Homework #7 Solution. Solutions: ΔP L Δω. Fig. 1

Transcription

1 Homework #7 Solution Aignment:. through.6 Bergen & Vittal. M Solution: Modified Equation.6 becaue gen. peed not fed back * M (.0rad / MW ec)(00mw) rad /ec peed ( ) (60) 9.55r. p. m ( 9.55) r. p. m The new peed i therefore:.. (opied directly from cla note) For the iolated generating tation with local load hown in Fig. below, it i oberved that Δ =0. bring about Δω=-0. in the teady-tate. Δ Δ Δω Fig. (a) Find /. Solution: We need the tranfer function between Δω and Δ. To get thi, write down Δω a a function of what i coming into it:

2 0 0 Now olve for Δω. Expanding: Bringing term in Δω to the left-hand-ide: Factoring Δω: Dividing: Multiply through by (+0)(+): 0 ) )( 0 ( ) ( 0 0

3 earrange the top and expand the bottom: 0 0( ) 0 ( 0 / ) Now we conider Δ =0, Δ =0., and aume it i a tep change. Therefore: Subtituting into (*), we get: 0 0( ) ( 0/ ) The above expreion i a alace function (i.e., in ). The problem give data for the teady-tate (in time). We may apply the final-value theorem to the above expreion to obtain: (*) lim ( t) t lim 0 lim 0 0 lim 0 0 0( ) ( 0/ ) 0( ) 0 ( 0/ ) 0/ that i, Solving for, we obtain: 0 0/ 3

4 0 0/ Subtituting Δ =0. and Δω=-0., we obtain: 0 0( 0.) 0 0(0.) ( 0.).5 The problem wa pecified with power in per-unit and Δω in rad/ec. eference to the block diagram indicate that the left-hand-ide umming junction output Δ -Δω/. To make thi um have commenurate unit, it mut be the cae that ha unit of (rad/ec)/pu power. The problem ak for /, which would be /.5=0.4 pu power/(rad/ec). One might alo expre and / in unit of pu frequency/pu power. Thi would be: pu =.5/60=0.047 / pu =4 ecalling the NE pecification that all unit hould have =0.05, then thi hould be adjuted upward. Quetion: What doe an =0.047 mean relative to an =0.05? Anwer: ecalling that pu =-Δω pu /Δ m,pu, we can ay that pu expree the teady-tate frequency deviation, a a percentage of 60 Hz, for which the machine will move by an amount equal to it full rating. So: if pu =0.05, then the teady-tate frequency deviation for which the machine will move by an amount equal to it full rating i 0.05*60=3hz. if pu =0.047, then the teady-tate frequency deviation for which the machine will move by an amount equal to it full rating i 0.047*60=.50hz. (b) Specify Δ to bring Δω back to zero (i.e., back to the teady-tate frequency ω=ω 0 ). 4

5 Solution: ecalling eq. (*): 0 0( ) 0 ( 0 / ) Now we have that and Δ =0. In thi cae, eq. (*) become: 0 0 Applying the final value theorem again: lim ( t) t lim lim 0 0 lim ( t) t lim lim lim 0 0 that i, Solving for Δ, we get: 0 ( 0 / 0 0 / / ( 0/ ) 0 ( 0 / ) 0 / (*) ( 0 / ) 0 ) 0 / 5

6 Having already computed /=0.4 in part (a), and with Δω=-0., we have 0.( 0 /.5) 0. 0 which indicate that for thi load increae of 0. which reult (from primary peed control) in a frequency deviation of -0. rad/ec, we need to adjut the peed-changer motor to increae plant output by 0. pu in order to correct the teady-tate frequency deviation back to 0. The change to the peed-changer motor would be accomplihed by the upplementary control..3 Uing the hint provided in the cla note: Hint on roblem.3: At the bottom of page 390, the text ay: The ~ ~ reader i invited to check that with i / Di and Ti Mi / Di, Figure.0 repreent (.) in block diagram form. In Figure.0 we have Δω i a an output and Δ Mi a an input and can cloe the power control loop by introducing the turbine-governor block diagram hown in Figure.4. Thi will reult in the following block diagram. Δ Δω Δ + - ( T )( G T T ) Δ M + - T Δδ Solution: / ~ i D i /0 00 6

7 From the above figure, uing the block diagram relation, we can derive the following equation: p T utting in known value: p p T ( T p G )( T T ) (.4) 0 ( ) To implify multiply the top and bottom by: [(+0)(+)]/0. To get: ( 5( ) 4 ) 0 0 Now implify the denominator o we can perform partial fraction expanion on it. Thi one i a bit ugly, on homework an eay approach i to et the denominator inide the parenthee equal to zero and olve on your calculator to olve for the root. (.55 5( ) j.33)(.55 j.33) After partial fraction expanion we get: 7

8 A (.55 j.33) * (.55 j.33) Solve for A and uing whatever method you deire (i.e. Heavyide over Up Method) A.439, Finally, to convert to the time-domain we take the aplace tranform which yield: ( t) e.55t co(.33t ).4 Firt we redraw the figure from.3 adding in the tie line. I alo lightly changed where Δω i defined (note: look different, but really i ame a.3) for implification purpoe. educing the portion inide the dahed line firt we get. Note for implicity I m putting number in right away. 8

9 With the dahed area reduced now to a ingle block the problem can now be olved in almot the exact ame fahion a (.4)( ) 000 ( ) To implify multiply the top and bottom by: [( )(+)]/0. To get: 5( ) Factor the bottom into 3 term: 5( ) (.96)(.044 j7.)(.044 j7.) After partial fraction expanion we get: A.96 (.044 j7.) * (.044 j7.) Once again ue your favorite method to olve for A and A. 0036, Finally, to convert to the time-domain we take the aplace tranform which yield: 9

10 ( t).036e.96t.694e.044t co(7.t 90.9) A compared with.3 Amplitude i approximately the ame Frequency i larger Damping i maller.. error = 0.5 Following Example., we have: f f o o Solving for Δ A we get:.05 A Where Δ A i in per unit.05 A.005pu.005* 50MW. 5MW 60*.05 Similarly:.05 B.005pu.005* 400MW. 0MW 60* pu D D f new (60) 59. 9Hz tie D pu.0 tie * MW 0