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SCR = O f ' O f ' ' size 12{ ital "SCR"= { {O { {f}} sup { ' }} over {O { {f}} sup { '' }} } } {} (5.37)

SCR = AFNL AFSC size 12{ ital "SCR"= { { ital "AFNL"} over { ital "AFSC"} } } {} (5.38)

Figure 5.13 Typical form of short-circuit load loss and stray load-loss curves.

R T R T = 234 . 5 + T 234 . 5 + t size 12{ { {R rSub { size 8{T} } } over {R rSub { size 8{T} } } } = { {"234" "." 5+T} over {"234" "." 5+t} } } {} (5.39)

R a , eff = short circuit load loss ( short circuit armature current ) 2 size 12{R rSub { size 8{a, ital "eff"} } = { {"short" - "circuit load loss"} over { \( "short" - "circuit armature current" \) rSup { size 8{2} } } } } {} (5.40)

§5.5 Steady-State Operating Characteristics

Figure 5.14 Characteristic form of synchronous-generator compounding curves.

Figure 5.15 Capability curves of an 0.85 power factor, 0.80 short-circuit ratio,

hydrogen-cooled turbine generator. Base MVA is rated MVA at 0.5 psig hydrogen.

Apprent power = P 2 + Q 2 = V a I a size 12{"Apprent power"= sqrt {P rSup { size 8{2} } +Q rSup { size 8{2} } } =V rSub { size 8{a} } I rSub { size 8{a} } } {}

Figure 5.16 Construction used for the derivation of a synchronous generator capability curve.

P jQ = V ˆ a + jX s I ˆ a size 12{P - ital "jQ"= { hat {V}} rSub { size 8{a} } + ital "jX" rSub { size 8{s} } { hat {I}} rSub { size 8{a} } } {} (5.41)

E ˆ af = V ˆ a + jX s I ˆ a size 12{ { hat {E}} rSub { size 8{ ital "af"} } = { hat {V}} rSub { size 8{a} } + ital "jX" rSub { size 8{s} } { hat {I}} rSub { size 8{a} } } {} (5.42)

P 2 + ( Q + V a 2 X s ) 2 = ( V a E af X s ) 2 size 12{P rSup { size 8{2} } + \( Q+ { {V rSub { size 8{a} } rSup { size 8{2} } } over {X rSub { size 8{s} } } } \) rSup { size 8{2} } = \( { {V rSub { size 8{a} } E rSub { size 8{ ital "af"} } } over {X rSub { size 8{s} } } } \) rSup { size 8{2} } } {} (5.43)

Figure 5.17 Typical form of synchronous-generator V curves.

§5.6 Effects of Salient Poles; Introduction to Direct-And

Quadrature-Axis Theory

§5.6.1 Flux and MMF Waves

Figure 5.18 Direct-axis air-gap fluxes in a salient-pole synchronous machine.

E 3, a = 2 V 3 cos ( e t + φ 3 ) size 12{E rSub { size 8{3,a} } = sqrt {2} V rSub { size 8{3} } "cos" \( 3ω rSub { size 8{e} } t+φ rSub { size 8{3} } \) } {} (5.44)

E 3, b = 2 V 3 cos ( 3 ( ω e 120 o ) + φ 3 ) = 2 V 3 cos ( e t + φ 3 ) size 12{E rSub { size 8{3,b} } = sqrt {2} V rSub { size 8{3} } "cos" \( 3 \( ω rSub { size 8{e} } - "120" rSup { size 8{o} } \) +φ rSub { size 8{3} } \) = sqrt {2} V rSub { size 8{3} } "cos" \( 3ω rSub { size 8{e} } t+φ rSub { size 8{3} } \) } {} (5.45)

E 3, c = 2 V 3 cos ( 3 ( ω e t 120 o ) + φ 3 ) = 2 V 3 cos ( e t + φ 3 ) size 12{E rSub { size 8{3,c} } = sqrt {2} V rSub { size 8{3} } "cos" \( 3 \( ω rSub { size 8{e} } t - "120" rSup { size 8{o} } \) +φ rSub { size 8{3} } \) = sqrt {2} V rSub { size 8{3} } "cos" \( 3ω rSub { size 8{e} } t+φ rSub { size 8{3} } \) } {} (5.45)

Figure 5.19 Quadrature-axis air-gap fluxes in a salient-pole synchronous machine.

Figure 5.20 Phasor diagram of a salient-pole synchronous generator.

§5.6.2 Phasor Diagrams for Salient-Pole Machines

Figure 5.21 Phasor diagram for a synchronous generator showing

the relationship between the voltages and the currents.

X d = X al + X ϕd size 12{X rSub { size 8{d} } =X rSub { size 8{ ital "al"} } +X rSub { size 8{ϕd} } } {} (5.46)

X q = X al + X ϕq size 12{X rSub { size 8{q} } =X rSub { size 8{ ital "al"} } +X rSub { size 8{ϕq} } } {} (5.47)

Figure 5.22 Relationships between component voltages in a phasor diagram.

o ' a ' oa = b ' a ' ba size 12{ { { { {o}} sup { ' } { {a}} sup { ' }} over { ital "oa"} } = { { { {b}} sup { ' } { {a}} sup { ' }} over { ital "ba"} } } {} (5.48)

o ' a ' = ( b ' a ' ba ) oa = I ˆ q X q I ˆ q I ˆ a = X q I ˆ a size 12{ { {o}} sup { ' } { {a}} sup { ' }= \( { { { {b}} sup { ' } { {a}} sup { ' }} over { ital "ba"} } \) ital "oa"= { { \lline { hat {I}} rSub { size 8{q} } \lline X rSub { size 8{q} } } over { \lline { hat {I}} rSub { size 8{q} } \lline } } \lline { hat {I}} rSub { size 8{a} } \lline =X rSub { size 8{q} } \lline { hat {I}} rSub { size 8{a} } \lline } {} (5.49)

E ˆ af = V ˆ a + R a I ˆ a + jX d I ˆ d + jX q I ˆ q size 12{ { hat {E}} rSub { size 8{ ital "af"} } = { hat {V}} rSub { size 8{a} } +R rSub { size 8{a} } { hat {I}} rSub { size 8{a} } + ital "jX" rSub { size 8{d} } { hat {I}} rSub { size 8{d} } + ital "jX" rSub { size 8{q} } { hat {I}} rSub { size 8{q} } } {} (5.50)

5.7 Power-Angle Characteristics Of Salient-Pole Machines

  • For the purposes of this discussion, it is sufficient to limit our discussion to the simple system shown in the schematic diagram of Fig.5.23a, consisting of a salient pole synchronous machine SM connected to an infinite bus of voltage V ˆ EQ size 12{ { hat {V}} rSub { size 8{ ital "EQ"} } } {} through a series impedance of reactance X EQ size 12{X rSub { size 8{ ital "EQ"} } } {} . Resistance will be neglected because it is usually small. Consider that the synchronous machine is acting as a generator. The phasor diagram is shown by the solid-line phasors in Fig.5.23b. The dashed phasors show the external reactance drop resolved into components due to I ˆ d size 12{ { hat {I}} rSub { size 8{d} } } {} and I ˆ q size 12{ { hat {I}} rSub { size 8{q} } } {} . The effect of the external impedance is merely to add its reactance to the reactances of the machine; the total values of the reactance between the excitation voltage E ˆ af size 12{ { hat {E}} rSub { size 8{ ital "af"} } } {} and the bus voltage V ˆ EQ size 12{ { hat {V}} rSub { size 8{ ital "EQ"} } } {} is therefore

X dT = X d + X EQ size 12{X rSub { size 8{ ital "dT"} } =X rSub { size 8{d} } +X rSub { size 8{ ital "EQ"} } } {} (5.50)

X qT = X q + X EQ size 12{X rSub { size 8{ ital "qT"} } =X rSub { size 8{q} } +X rSub { size 8{ ital "EQ"} } } {} (5.51)

  • If the bus voltage V ˆ EQ size 12{ { hat {V}} rSub { size 8{ ital "EQ"} } } {} is resolved into components its direct-axis component V d = V EQ sin δ size 12{V rSub { size 8{d} } =V rSub { size 8{ ital "EQ"} } "sin"δ} {} and quadrature-axis component V q = V EQ cos δ size 12{V rSub { size 8{q} } =V rSub { size 8{ ital "EQ"} } "cos"δ} {} in phase with I ˆ d size 12{ { hat {I}} rSub { size 8{d} } } {} and I ˆ q size 12{ { hat {I}} rSub { size 8{q} } } {} , respectively, the power P delivered to the bus per phase (or in per unit) is

P = I d V d + I q V q = I d V EQ sin δ + I q V EQ cos δ size 12{P=I rSub { size 8{d} } V rSub { size 8{d} } +I rSub { size 8{q} } V rSub { size 8{q} } =I rSub { size 8{d} } V rSub { size 8{ ital "EQ"} } "sin"δ+I rSub { size 8{q} } V rSub { size 8{ ital "EQ"} } "cos"δ} {} (5.52)

I d = E af V EQ cos δ X dT size 12{I rSub { size 8{d} } = { {E rSub { size 8{ ital "af"} } - V rSub { size 8{ ital "EQ"} } "cos"δ} over {X rSub { size 8{ ital "dT"} } } } } {} (5.53)

I q = V EQ sin δ X qT size 12{I rSub { size 8{q} } = { {V rSub { size 8{ ital "EQ"} } "sin"δ} over {X rSub { size 8{ ital "qT"} } } } } {} (5.54)

P = E af V EQ X dT sin δ + V EQ 2 ( X dT X qT ) 2X dT X qT sin size 12{P= { {E rSub { size 8{ ital "af"} } V rSub { size 8{ ital "EQ"} } } over {X rSub { size 8{ ital "dT"} } } } "sin"δ+ { {V rSub { size 8{ ital "EQ"} } rSup { size 8{2} } \( X rSub { size 8{ ital "dT"} } - X rSub { size 8{ ital "qT"} } \) } over {2X rSub { size 8{ ital "dT"} } X rSub { size 8{ ital "qT"} } } } "sin"2δ} {} (5.55)

Figure 5.23 Salient-pole synchronous machine and series impedance: (a) single-line diagram and (b) phasor diagram.

Figure 5.24 Power-angle characteristic of a salient-pole synchronous machine showing the fundamental component due to field excitation and the second-harmonic component due to reluctance torque.

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Source:  OpenStax, Electrical machines. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10767/1.1
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