Question

Problem 3: You have been requested to provide service to a new customer. The one-line drawing is shown below Peak Loac 250 kw,@ pf 0.89, VN 7808v Light Load: 95 kW, @ pf 0.96, VLN- 8200V Circuit Main Line Transformer 2.500 ft of 1/0 AL 3 13 87 96 kV-120/208 Y 300kVA phase with a neutral 0 167 10 139 ohm/ktt 1 What is the peak demand in kVA and kVAr? 2- What is the demand in kVA and kVAr at light load? 3- What is the primary current at both (peak and light) load-levels? 4 What is the voltage drop between the transformer and the sending terminal during both (peak and light) load- levels? 5- What is the voltage (line to neutral) at the sending terminal during both (peak and light) load- levels, assuming no voltage drop on the transformer?

Answer 1

Here loads are specified in High voltage. Assuming transformer High voltage connected to supply side and low voltage connected to Load side, questions 3,4,5 are solved.

1) During peak demand

250kW, pf = 0.89   ===>   KVA = 250kW / pf = 280.9 KVA,  

from power tringle, KVAR = sqrt[ (KVA)² - (KW)²] = 128.08 KVAR

2) During Light Load

95KW, pf = 0.96 ===> KVA = 95kw/pf = 98.96 KVA.

KVAR = sqrt[ (KVA)² - (KW)²] = 27.71 KVAR

3) primary current at peak load:

Transformer is star-star connected and primary is high voltage side and secondary is low voltage side.

primary line current, I_L = P/( √ 3 * V_L * cos Ф ) = 250*10³/(√ 3 *(√ 3* 7808)*0.89) = 12 A

taking load phase voltage as reference, current I_L lags V_L by cos^-1(0.89) = 27.12⁰

primary current at light load, I_L = P/( √ 3 * V_L * cos Ф ) = 95*10³/(√ 3 *(√ 3* 8200)*0.96) = 4.02 A

taking load phase voltage as reference, current I_L lags V_L by cos^-1(.96) = 16.26⁰

4)Voltage drop between transformer and sending end terminal

as transformer primary is star connected phase current and line currents are same, per phase voltage drop , Δ V = Z_line * I_L

Zline = (0.167+j0.139) * 2.500 Ω = 0.4175 + j 0.3475 Ω / phase

in case of peak load , Δ V = (0.4175 + j 0.3475)*12 ∠ -27.12 = 6.51 ∠ 12.65 V

in case of light load , Δ V = (0.4175 + j 0.3475)*4.02 ∠ -16.26 = 2.183 ∠ 23.511 V

5) phase voltage as sending end terminal , Vs = transformer primary phase voltage + Δ V.

transformer primary phase voltage is taken as reference

During peak load , Vs = 7808∠ 0 + 6.51 ∠ 12.65 = 7814.35∠0.01 V

During light load , Vs = 8200∠ 0 + 2.183 ∠ 23.511 = 8202∠6.08 V