Three phase system problem

[saw]

Hi there, I got stock with another problem. I can not get the right answer and it is frustrating me. I know there are lots of smart ones out there. Please help. Thanks

A three phase, three wire, 500V, 60HZ source supplies a three phase induction motor, a wye connected capacitor bank that draws 2 KVAR per phase, and a balanced three phase heater that draws a total of 10KW. The induction motor is operating at its rated 75hp and has an efficieny and power factor of 90.5 and 89.5 percent, respectively. Draw a one line diagram for the system, and determine a) the sysytem KW, B) the system KVAR, and c) the system KVA.

Answers: a) 71.82KW, B) 24.81 KVAR, c) 75.98KVA

 

[DK PE]

I'll give you some loose hints rather than all the math.... Draw power triangles for each load... the induction motor has to output 75 HP at the shaft, what is that equivalent in kW? then what is the input kW (real power) required given that efficiency? Draw the power triangle for INDUCTION motor (noting that lagging PF). You know the capacitors are leading PF and the heater is all real. Try to add all the reals and +/- jQ paying attention to signs and post back here if you still have difficulty.

 

[saw]

Thank you so much for your respond. If any one likes to see the answer, here is what I did. The only answer I can not get it right is apparent power.

output power= (75hp)X(746) = 55.95 KW

input power = (55.95KW) / .905 = 61.82 KW

Since the capacitor bank is per phase, we need to get the induction motor power per phase too.

p = (61.82KW) / 3 = 20.61 KW per phase

s = P / PF

S = 20.61 / .895 = 23.03 KVA

Q = sqr root of ( S2 - P2 ) = Sqr root of (23.03KVA)2- (20.61)2 = 10.28 KVR

QT per phase = -10.28 KVAR + 20KVAR = 8.28 KVAR leading

QT( 3phase) = 8.28 X3 = 24.83 KVAR

PT = (20.61 KW)X (3) + (10 KW) = 71.83 KW

ST = (71.83 KW)2 + (24.83 KVAR)2 = 5779 MVA

 

[saw]

My mistake the total needs to be sqr root. The answer for apperant power will be correct too. Thansk again for the hint.

 

[DK PE]

Thank you so much for your respond. If any one likes to see the answer, here is what I did. The only answer I can not get it right is apparent power.

Q = sqr root of ( S2 - P2 ) = Sqr root of (23.03KVA)2- (20.61)2 = 10.28 KVR

QT per phase = -10.28 KVAR + 20KVAR = 8.28 KVAR leading

QT( 3phase) = 8.28 X3 = 24.83 KVAR

PT = (20.61 KW)X (3) + (10 KW) = 71.83 KW

ST = (71.83 KW)2 + (24.83 KVAR)2 = 5779 MVA

Take a look at the part I've underlined above... I think there is typo but the total load is lagging not leading and convention is that is +Q. Think it should be:

QT per phase = 10.28 KVAR -2 KVAR (caps per phase) = +8.28 KVAR lagging (per phase)

 

[saw]

You are absolutly correct. The sign should be reversed and the system is lagging. Thank you.

 

[cconnawa]

Hi there, I got stock with another problem. I can not get the right answer and it is frustrating me. I know there are lots of smart ones out there. Please help. Thanks

A three phase, three wire, 500V, 60HZ source supplies a three phase induction motor, a wye connected capacitor bank that draws 2 KVAR per phase, and a balanced three phase heater that draws a total of 10KW. The induction motor is operating at its rated 75hp and has an efficieny and power factor of 90.5 and 89.5 percent, respectively. Draw a one line diagram for the system, and determine a) the sysytem KW, B) the system KVAR, and c) the system KVA.

Answers: a) 71.82KW, B) 24.81 KVAR, c) 75.98KVA

As reminder to all of us: Capacitors don't draw KVAR's, they supply them... so the system total KVAR will be the motor 3-phase KVAR minus the Capacitor 3-phase KVAR. Notice how the Capicitor KVAR is given in per phase... this requires conversion first to 3-phase KVAR to match the motor, or vice versa, otherwise the answer will be wrong...