Voltage at load of distribution network

[ClimberTom]

Please bare with me, this is my first post here (although I've read a lot on here over the past few months)
Hopefully I've used the correct formatting etc.
I've got to a sticking point on the final point of the question below (2iii) I'm pretty happy with my answers to 2i and 2ii so I've not shown them so as to not confuse the situation, the only parts carried over are the total current drawn (I_t) and the total load (S).
In case the attachment isn't very clear, the load at 'C' is 0.6pf lead.

I have omitted the volt drop part of the calculations as I cannot get the calculated current at the end to be close to the estimated current at the beginning. I am happy with the volt drop part once the currents make sense.

Details below and attached.

1. Homework Statement
Determine the load voltage levels for the following distribution feeder systems.
(see image attached for the sketches)

Homework Equations

$B=3.2 ∠-36.9° \text {MVA}\$
$B= 2.56-j1.92 \text {MVA}\$
$C=1.8 ∠53.1° \text {MVA}\$
$C=1.08+j1.44 \text {MVA}\$
$S=3.67 ∠-7.5° \text {MVA}\$
$S=3.64-j0.48 \text {MVA}\$
$I_t≈192.63∠-7.5° \text {A}\$
$I_t ≈ 192.78-j2.23 \text {A}\$

(I_t and S are both calculated from question 2i as shown on the attached sheet.

The Attempt at a Solution

Estimating Current drawn at points B and C:

$I_{BE} = {\frac{3.2*10^6}{\sqrt{3}*11*10^3}}$
$I_{BE}=167.96∠-36.9° \text {A}\$
$I_{BE}=134.37-j100.85 \text {A}\$

$I_{CE} = {\frac{1.8*10^6}{\sqrt{3}*11*10^3}}$
$I_{CE}=94.48∠53.1° \text {A}\$
$I_{CE}=56.69+j75.55 \text {A}\$

Assuming the lowest voltage will be at load 'B' (drawing the most current) therefore, 'B' should be fed from both sides as current will flow from highest potential to lowest potential.

Volt drop around ACBA = 0
Therefore:

$0=-2.7I_{z2}-5.4(I_{z2}-94.48∠53.1°)+4.6(192.63∠-7.5°-I_{z2})$
$=-2.7I_{z2}-5.4I_{z2}+306.126-j407.97+886.788-j10.258-4.6I_{z2}$
$=-12.7I_{z2}+1192.914-j418.228$
$12.7I_{z2}=1192.914-j418.228$
$=1264.10∠-19.3°$
$I_{z2}=99.54∠-19.3° \text {A}\$
$I_{z2}=93.95-j32.9 \text {A}\$

$I_{z1}= I_t-I_{z2}$
$=192.78-j2.23-93.95-j32.9$
$I_{z1}=98.83-j35.13 \text {A}\$
$I_{z1}=104.89∠-19.6° \text {A}\$

$I_{z3}=I_{z2}-I_ce$
$=93.95-j32.9-56.69+j75.55$
$I_{z3}=37.26+j42.65 \text {A}\$
$I_{z3}=56.63∠48.9° \text {A}\$

$I_b=I_{z1}+I_{z3}$
$=98.83-j35.13+37.26+j42.65$
$I_b=136.09+j7.52\text {A}\$##
$I_b=136.3∠3.16° \text {A}\$I feel sure that I have overlooked something simple here. I'm pretty confident that I'm correct in assuming B to be the lower voltage. That said, I have tried working the volt drop around the loop as ABCA = 0 and still not got the correct answer.

Can anyone offer any pointers on where I am going wrong?

Many thanks

 

[ClimberTom]

Ignore the above, After lots of tea and scrap paper, I realized that my equation for the voltage around the loop was wrong and also due to some rounding and conversion errors, other values were also incorrect.

I've now sussed it out and submitted it so just waiting for it to be marked.