Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor angle at this operation mode will be ________
a Explanation: VR = (11.68-10)*100/10 VR = 16.8 % Half the VR = 16.8/2 % Half the VR = 8.4% (10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1) I = 5000/(cosФr*10) …(2) Solving above equation Фr = 18.04°, lagging
a
See lessExplanation: VR = (11.68-10)*100/10
VR = 16.8 %
Half the VR = 16.8/2 %
Half the VR = 8.4%
(10.84-10)*1000 = |I|*(RcosФr + XsinФr) …(1)
I = 5000/(cosФr*10) …(2)
Solving above equation
Фr = 18.04°, lagging
A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. If the voltage regulation is reduced to 50%, then the power factor at this operation mode will be ________________
a Explanation: VR = (11.68-10)*100/10 VR = 16.8 % Half the VR = 16.8/2 % Half the VR = 8.4% (10.84-10)*1000 = |I|*( RcosФr + XsinФr ) …(1) I = 5000/(cosФr*10) …(2) Solving above eqaution Фr = 18.04° Cos Фr = 0.9508, lagging.
a
See lessExplanation: VR = (11.68-10)*100/10
VR = 16.8 %
Half the VR = 16.8/2 %
Half the VR = 8.4%
(10.84-10)*1000 = |I|*( RcosФr + XsinФr ) …(1)
I = 5000/(cosФr*10) …(2)
Solving above eqaution
Фr = 18.04°
Cos Фr = 0.9508, lagging.
A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. The new sending end voltage at the half the voltage regulation is _____________
a Explanation: VR = (11.68-10)*100/10 = 16.8 % Half the VR = 16.8/2 % = 8.4%
a
See lessExplanation: VR = (11.68-10)*100/10 = 16.8 %
Half the VR = 16.8/2 % = 8.4%
A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. Identify the transmission line and the voltage regulation
a Explanation: It is a short transmission line. Current, I = 5000/(10*0.8)=625 A Vs = |Vr|+|I|*(RcosФr + XsinФr) = 10000+625(0.39*0.8+3.96*0.6) = 11.68kV VR = (11.68-10)*100/10 = 16.8 %
a
See lessExplanation: It is a short transmission line.
Current, I = 5000/(10*0.8)=625 A
Vs = |Vr|+|I|*(RcosФr + XsinФr)
= 10000+625(0.39*0.8+3.96*0.6)
= 11.68kV
VR = (11.68-10)*100/10 = 16.8 %
A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is 11.68 kV. The voltage regulation will be
a Explanation: VR = (11.68-10)*100/10 = 16.8 %.
a
See lessExplanation: VR = (11.68-10)*100/10 = 16.8 %.
Taking a case study for the long line under no load condition, the receiving end voltage is ____________
a Explanation: Due to Ferranti effect, the voltage will be more at receiving end in a LTL.
a
See lessExplanation: Due to Ferranti effect, the voltage will be more at receiving end in a LTL.
A single phase 50 hz, generator supplies an inductive load of 5 MW at a power factor of 0.8 lagging using OHTL over 20 km. The resistance and reactance are 0.39Ω and 3.96 Ω. The voltage at receiving station is maintained at 10 KV. The sending end voltage is ______
a Explanation: Current, I = 5000/(10*0.8) =625 A Vs = |Vr|+|I|*(RcosФr + XsinФr) = 10000+625(0.39*0.8+3.96*0.6) = 11.68kV.
a
See lessExplanation: Current, I = 5000/(10*0.8)
=625 A
Vs = |Vr|+|I|*(RcosФr + XsinФr)
= 10000+625(0.39*0.8+3.96*0.6)
= 11.68kV.
For a transmission line under study of failure analysis, it is observed that the current at the receiving end is same as that of the sending end, then what can be concluded about the nature of the transmission line?
a Explanation: It is a short transmission line as the capacitance considered is zero and so the line charging current is also zero.
a
See lessExplanation: It is a short transmission line as the capacitance considered is zero and so
the line charging current is also zero.
The charging current of a 400 kV is _____ that of 220 kV line of the same length
a Explanation: Line charging current is proportional to voltage.
a
See lessExplanation: Line charging current is proportional to voltage.
If Xa is the armature reactance of a synchronous machine and Xl is the leakage reactance of the same machine, then the synchronous reactance is __________
a Explanation: Xs = Xa+Xl
a
See lessExplanation: Xs = Xa+Xl