problem # 513 ncees power

[ros]

hello,

prob 513..

i need help with this problem...

base (100mva and 12 kv)

i understand how they came up with Ibase. and then how they come up with 1 PU for Transformer... but i am having difficulty in calculating PU for 20 mile long transmission line...

given 0.145/1000ft... so for 20 miles... Z = J0.145*(20000/1000) = 2.9 ohms (IS THIS CORRECT)

SO NOW ZPU FOR TRANSMISSION LINE WILL BE = 2.9 / (Vbase/Ibase) = 2.9/2.49 pu...

can some one help me with this

 

[Dolphin P.E.]

hello,
prob 513..

i need help with this problem...

base (100mva and 12 kv)

i understand how they came up with Ibase. and then how they come up with 1 PU for Transformer... but i am having difficulty in calculating PU for 20 mile long transmission line...

given 0.145/1000ft... so for 20 miles... Z = J0.145*(20000/1000) = 2.9 ohms (IS THIS CORRECT)

SO NOW ZPU FOR TRANSMISSION LINE WILL BE = 2.9 / (Vbase/Ibase) = 2.9/2.49 pu...

can some one help me with this

1 Mile = 5280 ft

 

[ros]

ok ... so Z for transmission line will be (0.145*20*5280) / 1000 = 15.312 ohm

now Zbase is 2.49 ....

so Z pu = 15.312/2.49 is this correct ?

if not .. help me solve this problem...

thanks

 

[ros]

finally i am able to solve this problem.... but still i have confusion in finding short ckt current

i found that Z trans = 1 PU and Z trans line = 10.6 pu

but why SC = I base / ( 1 + 10.6) , where I base = 100mva/(sq 3 * 12 kv)

any help on that

thanks

 

[Dolphin P.E.]

finally i am able to solve this problem.... but still i have confusion in finding short ckt current
i found that Z trans = 1 PU and Z trans line = 10.6 pu

but why SC = I base / ( 1 + 10.6) , where I base = 100mva/(sq 3 * 12 kv)

any help on that

thanks

See Attached file

413.pdf

 

[ros]

finally i am able to solve this problem.... but still i have confusion in finding short ckt current
i found that Z trans = 1 PU and Z trans line = 10.6 pu

but why SC = I base / ( 1 + 10.6) , where I base = 100mva/(sq 3 * 12 kv)

any help on that

thanks

See Attached file

thanks a lot ... but i am not able to understand the concept of Isc = Ib/Zpu

any help on that

thanks

 

[Dolphin P.E.]

finally i am able to solve this problem.... but still i have confusion in finding short ckt current
i found that Z trans = 1 PU and Z trans line = 10.6 pu

but why SC = I base / ( 1 + 10.6) , where I base = 100mva/(sq 3 * 12 kv)

any help on that

thanks

See Attached file

thanks a lot ... but i am not able to understand the concept of Isc = Ib/Zpu

any help on that

thanks

You are looking for the actual current at the fault. I actual = I base * I pu. = I base / Zpu. since Ipu = Vpu/Zpu note, Vpu=1

same way as S fault = S base / Zpu. I hope I didn't confuse you.

 

[ros]

no u didnt confuse me.. u actually made my concept clear...thank you

 

[willsee]

I solved this using MVA Method as well

z_line=j0.145/1000ft * 5280ft/mi * 20 mi = 15.312 ohms

7.5MVA/.075 = 100MVA

(12kv)^2 / 15.312 ohms = 9.4 MVA

Both are in series

100*9.4 / (100+9.4) = 8.59 MVA

I_line = 8.59MVA / (sqrt(3) * 12kv)) = 413A

 

[dianevp]

I solved this using MVA Method as well
z_line=j0.145/1000ft * 5280ft/mi * 20 mi = 15.312 ohms

7.5MVA/.075 = 100MVA

(12kv)^2 / 15.312 ohms = 9.4 MVA

Both are in series

100*9.4 / (100+9.4) = 8.59 MVA

I_line = 8.59MVA / (sqrt(3) * 12kv)) = 413A

Why are the MVAsc's calculated in parallel? And if following this logic, would parallel MVA's be calculated in as series MVAsc?

Please advice...I'm not very familiar with this method...it does look quicker.

Thanks!

 

[Dolphin P.E.]

I solved this using MVA Method as well
z_line=j0.145/1000ft * 5280ft/mi * 20 mi = 15.312 ohms

7.5MVA/.075 = 100MVA

(12kv)^2 / 15.312 ohms = 9.4 MVA

Both are in series

100*9.4 / (100+9.4) = 8.59 MVA

I_line = 8.59MVA / (sqrt(3) * 12kv)) = 413A

Why are the MVAsc's calculated in parallel? And if following this logic, would parallel MVA's be calculated in as series MVAsc?

Please advice...I'm not very familiar with this method...it does look quicker.

Thanks!

Yes, you actually add MVAsc as a capacitors.

 

[Dolphin P.E.]

I solved this using MVA Method as well
z_line=j0.145/1000ft * 5280ft/mi * 20 mi = 15.312 ohms

7.5MVA/.075 = 100MVA

(12kv)^2 / 15.312 ohms = 9.4 MVA

Both are in series

100*9.4 / (100+9.4) = 8.59 MVA

I_line = 8.59MVA / (sqrt(3) * 12kv)) = 413A

Why are the MVAsc's calculated in parallel? And if following this logic, would parallel MVA's be calculated in as series MVAsc?

Please advice...I'm not very familiar with this method...it does look quicker.

Thanks!

Yes, you actually add MVAsc as a capacitors.

take a look at this thread. It has the link to the MVA method papers.

http://engineerboards.com/index.php?showtopic=11332