= 525 \cdot 18.35 \cdot 0.444 \approx 4,072.34 W
\]
Now using the relation \( W_1 + W_2 = P \):
\[
W_2 = P - W_1 \approx 4,072.34 W - W_1
\]
However, because we have a balanced load, \( W_1 \) and \( W_2 \) will not differ much and will share the total power evenly. Since the system is balanced:
\[
W_1 = W_2 \approx \frac{P}{2} \approx \frac{4,072.34 W}{2} \approx 2,036.17 W
\]
4. **Final Result**:
Thus, the readings on the two wattmeters are approximately:
\[
W_1 \approx 2,036.17 W
\]
\[
W_2 \approx 2,036.17 W
\]
1. **Calculate the phase current (I)**:
The phase voltage \( V_{ph} \) for a star-connected load is given by:
\[
V_{ph} = \frac{V_{line}}{\sqrt{3}} = \frac{525V}{\sqrt{3}} \approx 303.12V
\]
The phase current \( I \) can be calculated using Ohm's Law:
\[
I = \frac{V_{ph}}{Z}
\]
where \( Z = 7 + j15 \, \Omega \) and its magnitude \( |Z| \) is calculated as:
\[
|Z| = \sqrt{7^2 + 15^2} = \sqrt{49 + 225} = \sqrt{274} \approx 16.52 \, \Omega
\]
Now, using the phase voltage:
\[
I = \frac{303.12}{16.52} \approx 18.35 \, A
\]
2. **Calculate the total active power (P)**:
The active power in a three-phase system is calculated as:
\[
P = 3 \cdot V_{ph} \cdot I \cdot \cos(\phi)
\]
where \( \phi \) is the phase angle. The phase angle \( \phi \) can be determined from the impedance:
\[
\tan(\phi) = \frac{15}{7} \quad \Rightarrow \quad \phi \approx \tan^{-1}(2.14) \approx 64.62^\circ
\]
Hence, the cosine of the phase angle is:
\[
\cos(\phi) \approx \cos(64.62^\circ) \approx 0.444
\]
Now substituting the values:
\[
P = 3 \cdot 303.12V \cdot 18.35A \cdot 0.444 \approx 3 \cdot 303.12 \cdot 18.35 \cdot 0.444 \approx 3 \cdot 303.12 \cdot 18.35 \cdot 0.444 \approx 4,072.34 W \quad (approximately)
\]
3. **Determine the wattmeter readings**:
Using the two-wattmeter method:
\[
W_1 = V_{line} \cdot I \cdot \cos(\phi) = V_{line} \cdot I_A \cdot \cos(\phi)
\]
\[
W_2 = V_{line} \cdot I \cdot \cos(\phi - 60^\circ)
\]
Now we can compute \( W_1 \) and \( W_2 \):
Using \( V_{line} = 525V \):
\[
W_1 = 525V \cdot 18.35A \cdot \cos(64.62^\circ)
\]
\[
= 525 \cdot 18.35 \cdot 0.444 \approx 4,072.34 W
\]
Now using the relation \( W_1 + W_2 = P \):
\[
W_2 = P - W_1 \approx 4,072.34 W - W_1
\]
However, because we have a balanced load, \( W_1 \) and \( W_2 \) will not differ much and will share the total power evenly. Since the system is balanced:
\[
W_1 = W_2 \approx \frac{P}{2} \approx \frac{4,072.34 W}{2} \approx 2,036.17 W
\]
4. **Final Result**:
Thus, the readings on the two wattmeters are approximately:
\[
W_1 \approx 2,036.17 W
\]
\[
W_2 \approx 2,036.17 W
\]