Increases ω_m,nl and f_e,gen@nl.

This will then in hand increase E_A because E_A=KΦ_magω_m,nl.

(a)f_e,gen@op = f_e,bus and ω_m drops to ω_bus. E_A is fixed and PG increases

(b) V_T,gen=V_T,bus V_{φ is fixed.}

P_G increases because P_G=3V_φE_Asinδ/ Xs

Therefore sinδ increases

[image]

The power P_G must remain constant when changing I_F

because (1) generator's speed cannot be changed when connected to the bus(f_e cannot change)

(2) the power into the generator is constant(P_in=τ_indω_m) τ_ind is controlled by the govenor set point no I_F 

(3) Implies that I_AcosΘ and E_Asinδ do not change when adjusting I_F

the magnetic flux φ_max increases. This in turn increases E_A=Kφ_maxω_m. I_AcosΘ and E_Asinδ

V_φ is constant so the angle of jX_sI_A changes and I_AsinΘ increases(higher positive Q)

[image]

(1) frequency and terminal volatge are controlled by the system to which it is connected.

(2) Govenor set points of the generator controll the real power supplied by the generator to the system

(3) Field current controls the reactive power suppled by the generator 

(1) Adjusting the field resistance R_F( adujusting φ_s)

(2) Adjusting Armature Voltage V_A without changing V_T

(3) Inserting a resistor in series with armature circuit

KNOW:

ω_m=V_A/Kφ_s - R_A/K2φ_s²τ_ind

This is low torque. There is a decrease in Flux and E_A( because E_a=Kφ_sω_m).ω_mwill increase because I_A increases a lot more than φ_s decreases. After ω_m Increases, I_A2 will decrease τ_ind until τ_ind= τ_Load.

ω_m↑ = V_A/Kφ_s↓ - R_A/K²(φ_s²↓) τ_ind

Graoh showing the behavior at High speed           (Slope of -φ_s²)

[image] 

This is high torque. There is a decrease in Flux and E_A( because E_a=Kφ_sω_m)but ω_mwill decrease( slow down) because I_A increases less than φ_s decreases. After ω_m decreases, I_A2 will increase τ_ind until τ_ind= τ_Load at a slower speed of ω_m .

ω_m↓ = V_A/Kφ_s↓ - R_A/K²(φ_s²↓) τ_ind

This is the final plot, the top is at high speed , 3/4 the way is middle and about the bottom is low

[image]

V_T is still constant.

I_A=(V_A-E_A)/R_A

When I_A increases V_A increases and when I_A decreases E_A increases until T_ind = T_load

ω_m↑ = V_A↑/Kφ_s - R_A/K²(φ_s²) τ_ind

[image]

ω_m↓ = V_A/Kφ_s - R_A↑/K²(φ_s²) τ_ind

At Same T_load = Kφ_sI_A, more power wasted

ω_m@Tind=0 = fixed

[image]

I_base=S_base/V_base  Z_base=(V_base)²/S_base

V_base2=V_base1/a

I_base2=I_base1a

Z_base2=Z_base1/(a)²

S_base1=S_base2

V_x,pu=V_x/V_base1

Same with all other terms 

(1) Find I of the system I = V/(Z_Tot)

(2) Find the real power P = real( |I|²Z)

(3) Convert back to actual power P=P_puS_base

(1) Find I I=V/Z

(2) Solve for V_p in V_p=V_s,pu +R_eq,puI_pu + jX_eq,puI

(3)VR = (V_s-V_p)/V_p

η= P_out/P_in

=P_out/(P_out + P_loss)

V_T=Vφ√3 (θ +/- 30^o)

+ for ABC sequence - for ACB sequence

V_φ= E_A- jX_SI_A - R_AI_A

[image]

T=P/ω = 9.55O/n_r where n_r is in RPM

P_conv=3E_AI_Acosγ

<γ = <E_A - <I_A

P_out=3V_φI_Acosθ

Q_out=3V_φI_Asinθ

θ= phase of the load

[image]

P_out=3V_φE_Asinδ /X_s

=P_maxsinδ

δ= γ-Θ ( Phase between E_A and V_φ)

γ = the phase between E_A and I_A

X_s>>>>R_A

Static power limit is when δ=90^o or sin(δ) = 1

Therefore static power limit is when 

P_out=P_Max = 3V_φE_A /X_S

Is equal to KB_RxB_s

Which is equal to T_ind= P_out/ω_m

Where P_out = 3V_φE_Asinδ/ X_S

For Y connection I_A=I_L

For Delta connection I_A =√3I_L

and we also know that S=√3V_TI_L^*

(I^* represents the complex conjugate of I and therefore means it has a -ve phase i.e -30^o)

So if we rearrange for I_L, we get:

I_L=S/√3V_T

(1)Solve for I_A

(2) Find V_φ (For Y-connection V_T/√3. In Delta V_φ=V_T)

(3) Sub all values into the E_A equation:

E_A=V_φ + R_AI_A + jX_SI_A

P_out=S*PF

P_CLA=3I_A²R_A

P_CLF,dc=I_FV_F

P_IN= P_out+P_Losses

(1) Find the RPM usinf n_m=120f_e/P

(2)Use the Equations T_App=P_in/ω_m

n_slip=n_sync-n_m

s(slip)=n_slip/n_sync

P_AG=P_in-P_SCL-P_core = 3I₂²(R₂/s)

Where P_in=3•Re(V_φI₁^*)

P_SCL=3I₁²R₁

P_Core=3E_A2/R_C

P_AG=P_RCL+P_conv

Where Rotor Copper Losses P_RCL=s(3I₂²(R₂/s)) = sP_AG

Converted Power: P_Conv=P_AG-P_RCL = P_AG -sPAG  =3I₂²R₂((1-s)/s)

Output Power: P_out= P_Conv-P_f&W&misc

T_{ind_developed(mech)}= P_conv/ω_m =P_AG/ω_sync = T_{ind_Magnetic(elec)}

P_AG/ω_sync= 3I₂²(R₂/s)/ω_sync

I₂=V_TH/√(R_TH+R₂/s)² + (X_TH+X₂)²

 sub this into the Torque equation to get:

T_ind=(3V_TH²R₂/s)/ω_sync((R_TH+R₂/s)² + (X_TH+X₂)²)

[image]

V_TH≈(X_M/X₁+X_M)V_φ

Z_TH≈ R₁(X_M/X₁+X_M)² +jX₁

[image]