Some practice problems

At end of the module G1 student text aresome problems. I suggest that you workon problems 1, 4, 6, and 7. This will provideyou with good practice for this module.I will make the solutions available to youlater. It is also possible that I may assignone or more of these problems as an in-classgroup exercise sometime next week.

The power angle

astf IjXVE

Let’s define the angle that Ef makes with Vt as

|| ff EE

For generator operation (power supplied by machine),this angle is always positive.

For motor operation, this angle is negative.

The power angle

astf IjXVE

|| ff EE

Vt

Ia

XsIa

jXsIa

jXsIa

Ef

This is forgenerator operation.

You should seeif you can constructthe phasor diagramfor motor operation,lagging operation(just reverse the arrow on the current)

The power angleis the angle betweenVt and Ef, and isthe angle of Efif Vt is reference

Example

A 10 MVA, 3 phase, Y-connected, two pole, 60 Hz,13.8 kV (line to line) generator has a synchronousreactance of 20 ohms per phase. Find the excitationvoltage if the generator is operating at rated terminalvoltage and supplying (a) 300 Amperes at 30 degreeslagging, (b) 300 Amperes at 30 degrees leading

Example (cont’d)Solution

(a)

(b)

Note: 1) 2) lag case referred as overexcited operation 3) lead case referred as underexcited operation

VkV

kV Vt t 1383

7 97 7 97 10 03.. .

E V jX If t s a

I E

kV

a f

300 30 7 97 10 0 20 90 300 30

1214 2534

3. ( )( )

. .

I E

kV

a f

300 30 7 97 10 0 20 90 300 30

719 46 27

3. ( )( )

. .E Ef lag f lead

It is usually the case that excitation voltage Ef is

lower in magnitude for leading operation as

compared to lagging operation.

VtVtIa

Ia jXsIa

EfEf jXsIa

Implication: If |Ef| is large (small), relative to |Vt|,machine is probably lagging (leading).

It is usually the case that excitation voltage Ef is

lower in magnitude for leading operation as

compared to lagging operation, but not always.

VtVt Ia

Ia jXsIa

jXsIa

Ef

EfHere note that |Ef| forleading case is larger thanfor lagging case. This is due to much larger Ia inthe leading case.

So how can we tell, using Ef and Vt, whether the generator is lagging or leading?

We need to look at power relations for the answer.

We will derive power relations that do not require knowledge of the current.

Power relationshipsRecall the power angle, , as the angle at which theexcitation voltage, , leads the terminalvoltage, . Therefore, from the circuit….

E Ef f | | V Vt t | | 0

Ef

jXs

ZloadVt

Ia

Power relationships

IE V

jX

E j E V

jX

E V

jX

j E

jX

af t

s

f f t

s

f t

s

f

s

| | | |cos | |sin

| |cos | |sin

0

| |sin | |cosE

XjE V

Xf

s

f t

s

sin||cos|| aIjaIaI But

Power relationships (cont’d)Equating real and imaginary parts of equ. 1 and 2and multiplying both sides of the equations by : (3)

(4)

Note: reactive power is positive when the machine is operated overexcited, i.e., when it is lagging

3Vt

sXfEtV

aItVoutP

sin||3

cos||3

sXtV

sXfEtV

aItVoutQ23cos||3

sin||3

ExampleFind Pout and Qout for the conditions (a) and (b)described in the previous example.

Solution:

(a)

2534 7 97 1214. , . ,| | .V kV E kVt f

Pout3 7 97 10 1214 10 2534

206 21

3 3( . )( . )sin ..

59.320

2)31097.7(3

20

34.25cos)31014.12)(31097.7(3

outQ

MW

MVARS

Example (cont’d)

(b) 46 27 7 97 719. , . ,| | .V kV E kVt f

Pout3 7 97 10 719 10 46 27

206 21

3 3( . )( . )sin ..

Qout3 7 97 10 719 10 46 27

203 7 97 10

20359

3 3 3 2( . )( . )cos . ( . ).

MW

MVARS

Example (cont’d)Consider:

• Why is real power the same under the two conditions?

• When the generator is operating lagging, is it absorbing vars from or supplying vars to the network? What about when the generator is operating leading?

• For a particular angle , how are the terms “lagging” and “leading” meaningful with respect to real power? With respect to reactive power?

1.0 power factor condition in terms of Ef, Vt, and delta

We have seen that gen lagging operation

results in +Qout and leading operation

results in -Qout. So 1.0 power factor implies

Qout=0. Applying this to reactive power equation:

1.0 power factor condition

tVfEsXtV

sXfEtV

sXtV

sXfEtV

aItVoutQ

cos||

23cos||3

23cos||3sin||30

This is the condition for which the generator is operatingat 1.0 power factor.

What about leading/lagging operation?

In the final equation on the previous page, just replace the equals sign by > (for lagging) or< (for leading).

0)Qoutor leading(for cos||

0)Qoutor lagging(for cos||

tVfE

tVfE

tVfE cos||

Let’s look at the phasor diagrams.

tVfE cos||

VtIa

jXsIa

Ef

VtIa

EfjXsIa

tVfE cos||

VtIa

Generation Operation: Generator pull-out power

From equ. 3, the plot of the electric power against

the power angle

Pout

PV E

Xt f

smax

3

Generation Operation: Generator pull-out power (cont’d)

• For simplicity neglect all real power losses associated with:– windage and heat loss in turbine– friction in turbine– friction in generator bearings– winding resistances

• when operating in steady-state• if then decreases while remains

unchanged which causes machine to accelerate beyond synchronous speed-i.e. the machine has pulled out or lost synchronism

P Pmech out

90 Pout Pmech

P Pmech out

ExampleCompute the pull-out power for the two conditionsdescribed in Example 2

Solution(a) Overexcited case (lagging):

(b) Underexcited case (leading):

Note: The limit is lower when the generator is underexcited because is lower

Pmax( . )( . )

. 3 7 97 10 1214 1020

14 513 3

Pmax( . )( . )

. 3 7 97 10 719 1020

8 63 3

MW

MW

E f