Question

2) Take the following system. It represents the frequency control of a steam-turbine gener- ator unit Assume TRH = 6, FHP = 0.3, Tg = 0.1, and M = 10, D = 1.0 (Both in p.u.) Ape Load ref. Tos+1 11+TRE PS 1+TRHS DUM (Note that AP represents the change in electric load. - Assume that the load APe, increases by 0.01 p.u. What is the effect on the frequency, if the droop R was infinity, i.e., what is the frequency deviation if no governors were present? -Plot the root-locus as k = 1, varies from zero to infinity. -Set R=0.05, Assume that the load changes as a step function AP = 0.01u(t) p.u. What is the effect of the change in the load on the frequency?

Answer 1

5 * Gruen TRA = 6; Fap = 0.3; TG=ool & M= 10; D=1 PE=0.0ipu Δώ. 1 + 6x0.33 1+68 0.13+1 THIOS Black diagram T YRI reduction technique :- Gurin AP-0.01 ult). in s-domain = 0.01% Load 1 +1. 85 (0.15+1)(1+65) Itos YR = 76.05= 20 1 +1.83 1 +10s 20 201

Gls) I (59+1) (1+51:0) 10:0-( 2081 +0) (89+1) (1+510)s (S01+1) 20 2014 HIS) (9,49 (99+1)(10.0+ 8€_9140)=288'17 5 (soitt) (99+1)(1+510)s = 1.852 + Backsboa S – 1 610-39 +1-6x16-) -0.01 -0.068 10.182+3)(1 +108 +68+6052) - 1.882 +8 (1-14162_6x10-3-0.06) -0.01 10.152+)(1+168 +603>) a 1.882 +0.9338 -0.01 0.182 +1.682 +684 +3 +1682+bos3 1.882 +0.9338 -0.01 71 to 654 161.683 + 16.182 +8 it as a When R o, then as the feedback path ů VR ie, Yo =0; Hence the destem becomes as a open-loop system.

When the System is open-loop, then the effect on the frequeny s dretty controlled by their load Value APE . Loud Aww) KOY Ref. - 1.89270.9338-0.01 684 +6.653416.18278 Thouefore the transfer function be, SWC) 1.882 +0.9233 – 0.01 > 684 +61.683 +16.182 + Frequency AW(S) 1.832 +0.9333 -0.01 - XRG) 684+61.683 +16.182+ devrå kon Load reference as input Part -Bi- MATLAB :- >> num = [1.8 0.9333 -0.01] » den - [6 61.6 16.) i o] » Lys = tf(num, den) » rlows (lys) ^ - À

octave: 2 source("almy_scriptasafsadasd.m") num = 1.800000 0.933300 -0.010000 den = 6.00000 61.60000 16.10000 1.00000 0.00000 Transfer function sys' from input 'ul' to output ... 1.8 s^2 + 0.9333 S - 0.01 l : -- --------- --------- ----------- 6 3^4 + 61.6 5^3 + 16.1 5^2 + S Continuous-time model. Root Locus of sys ---- asymptotes locus * open loop poles O zeros Imaginary Axis -10 -8 - 0 -6 Real Axis 4 2 gain = [0, 875.407]

Part co Refer From part-a for the transfer function be > Rai = 1.852 +0.9338-0:0) 684 +61.683+16.18275 664 +6.69236 ***** aone When including the value of R=0.05, .. then the feedback faith having the magnitude be, HIS) = ☆ a dos = 207 - Thus, the deptem become closed loop System do that there is change is load refence definitely will cause the change in frequency for that respective load and also meet to run with respective load reference

Output: 1.8000 0.9333 -0.0100 den = 6.0000 61.5000 16.1000 1.0000 0 sys = 1.8 s^2 + 0.9333 s - 0.01 --------- 6 5^4 + 61.6 5^3 + 16.1 5^2 + 5 Continuous-time transfer function. ans = RiseTime: NaN SettlingTime: NaN Settlingkin: NaN SettlingMax: NaN Overshoot: NaN Undershoot: NaN Peak: Inf PeakTime: Inf Output: Step Response andum 2000 2500 1000 1500 Time (seconds)

MATLAB code:

num=[1.8 0.9333 -0.01]
den=[6 61.6 16.1 1 0]
sys=tf(num,den)
step(sys)
stepinfo(sys)