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3) A generator rated 75 MVA is' delivering 0.8 p.u power to a motor through a transmission line of reactance 0.2 p.u. The terminal voltage of the generator is 1.0 p.u and that of the motor is also 1.0p.u. Determine the generator E.M.F behind transient reactance. Find also the maximum power that can be transferred.?
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3) A generator rated 75 MVA is' delivering 0.8 p.u power to a motor th...
Generator E.M.F Calculation
To find the E.M.F of the generator behind the transient reactance, we need to apply the formula for the voltage drop across the reactance of the transmission line:
- Given:
- Generator rating: 75 MVA
- Power factor (PF): 0.8 p.u
- Transmission line reactance (X): 0.2 p.u
- Terminal voltage of the generator (Vg): 1.0 p.u
- Terminal voltage of the motor (Vm): 1.0 p.u
- The power delivered to the motor in p.u is:
- P = 0.8 p.u * 75 MVA = 60 MW
- The current (I) can be calculated using:
- I = P / (Vg * PF) = 60 MW / (1.0 p.u * 0.8) = 75 p.u
- The voltage drop across the reactance (Vdrop) is:
- Vdrop = I * X = 75 p.u * 0.2 p.u = 15 p.u
- The E.M.F (Eg) behind the transient reactance is:
- Eg = Vg + Vdrop = 1.0 p.u + 0.15 p.u = 1.15 p.u
Maximum Power Transfer Calculation
The maximum power that can be transferred through a transmission line occurs when the load is matched to the source's E.M.F. This can be calculated using the following:
- The maximum power transfer equation is:
- Pmax = (Vg * Vm) / X
- Substituting the values:
- Pmax = (1.0 p.u * 1.0 p.u) / 0.2 p.u = 5 p.u
- Converting to MW at the generator's rating:
- Pmax = 5 p.u * 75 MVA = 375 MW
Summary
- The E.M.F behind the generator's transient reactance is 1.15 p.u.
- The maximum power that can be transferred is 375 MW.
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3) A generator rated 75 MVA is' delivering 0.8 p.u power to a motor through a transmission line of reactance 0.2 p.u. The terminal voltage of the generator is 1.0 p.u and that of the motor is also 1.0p.u. Determine the generator E.M.F behind transient reactance. Find also the maximum power that can be transferred.?
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3) A generator rated 75 MVA is' delivering 0.8 p.u power to a motor through a transmission line of reactance 0.2 p.u. The terminal voltage of the generator is 1.0 p.u and that of the motor is also 1.0p.u. Determine the generator E.M.F behind transient reactance. Find also the maximum power that can be transferred.? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 3) A generator rated 75 MVA is' delivering 0.8 p.u power to a motor through a transmission line of reactance 0.2 p.u. The terminal voltage of the generator is 1.0 p.u and that of the motor is also 1.0p.u. Determine the generator E.M.F behind transient reactance. Find also the maximum power that can be transferred.? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 3) A generator rated 75 MVA is' delivering 0.8 p.u power to a motor through a transmission line of reactance 0.2 p.u. The terminal voltage of the generator is 1.0 p.u and that of the motor is also 1.0p.u. Determine the generator E.M.F behind transient reactance. Find also the maximum power that can be transferred.?.
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