saturated synchronous reactance Xs at the rated voltage. Express the

EE3PI4

1

_Mid

_‐

_Term

_Examination

₍₂

_Pages)

Note:Q1toQ4areallmandatory.

1. (36 Points) Fig. 1 shows a 60 Hz, three‐phase power system with two loads. The source is a three‐phase, Δ‐connected generator in ACB sequence, which produces a line‐to‐line voltage |VL| of 480 V. The line impedance Zline in the power system is 1 + j2 . Load 1 is Y‐connected, with its phase impedance Z1 of 40∠36.87° . Load 2 is Δ‐connected with its phase impedance Z2 of 90∠30° .

Fig. 1: Schematic of a three‐phase power system. (a) What is the line‐to‐line voltage |Vload| across these loads?

(b) Find the real and reactive power losses in the transmission line.

(c) Find the real & reactive powers, and power factor (PF) supplied by the generator. (d) If we want to improve the power factor of the generator to unity by using a Y‐ connected capacitor bank (load 3 in Fig. 1), calculate the required equivalent bank capacitance C3 per phase.

2. (34 Points) A 13.8‐kV, 50‐MVA, 0.9‐power‐factor‐lagging, 60‐Hz, six‐pole Y‐ connected synchronous generator has a synchronous reactance of 2.5  and an armature resistance of 0.2 . At 60 Hz, its friction and windage losses are 2 MW, its core losses are 1.5 MW, and assuming there is no stray loss.

(a) Calculate the rotor speed in RPM.

Page 2 of 2 first one. What happens to terminal voltage of the generator (using the phasor diagram to justify your answer)?

(f) Calculate the new terminal voltage after the load has been added to confirm the expectation in (e).

(g) What must be done to restore the terminal voltage to its original value?

3. (22Points) A 67.5‐MVA (Srated), 14.8‐kV (VT,rated), three‐phase, ‐connected, 60‐Hz synchronous generator was tested by the open‐circuit, short‐circuit, and dc tests, and its air‐gap voltage was extrapolated with the following results:

Open‐Circuit Test

Field current (IF), A 275 320 365 380 475 570 Line-to-line voltage (VT,OC), kV 13.75 14.0 14.8 15.1 16.2 17.0

Extrapolated air-gap voltage (VT,OC AG), kV 13.75 16.0 18.25 19.0 23.75 28.5

Short‐Circuit Test

Field current (IF), A 275 320 365 380 475 570 Line current (IL,SC), A 1524 1774 2023 2106 2633 3159

DC Test:

When applying a dc voltage of 4 V to two terminals, a current of 10 A was measured. (a) Find the armature resistance RA and the unsaturated synchronous reactance Xsu

of this generator in ohms perphase.

(b) Find the saturated synchronous reactance Xs at the rated voltage. Express the answer in ohms per phase.

(c) Find the short‐circuit ratio for this generator.

4. (8Points) Please answer the following short questions:

(a) What causes the armature reactance in a generator, and how to model this effect in the equivalent circuit?

(b) What effects are included in the synchronous reactance jXs of a generator? Is it a constant? If not, how does it behave?

(c) How to obtain the short‐circuit ratio of a generator?

(d) What is the physical meaning of the reactive power Q in a power system? How to interpret the negative real power P?

(e) Why we want to improve the power factor of a generator as close to unity as possible?

Solution of EE 3PI4 1st Mid-Term Exam, 2013-2014 Term 2

SolutionforEE3PI41st_{Mid‐TermExam,2013~2014} (FullScore:100Points)

Q1.(36Points)

(a)To solve this problem, we first convert the ‐connected load Z2 to its equivalent Y‐ connected load Z2Y, and get the per‐phase equivalent circuit as follows (2pts):

where V and Z2Y are given by

480

(0 30 )= 277 30 [V] for AC sequence

3 B

V_       (2 pts)

and

2 2

90 30

30 30 [ ]

3 3

Y

Z

Z_ 



 

     . (2 pts)

So the equivalent impedance seen by the generator can be obtained by

1 2

1 2

15.41 11.34 19.13 36.34 [ ]

Y

eq line

Y

Z Z

Z Z j

Z Z

 

 

 

      

 , (2 pts)

and the line current Iline from the generator can be calculate by

14.48 6.34 14.40 1.60 [A]

line eq

V

I j

Z

 

      . (2 pts)

So the phase load voltage V,load can be calculated by

,load line line 222.4 111.37 248.73 26.6 [V]

V_ V_ I Z   j    (2 pts)

and the line‐to‐line load voltage Vload can be obtained by

,

3 ( 30 ) 430.82 3.4 [V]

load load

V  V_        .

Therefore |Vload| = 430.82 V(2 pts)

3 * 12.04 36.34 9.7 7.14 [kVA]

line

S_generator   V_ I      j .

Therefore, the real power supplied by the generator is Pgenerator = 9.7 kW (2pts)and the reactive power supplied by the generator is Qgenerator =7.14 kVAR (2 pts). The power factor is calculated by

cos(36.34 ) 0.81

 

P

F (2 pts) lagging. (2 pts)

Solution of EE 3PI4 1st Mid-Term Exam, 2013-2014 Term 2

Q2.(34Points)

(a)The rotor speed in RPM is calculated by (2pts)

120 120 60

1200 RPM. 6

e m

f n

Pole

 

  

(b)When the field current is 5.0 A, and the open‐circuit terminal voltage is 16,500 V, and the open‐circuit phase voltage in the generator (and hence EA) will be given by

16, 500 / 3 9526 V A

E   (2 pts). The load is ‐connected with three impedances of 2425 . From the Y‐ transformation, this load is equivalent to a Y-connected load with three impedances of 825  (2 pts). The resulting per-phase equivalent circuit is shown below (2pts):

The magnitude of the phase current flowing in this generator is (2pts)

9526 V

1004 25 A 0.2 2.5 8 25

A A

A S

E I

R jX Z j

 

    

    

Therefore, the magnitude of the phase voltage is (1pt)

(1004 V)(8 ) 8032 V A

V_ I Z   

and the terminal voltage is (1pt)

3 3(8032 V) 13, 910 V. T

V  V_  

(c)Based on the equivalent circuit diagram, the internally generated voltage is (2pts)

(8032 0 V) (0.2 )(1004 25 A) (2.5 )(1004 25 A)

9530 13.3 V

A A A S A

E V R I jX I

j



  



  

         

 

2 2

3 3(1004 A) (0.2 ) 605 kW

cu A A

P  I R    .

& 2 MW

F W

P  .

1.5 MW core

P  .

0 (assumed) stray

P  .

& 26 MW

in out cu F W core stray

P P P P P P  (2pts).

21.9 MW

100% 100% 84%

26 MW out

in P Efficiency

P 

      (2pts).

(e)To get the basic idea of what happens, we will ignore the armature resistance for the moment. If the field current and the rotational speed of the generator are constant, then the magnitude of EA (= k) is constant. The quantity jXSIA increases in length at the same angle, while the magnitude of EA must remain constant. Therefore, EA “swings” out along the arc of constant magnitude until the new jXSIA fits exactly between V and EA(2pts).

Therefore, from the phasor diagram shown above, the terminal voltage drops (2pts). (f) The new impedance per phase will be half of the old value, and it is Z = 425(2pts). The

magnitude of the phase current flowing in this generator becomes

9526 V 9526 V

1680 A 5.67

0.2 2.5 4 25 A

A

A S

E I

R jX Z j 

   

      (2pts).

Therefore, the magnitude of the phase voltage is (2pts)

(1680 V)(4 ) 6720 V A

V_ I Z   

and the terminal voltage is (2pts)

3 3(6720 V) 11, 640 V. T

V  V  

Solution of EE 3PI4 1st Mid-Term Exam, 2013-2014 Term 2

Q3.(22Points)

(a)For the armature resistance, since the generator is in‐connection, the dc resistance measured between any two terminals is given by

2

2 2 4 V

( ) 0.4 0.6 .

2 3 10 A

A

eq A A A A A

A A

R

R R R R R R

R R

         



 (2 pts)

For the unsaturated synchronous reactance of this generator, it is given by

, , 2 2 2 2

,

0.6 / 3

OC T OC

A Su Su

A SC L

V V

R X X

I I

 _{     (1)}

where V,OC and IA,SC are the phase voltage in the open‐circuit test and phase current in the short‐circuit test under the same field current IF, respectively. If we use the extrapolated air‐gap voltage vs. field current characteristic for V,OC, we can read IL at any field current

IF. So we will look at it at a field current of 380 A, for example. The extrapolated air‐gap voltage at this point is 19.0 kV (2pts), and the short‐circuit current is 2106 A (2pts). Therefore, using (1), the unsaturated synchronous reactance is





2 19.0 kV 3

0.6 15.61 . 2106 A  _            Su

X (2 pts)

(b)For saturated synchronous reactance, since the field current IF at the rated voltage (VT =

VT,rated = 14.8 kV) is 365 A (2pts), the corresponding short‐circuit current IL is 2023 A (2

pts). Using (1) again, we can calculate the saturated synchronous reactance by





2 14.8 kV 3

0.6 12.66 . 2023 A              S

X (2 pts)

(c) The short‐circuit ratio (SCR) is defined as SCR = IF@V,rated/IF@IA,rated =IF@V,rated/IF@IL,rated (2 pts). Since V,rated= 14.8 kV, from the OCC result, its corresponding IF@V,rated= 365 A. The

next step is to find the rated current IL,rated by





,

,

67.5 MVA

2633 A

3 3 14.8 kV

    rated L rated T rated S I

V (2 pts)

and from the SCC results, its corresponding IF@IL,ratedis about 475 A (2 pts). Therefore,

Q4.(8Points)

(a)The armature reactance in a generator is caused by the Estatfrom the magnetic flux due to the armature current (1pt). It is modeled by an inductor in the equivalent circuit (1pt). (b)The synchronous reactance jXs of a generator includes armature reactance and the self‐

inductance of the armature windings (1pt). It is NOT a constant, and it decreases when

field current increases (1pt).

(c)The short‐circuit ratio (SCR) of a generator is calculated by the field current obtained at the rated voltage divided by the field current obtained at the rated armature current (1 pt).

(d)The physical meaning of the reactive power Q in a power system is the power bouncing

back and forth between the source and load but is never consumed by the load (1pt). The

negative real power P for a generator means the power being consumed by the generator, while for the load, it means the power generated from the load. (1pt)