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You’re the operator of a 15000Vrms, 60HZ electrical substation. When you get to work one day, you see that the station is delivering 6MWof power with a power factor of0.90.

a. What is the rms current leaving the station?

b. How much series capacitance should you add to bring the power factor up to 1.0?

c. How much power will the station then be delivering?

Short Answer

Expert verified

(a).The rms current leaving at station is 0.44kA

(b). The value of capacitance should be added is 1.80x10-4F

(c). The power leaving by station is7.4MW

Step by step solution

01

Part (a) Step 1: Given information 

We are given that you’re the operator of a 15000Vrms/60HZ electrical substation. When you get to work one day, you see that the station is delivering 6MWof power with a power factor of 0.90.

We need to find that What is the rms current leaving the station?

02

Part (a) Step 2: Explanation 

We know that the avg power is

P=IrmsErmscosIrms=PErmscosIrms=6x10615x103x0.9Irms=0.44kA

03

Part (b) Step 1: Given information 

We are given that you’re the operator of a 15000Vrms/60HZ electrical substation. When you get to work one day, you see that the station is delivering 6MWof power with a power factor of 0.90.

We need to find that How much series capacitance should you add to bring the power factor up to 1?

04

Part (b) Step 2: Explanation 

In the LCR circuit the value of Z=ErmsIrms

The power factor is given by role="math" localid="1650738938667" cos=0.9so the phase angle is 25.840{"x":[[5,4,16,30,35,27,4,4,35],[73,45,45,44,45,54,66,72,73,72,67,48,43],[85],[112,100,100,118,127,122,102,97,101,125,122,112],[154,154,155,133,133,163],[185.3333740234375,184.3333740234375,181.3333740234375,179.3333740234375,178.3333740234375,178.3333740234375,177.3333740234375,177.3333740234375,176.3333740234375,176.3333740234375,175.3333740234375,173.3333740234375,172.3333740234375,172.3333740234375,171.3333740234375,171.3333740234375,171.3333740234375,171.3333740234375,171.3333740234375,171.3333740234375,171.3333740234375,171.3333740234375,172.3333740234375,172.3333740234375,174.3333740234375,175.3333740234375,175.3333740234375,176.3333740234375,177.3333740234375,177.3333740234375,179.3333740234375,181.3333740234375,183.3333740234375,184.3333740234375,185.3333740234375,185.3333740234375,185.3333740234375,186.3333740234375,186.3333740234375,186.3333740234375,187.3333740234375,188.3333740234375,188.3333740234375,188.3333740234375,188.3333740234375,189.3333740234375,189.3333740234375,190.3333740234375,190.3333740234375,190.3333740234375,190.3333740234375,190.3333740234375,190.3333740234375,190.3333740234375,190.3333740234375,190.3333740234375,190.3333740234375,190.3333740234375,189.3333740234375,189.3333740234375,188.3333740234375,187.3333740234375,186.3333740234375]],"y":[[30,16,8,11,25,51,116,116,116],[9,9,9,51,51,48,51,63,88,107,116,116,97],[115],[9,12,47,52,85,114,117,85,58,44,16,10],[116,9,9,85,85,85],[-13.388885498046875,-13.388885498046875,-13.388885498046875,-13.388885498046875,-13.388885498046875,-12.388885498046875,-11.388885498046875,-11.388885498046875,-11.388885498046875,-10.388885498046875,-8.388885498046875,-5.388885498046875,-1.388885498046875,-0.388885498046875,0.611114501953125,0.611114501953125,2.611114501953125,3.611114501953125,4.611114501953125,5.611114501953125,6.611114501953125,7.611114501953125,8.611114501953125,9.611114501953125,9.611114501953125,9.611114501953125,10.611114501953125,10.611114501953125,10.611114501953125,11.611114501953125,11.611114501953125,11.611114501953125,11.611114501953125,11.611114501953125,11.611114501953125,11.611114501953125,10.611114501953125,10.611114501953125,9.611114501953125,8.611114501953125,8.611114501953125,7.611114501953125,6.611114501953125,3.611114501953125,2.611114501953125,0.611114501953125,-0.388885498046875,-1.388885498046875,-3.388885498046875,-5.388885498046875,-6.388885498046875,-7.388885498046875,-7.388885498046875,-8.388885498046875,-9.388885498046875,-10.388885498046875,-11.388885498046875,-12.388885498046875,-12.388885498046875,-14.388885498046875,-14.388885498046875,-15.388885498046875,-15.388885498046875]],"t":[[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0,0],[0],[0,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0],[1650738973683,1650738973904,1650738973918,1650738973935,1650738973951,1650738973967,1650738973987,1650738974017,1650738974040,1650738974051,1650738974069,1650738974088,1650738974103,1650738974119,1650738974137,1650738974157,1650738974202,1650738974217,1650738974235,1650738974251,1650738974281,1650738974301,1650738974319,1650738974335,1650738974351,1650738974368,1650738974389,1650738974402,1650738974417,1650738974434,1650738974451,1650738974468,1650738974487,1650738974500,1650738974551,1650738974572,1650738974584,1650738974601,1650738974625,1650738974634,1650738974651,1650738974669,1650738974684,1650738974701,1650738974718,1650738974734,1650738974751,1650738974769,1650738974784,1650738974803,1650738974817,1650738974835,1650738974851,1650738974891,1650738974910,1650738974928,1650738974984,1650738975001,1650738975034,1650738975085,1650738975119,1650738975151,1650738975176]],"version":"2.0.0"}, to make power factor is equal to 1 so

Xc=ZsinXc=14.72ohm

And we know that

Xc=12πfCC=12πfXcC=12x3.14x60x14.72C=1.80x10-4F

05

Part (c) Step 1: Given information 

you’re the operator of a 15000Vrms/60HZ electrical substation. When you get to work one day, you see that the station is delivering6MW of power with a power factor of 0.90.

We need to find How much power will the station then be delivering?

06

Part (c) Step 2: Explanation

So power dissipated by resistor is

P=Vrms2R=Vrms2ZcosP=15000233.78xcos(25.84)P=7.4MW

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