Solution Electrical Numerical 3

Numerical 3: A 100 KVA transformer has 400 turns on the primary and 80 turns on the secondary. The primary and secondary resistances are 0.3 Ω and 0.01 Ω respectively, and the corresponding leakage reactance’s are 1.1 Ω and 0.035 Ω respectively. The supply voltage is 2200 V. Calculate: (i) The equivalent impedance referred to the primary circuit; (ii) The voltage regulation and secondary terminal voltage for full load having a power factor of (a) 0.8 lagging and (b) 0.8 leading. Solution: Given:

Transformer rating = 100 KVA

Turn ratio = 400:80

Resistance Primary Coil $\displaystyle \mathrm{R_{p}}$  = 0.3 Ω

Resistance Secondary coil $\displaystyle \mathrm{R_{s}}$ = 0.01 Ω

(i) Equivalent impedance referred to primary circuit; 

$\displaystyle \mathrm{R_{e1}\ =\ R_{1} + R_{2} \left (\frac{N_1}{N_2} \right )^2}$

$\displaystyle \mathrm{R_{e1}\ =\ 0.3 + 0.01 \left (\frac{400}{80} \right )^2\ =\ 0.55}$ Ω

$\displaystyle \mathrm{X_{e1}\ =\ X_{1} + X_{2} \left (\frac{N_1}{N_2} \right )^2}$ 

$\displaystyle \mathrm{X_{e1}\ =\ 1.1 + 0.035 \left (\frac{400}{80} \right )^2\ =\ 1.975}$ Ω

$\displaystyle \mathrm{Z_{1}\ =\ \sqrt{R_{1}^{2}+X_{1}^{2}}}$ 

$\displaystyle \mathrm{Z_{1}\ =\ \sqrt{0.55^{2}+ 1.975^{2}}=\ 2.05}$ Ω

(ii) (a) 0.8 Leading

 Cosф =0.8 

 thus, Sinф =0.6

$\displaystyle \mathrm{I_{1f1}\ =\ \frac{KVA\times 10^3}{V}}$ 

$\displaystyle \mathrm{I_{1f1}\ =\ \frac{100\times 10^3}{2200}\ =\ 45.45\ A}$ 

$\displaystyle \mathrm{E\ =\ I_{1f1}\frac{R_{e1}Cos\phi + X_{e1}Sin\phi}{V}}$ Per Unit (Pu)

$\displaystyle \mathrm{E\ =\ 45.45\times \frac{0.55\times 0.8 + 1.975\times 0.6}{2200}\ =\ 0.0336 Pu}$ 

The Secondary No-Load Voltage:

$\displaystyle \mathrm{V_{2n}\ =\ V\times \frac{N_2}{N_1}\ =\ 2200\times \frac{80}{400}\ =\ 440\ V}$ 

The secondary Terminal voltage:

$\displaystyle \mathrm{V_{2}\ =\ V_{2n}\times (1-E) =\ 440\times (1-0.0336) =\ 425.2\ V}$ 

(b) 0.8 lagging

$\displaystyle \mathrm{E\ =\ I_{1f1}\frac{R_{e1}Cos\phi - X_{e1}Sin\phi}{V}}$ 

$\displaystyle \mathrm{E\ =\ 45.45\times \frac{0.55\times 0.8 - 1.975\times 0.6}{2200}\ =\ 0.0154 Pu}$ 

The secondary Terminal voltage:

$\displaystyle \mathrm{V_{2}\ =\ V_{2n}\times (1-E) =\ 440\times (1-0.0154) =\ 433.22\ V}$ 

Maximum Voltage Regulation

$\displaystyle \mathrm{E_{max}\ =\ \frac{I_{1f1}\times Z_e}{V}}$ 

$\displaystyle \mathrm{E_{max}\ =\ \frac{45.45\times 2.05}{2200}}$ = 0.0425 Pu