ECE 422 Homework 6 Power System Controls 1. Problem 12.2 on page 683 of the course textbook. The automatic voltage regulator (AVR) system of a generator is represented by the simpified block diagram shown in Figure 12.20, in which the sensor is modeled by a simple first-order transfer function. The voltage is sensed through a voltage transformer and then rectified through a bridge rectifier. Parameters of the AVR system are given as follows: Element Amplifier Gain KA Time Constant τA=0.1 Exciter KE=1 τE=0.4 Generator KG=1 τG=1.0 Sensor KR=1 τR=0.05 a. Determine the open loop transfer function of the block diagram and the close loop transfer function relating the generator terminal voltage Vt(s) to the reference voltage Vref(s). The open loop gain can be interpreted as either of two things: the forward path gain or the loop gain. The forward path gain is KE KA KG KA  1 1 GF( s ) =   = 1  τA  s 1  τE s 1  τG s ( 1  0.1 s )  ( 1  0.4 s )  ( 1  1.0 s ) The loop gain is KE KA KG KR KA  1 1 GOL( s ) =    = 1  τA  s 1  τE s 1  τG s 1  τR s ( 1  0.1 s )  ( 1  0.4 s )  ( 1  1.0 s )  ( 1  0.05 s )   The closed loop transfer function is K A  1 1 ( 1 0.1 s )  ( 1 0.4 s)  ( 1 1.0 s ) GCL( s) = 1 KA 1  1  1 ( 1  0.1 s)  ( 1  0.4 s)  ( 1  1.0 s)  ( 1  0.05 s) KA ( 1  0.05 s) GCL( s) = [ ( 1  0.1 s)  ( 1  0.4 s)  ( 1  1.0 s)  ( 1  0.05 s) ]  KA GCL( s) = KA ( 1  0.05 s)  0.002  s4  0.067  s3  0.615  s2  1.55 s  K  1.0 A   25 KA s  500  KA GCL( s) = 4 3 2 s  33.5 s  307.5  s  775.0  s  500.0  KA  500.0 b. For the range of KA from 0 to 12.16, comment on the stability of the system. Our characteristic polynomial is 4 3 2 s  33.5 s  307.5  s  775.0  s  500.0  KA  500.0 = 0 KA  0 For our polynomial roots are found as v  1 33.5 307.5 775  For KA  12.15 500 KA  500 T our polynomial roots are found as  1  0.4  polyroots( v )    0.1   0.05     0.059  6.848i  10 3  v  1 33.5 307.5 775  500  KA  500    3   0.059  6.848i  10   polyroots( v )    5  2.775  10  0.208i  our polynomial roots are found as For KA  12.16  2.775  10 5  0.208i     0.059  6.862i  10 3  T v  1 33.5 307.5 775  500  KA  500      3 0.059  6.862i  10   polyroots( v )    5  0.208i 1.069  10   MATLAB commands to get a rootlocus for this function are  1.069  10 5  0.208i    T >> sys=tf([1],[0.002 0.0667 0.615 1.55 1]) sys = 1 ----------------------------------------------0.002 s^4 + 0.0667 s^3 + 0.615 s^2 + 1.55 s + 1 Continuous-time transfer function. >> K=linspace(0,12.16,500); >> rlocus(sys,K) >> Therefore, we find that the sytem is stable for values of KA up to 12.15 and gives roots in the right half plane for KA=12.16 or greater. 6 4 Root Locus Root Locus 6 Imaginary Axis (seconds-1) 4 2 0 -2 -4 -6 -25 -20 -15 -10 -5 0 5 -1 Real Axis (seconds ) Routh - Hurwitz criterion: 4 S 3 s 2 s 1 307.5 33.5 775   1 307.5     33.5 775   284.366  33.5  1 s 0 s  0  1 500  500  KA    0  33.5  = 500  500  K A 33.5 775  33.5   284.366 500  500  K  A  = 716.10  58.90  KA 284.366 284.366 500  500  KA     716.10  58.90  KA  0 716.10  58.90  KA 500  500  KA = 500  500  KA   33.5 0     284.366 0   0 284.366 0 From the s 1 row, 716.10  58.90  KA = 0 KA = 12.16 From the s 0 row, 500  500  KA = 0 KA = 1 Therefore, to avoid sign changes in the array, 1  KA  12.16 c. For KA=10, evaluate the steady state step response and steady state error. 500  KA  10 4 GCL( s) = 4 3 2 s  33.5 s  307.5  s  775.0  s  500.0  KA  500.0 Steady state step response is GCLss( s) =  s G ( s)  1   CL  s s0 lim Recall 25 KA s  500  KA GCL( s) = 4 3 2 s  33.5 s  307.5  s  775.0  s  500.0  KA  500.0 KA  10 Evaluate 25 KA s  500  KA    1 s    s  0   s4  33.5 s3  307.5  s2  775.0  s  500.0  K  500.0  s  A     lim Steady state output (for a unit step input) is 500.0  KA 500.0  KA  500.0 KA  10 500.0  KA GSS KA  500.0  KA  500.0     GSS KA  0.909 Steady state error is the difference between the steady state input and the steady state feedback, Xin  1.0   Xfb  GSS KA  Xin  0.909 ESS  Xin  Xfb  0.091 Problem 2. 12.5 on page 684 of the course text book. An area of an interconnected 60-Hz power system has three turbine generator units rated 200, 300 and 500 MVA. The regulation constants of the units are 0.03, 0.04 and 0.06, respectively, based on their ratings. Each unit is initially operating at one half of its own rating when the load suddenly decreases by 150MW. a. Determine the unit area frequency response characteristic β on a 100MVA base b. Determine the steady state increase in area frequency c. Determine the MW decrease in the mechanical power output of each turbine. Assume that the reference power setting of each turbine governor remains constant. Neglect losses and the dependence of load on frequency. a. Determine the unit area frequency response characteristic β on a 100MVA base 6 MVA  10  V A Sbase  100  MVA fbase  60 Hz S1  200  MVA S2  300  MVA S3  500  MVA R1x  0.03 R2x  0.04 R3x  0.06 Sbase R1  R1x S1 R1  0.015 Sbase R2  R2x S2 R constants are inversely related to MVA. R2  0.013 Sbase R3  R3x S3 R3  0.012 Add the inverse R constants to get β. β  1 R1  1 R2  1 β  225 R3 b. Determine the steady state increase in area frequency Δpm  150  MW Δpm  1.5 Sbase 1 Δf    Δpm β 3 Δf  6.667  10 ΔfHz  Δf  fbase ΔfHz  0.4 Hz c. Determine the MW decrease in the mechanical power output of each turbine. Let ∆pref=0. Δpm = Δpref  β Δf Calculate each change in power output Δpm1   Δpm2   Δpm3   1 R1 1 R2 1 R3  Δf Δpm1  0.444 Δpm1 Sbase  44.444 MW  Δf Δpm2  0.5 Δpm2 Sbase  50 MW  Δf Δpm3  0.556 Δpm3 Sbase  55.556 MW Problem 3. 12.10 on page 685 of the text book. Open PowerWorld simulator case Problem 12.10. The case models the system from Example 12.5 except the following: 1) the load increase is a 50% rise at bus 6 for a total increase of 250MW (from 500MW to 750MW) 2) the value of R for generator 1 is changed from 0.05 to 0.04 per unit. Repeat Example 12.4 for these modified values. Problem 3. 12.11 on page 685 of the text book. Open Power World simulator case problem 12.11, which includes a transient stability representation of the system from example 6.13. Each generator is modeled using a two-axis model, an IEEE Type 1 exciter and a TGOV1 governor with R=0.05 per unit. A summary of the generator models is available by selecting either Stability Case Info, Transient Stability Generator Summary, which includes the generator MVA base, or Stability Case Info, Transient Stability Case Summary. The contingency is the loss of tthe generator at Bus 50, which intially has 42.1MW of generation. Analytically determine the steady state frequency error in Hz following this contingency. Use PowerWorld Simulator to confirm this result. Also determine the magnitude and time of the largest bus frequency deviation. See next page for answers to these two problems. Problem 4. 12.14 and 12.16 on page 685 of the text book. 12.14 A 60 Hz power sytem consists of two interconnected areas. Area 1 has 1200 MW of generation and an area frequency response characteristic β1 = 600MW/Hz. Area 2 has 1800 MW of generation and β2 =800MW/Hz. Each area is initially operating at one half of its total generation, at ∆ptie1 =∆ptie2 =0 and at 60 Hz when the load in Area 1 suddenly increases by 400MW. Determine the steady state frequency error and the steady state tie line error ∆ptie of each area. Assume that the reference power settings of all turbine governors are fixed. That is, LFC is not employed in any area. Neglect losses and the dependence of load on frequency. 12.16 Repeat Problem 12.14 if LFC is employed in both areas. The frequency bias coefficients are Bf1 =β1 =600MW/Hz and Bf2 =β2 =800MW/Hz. Δpm1  400  MW MW β1  600  Hz Δpm2  0  MW MW β2  800  Hz The overall frequency - power droop relation is     Δpm1  Δpm2 = Δpref1  Δpref2  β1  β2  Δf Solve this for the frequency Δf   Δpm1  Δpm2 β1  β2 Δf  0.286  Hz Δptie2  β2  Δf Δptie2  228.571  MW Δptie1  Δptie2 Δptie1  228.571  MW Checking,  Δptie1  Δpm1  β1  Δf  Δptie1  228.571  MW Problem 12.16. Area Control restricts the net tie line changes to zero. Area 1 must pick up its own generation increase without any change in system frequency in steady state. Problem 5. 12.19, 12.20, and 12.21 on page 686 of the text book. 12.16 The fuel cost curves for two generators are given as follows:   2   2 C1 P1  600  15 P1  0.05 P1 C2 P2  700  20 P2  0.04 P2 Assuming the system is lossless, calculate the optimal dispatch values of P1 and P2 for a total load of 1000MW, the incremental operating cost, and the total operating cost. PT  1000 The incremental cost functions are   IC1 P1  15  0.10 P1   IC2 P2  20  0.08 P2 IC1 = d C1 P1 dP1 IC2 = d C2 P2 dP2     Set the incremental costs equal and solve for the power values. IC1 = IC2 15  0.10 P1 = 20  0.08 P2 0.10 P1  0.08 P2 = 5 P1  P2 = 1000  P1   0.10 0.08   1  5       P2   1 1   1000   P1   472.222    P2   527.778  P units are MW. λ values are the incremental costs. Solve for both as a check.   λ  IC1 P1   λ  62.222 λ units are dollars/MWh. λ  IC2 P2  62.222 Total cost is the sum of the cost functions.     C1 P1  600  15 P1  0.05 P1 2 C1 P1  18833.02   Ctotal P1 P2   C1  P1   C2  P2  2 C2 P2  22397.531 C2 P2  700  20 P2  0.04 P2 C units are dollars/hour.     Ctotal P1 P2  41230.56 12.20 Rework Problem 12.19 except assume that the limit ouptut are subject to the following inequality constraints: 200 < P1 < 800 MW 100 < P2 <500 MW P2 is over this limit. Set P2 to its limit and solve the problem. P1 is the difference between Ptotal and P2. P2  500 PT  1000 P1  PT  P2 P1  500 The units on P are MW. System λ is the λ of the unlimited unit, P1,   λ  65 The λ units are dollars/MWh. λ  IC1 P1 Total cost is the sum of the cost functions.   C1 P1  600  15 P1  0.05 P1 2   Ctotal P1 P2   C1  P1   C2  P2  2 C2 P2  700  20 P2  0.04 P2   C1 P1  20600 C units are dollars/hour.   C2 P2  20700   Ctotal P1 P2  41300 12.21 Rework problem 12.19 except assume the 1000MW value also includes losses and that the penalty factor on the first unit is 1.0 and the second 0.95. L1  1.0 L2  0.95 Include the losses in the 1000MW figure. PT = P1  P2  PL PT  PL = P1  P2 = 1000 λ is the product of incremental cost and penalty factors. λ = L1  IC1 = L2  IC2 Calculate the incremental costs and multiply by the penalty factors.   IC1 P1  15  0.10 P1   IC2 P2  20  0.08 P2 IC1 = d C1 P1 dP1 IC2 = d C2 P2 dP2 L1  IC1 = L2  IC2        1.0 15  0.01 P1 = 0.95 20  0.08 P2  Write the equations from this information. 0.95 0.08  0.076 0.01 P1  0.95 0.08 P2 = 0.95 20  1.0 15 0.95 20  1.0 15  4 0.01 P1  0.076  P2 = 4 P1  P2 = 1000 Solve for the power values.  P1    P2  1  0.1 0.076    4      1  1  1000   P1   454.545    P2   545.455  P units are MW. Solve for the λ value; check with the other generator.   λ  L1  IC1 P1     λ  L2  IC2 P2 IC1 P1  15  0.10 P1 λ  60.455 λ units are dollars/MWh.   IC2 P2  20  0.08 P2 λ  60.455 Total cost is the sum of the cost functions.   C1 P1  600  15 P1  0.05 P1 2 2   Ctotal P1 P2   C1  P1   C2  P2  C2 P2  700  20 P2  0.04 P2   C1 P1  17748.76 C units are dollars/hour.   C2 P2  23509.917   Ctotal P1 P2  41258.68