A 10 kVA, 2500/250 V, single phase transformer gave the following results:
Open circuit Test: 250 V, 0.8 A, 50 W
Short circuit Test: 60 V, 3A, 45 W
Determine
(a) Equivalent circuit parameters referred to LV side
(b) The efficiency at full load and 0.8 power factor lagging
(c) Load (kVA) at which the maximum efficiency occurs
(d) Voltage regulation at rated load and 0.8 pf leading
(e) Secondary terminal voltage at rated load and 0.8 pf lagging.
Question : A 10 kVA, 2500/250 V, single phase transformer gave the foll...
Step 1:
To determine the equivalent circuit parameters referred to the LV side, we need to use the open circuit and short circuit test results.
Step 2:
From the open circuit test, we have the following values: V2 = 250 V, I0 = 0.8 A, and W0 = 50 W.
Step 3:
Using these values, we can calculate the exciting current component: Iw = W0 /V2= 50 / 250 = 0.2 A.
Step 4:
We can also calculate the magnetizing current component: Imag = sqrt(I0^2 - Iw^2) = sqrt(0.8^2 - 0.2^2) = 0.7746 A.
Step 5:
The exciting resistance is calculated using R0 = V2 / Iw = 250 / 0.2 = 1250 Ω.
Step 6:
The exciting reactance is calculated using X0 = V2 / Imag = 250 / 0.7746 = 323 Ω.
Step 7:
From the short circuit test, we have the following values: V1(sc) = 60 V, I1(sc) = 3 A, and Wc = 45 W.
Step 8:
Using these values, we can calculate the equivalent impedance: Zep = V1(sc) / I1(sc) = 60 / 3 = 20 Ω.
Step 9:
We can also calculate the equivalent resistance: Rep = (Wc / (I1(sc))^2) = (45 / (3)^2) = 5 Ω.
Step 10:
Finally, we can calculate the equivalent reactance: Xep = sqrt(Zep^2 - Rep^2) = sqrt(20^2 - 5^2) = 19.36 Ω.
Step 11:
The transformation ratio K is calculated as the ratio of the secondary voltage to the primary voltage: K = V2 / V1 = 250 / 2500 = 0.1.
Step 12:
To find the equivalent circuit parameters referred to the LV side, we use the following formulas:
Res = K^2 × Rep = (0.1)^2 ×5 = 0.05 Ω.
Xes = K^2 × Xep = (0.1)^2 × 19.36 = 0.1936 Ω.
Step 13:
To calculate the efficiency at full load and 0.8 power factor lagging, we need to use the following formulas:
Full-load current on HV side: I1(fl) = (10 ×1000) / V1 = (10 × 1000) / 2500 = 4 A.
Copper loss at full load: Pc = (I1(fl) / I1(sc))^2 Wc = (4/ 3)^2 × 45 = 80 W.
Iron loss: Pi = W0 = 50 W.
Efficiency at full load: η = (kVA ×1000 cosϕ) (kVA × 1000 cosϕ + Pi + Pc) = (10 ×1000 × 0.8) (10 × 1000 × 0.8 + 50 + 80)100 = 98.4%.
Step 14:
To determine the load at which the maximum efficiency occurs, we need to calculate the fraction of load at which efficiency is maximum:
x = √Pi / Pc = √50 / 80 = 0.79.
Load in kVA at which the efficiency is maximum: x rated kVA = 0.79 ×10 = 7.9 kVA.
Step 15:
To calculate the voltage regulation at rated load and 0.8 power factor leading, we need to use the following formulas:
I2(fl) = (10 ×1000) / V2 = (10 × 1000) / 250 = 40 A.
V2 = E2 - I2(fl) Res cosϕ + I2(fl) Xes sinϕ.
Step 16:
Substitute the given values into the equation and solve for V2:
V2 = 250 - 40 × 0.05 × 0.8 + 40 × 0.1936 × 0.6 = 253 V.
Step 17:
Calculate the voltage regulation using the formula: %Reg = (E2 - V2) / E2 * 100.
%Reg = (250 - 253) / 250 * 100 = -1.2.
Step 18:
To calculate the secondary terminal voltage at rated load and 0.8 power factor lagging, follow a similar process as in Step 15:
V2 = 250 - 40 × 0.05 × 0.8 - 40 × 0.1936 × 0.6 = 243.75 V.
This is the final answer.
Final Answer
Transformer rating =10\: kVA;\: E_{1}=2500\: V;\: E_{2}=250\: V
As per data, open circuit test is performed on LV side and short circuit test is performed on HV side.
Open circuit test (LV side); V_{2}=250\: V;\: I_{0}=0.8\: A;\: W_{0}=50\: W
Short circuit test (HV side); V_{1(sc)}=60\: V;\: I_{1(sc)}=3\, A;\: W_{c}=45\: W
(a) Open circuit test performed on LV side given;
I_{0}=0.8\: A;\: I_{w}=\frac{W_{0}}{V_{2}}=\frac{50}{250}=0.2\: AI_{mag}=\sqrt{I^{2}_{0}-I^{2}_{w}}=\sqrt{(0.8)^{2}-(0.2)^{2}}=0.7746\: A
Exciting resistance, R_{0}=\frac{V_{2}}{I_{w}}=\frac{250}{0.2}=1250\: \Omega
Exciting reactance, X_{0}=\frac{V_{2}}{I_{mag}}=\frac{250}{0.7746}=323\: \Omega
Short circuit test performed on HV side gives;
Z_{ep}=\frac{V_{1(sc)}}{I_{1(sc)}}=\frac{60}{3}=20\: \OmegaR_{ep}=\frac{W_{c}}{(I_{1(sc)})^{2}}=\frac{45}{(3)^{2}}=5\: \Omega
X_{ep}=\sqrt{Z^{2}_{ep}-R^{2}_{ep}}=\sqrt{(20)^{2}-(5)^{2}}=19.36\: \Omega
Transformation ratio, K=\frac{E_{2}}{E_{1}}=\frac{250}{2500}=0.1
Parameters, when referred to LV side;
R_{es}=K^{2}*R_{ep}=(0.1)^{2}*5=0.05\: \OmegaX_{es}=K^{2}*X_{ep}=(0.1)^{2}*19.36=0.1936\: \Omega
(b) Full-load current on HV side, I_{1(fl)}=\frac{10*1000}{2500}=4\: A
Copper loss at full-load, P_{c}=\left ( \frac{I_{1(fl)}}{I_{1(sc)}} \right )^{2}*W_{c}=\left ( \frac{4}{3} \right )*45=80\: W
Iron loss, P_{i}=W_{0}=50\: W
Efficiency at full-load, 0.8 p.f. lagging;
\eta =\frac{kVA*1000*cos\: \phi }{kVA*1000*cos\: \phi +P_{i}+P_{c}}=\frac{10*1000*0.8}{10*1000*0.8+50+80}*100=98.4\, \%
(c) Fraction of load at which the efficiency is maximum;
x=\sqrt{\frac{P_{i}}{P_{c}}}=\sqrt{\frac{50}{80}}=0.79
Load in kVA at which the efficiency is maximum
= x * rated kVA = 0.79 × 10 = 7.9 kVA
(d) When p.f., cos\: \phi =0.8\: leading;\: sin\: \phi =sin\: cos^{-1}\: 0.8=0.6
I_{2(fl)}=\frac{10*1000}{250}=40\: A
V_{2}=E_{2}-I_{2}R_{es}\: cos\: \phi +I_{2}X_{es}\: sin\: \phi
= 250 – 40 × 0.05 × 0.8 + 40 × 0.1936 × 0.6
= 250 – 1.6 + 4.6464 = 253 V
\%\: Reg=\frac{E_{2}-V_{2}}{E_{2}}*100=\frac{250-253}{250}*100=-1.2\: %
(e) When p.f. cos\: \phi =0.8\: lagging;\: sin\: \phi =sin\: cos^{-1}\, 0.8=0.6
V_{2}=E_{2}-I_{2}R_{es}\: cos\: \phi -I_{2}X_{es}\: sin\: \phi
= 250 – 40 × 0.05 × 0.8 + 40 × 0.1936 × 0.6
= 250 – 2.6 – 4.6464 = 243.75 V