Quiz. Assess the following concerns. Show your remedy and encircle your last answer. 1 . In a 25-kVA, 2000/200 V, single phase transformer, the iron and full load copper losses are 350 W and 400 T respectively. Compute the performance at unity power aspect on a) full load, b) fifty percent full-load. installment payments on your If P1 and P2 be the iron and copper loss of the transformer on total load, locate the ratio of p1 and P2 such tat maximum effectiveness occurs for 75% total load.

three or more. A 5-kVA, 2300/230 Sixth is v, 50 Hz transformer was tested pertaining to the straightener losses with normal excitation and copper mineral losses at full-load and these were located to be forty five W and 112 T respectively. Estimate the efficiencies of the transformer at 0. 8 l. f. to get the following results: a) 1 . 25 kVA, b) 2 . 5kVa, c) 5 kVA 4. A 200-kVa transformer has an efficiency of 98% at complete load. In the event the maximum efficiency occurs for three quarters of full insert, calculate the efficiency by half load. Assume minimal magnetizing current and s. f. 0. 8 by any means loads.

5. A 25-kVA single transformer 2200/220 volts, has primary resistance of 0. 1О© and the second resistance of 0. 01О©. Find the same secondary amount of resistance and the full-load efficiency at 0. almost 8 p. n. if the flat iron loss of the transformer can be 80% belonging to he total load water piping loss. 6th. Consider a 4-Kva 200/400 Sixth is v single period transformer providing full insert current in 0. almost eight lagging electricity factor. The open ckt. And brief ckt test results are the following: O. C.: 200V

0. 8A

70W

S. C.: 20V

10A

60W

six. A 10-kVA 5000/440 Versus, 25 Hz, single phase transformer offers copper, turn current and hysteresis deficits of 1. a few, 0. 5 and 0. 6 percent of outcome on complete load. What is going to be the percentage losses in the event the transformer is employed on a 10-kVA, 50 Hz system keeping the full fill current regular? Assume oneness power element operation. Review the full load efficiencies intended for the two situations. 8. A 10-kVa 500/250 V single phase transformer gave the next test effects: S. C.: 60V, 20A, 150W

The most efficiency arises at unanimity power element and at 1 . 20 occasions full insert current. Decide the full-oad efficiency in 0. 80 p. farreneheit.. Also calculate the maximum effectiveness. 9. A 100-kVA light transformer contains a full-load lack of 3 kW, the failures being equally divided between iron and copper. During a day, transformer operates in full-load intended for 3 several hours, one-half weight for 4 hours, the output staying negligible to get the remainder during. Calculate the all-day productivity. 10. A 5-kVA distribution transformer has a full-load efficiency at oneness p. farrenheit. of 95%, the birdwatcher and straightener losses after that being equal. Calculate its all-day efficiency if it is packed throughout the a day as follows: No load for 10 several hours

Full load for two hours

Half weight for five hours

Quarter load intended for 7 hours

Assume fill p. farrenheit. of oneness.

2 .

1 )

So= 25 KVA

KVA Load О·M = Complete Load KVA в€љ

Pco= 350 W

PcuFL= 500 W

@ О·M

COMPUTER = Pcux

p. farreneheit. = you

P1 = core

P2 = PcuFL

a. О· @ Total Load

О·FL =

zero. 75 Full Load sama dengan Full Load в€љ

=

=в€љ

О·FL= 97. 09%

b. О· @ 50 percent Load

a couple of

в€љ

[

Pcux = back button PcuFL

= (0. 5)2(400)

(0. 75)2 =

= 100 Watts

О·=

О· = ninety six. 53%

sama dengan 0. 56

=

]2

a few.

4.

Public carriage office = forty W

PcuFL = 112 W

g. f. = 0. almost 8

@ 98%

=

a. So sama dengan 1 . twenty-five KVA

=

x=

sama dengan 0. 25

Plosses = 3265. 31 W

a couple of

Pcux sama dengan (0. 25) (112)

3265. 31 sama dengan Pco + PcuFL frequency. 1

=7W

@ Maximum Efficiency

sama dengan

Pco = Pcux

Pcux = ( вЃ„ )2(PcuFL)

О·=

Pcux = 0. 5625 PcuFL eq. a couple of

О· = 95. 51%

Subst. frequency. 2 to eq. one particular

b. Thus = 2 . 5 KVA

x=

= 0. five

Pcux = (0. 5)2(112)

3265. thirty-one = zero. 5625 PcuFL + PcuFL

PcuFL sama dengan 2089. 798 W

Coming from Eq. 2

= twenty-eight W

Pcux = 0. 5625(2089. 798)

О·=

Pcux = 1175. 51 Watts = Pco

О· = 96. 71%

Pcux = (0. 5)2(2089. 798)

c. So sama dengan 5 KVA

x=

=1

= 522. 45 Watts

О·=

a couple of

Pcux = (1) (112)

= 112 W

О·=

О· sama dengan 96. 34%

О· = 97. 92%

5.

6th.

Vp =2200 V

rp = zero. 1 О©

Isc = 10 A

Vs = 220 Versus

rs sama dengan 0. 01 О©

Vsc = 20 V

And so = 25 KVA

a=

=

Available ckt. = Pco sama dengan 70 T

Short ckt. = psc = 70 W

l. f. = 0. almost 8

Ip sama dengan

= 15

Оё =...

### Tags:

full, current, transformer, pcux, pcufl, deficits, copper, performance, hours, load, full-load, iron, main loss, birdwatcher loss-
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