# Feeder with transformer and loads

#### Jess_88

Joined Apr 29, 2011
174
Hey guys I'm having a little trouble with this problem for the circuit given above I need to determine

a) Compute ~IL load1 (magnitude and phase).
b) Compute IL load2 (magnitude and phase).
c) Determine the total load current (magnitude and phase).
d) Compute the voltage at the sending end of the feeder Vsend (magnitude and phase) if the voltage at the load ~(receiving) end is maintained at Vload 2400<0° V .
e) What is the real power Psend and reactive power Qsend at the sending end of the feeder?
f)Calculate the voltage regulation VR

given
Two single-phase loads are supplied through a 35 kV feeder whose impedance is Zf=(Rf+jXf)=(100+j300) and a ￼ VH / VL ￼ =35000V/2400V single-phase transformer whose equivalent (series-branch) impedance is Z L =( R Low + jX Low ) =(0.2+j1.0) referred to its low-voltage (secondary) side. Note H and ￼￼￼￼L stand for high side and low side, respectively.

I feel like I must be looking at this question wrong... seems a maybe a bit 'easy' to calculate the currents.

for parts a,b,c would I do the following

S = VI
I = S/V

could I just use I = P/V???
or would I work out the power at load2 and use that to determine the current in load 1

part 'c' I just add them together right?

Last edited:

#### t_n_k

Joined Mar 6, 2009
5,447
To find the current in each load you divide the apparent power (S) by the voltage Vload.

No you can't just divide P1/Vload to give Iload1. You must use the apparent power for this case as well.

Also remember the two currents Iload1 & Iload2 are complex values so they must be added accordingly to derive the total current Iload - i.e. not just as a simple algebraic addition.

#### Jess_88

Joined Apr 29, 2011
174
ok I think I have it.

(a)
I = P/(Vcosθ)
=69.44

θ = cos^(-1) (0.6)

I = 69.44 < -53.13

(b)
I = S/V

θ = cos^(-1)(0.3)

I = 41.66 <72.54

(c)
I1 + I2 = 56.42< -16.262

(d)
Referring Xeq and Req to primary (x(35k/2400)^(2))

Zeq = 42.534 + j212.673

I(send) = (1/a)*IL
= [1/(35k/2400)]*(54.164<-16.262)
=3.714<-16.262 A

voltage drop across lines
Vd = Isend*(Zf + Zeq)

= 1976.286<58.2

Vsend = VH + Vd
= 35000<0 + 1976.286<58.2
= 36 080.53<2.668 V

(e)
S=V*I(conjugate)
= (36 080.53<2.668 V)(3.714<16.262 A)
=126 755.62 + j43 472.31

P = 126 755.62 W
Q = 43 472.31 VAR

(f)
VL = 2400<0 + (56.41<-16.26)(1.02<80.25)
=2425.23<1.22

VH = VL*(35k/2400)
=35 364.5<1.22 V

VR = [(VH - 35k)/35k]*100
=1.04%

Hows that look?