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Answer
Step 1 > To determine
The voltage at the power source of the system.
Step 2 > Answer
The voltage at the power source is Vsource = 15.5∠3.00 kV .
Step 3 > Explanation
Given Information:
Transformer rating is 100KVA, voltage ratio is 14/2.4KV.
Load is 90KW at a voltage of 2300V with power factor 0.8Lag.
Calculation:
The circuit to the secondary side is used for the solution. The feeder’s impedance referred to the secondary side is,
2
Zline ' =( 2.4kV
14kV ) (38.2Ω+j140Ω)
= 1.12 + j4.11Ω
The secondary current Isis,
Is =
90kW
(2400V)(0.8)
= 46.88A
The power factor is 0.80 lagging, so the impedance angle is,
θ = cos−1 (0.8)
=36.87o
Hence, the phasor current is,
Is = 46.88∠ −36.87o
The voltage at the power source of this system (referred to the secondary side) is,
Vsource ' = Vs + Is Zline ' + Is ZEQ
Vsource ' = 2400∠00 V +(46.88∠ −36.870 A)(1.12 + j4.11Ω) + (46.88∠ −36.870 A)(0.10 + j0.40Ω)
= 2576∠3.00 V
Hence, the voltage at the power source is,
14kV
Vsource = (2576∠3.00 V) 2.4kV
Vsource = 15.5∠3.00 kV
Step 1 > To determine
The voltage regulation of the transformer.
Step 2 > Answer
The voltage regulation of the transformer is V R = 0.63o/o .
Step 3 > Explanation
Given Information:
Transformer rating is 100KVA, voltage ratio is 14/2.4KV.
Load is 90KW at a voltage of 2300V with power factor 0.8Lag.
Calculation:
The voltage at the primary side of the transformer (referred to the secondary side) under these conditions:
Vp ' = Vs + Is ZEQ
Vp ' = 2400∠00 V+(46.88∠ −36.870 A)(0.10 + j0.40Ω)
= 2415∠3.0o V
There is a voltage drop of 15 V under these load conditions. Hence, the voltage regulation of the transformer is,
VR =
2415−2400
2400
× 100
V R = 0.63%
Step 1 > To determine
The efficiency of the power system.
Step 2 > Answer
The efficiency of the power system is η = 97.1% .
Step 3 > Explanation
Given Information:
Transformer rating is 100KVA, voltage ratio is 14/2.4KV.
Load is 90KW at a voltage of 2300V with power factor 0.8Lag.
Calculation:
The overall efficiency of the power system will be the ratio of the output power to the input power.
The output power supplied to the load is Pout = 90 kW. The input power supplied by the source is,
Pin = Pout + Ploss
= Pout + I 2 R
= 90kW +(46.88A)2 (1.22Ω)
= 92.68kW
Pin = Vsource 'Is cos θ
= (2415V)(46.88A) cos(36.57o )
= 90.93kW
Hence, the efficiency of the power system is,
η=
η=
Pout
Pin × 100
90kW
92.68kW × 100
η = 97.1%