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The sending-end and receiving-end voltages of a three-phase transmission line at a 200 MW load are equal at 230 kV. The per phase line impedance is 14 ohm. The maximum steady-state power that can be transmitted (MW/phase) over the line would be _______. (Answer up to one decimal place)
Correct answer is '1259.5'. Can you explain this answer?
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Given data:
- Sending-end and receiving-end voltages: 230 kV
- Per phase line impedance: 14 ohm
- Load: 200 MW
To find:
- Maximum steady-state power that can be transmitted (MW/phase)
Formula:
The maximum steady-state power that can be transmitted over a transmission line is given by the formula:
\[
P_{\text{max}} = \frac{{V^2}}{{Z}}
\]
where:
- \(P_{\text{max}}\) is the maximum power that can be transmitted
- \(V\) is the sending/receiving voltage
- \(Z\) is the line impedance
Calculation:
Substituting the given values into the formula, we have:
\[
P_{\text{max}} = \frac{{(230 \text{ kV})^2}}{{14 \Omega}}
\]
Converting the voltage to volts and simplifying, we get:
\[
P_{\text{max}} = \frac{{(230 \times 10^3 \text{ V})^2}}{{14 \Omega}}
\]
\[
P_{\text{max}} = \frac{{52900 \times 10^6 \text{ V}^2}}{{14 \Omega}}
\]
\[
P_{\text{max}} = 3785.71 \times 10^6 \text{ W}
\]
Converting the power to megawatts, we have:
\[
P_{\text{max}} = 3785.71 \text{ MW}
\]
Rounding off the answer to one decimal place, the maximum steady-state power that can be transmitted over the line is approximately 3778.5 MW.
- Sending-end and receiving-end voltages: 230 kV
- Per phase line impedance: 14 ohm
- Load: 200 MW
To find:
- Maximum steady-state power that can be transmitted (MW/phase)
Formula:
The maximum steady-state power that can be transmitted over a transmission line is given by the formula:
\[
P_{\text{max}} = \frac{{V^2}}{{Z}}
\]
where:
- \(P_{\text{max}}\) is the maximum power that can be transmitted
- \(V\) is the sending/receiving voltage
- \(Z\) is the line impedance
Calculation:
Substituting the given values into the formula, we have:
\[
P_{\text{max}} = \frac{{(230 \text{ kV})^2}}{{14 \Omega}}
\]
Converting the voltage to volts and simplifying, we get:
\[
P_{\text{max}} = \frac{{(230 \times 10^3 \text{ V})^2}}{{14 \Omega}}
\]
\[
P_{\text{max}} = \frac{{52900 \times 10^6 \text{ V}^2}}{{14 \Omega}}
\]
\[
P_{\text{max}} = 3785.71 \times 10^6 \text{ W}
\]
Converting the power to megawatts, we have:
\[
P_{\text{max}} = 3785.71 \text{ MW}
\]
Rounding off the answer to one decimal place, the maximum steady-state power that can be transmitted over the line is approximately 3778.5 MW.
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The sending-end and receiving-end voltages of a three-phase transmission line at a 200 MW load are equal at 230 kV. The per phase line impedance is 14 ohm. The maximum steady-state power that can be transmitted (MW/phase) over the line would be _______. (Answer up to one decimal place)Correct answer is '1259.5'. Can you explain this answer?
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The sending-end and receiving-end voltages of a three-phase transmission line at a 200 MW load are equal at 230 kV. The per phase line impedance is 14 ohm. The maximum steady-state power that can be transmitted (MW/phase) over the line would be _______. (Answer up to one decimal place)Correct answer is '1259.5'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The sending-end and receiving-end voltages of a three-phase transmission line at a 200 MW load are equal at 230 kV. The per phase line impedance is 14 ohm. The maximum steady-state power that can be transmitted (MW/phase) over the line would be _______. (Answer up to one decimal place)Correct answer is '1259.5'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sending-end and receiving-end voltages of a three-phase transmission line at a 200 MW load are equal at 230 kV. The per phase line impedance is 14 ohm. The maximum steady-state power that can be transmitted (MW/phase) over the line would be _______. (Answer up to one decimal place)Correct answer is '1259.5'. Can you explain this answer?.
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