ACTIVITY 1: PARALLEL OPERATION OF ALTERNATORS
EE133L POWER PLANT ENGINEERING LABORATORY
LEARNING OUTCOMES:
To be able to solve problem involving parallel operation of alternators.
DISCUSSION
Reasons for parallel operation of alternators:
1. Several generators can supply a bigger load than one machine by itself.
2. Having many generators increases the reliability of the power system, since the failure of
any one of them does not cause a total power loss to the load.
3. Having many generators operating in parallel allows one or more of them to be removed for
shutdown and preventive maintenance.
Conditions required for paralleling:
1.
2.
3.
4.
Equal rms line voltages.
Equal phase angles of the two phases.
Same frequency.
Same phase sequence. (for 3-phase systems).
General Procedure for Paralleling Generators:
1. Using voltmeters, the field current of the oncoming generator should be adjusted until its
terminal voltage is equal to the line voltage of the running system.
2. The phase sequence of the oncoming generator must be compared to the phase sequence
of the running system.
3. The frequency of the oncoming generator is adjusted to be slightly higher than the
frequency of the running system.
When a generator is operating by itself supplying the system loads, then
1. The real and reactive power supplied by the generator will be the amount demanded by the
attached load.
2. The governor set points of the generator will control the operating frequency of the power
system.
3. The field current (or the field regulator set points) controls the terminal voltage of the
power system.
This is the situation found in isolated generators in remote field environments.
OPERATION OF GENERATORS IN PARALLEL WITH LARGE POWER SYSTEMS
When a generator is operating in parallel with an infinite bus:
1. The frequency and terminal voltage of the generator are controlled by the system to which
it is connected.
2. The governor set points of the generator controls the real power supplied by the generator
to the system.
3. The field current in the generator controls the reactive power supplied by the generator to
the system.
OPERATION OF GENERATORS IN PARALLEL WITH OTHER GENERATORS OF THE SAME SIZE
When two generators are operating together, an increase in governor set points on the one of them
1. Increases the system frequency.
2. Increases the power supplied by that generator, while reducing the power supplied by the
other one.
When two generators are operating together and the field current of G2 is increased,
1. The system terminal voltage is increased.
2. The reactive power Q supplied by that generator is increased, while the reactive power
supplied by the other generator is decreased.
If the slopes and no-load frequencies of the generator’s speed droop (frequency-power) curves are
known, then the powers supplied by each generator and the resulting system frequency can be
determined quantitatively.
FOR TWO GENERATORS OPERATING TOGETHER:
1. The system is constrained in that the real power supplied by the two generators together
must equal the amount consumed by the load. Neither fsys nor VT is constrained to be
constant.
2. To adjust the real power sharing between generators without changing fsys, simultaneously
increase the governor set points on one generator while decreasing the governor set points
on the other. The machine whose governor set point was increased will assume more of the
load.
3. To adjust fsys without changing the real power sharing, simultaneously increase or decrease
both generators’ governor set points.
4. To adjust the reactive power sharing between generators without changing V T,
simultaneously increase the field current on the other. The machine whose field current was
increased will assume more of the reactive load.
5. To adjust VT without changing the reactive power sharing, simultaneously increase or
decrease both generators’ field currents.
ILLUSTRATIVE PROBLEMS:
1. There are two alternators, 100 kW, 3-ф in parallel are driven by shunt motors whose speedload characteristics are as follows: Alternator A: NNL= 600 rpm and NFL=530 rpm; while
alternator B: NNL= 590 rpm and NFL=550 rpm. What is the greatest load that can be delivered
without overloading either alternator? (PT=171.428kW)
Solution:
Alternator A: Pfla = 100 kW: NNLa= 600 rpm, NFLa=530 rpm
Alternator B: Pflb = 100 kW: NNLb= 590 rpm, NFLb=550 rpm
When plotting the speed-power curves of the 2 alternators back-to-back, the alternator with the lower
slope (droop) will attain 100 kw first. Alternator B has the lower droop. Thus assuming alternator B is
operating at 100 kW, Nb = Nflb = 550 rpm = Na.
By ratio and proportion on the speed-power curve of alternator A,
Pfla/(Nnla – Nfla) = Pa/(Nnla – Na)
Pa = [(600-550)/(600-530)]x100 = 71.43 kW
Total load = Pa + Pb = 71.43 + 100 = 171.43 kW
2. Two alternators are operating in parallel, supplying a load drawing 1000 kVA at 0.80pf
lagging. If alternator A contributes 500 kVA at 0.60pf lagging, determine the pf of alternator
B. (0.928 lagging)
Solution:
Let: Sa, Sb = the complex power supplied by alternator A and B
Sl = complex power received by load
Sa + Sb = Sl
500 <-arccos(0.6) + Sb = 1000 <-arccos(0.8)
Sb = 538.52 <-21.8 kVA
Pfb = cos(-21.8) = 0.9285 lagging
3. Alternator A (100kVA, 3-ф, 240 v, 60hz, 1800 rpm) is operating in parallel with alternator B
(125 kVA, 3-ф, 240 v, 60hz, 1800 rpm). The load of alternator A is 60kW at 90% pf lagging
and the load of alternator B is 80kW at 70% pf lagging. Determine the pf of the load. (0.784
lagging)
Solution:
Let: Sa, Sb = the complex power supplied by alternator A and B
Sl = complex power received by load
Sa + Sb = Sl
(60/0.9) <-arccos(0.9) + (80/0.7) <-arccos(0.7) = Sl
Sl = 178.46 <-38.33
Pfl = cos(-38.33) = 0.784 lagging
4. Two alternators are connected in parallel. The total load is 4250 kW at 0.85 pf lagging.
Alternator A operates with a load of 2125kW at 0.707 pf lagging. Determine the kVA load of
alternator B. (2235kVA)
Solution:
Let: Sa, Sb = the complex power supplied by alternator A and B
Sl = complex power received by load
Sa + Sb = Sl
2125 <-arccos(0.707) + Sb = 4250 <-arccos(0.85)
Sb = 2235 kVA
5. Two 3- φ 4160V, 60Hz alternators are operated in parallel. The total load of the system as
1050 W at 75% lagging pf. If alternator A is carrying 700kW at 80% pf lagging, determine
kVAR of alternator B. (401kVAR)
Solution similar to problem 3 & 4
6. Two Y-connected alternators A and B running in parallel supply the following loads at 3.3kV:
Load 1: 800 kW at unity pf
Load 2: 600 kW at 0.8 pf lagging
Load 3: 400 kW at 0.707 pf lagging
If alternator A is adjusted to carry an armature current of 150A at 0.85pf lagging, what is the
armature current of alternator B?
Solution:
Total load = 800 + (600/0.8) <-arccos(0.8) + (400/0.707) <-arccos(0.707) = 1990.65 <-25.28
Sa = sqrt(3) (3.3)(150) <-arccos(0.85) = 857.365 <-31.79 kva
Sa + Sb = Sl
Sb = 1990.65 <-25.28 - 857.365 <-31.79 = 1142.95 <-20.4 kva
1142.95 = sqrt(3) x 3.3 x Ib (apparent power formula for 3-phase)
Ib = 200 A
7. Two generators connected in parallel is supplying a load. Generator 1 has a no-load
frequency of 61.5Hz and a slope sP1 of 1MW/Hz. Generator 2 has a no-load frequency of
61Hz and a slope sP2 of 1MW/Hz. The two generators are supplying a real load totaling
2.5MW at 0.8 pf lagging.
(a) At what frequency is this system operating, and how much power is supplied by
each of the two generators? (60Hz , 1.5MW and 1MW)
(b) Suppose an additional 1-MW load were attached to this power system. What would
the new system frequency be,, and how much power would G1 and G2 supply now.
(59.5Hz, 2MW and 1.5MW)
(c) With the system in configuration described in part b, what will the system frequency
and generator powers be if the governor set points on G2 are increased by 0.5Hz?
(59.75Hz, 1.75MW)
Solution:
a)
Let Fsys = operating frequency of the 2 generators
F1, P1 = frequency and power of generator 1
F2, P2 = frequency and power of generator 2
The frequency-power curve of the 2 generators can be drawn back-to-back.
F1 = (-1/Sp1)P1 + F1nl
F1 = (-1/1)P1 + 61.5 = -P1 + 61.5 (where P1 is in MW)
F2 = (-1/Sp2)P2 + 61 = -P2 + 61 (where p2 is in MW)
At the operating frequency F1 = F2 = fsys
-P1 + 61.5 = -P2 + 61
P1 – P2 = 0.5 MW (eq. 1)
But;
P1 + P2 = 2.5 MW (eq. 2)
Adding the 2 equations:
2 P1 = 3
P1 = 1.5 MW
P2 = 2.5 – 1.5 = 1 MW
Fsys = -P1 + 61.5 = -1.5 + 61.5 = 60 Hz.
b) For additional 1 MW, F1 = F2 = Fsys
P1 – P2 = 0.5 MW (eq. 1)
P1 + P2 = 3.5 MW (eq. 2)
2 P1 = 4
P1 = 2 MW
P2 = 3.5 – 2 = 1.5
Fsys = -P1 + 61.5 = -2 + 61.5 = 59.5 Hz.
c) If the governor setpoint of G2 is changed such that the no load frequency increase by 0.5 Hz,
F2 = -P2 + 61.5
F1 = -P1 + 61.5 (still the same equation)
F2 = F1 = Fsys
-P2 + 61.5 = -P1 + 61.5
P1 = P2 = 3.5/2 = 1.75 W
Fsys = F1 = -1.75 + 61.5 = 59.75 Hz.
GENERAL INSTRUCTIONS:
a) Answer the following problems in a letter size bond paper.
b) Solve neatly. Show all the details of your solution, including circuit diagram(s) if necessary.
Box your final answer(s).
c) There will be corresponding deductions from your accumulated scores for not following the
general instructions and submitting on time.
1. Two, 1,250kVA, 3-phase generators operating in parallel supply the following 3-phase loads: 600kW,
0.866pf lagging; 400kW, unity pf; 500kW, 0.9pf lagging. One alternator has a load of 500kW at 0.887
pf lagging, Determine the pf of the other alternator.
2. Two, 3- phase generators operate in parallel supplying a load of 2,400kW at 13.8kV and 0.92 pf
lagging. Alternator 1 has line current of 80A and pf of 0.8 lagging. Determine the line current of
alternator 2 in amperes and its power factor.
3. Two, 100kw alternators are driven by shunt motors having speed-load droop characteristics of 3%
and 4% respectively. The alternators are in parallel and each carrying 50kW. An additional 75kW
load is switched on. What are the resulting loads of the alternators assuming that the speed-load
control of each is not adjusted?
4. Two, identical, 2,000kW, 3-phase alternators operate in parallel. The governor of the first machine is
such that the frequency drops uniformly from 61.2Hz at no-load to 60Hz at full load. The
corresponding uniform speed drop of the second machine is 61.2Hz at no-load to 59.2Hz at full load.
What is the maximum kW loading that can be supplied by the combination without overloading
either machine?
5. Two 3-phase alternators connected in parallel are driven by hydraulic turbines whose speed-load
characteristics are as follows: The speed of the first falls uniformly from 624rpm at no-load to
600rpm at full load of 1,000kW. The speed of the second falls uniformly from 630 rpm at no-load to
600rpm at full load of 1,000kW. If the total load is 1,00kW at 0.8pf lagging, determine the kW load
of each alternator.
6. Three physically identical synchronous generators are operating in parallel. They are all rated for a
full load of 3MW at 0.8pf lagging. The no-load frequency of Generator A is 61Hz and its speed droop
is 3.4%. The no-load frequency of Generator B is 61.5Hz and its speed droop is 3%. The no-load
frequency of Generator C is 60.5Hz, and its speed droop is 2.6%. If a total load consisting of 7Mw is
being supplied by this power system, what will the system frequency be and how will the power be
shared among the three generators?
References:
Electric Machinery Fundamentals, 3rd Edition
By: Stephen J. Chapman
Electrical Engineering Reviewer Series Volume 1
By: Engr. Eldorado L. Corpuz