Worked Problems on Synchronous Machines
EEEN20090 – Electrical Energy Systems
Problem 1
A 3-phase synchronous generator has the following synchronous reactance Xs =
5 Ω and Ra = 0. It is connected to a grid with V = 6, 600 V. The e.m.f. of
the machine is E = 6, 000 V. Determine the maximum active power that the
machine can deliver to the grid.
Problem 2
A 3-phase synchronous machine has the following nominal data:
SN = 5, 000 kVA
VN = 6, 600 V
The e.m.f. is a function of the excitation current Ie , as follows:
E=
7, 400Ie
,
85 + Ie
(1)
where E is expressed in V and Ie in A. E in (1) is the phase-to-neutral voltage.
Moreover, the machine has: Ra = 0.2 Ω and Xd = 1 Ω. The mmf of the
armature reaction is equivalent to a current Ic = 20 A.
Determine:
a. Variation of the current to obtain the nominal voltage of 6, 600 V from
open-circuit to full load with cos ϕ = 0.6 lagging.
b. Efficiency of the machine at full load with cos ϕ = 0.6 lagging.
Note: Iron losses and mechanical losses are 100 kW and Ve = 200 V is the
exitation dc voltage.
1
Problem 3
A salient-pole generator has the following data:
SN = 1500 kVA
VN = 3000 V
fN = 50 Hz
Xd = 2 Ω
Xq = 1.5 Ω
Determine:
a. The emf at full load with cos ϕ = 0.8 lagging.
b. Maximum active power that the machine can deliver to the grid.
2
Solution of Problem 1
The phase-to-neutral voltage and emf are:
6600
V = √ = 3810.5 V
3
6000
E = √ = 3464.1 V
3
The active power provided by the generator is:
Pe =
This power is maximized for δ =
π
2,
3EV
sin δ
Xs
hence:
√
√
3EV
3 · 6, 000/ 3 · 6, 600/ 3
=
Xs
Xs
6, 000 · 6, 600
= 7.92 MW
=
5
Pemax =
One can solve the same problem by imposing:
Ē = V̄ + jXs I¯
⇒ 3, 464.1∠90◦ = 3, 810.5∠0◦ + j5I∠ϕ
⇒ I sin ϕ = 762.1 A and I cos ϕ = 692.8 A
I sin ϕ
⇒ tan ϕ =
= 1.1
I cos ϕ
⇒ ϕ = 47.74◦
⇒ cos ϕ = 0.673 (leading)
⇒ I · 0.673 = 692.8 A
⇒ I = 1, 029.95 A
√
Pmax = 3V I cos ϕ = 7.92 MW
Solution of Problem 2
a. The emf in open-circuit is:
6, 600
E = √ = 3, 810.5 V
3
Then:
E=
7, 400Ie
85 + Ie
⇒ Ie = 90.23 A
At full-load, the nominal current is:
SN
5, 000, 000
IN = √
= 437.39 A
=√
3 VN
3 · 6, 600
3
Then, one has:
Ēr = V̄ + (Ra + jXd )I¯
6, 600
Ēr = √ ∠0◦ + (0.2 + j1)437.39∠ − 53.13◦
3
= 4217.3∠2.62◦ V
Hence the Ir required to produce Er is:
7, 400Ir
85 + Ir
7, 400Ir
⇒ 4, 217.3 =
85 + Ir
⇒ Ir = 112.63 A
Er =
The excitation current is given by:
I¯
I¯e = I¯r −
KP
where
I¯r = Ir (∠Ēr + 90◦ )
= 112.63∠2.62◦ + 90◦
= 112.63∠92.62◦ A
and
I¯
I¯c =
= 20∠ − 53.13◦
Kp
⇒ cos ϕ = 0.6 (lagging)
Hence:
I¯
I¯e = I¯r −
Kp
= 112.63∠92.62◦ − 20∠ − 53.13◦
= 129.64∠97.6◦ A
Finally:
Ie (open-circuit) = 90.23 A
Ie (full-load, cos ϕ = 0.6 lagging) = 129.64 A
⇒
∆I ≈ 40 A
4
b. Let’s compute the efficiency:
Pe = SN cos ϕ = 5000 · 0.6 = 3000 kW
PFe + Pm,losses = 100 kW
Pex = Ve Ie = 200 · 129.6 = 25.92 kW
Pj,a = 3Ra I 2 = 3 · 0.2 · 437.392 = 114.8 kW
Pm = Pe + PFe + Pj,a + Pm,losses
Ptot = Pm + Pex = 3, 240.72 kW
Finally:
η=
Pe
3, 000
100 = 92.6%
=
Ptot
3, 240.72
Solution of Problem 3
a. Blondell Model ⇒ Ē = V̄ + jXd I¯d + jXq I¯q
Let’s use Ē = Ē − j(Xd − Xq )I¯d
Then we can write:
Ē = V̄ + jXq I¯
At full load:
SN
1, 500, 000
I=√
=√
= 288.7 A
3V
3 · 3, 000
3, 000
V̄ = √ ∠0◦ = 1, 732.1∠0◦ V (phase reference)
3
I¯ = 288.7∠ − 36.87◦ A (cos ϕ = 0.8 lagging)
Then
Ē = V̄ + jXq I¯
= 1732.1∠0◦ + j1.5 · 288.7∠ − 35.87◦
= 2022∠9.9◦
⇒ δ = 9.9◦
ψ = δ + ϕ = 9.9◦ + 36.87◦ = 46.77◦
Id = I sin ψ = 288.7 sin 46.77◦ = 198 A
Iq = I cos ψ = 288.7 cos 46.77◦ = 210 A
Then
(Xd − Xq )Id = (2 − 1.5) · 198 = 99 V
E = E + (Xd − Xq )Id = 2, 022 + 99 = 2, 121 V
Ē = 2, 121∠9.9◦ V
√
→ E(phase-to-phase) = 3 · 2, 121 = 3, 674 V
5
b. To determine the maximum active power, we know that:
1
3EV
3V 2
1
Pem =
sin δ +
−
sin 2δ
Xd
2
Xq
Xd
∂Pem
max
|δ =0
(δM ) ⇒
Pem
∂δ M
1
3EV
∂Pem
1
2
=0=
cos δ + 3V
−
cos 2δ
∂δ
Xd
Xq
Xd
Substituting the values of E, V , Xd and Xq :
0 = 5.51 cos δ + 1.5 cos 2δ
Since
cos 2δ = cos2 δ − sin2 δ = 2 cos2 δ − 1
Equation (2) becomes:
3 cos2 δ + 5.51 cos δ − 1.5 = 0
⇒ 2 solutions with known cos δ = y:
3 y 2 + 5.51 y − 1.5 = 0
⇒ δ = 76◦
Hence:
max
Pem
= 35.31 sin(76◦ ) + 0.75 sin(152◦ ) = 5.7 MW
6
(2)