APPROXIMATE EXPRESSION FOR VOLTAGE REGULATION

The approximate expression for the no-load secondary voltage is derived in Section 16.1.

For inductive load

 

E₂ = V₂ + I₂R_es cosɸ₂ + I₂X_es sinɸ₂

or

 

E₂ − V₂ = I₂R_es cosɸ₂ + I₂X_es sinɸ₂

or

where,

∴

 

% Reg = % resistance drop × cos ɸ₂ + % reactance drop × sin ɸ₂

Similarly

(ii) For resistive load: % Reg = % resistance drop

(iii) For capacitive load

∴

 

% Reg = % resistance drop × cos ɸ₂ − % reactance drop × sin ɸ₂

Example 10.17

A 10 kVA, 2000/400 V, single-phase transformer has resistance and leakage reactance as follows:

Primary winding: Resistance = 5.5 Ω Reactance = 12 Ω

Secondary winding: Resistance = 0.2 Ω, Reactance = 0.45 Ω

Determine the value of the secondary voltage at full load, 0.8 p.f. lagging, when the primary supply voltage is 2000 V.

Solution:

Transformer rating = 10 kVA = 10 × 10³ VA

Primary induced voltage, E₁ = 2000 V

Secondary induced voltage, E₂ = 400 V

Primary resistance, R₁ = 5.5 Ω; Primary reactance, X₁ = 12 Ω

Secondary resistance, R₂ = 0.2 Ω; Secondary reactance, X₂ = 0.45 Ω

Load p.f.,

 

cos ɸ₂ = 0.8 lagging

Transformation ratio,

Primary resistance referred to secondary side,

 = K²R₁ = (0.2)² × 5.5 = 0.22Ω

Equivalent resistance referred to secondary side,

R_es = R₂ +  = 0.2 + 0.22 = 0.42Ω

Primary reactance referred to secondary side,

 = K²X₁ = (0.2)² × 12 = 0.48Ω

Equivalent reactance referred to secondary side,

X_es = X₂ +  = 0.45 + 0.48 = 0.93Ω

Load p.f.,

 

cosɸ₂ = 0.8 ∴ sinɸ₂ = sincos⁻¹ 0.8 = 0.6

Full-load secondary current,

As the primary supply voltage,

 

V₁ = E₁ = 2000V

Secondary induced voltage,

 

E₂ = KE₁ = 0.2 × 2000 = 400V

Using the expression;

 

E₂ = V₂ + I₂R_es cosɸ₂ − I₂X_es sinɸ₂

Secondary terminal voltage,

 

V₂ = E₂ − I₂R_es cosɸ₂ − I₂X_es sinɸ₂

= 400 − 25 × 0.42 × 0.8 − 25 × 0.93 × 0.6 = 400 − 8.4 − 13.95 = 377.65 A

Example 10.18

The ratio of turns of a single-phase transformer is 8, the resistance of the primary and the secondary windings are 0.85 Ω and 0.012 Ω, respectively, and the leakage reactance of these windings are 4.8 Ω and 0.07 Ω, respectively. Determine the voltage to be applied to the primary to obtain a current of 150 A in the secondary when the secondary terminal are short circuited. Ignore the magnetizing current.

Solution:

Ratio of turns,

Primary resistance,

 

R₁ = 0.85;

Primary reactance,

 

X₁ = 4.8Ω

Transformation ratio,

Secondary resistance

 

R₂ = 0.012Ω

Secondary reactance,

 

X₂ = 0.07Ω

Secondary resistance referred to primary, 

Equivalent resistance referred to primary, R_ep = R₁ +  = 0.85 + 0.768 = 1.618Ω

Secondary reactance referred to primary, 

Equivalent reactance referred to primary, X_ep = X₁ +  = 4.8 + 4.48 = 9.28Ω

Equivalent impedance referred to primary, 

= 9.42 Ω

Short circuit current referred to primary, I_1(SC) = KI_2(SC) =  × 150 = 18.75 A

Voltage applied to the primary under short circuit condition,

 

V_1(SC) = I_1(SC) × Z_ep =18.75 × 9.42 = 176.625 V