Transcript IES PE Short Course

Some practice problems
At end of the module G1 student text are
some problems. I suggest that you work
on problems 1, 4, 6, and 7. This will provide
you with good practice for this module.
I will make the solutions available to you
later. It is also possible that I may assign
one or more of these problems as an in-class
group exercise sometime next week.
The power angle
E f  Vt  jX s I a
Let’s define the angle that Ef makes with Vt as 
E f | E f | 
For generator operation (power supplied by machine),
this angle is always positive.
For motor operation, this angle is negative.
The power angle
E f  Vt  jX s I a
The power angle E f | E f | 
is the angle between
Ef
Vt and Ef, and is
the angle of Ef
jXsIa
if Vt is reference
jXsIa

Vt
Ia
XsIa
This is for
generator
operation.
You should see
if you can construct
the phasor diagram
for motor operation,
lagging operation
(just reverse the
arrow on the current)
Example
A 10 MVA, 3 phase, Y-connected, two pole, 60 Hz,
13.8 kV (line to line) generator has a synchronous
reactance of 20 ohms per phase. Find the excitation
voltage if the generator is operating at rated terminal
voltage and supplying (a) 300 Amperes at 30 degrees
lagging, (b) 300 Amperes at 30 degrees leading
Example (cont’d)
Solution
138
. kV
Vt 
 7.97kV  Vt  7.97  103 0
3
E f  Vt  jX s Ia
(a)
I a  300  30 E f  7.97  103 0( 2090 )(300  30 )
 1214
. kV25.34
3
(b) I a  30030 E f  7.97  10 0( 2090 )(30030 )
 719
. kV46.27
Note: 1) E f lag  E f lead
2) lag case referred as overexcited operation
3) lead case referred as underexcited operation
It is usually the case that excitation voltage Ef is
lower in magnitude for leading operation as
compared to lagging operation.
Ef
Ef
jXsIa
jXsIa
Ia
Vt
Ia
Vt
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.
jXsIa
Here note that |Ef| for
leading case is larger than
for lagging case. This is
due to much larger Ia in
the leading case.
Ef
Ia
jXsIa
Vt
Ia
Vt
Ef
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 relationships
Recall the power angle,  , as the angle at which the
excitation voltage, E f | E f |  , leads the terminal
voltage, Vt |Vt | 0. Therefore, from the circuit….
Ia
jXs
Ef
Vt
Zload
Power relationships
Ia 
| E f |   Vt 0
jX s



But
| E f |cos   j| E f |sin   Vt
jX s
| E f |cos   Vt
jX s
| E f |sin 
Xs

j| E f |sin 
jX s
| E f |cos   Vt 
 j

X


s
I a | I a | cos   j | I a | sin 
Power relationships (cont’d)
Equating real and imaginary parts of equ. 1 and 2
and multiplying both sides of the equations by 3Vt :
Pout  3Vt | I a | cos  
3Vt | E f | sin 
(3)
Xs
3Vt | E f | cos  3Vt 2
Qout  3Vt | I a | sin  

(4)
Xs
Xs
Note: reactive power is positive when the machine is
operated overexcited, i.e., when it is lagging
Example
Find Pout and Qout for the conditions (a) and (b)
described in the previous example.
Solution:
(a)   25.34 ,Vt  7.97kV ,| E f |  1214
. kV
3( 7.97  103 )(1214
.  103 )sin 25.34
 Pout 
 6.21 MW
20
3(7.97  103 )(12.14  103 ) cos 25.34 3(7.97  103 ) 2
 Qout 

 3.59
20
20
MVARS
Example (cont’d)
. kV
(b)   46.27 ,Vt  7.97kV ,| E f |  719
 Pout
 Qout
3( 7.97  103 )( 719
.  103 )sin 46.27

 6.21 MW
20
3( 7.97  103 )( 719
.  103 )cos 46.27 3( 7.97  103 )2


 359
. MVARS
20
20
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
Qout  0  3Vt | I a | sin  

3Vt | E f | cos 
Xs
| E f | cos   Vt
3Vt | E f | cos 
Xs
3Vt 2

Xs
3Vt 2

Xs
This is the condition for which the generator is operating
at 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).
| E f | cos   Vt (for lagging or Qout  0)
| E f | cos   Vt (for leading or Qout  0)
Let’s look at the phasor diagrams.
Ef
Ef
jXsIa
jXsIa
Ia
Vt
Ia
Vt
| E f | cos   Vt
| E f | cos   Vt
Ia
Vt
| E f | cos   Vt
Generation Operation:
Generator pull-out power
From equ. 3, the plot of the electric power Pout against
the power angle 
Pmax 
3Vt E f
Xs
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
• Pmech  Pout when operating in steady-state
• if   90 then Pout decreases while Pmech remains
unchanged  Pmech  Pout which causes machine
to accelerate beyond synchronous speed-i.e. the
machine has pulled out or lost synchronism
Example
Compute the pull-out power for the two conditions
described in Example 2
Solution
(a) Overexcited case (lagging):
Pmax
3( 7.97  103 )(1214
.  103 )

 14.51
20
(b) Underexcited case (leading):
Pmax
3( 7.97  103 )( 719
.  103 )

 8.6
20
Note: The limit is lower when the generator is
underexcited because E f is lower
MW
MW