Determine maximum permissible load which a 15 MVA, ONAN cooled transformer to IS : 2026 can take for 6 hr, if the initial load on it was 12 MVA. The weighted ambient temperature is 20˚C. Assume the permissible load kVA as a fraction of rated kVA is 1.25.

Maximum permissible load = Rated MVA rating*fraction of rated kVA
Maximum permissible load = 15*1.25
Maximum permissible load = 18.75 MVA

A balanced delta connected load of 8+j6 Ω per phase is connected to a 415 V, 50 Hz, 3-phase supply lines. If the input power factor is improved to 0.9 by connecting a star connected capacitor bank, then the required kVAR of the bank is

Current flow through the circuit,
I = V/Z
= 400/10
= 41.5 A
Power factor = 8/10
= 0.8
After connecting a capacitor bank, the reactive power component of current changes but the active power component of current is unchanged.
I1*cosφ1 = I2*cosφ2
41.5*0.8 = I2*0.9
I2 = 36.88 A
Q1 (before connecting capacitor) = √3*VL*IL*sinφ1
= √3*415*41.5*sin(36.86)
= 26.847 kVAR
Q2 (after connecting capacitor) = √3*VL*IL*sinφ2
= √3*415*36.88*sin(25.84)
= 15.867 kVAR
kVAR supplied by capacitor bank = Q1 - Q2
= 26.847 - 15.867
= 10.98 kVAR

માણસ: ઘર :: પશુ :

___ is the short time reduction in the rms voltage between 0.1 to 0.9 p.u for a duration of ___

Voltage sag:
It is usually associated with system faults but can also be caused by energization of heavy loads or starting of large motors.
Voltage sag is the short reduction in the RMS voltage between 0.1 to 0.9 pu for a duration of 0.5 cycle to 1 minute

Which of the following features of Microsoft Word automatically moves to the next line while reaching the end of current line?

In a 3-phase, 4-wire, 400/230 V system, a lamp (L1) of 100 W is connected to one phase and neutral and a lamp (L2) of 150 W is connected to the second pahse and neutral. If the neutral wire is disconnected accidently, what will be the voltage across a 100 W lamp (L1)?

When neutral disconnected Lamp are in series connection,
So, for series connection current is same for both Lamp,
Resistance of Lamp (L1) = (230)²/P
= (230)²/100
= 529 ohm
Resistance of lamp (L2) = (230)²/150
= 352.67 ohm
So, current through the lamp = E/(R1 + R2)
= 400/(529 + 352.67)
= 0.454 A
Voltage across lamp (L1) = I*R1
= 0.454*230
= 240 V