Dublin Institute of Technology, Kevin Street, Dublin 8

Dublin Institute of Technology, Kevin Street, Dublin 8. School of Electrical Engineering Systems

SOLUTION SHEET

Course: DT704/705 Subject Power Session: Sitting: Systems Analysis 2 2009/2010 May Examiners: Page: 1 of 3 Kevin Gaughan

Question Number: 1 ( Please begin new solution/answer on a separate page) Section Marks (c) First convert percentage droop into Hz/puMW for calculation purposes R1=3%=0.03*50Hz = 1.5Hz/puMW R2=6%=0.06*50Hz = 3Hz/puMW

Note that the base MVA is the same on both systems so change of base is not required.

The combined Area Frequency Response [5 marks] βtotal=1/R1+1/R2+D1+D2=1/1.5+1/3+0.015+0.03=1.045 puMW/Hz

The step power change of 100MW = 0.1 pu will cause a steady state frequency change of ∆f=0.1pu/ β=0.1/1.045=0.096Hz (negative for a load increase).

This ∆f will cause a net change in power in Region 2 of ∆f*(1/R2+D2) = 0.096*(1/3+0.03)=0.035 puMW. Since there is no change in demand in region 2 this [5 marks] must mean that the tie line power flow changes by 0.035 puMW.

(d) (i) Operating Reserve refers to spare generating capacity that is available at very short [3 marks] notice to deal with unforeseen demand variations or supply disruptions.

(ii) The main difference relates to the time frame within which a given category of reserve will be available. Primary Operating reserve must be available within 15 seconds. Secondary operating reserve must be available from 15 seconds to 90 seconds. Tertiary operating reserve is available after 90 seconds. (90s to 5 minutes for TOR1, 5 minutes to 20 minutes for TOR2). Primary and secondary operating reserve are sometimes known as "Spinning Reserve" because these response times can only be achieved by generators that are already spinning and synchronised to the [4 marks] grid.

(iii) Reserve as a service refers to contracts between the transmission system operator and generators who agree to be available to provide reserve if it is needed. The reserve providers are paid for being available regardless of whether or not they are called upon. In general the faster the response time promised the higher the payment. Primary Operating reserve will cost more than secondary and so on. [4 marks] Dublin Institute of Technology, Kevin Street, Dublin 8. School of Electrical Engineering Systems

SOLUTION SHEET

Question Number: 3 ( Please begin new solution/answer on a separate page) Section Marks (a) Through block diagram manipulation of the control loop the student should first be able to produce the following single loop negative feedback system (noting that Ggt(s) = 1.0) :

[6 marks]

The closed loop transfer function of the loop may now be readily extracted

Closed looptransfer function with forward gainG and feedback gain H is :

G 2K p  1 GH  1    s(1 p s)  2K p K1 1 s  2K1R  [5 marks] The characteristic equation is got by making the denominator of the loop transfer function equal to 0:  1    s(1 p s)  2K p K1 1 s  0  2K1R  simplifying  K  2  p   p s  1 s  2K p K1  0  R  [4 marks] Dublin Institute of Technology, Kevin Street, Dublin 8. School of Electrical Engineering Systems

SOLUTION SHEET

Question Number: 3 continued ( Please begin new solution/answer on a separate page) Section Marks (b) For stability we must ensure that none of the roots of the characteristic equation occur in the right half plane (positive real parts). A Routh analysis of the coefficients will quickly determine if any roots are in the right half plane

2 a2=the coefficient of s =p=20 a1=the coefficient of s = (1+Kp/R)=(1+100/2.5)=41 a0=the remainder = 2πKpK1=2π*100*0.7=440

Routh Table: Col 1 Col 2 Col 3

Row 2 a2=20 a0=440

Row 1 a1=41 0

Row 0 b1=a1a0-a2.0=18,040

The number of right half plane zeros equals the number of sign changes in column 1 = none. Therefore there are no right half plane zeros and the system is stable. [10 marks]

(c) However if a voltage regulator is fitted it will pick up the small increase in voltage caused by a small increase in δ and reduce E' to compensate. Reducing E' will reduce transmitted electrical power Pe leading to a further destabilising increase in δ.

The relative importance of these two effects depends on the relative values of K1 and K2 which depend on the cos and sine of the operating power angle respectively. [5 marks]

We can conclude that careful tuning will be required to ensure that a voltage regulator does not destabilise the system.