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maquinas eletricas - transformadores, Exercícios de Máquinas Elétricas

Pontifícia Universidade Católica de Goiás (PUC-GO)Máquinas Elétricas

exercicios de maquinas eletricas sobre transformadores

Tipologia: Exercícios

2020

Compartilhado em 16/04/2020

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Transformer sheet
1 A single-phase 100 kVA, 1000/ 100 V transformer gave the following test results:
open-circuit test 100 V, 6.0 A, 400 W short-circuit test 50 V, 100 A, 1800 W
(a) Determine the rated voltage and rated current for the HV and LV sides.
(b) Derive an approximate equivalent circuit referred to the HV side.
(c) Determine the voltage regulation at full load, 0.6 PF leading.
(d) Draw the phasor diagram for condition (c).
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pf4
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pfa
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Transformer sheet

1 A single-phase 100 kVA, 1000/ 100 V transformer gave the following test results:

open-circuit test 100 V, 6.0 A, 400 W short-circuit test 50 V, 100 A, 1800 W

(a) Determine the rated voltage and rated current for the HV and LV sides.

(b) Derive an approximate equivalent circuit referred to the HV side.

(c) Determine the voltage regulation at full load, 0.6 PF leading.

(d) Draw the phasor diagram for condition (c).

2 A 1 single-phase, 25 kVA, 220/440 V, 60 Hz transformer gave the following test results.

Open circuit test : 220 V, 9.5 A, 650 W Short-circuit test : 37.5 V, 55 A, 950 W

(a) Derive the approximate equivalent circuit in per-unit values.

(b) Determine the voltage regulation at full load, 0.8 PF lagging.

(c) Draw the phasor diagram for condition (b).

No-load test results: Voc , Ioc , Poc Short-circuit test results: Vsc , Isc , Psc

(b) The HV winding of the transformer is connected to the 11 kV supply and a load,

o

Z L 15 90 is connected to the low voltage winding. Determine:

(i) Load voltage. (ii) Voltage regulation.

Solution:

a = =

2 2 Reqp Rp a Rs

2 2 Xeqp Xp a Xs

R (^) cp = a Rcs = 5 * 2. 4 = 60 k Ω

2 2

X (^) mp = a Xms = 5 * 0. 8 = 20 k Ω

2 2

For open circuit test on the primary side:

V oc = 11 kV Voltameter eadings

A

R

V

I

c

oc c^0.^183 60000

A

X

V

I

m

oc m^0.^55 20000

I (^) oc Ic Im 0. 183 0. 55 0. 58 A

2 2 2 2

= + = + = Ameter readings

The wattmeter readings is Poc IcRc 0. 183 * 60000 2009 W

2 2 = = =

For short circuit test on the primary side:

A

V

VA

I

r

r^22.^73 11 * 10

3

3

1

1 =^ = = (Ameter readings)

The voltameter readins is I (^) sc * ( Reqp + jXeqp ) = 22. 73 *( 2. 55 + j 8. 5 ) = 201. 71 V

The wattmeter readings is I (^) sc * ( R (^) eqp ) 22. 73 *( 2. 55 ) 1317. 51 W

2 2 = =

(b) = ∠− Ω

o Z (^) L 15 90

o o Z (^) L a * ZL 5 * 15 90 375 90

2 2

( )

o o

o o

L eqp

L

Z Z j

Z

V V 11270 0. 4

Then load voltage = V 2 = V 2 ′/ a = 11270 / 5 = 2254 V

2

1 2 =−

V

V V

VR

5 A 1 φ , 10 kVA, 460/ 120 V, 60 Hz transformer has an efficiency of 96% when

it delivers 9 kW at 0.9 power factor. This transformer is connected as an

autotransformer to supply load to a 460 V circuit from a 580 V source.

(a) Show the autotransformer connection.

(b) Determine the maximum kVA the autotransformer can supply to the 460 V circuit.

(c) Determine the efficiency of the autotransformer for full load at 0.9 power factor.

6 Reconnect the windings of a 1 φ , 3 kVA, 240/120 V, 60 Hz transformer so that it can supply

a load at 330 V from a 110 V supply. (a) Show the connection.

(b) Determine the maximum kVA the reconnected transformer can deliver.

8 A 1 φ 200 kVA, 2100/210 V, 60 Hz transformer has the following characteristics. The

impedance of the high-voltage winding is 0.25 + j 1.5 Ω with the lowvoltage winding

short-circuited. The admittance (i.e., inverse of impedance) of the low-voltage winding is 0.025 - j

O.075 mhos with the high-voltage winding open-circuited.

(a) Taking the transformer rating as base, determine the base values of power, voltage, current,

and impedance for both the high-voltage and low-voltage sides of the transformer.

(b) Determine the per-unit value of the equivalent resistance and leakage reactance of the

transformer. (c) Determine the per-unit value of the excitation current at rated voltage.

(d) Determine the per-unit value of the total power loss in the transformer at full-load output

condition.

9- A 460 kVA, (4600/460) V, 60 Hz single phase transformer has the following parameters:

A

E

I 7. 58 1. 233

V (^) 2 ′ = E − I 2 ′* ( R 2 ′+ JX 2 ′) = 4565. 17 ∠− 0. 59 V − 7. 58 ∠− 1. 233 *( 2. 6 + j 6 ) = 4545. 19 ∠− 1. 16 V

( )

( )

( )

( )

4600 * 8. 94 *cos 13. 43

  1. 19 * 7. 58 *cos 0. 076

cos

  • cos

11

2 2 = =

inp

L

VI

V I

φ

φ η

At no load the equivalent circuit becomes:

Noload equivalent circuit

J

Z (^) excitation

Then V j

V NL * 2057. 98 59. 04 4585. 56 0. 024

Then voltage regulation can be obtained as following:

2

2 2

V

V V

VR

NL

10- A single-phase, 10 kVA, 2000/200 V, 60 Hz distribution transformer has the following

characteristics:

Core loss at full voltage =120 W

Copper loss at half load =80 W

i. Determine the efficiency of the transformer when it delivers full load at 0.8 power factor

lagging.

ii. Determine the rating at which the transformer efficiency is a maximum. Determine the

efficiency if the load power factor is 0.

iii. Determine the rating of transformer at 92% efficiency and 0.8 power factor.

iv. The transformer has the following load cycle:

No load for 6 hours

66% full load for 10 hours at 0.8 PF

85% full load for 8 hours at 0.9 PF

Determine the all day efficiency of the transformer

2- (i) Pout = 10 * 0. 8 = 8 kW

Pcore = 120 W

2

,

x P

P

cuFL

cu (^) = P W cu FL^320

, (^2)

(ii) Maximum efficiency occurs at Pcore = Pcu

Then Pcu x Pcu , FL

2 = 120 = max

,

max = = = cu FL

core

P

P

x

(iii) 92 %

2

x x

x

η

2

8000 * x = 0. 928000 * x + 120 + 320 * x

2

x − x + =

2 = ±

x =

Then x = 1. 985 refused or x = 0. 189

(iv) E 0 10 * 0. 66 * 0. 8 * 10 8 * 0. 9 * 0. 85 * 10 114 kWh

24

E kWh

core

3

−

E kWh

cu

2 2

Then * 100

24

24 _

core cu

all day

E E E

E

_

all day

η

11- A 6kVA, 250/500 V, transformer gave the following test results

short-circuite 20 V ; 12 A, 100 W and Open-circuit test : 250 V, 1 A, 80 W

i. Determine the transformer equivalent circuit.

ii. calculate applied voltage, voltage regulation and efficiency when the output is 10 A at 500

volt and 0.8 power factor lagging.

iii. Maximum efficiency, at what percent of full load does this maximum efficiency occur? (At

0.8 power factor lagging).

iv. At what percent of full load does the effeciency is 95% at 0.8 power factor lagging.

3- (I) From O.C. Test

  1. 32 250 * 1. 0

80 cos

*cos

∴ = = =

=

o o

o o

o o o o

V I

P

P V I

ϕ

ϕ

So, = = = 0. 4167 Ω

1

1 1 sc

sc eq I

V

Z

Also, 1

2 Psc = I 1 scReq

Then, = = 0. 174 Ω

eq (^12)

R

Then, = − = 0. 4167 − 0. 174 = 0. 3786 Ω

2 2 2 1

2 Xeq 1 Zeq 1 Req

It is clear the second method gives the same results easly.

(II) Output KVA = 10 * 500 * 0. 8 = 4 kVA

Now, from the aproximate equivalent circuit refeared to secondery :

1 2 0 2 * eq 1

o o o V ∠δ = V ′∠ + I ′ ∠ ϕ Z

Then,

( )

o

o o o V j

1 250 0 20 36.^87 *^0.^1740.^3786

2

1 2

V

V V

V R

Pout = 10 * 500 * 0. 8 = 4 kW ,

Pi = Poc = 80 W , and ,

Pcu 10 * Req 2 100 * 0. 694 69. 4 W

2

= = = or

W

I

I

P P

SC

cu sc^69.^4 12

(^22)

2

out i cu

out

P P P

P

(III) maximum effeciency ocures when Pc = Pcu = 80 W

the

The percent of the full load at which maximum efficiency occurs is :

,

cu FL

c

P

P

X

Then, the maximum efficiency is :

(IV)

2

x x

x

P P P

P

out i cu

η out

Then,

2 x − x + =

Then, x = 2. 155 (Unacceptable)

Or x = 0. 3712

Then to get 95% efficiency at 0.8 power factor the transformer must work at 37.12% of full load.

12- A single phase, 50 kVA, 2400/460 V, 50 Hz two-winding transformer has an efficiency of 0.95%

when it delivers 45kW at 0.9 power factor. This transformer is connected as an auto-transformer to supply

load to a 2400 V circuit from 2860 V source.

(a) Show the transformer connection.

(b) Determine the maximum kVA that autotransformer can supply to 2400 V circuit.

(c) Determine the efficiency of the autotransformer for full load at 0.9 power factor.

Solution:

(a)

(b) I (^) s w 108. 7 A 2460

50 * 10

3

, 2 = =

Then, kVA ) kW Auto

= 108. 782860 = 310. 87

(c) 0. 95

50 * 10 * 0. 9

50 * 10 * 0. 9

,

3

3

2 =

=

i cu FL

w P P

η

Then, Pi^ + Pcu , FL =^2368.^42 W

  1. 61 % 310870 * 0. 9 2368. 42

310870 * 0. 9

η Auto =

13- Three single phase, 30 kVA, 2400/240 V, 50 Hz transformers are connected to form 3 ϕ, 4160/240 V

transformer bank. The equivalent impedance of each transformer referred to the high voltage side is 1.5+j2Ω.

The transformer delivers 60 kW at 0.75 power factor (leading). (a) Draw schematic diagram showing the

transformer connection. (b) Determine the transformer winding current (c) Determine the primary voltage.

(d) Determine the voltage regulation. (e) determine the maximum efficiency if the maximum effeciency

occurred at 90% of full load at 0.9 power factor.

2860

460

2400

14- Three single-phase, 50 kVA, 2300/230 V, 60 Hz transformers are connected to form a three-phase,

4000/230 V transformer bank. The equivalent impedance of each transformer referred to low voltage is

0.012 + j0.016 Ω. The three-phase transformer supplies a three-phase 120 kVA, 230 V, 0.85 PF (lag) load.

(a) Draw a schematic diagram showing the transformer connection.

(b) Determine the transformer winding currents.

(c) Determine the primary voltage (line-to-line) required.

(d) Determine the voltage regulation.