3. A 5-kVA, 2300/230 V, 50 Hz transformer was tested for the iron losses with normal excitation and copper losses at full-load and these were found to be 40 W and 112 W respectively. Calculate the efficiencies of the transformer at 0.8 p.f. for the following outputs: a) 1.25 kVA, b) 2.5kVa, c) 5 kVA 4. A 200-kVa transformer has an efficiency of 98% at full load. If the maximum efficiency occurs at three quarters of full load, calculate the efficiency at half load. Assume negligible magnetizing current and p.f. 0.8 at all loads.

5. A 25-kVA single transformer 2200/220 volts, has primary resistance of 0.1Ω and a secondary resistance of 0.01Ω. Find the equivalent secondary resistance and the full-load efficiency at 0.8 p.f. if the iron loss of the transformer is 80% of the full load copper loss. 6. Consider a 4-Kva 200/400 V single phase transformer supplying full load current at 0.8 lagging power factor. The open ckt. And short ckt test results are as follows: O.C.: 200V

0.8A

70W

S. C.: 20V

10A

60W

7. A 10-kVA 5000/440 V, 25 Hz, single phase transformer has copper, eddy current and hysteresis losses of 1.5, 0.5 and 0.6 percent of output on full load. What will be the percentage losses if the transformer is used on a 10-kVA, 50 Hz system keeping the full load current constant? Assume unity power factor operation. Compare the full load efficiencies for the two cases. 8. A 10-kVa 500/250 V single phase transformer gave the following test results: S.C.: 60V, 20A, 150W

The maximum efficiency occurs at unity power factor and at 1.20 times full load current. Determine the full-oad efficiency at 0.80 p.f.. Also calculate the maximum efficiency. 9. A 100-kVA lighting transformer has a full-load loss of 3 kW, the losses being equally divided between iron and copper. During a day, transformer operates on full-load for 3 hours, one-half load for 4 hours, the output being negligible for the remainder of the day. Calculate the all-day efficiency. 10. A 5-kVA distribution transformer has a full-load efficiency at unity p.f. of 95%, the copper and iron losses then being equal. Calculate its all-day efficiency if it is loaded throughout the 24 hours as follows: No load for 10 hours

Full load for 2 hours

Half load for 5 hours

Quarter load for 7 hours

Assume load p.f. of unity.

2.

1.

So= 25 KVA

KVA Load ηM = Full Load KVA √

Pco= 350 W

PcuFL= 400 W

@ ηM

PC = Pcux

p.f.= 1

P1 = core

P2 = PcuFL

a. η @ Full Load

ηFL =

0.75 Full Load = Full Load √

=

=√

ηFL= 97.09%

b. η @ Half Load

2

√

[

Pcux = x PcuFL

= (0.5)2(400)

(0.75)2 =

= 100 W

η=

η = 96.53%

= 0.56

=

]2

3.

4.

Pco = 40 W

PcuFL = 112 W

p.f. = 0.8

@ 98%

=

a. So = 1.25 KVA

=

x=

= 0.25

Plosses = 3265.31 W

2

Pcux = (0.25) (112)

3265.31 = Pco + PcuFL eq.1

=7W

@ Maximum Efficiency

=

Pco = Pcux

Pcux = ( ⁄ )2(PcuFL)

η=

Pcux = 0.5625 PcuFL eq. 2

η = 95.51%

Subst. eq. 2 to eq. 1

b. So = 2.5 KVA

x=

= 0.5

Pcux = (0.5)2(112)

3265.31 = 0.5625 PcuFL + PcuFL

PcuFL = 2089.798 W

From Eq. 2

= 28 W

Pcux = 0.5625(2089.798)

η=

Pcux = 1175.51 W = Pco

η = 96.71%

Pcux = (0.5)2(2089.798)

c. So = 5 KVA

x=

=1

= 522.45 W

η=

2

Pcux = (1) (112)

= 112 W

η=

η = 96.34%

η = 97.92%

5.

6.

Vp =2200 V

rp = 0.1 Ω

Isc = 10 A

Vs = 220 V

rs = 0.01 Ω

Vsc = 20 V

So = 25 KVA

a=

=

Open ckt. = Pco = 70 W

Short ckt. = psc = 60 W

p.f. = 0.8

Ip =

= 10

θ =...