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Example 1. A 208 V 45 kVA 0.8 PF leading, Δ connected 60 Hz synchronous machine has a synchronous reactance of 2.5Ω and negligible armature resistance. Its friction and windage losses are 1.5 kW, and its core losses are 1.0 kW. Initially, the shaft is supplying a 15-hp load and the motor’s power factor is 0.80 leading.
(a) Find the values of , ,A L AI I E (15 )(0.746 / ) 11.19outP hp kW Hp kW= =
11.19 1.5 1.0 0 13.69in out mechloss coreloss elec lossP P P P P
kW kW kW kW kW
= + + +
= + + + = Since the power factor is 0.8 leading,
/( 3 cos ) 13.69 /( 3 208 0.8) 47.5L in TI P V kW Aθ= = ⋅ ⋅ = The phase current is then
(b) Assume that the load is increased to 30 hp. Sketch the
phasor diagram and find , ,A L AI I E , and power factor.
(30 )(0.746 / ) 1.5 1.0 0 24.88in out mechloss coreloss elec lossP P P P P
hp kW hp kW kW kW kW
= + + +
= + + + =After the load changes, the shaft slows down momentarily.
4-15/34
This makes AE lagging φV more but its magnitude remains constant.
1
1
3sin
sin3
2.5 24.88sin 233 208 255
255 23
A
S
S
A
A
V EP
XPXV E
kW
V
φ
φ
δ
δ −
−
=
=
⋅= = °
⋅ ⋅= ∠ − °E
( ) / (208 0 255 23 ) / 2.5 41.2 15
3 41.2 (15 30) 71.4 15cos15 0.966
A A S
L
jX j A
APF leading
φ= − = ∠ ° − ∠− ° = ∠ °
= ⋅ ∠ − ° = ∠ − °= ° =
I V E
I
4-16/34
Example 2. A 208 V 45 kVA 0.8 PF leading, Δ connected 60 Hz synchronous machine has a synchronous reactance of 2.5Ω and negligible armature resistance. Its friction and windage losses are 1.5 kW, and its core losses are 1.0 kW.
(a) If the motor is supplying a 15-hp load at 0.85 PF lagging and the field current is 4.0 A. Find the values of ,A AI E
From the previous example, the input power is 13.69 kW.
1
13.69 25.83 cos 3 208 0.85
25.8 cos 0.85 25.8 31.8
inA
A
P kWI AV
Aφ θ
−
= = =⋅ ⋅
= ∠ − = ∠− °I
and 208 0 ( 2.5)(25.8 31.8 ) 182 17.5A S AjX j Vφ= − = ∠ ° − ∠ − ° = ∠ − °E V I
4-17/34
(b) If the flux is increased by 25%, what are ,A AI E and the power factor of the motor now?
( ) ( )'
1 1
1.25 1.25 182 227.5 ; sin sin
sin sin / sin 182sin( 17.5 ) / 227.5 13.9227.5 13.9
A A A A
A A
A
E E V E E
E EV
δ δ
δ δ− −
′ ′= ⋅ = ⋅ = =
′ ′= = − ° = − °
′ = ∠ − °E
( ) / (208 0 227.5 13.9 ) / 2.5 22.5 13.2
cos(13.2 ) 0.974A A SjX j A
PF leadingφ′ ′= − = ∠ ° − ∠ − ° = ∠ °
= ° =
I V E
4-18/34
Example 3. The infinite bus operates at 480 V. Load#1 is an induction motor consuming 100kW at 0.78 PF lagging, and load#2 is an induction motor consuming 200kW at 0.8 PF lagging. Load#3 is a synchronous motor whose real power consumption is 150kW.
4-19/34
(a) If the synchronous motor is adjusted to operate at 0.85
PF lagging, what is the transmission line current then?
11 1
12 2
13 3
tan (100 ) tan(cos 0.78) 80.2
tan (200 ) tan(cos 0.80) 150
tan (150 ) tan(cos 0.85) 93
Q P kW kVar
Q P kW kVar
Q P kW kVar
θ
θ
θ
−
−
−
= = =
= = =
= = =
1 2 3
1 2 3
1
1
100 200 150 45080.2 150 93 323.2
cos cos(tan / )
cos(tan 323.2 / 450 ) cos35.7 0.812450 667
3 cos 3 480 0.812
total
total
total total
totalL
L
P P P P kWQ Q Q Q kVar
PF Q P
kVar kW laggingP kWI AV
θ
θ
−
−
= + + = + + =
= + + = + + =
= =
= = ° =
= = =⋅ ⋅
4-20/34
(b) If the synchronous motor is adjusted to operate at 0.85
PF leading what is the transmission line current then?