SME
Power Systems · Problem type
Synchronous generator internal EMF
Use this for “A generator delivers 100 A at pf 0.8 lagging and 277 V per phase, with Xₛ = 1 Ω. Find E and δ.”
Diagram
V = 277∠0° V, I = 100∠−36.87° A, E = 346.4∠13.35° V.
jXₛI = 100∠53.13° V, tip to tail: E = V + jXₛI.
Phasors in A are drawn to their own scale.
δ = 13.35° between V and E.
This is an example. Type your own values to start a new problem.
E (V)
∠
δ
°
=
V (V)
∠0°
+
j
Xₛ (Ω)
×
I (A)
∠−
θ
°
pf
Power factor
example
XₛI
Reactance drop
worked out (not typed)
Eᵣₑ
Real part of E
worked out (not typed)
Eᵢₘ
Imaginary part of E
worked out (not typed)
Formulas
pf
=
cos(
θ
)
0.8 = cos(36.8699°)
XₛI
=
Xₛ
×
I
100 = 1 × 100
Eᵣₑ
=
V
+
XₛI
×
sin(
θ
)
337 = 277 + 100 × sin(36.8699°)
Eᵢₘ
=
XₛI
×
cos(
θ
)
80 = 100 × cos(36.8699°)
E
=
√
Eᵣₑ
2
+
Eᵢₘ
2
346.3654
=
√
337
2
+
80
2
δ
=
atan2(Eᵢₘ,
Eᵣₑ)
13.3542 = atan2(80, 337)
Assumptions
•
Per phase, with V the reference at 0°; j is the imaginary unit (j² = −1).
•
The load is lagging: the current trails V by θ = cos⁻¹(pf), so it is I∠−θ.
•
Armature resistance is left out, so E = V + jXsI, and δ is how far E leads V.
Step-by-step
WE KNOW
V = 277 V, I = 100 A, pf = 0.8, Xₛ = 1 Ω
FIND
power-factor angle (θ), reactance drop (XₛI), real part of E (Eᵣₑ), imaginary part of E (Eᵢₘ), internal EMF (E), power angle (δ)
Step 1 · Find power-factor angle (θ)
Use pf = cos(θ)
Undo the cosine; the current lags by this angle.
θ
=
cos⁻¹(pf)
θ
=
cos⁻¹(0.8)
θ
=
36.8699°
Step 2 · Find reactance drop (XₛI)
Use XₛI = Xₛ × I
The size of the drop across the synchronous reactance.
XₛI = 1 × 100
XₛI = 100 V
Step 3 · Find real part of E (Eᵣₑ)
Use Eᵣₑ = V + XₛI × sin(θ)
jXsI
points
90°
ahead
of
I
,
at
90°
−
θ
,
so
its
real
part
is
XsI
sin
θ
.
Eᵣₑ = 277 + 100 × sin(36.8699°)
Eᵣₑ = 277 + 100 × 0.6
Eᵣₑ = 277 + 60
Eᵣₑ = 337 V
Step 4 · Find imaginary part of E (Eᵢₘ)
Use Eᵢₘ = XₛI × cos(θ)
Its
imaginary
part
is
XsI
cos
θ
;
V
adds
nothing
here.
Eᵢₘ = 100 × cos(36.8699°)
Eᵢₘ = 100 × 0.8
Eᵢₘ = 80 V
Step 5 · Find internal EMF (E)
Use E = √(Eᵣₑ² + Eᵢₘ²)
The
two
parts
are
at
right
angles:
the
size
of
E
is
the
hypotenuse.
E
=
√
337
2
+
80
2
E
=
√
113,569 + 6,400
E
=
√
119
,969
E
=
346.3654
V
Step 6 · Find power angle (δ)
Use δ = atan2(Eᵢₘ, Eᵣₑ)
The
angle
of
E
above
the
real
axis,
where
V
lies.
δ
=
atan2(80,
337)
(337,
80)
is
in
quadrant
I
:
tan⁻¹
gives
the
angle
as
it
is
δ
=
tan⁻¹(
80
337
)
=
13.3542°
δ
=
13.3542°
Check
Put the numbers back in. Both sides should match.
0.8 = cos(36.8699°) ✓
100 = 1 × 100 ✓
337 = 277 + 100 × sin(36.8699°) ✓
80 = 100 × cos(36.8699°) ✓
346.3654
=
√
337
2
+
80
2
✓
13.3542 = atan2(80, 337) ✓
Topic check
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