(McGraw-Hill) (Instructors Manual) Electric Machinery Fundamentals 4th Edition Episode 1 Part 7 doc

115

(a) This generator is Y-connected, so L A I I = . At rated conditions, the line and phase current in this

generator is

( )

1000 kVA 251 A

3 3 2300 V A L

L

P I I

V == = = at an angle of –36.87°

The phase voltage of this machine is V V φ = = T / 3 1328 V . The internal generated voltage of the machine

is

EV I I A AA SA =+ + φ R jX

EA = ∠ °+ Ω ∠− ° + Ω ∠− ° 1328 0 0.15 251 36.87 A 1.1 251 36.87 A ( )( ) j( )( )

1537 7.4 V EA = ∠°

The input power to this generator is equal to the output power plus losses. The rated output power is

POUT = = ( )( ) 1000 kVA 0.8 800 kW

( )( ) 2 2

CU P IR = = Ω= 3 3 251 A 0.15 28.4 kW A A

PF&W = 24 kW

core P = 18 kW

stray P = (assumed 0)

IN OUT CU F&W core stray PP PP P P = ++ + + = 870.4 kW

OUT

IN

800 kW 100% 100% 91.9%

870.4 kW

P

P η =× = × =

(b) If the generator is loaded to rated kVA with lagging loads, the phase voltage is 1328 0 V Vφ = ∠ ° and

the internal generated voltage is 1537 7.4 V EA = ∠° . Therefore, the phase voltage at no-load would be

Vφ = ∠° 1537 0 V . The voltage regulation would be:

1537 1328 VR 100% 15.7%

1328

− = ×=

(c) If the generator is loaded to rated kVA with leading loads, the phase voltage is 1328 0 V Vφ = ∠ ° and

the internal generated voltage is

EV I I A AA SA =+ + φ R jX

EA = ∠ °+ Ω ∠ ° + Ω ∠ ° 1328 0 0.15 251 36.87 A 1.1 251 36.87 A ( )( ) ( )( ) j

1217 11.5 V EA = ∠°

The voltage regulation would be:

1217 1328 VR 100% 8.4%

1328

− = × =−

(d) If the generator is loaded to rated kVA at unity power factor, the phase voltage is

Vφ =1328 0 V ∠ ° and the internal generated voltage is

EV I I A AA SA =+ + φ R jX

116

EA = ∠ °+ Ω ∠ ° + Ω ∠ ° 1328 0 0.15 251 0 A 1.1 251 0 A ( )( ) ( )( ) j

1393 11.4 V EA = ∠°

The voltage regulation would be:

1393 1328 VR 100% 4.9%

1328

− = ×=

(e) For this problem, we will assume that the terminal voltage is adjusted to 2300 V at no load

conditions, and see what happens to the voltage as load increases at 0.8 lagging, unity, and 0.8 leading

power factors. Note that the maximum current will be 251 A in any case. A phasor diagram representing

the situation at lagging power factor is shown below:

I

A

Vφ

EA

θ

δ

θ

θ

I

A

R

A

jX S

I

A

By the Pythagorean Theorem,

( ) ( ) 2 2 2 cos sin cos sin E V RI XI XI RI A AA S A S A AS =+ + + − φ θ θ θθ

( )2 2 cos sin cos sin V E XI RI RI XI φ =− − − − A S A AS AA S A θθ θ θ

A phasor diagram representing the situation at leading power factor is shown below:

I

A

Vφ

EA

δ θ θ

θ

I

A

R

A

jX S

I

A

By the Pythagorean Theorem,

( ) ( ) 2 2 2 cos sin cos sin E V RI XI XI RI A AA S A S A AS φ =+ − + + θ θ θθ

( )2 2 cos sin cos sin V E XI RI RI XI φ =− + − + A S A AS AA S A θθ θ θ

A phasor diagram representing the situation at unity power factor is shown below:

I

A Vφ

EA

δ

I

A

R

A

jX S

I

A