Transformer no - load
voltage and full-load conditions of voltage and full load voltage difference is
ratio Voltage Regulation. The percentage rate of the ratio. Such as: -
Regulation of the
voltage transformer must be less than that amount, the lower the better.
The regulation voltage
transformer coil voltage detection, both with full load terminal voltage drops
to the no - load voltage is determined. The no - load voltage is determined as
the secondary.
In other words,
Vp / a = Vs + Zes
Here,
Vp / a = no - load
voltage secondary
Vs = secondary
full-load terminal Voltage.
Is Zes = transformer
coil in both the total voltage drop (secondary side)
The Vp / a = VNL and Vs = VFL in
Transformer
regulation,
Vreg = VNL - VFL ÷ VFL × 100%
Vs-a reference to the
Vp / a = √ (Vs Cos θ +
IsRes) ² + √ (Vs Sin θ + Is Xes) ²
Or Vs-a reference to
the
Vp / a = Vs + Is (Cos
θ - j Sin θ) (Res + j Xes)
Methods and equations
can be used.
When, Cos θ = load
power factor
If the power factor
leading θ-value is negative.
1. 25 KVA, 2300/230 V
Transformers Rp = 0.8Ω Xp = 3.2Ω, Rs = 0.009Ω, Xs = 0.03Ω equal to transformer
of regulation. When -
1. Power Factor unit.
2. Power factor 0.8
lagging
3. Power factor
leading 0.866.
Ans: - a = 2300/230
=
10
Is = 25 × 1000 ÷ 230 =
108.7 A
Secondary terms
Res = Rp / a ² + Rs =
0.8 / (10) ² + 0.009 = 0.017 Ω
Xes = Xp / a ² + Xs =
3.2 / (10) ² + 0.03 = 0.062 Ω
IsRes = 108.7 × 0.017
= 1.85 V
IsXes = 108.7 × 0.062
= 6.74 V
1. V / a = VNL = √ (VFL cosθ +
IsRes) ² + √ (VFLsinθ + IsXes) ²
= √ (230 + 1.85) ² + √
(0 + 6.74) ²
= 231.95 V.
Vreg = VNL - VFL ÷ VFL × 100%
= 231.95 - 230 ÷ 230 ×
100%
= 0.84%
2. Here Cos θ = 0.8,
Sin θ = 0.6
V / a = VNL = √ (VFL cosθ +
IsRes) ² + √ (VFLsinθ + IsXes) ²
= √ (230 × 0.8 + 1.85)
² + √ (230 × 0.6 + 6.74) ²
= 235.57 V.
Vreg = VNL - VFL ÷ VFL × 100%
= 235.57 - 230 ÷ 230 ×
100%
= 2.82%
3. Here Cos θ = 0.866
Sin θ = - 0.5
(Θ-'s value because of
the negative power factor leading)
V / a = VNL = √ (VFL cosθ +
IsRes) ² + √ (VFLsinθ + IsXes) ²
= √ (230 × 0.8 + 1.88)
² + √ (230 × (- 0.5 + 6.74) ²
= 228.32 V.
Vreg = VNL - VFL ÷ VFL × 100%
= 228.32 - 230 ÷ 230 ×
100%
= - 0.73%
2. Short-Circuit Test of a 10KVA 2300/230V Transformers were found to have the following information, Ese = 137V, Psc = 192 W, Isc = 4.34A, 0.8 power factor of percent voltage regulation lagging to out now.
2. Short-Circuit Test of a 10KVA 2300/230V Transformers were found to have the following information, Ese = 137V, Psc = 192 W, Isc = 4.34A, 0.8 power factor of percent voltage regulation lagging to out now.
Ans: -
That have been invoked
Rating = 10KVA. % Reg
= ?
Ep = 2300V Pf = 0.8
lagging
Es = 230V
Esc = 137V
Psc = 192W
Isc = 4.34A
Re = Psc / I ² sc
= 192 / (4.34) ²
= 10.19 Ω
Ze = Esc / Isc
= 137 / 4.34 = 31.57 Ω
Xe = √ Ze - √ R ² e
= √ (31.57) - √
(10.19) ²
= 29.88 Ω
I = KVA × 1000 / Ep
= 10 × 1000/2300
= 4.35A
IRe = 4.35 × 10.19 =
44.3V
IXe = 4.35 × 29.98 =
129.98V
Cosθ = 0.8 & sinθ
= 0.6
Vp = √ (V cosθ + IRe)
² + √ (Vsinθ + IXe) ²
= √ (2300 × 0.8 +
44.3) - √ (2300 × 0.6 + 129.98) ²
= 2414.8 V
Vreg = VNL - VFL ÷ VFL × 100%
= 2414.8 - 2300 ÷ 2300
× 100%
= 4.99%
