. Introducton AGC 3 The prmary controller response to a load/generaton mbalance results n generaton adjustment so as to mantan load/generaton balance. However, due to droop, t also results n a non-zero steady-state frequency devaton. Ths frequency devaton must be corrected. Also, the net scheduled export must be mantaned accordng to the rchase agreements. Prmary control does nothng to correct the steady-state frequency error or the net scheduled export. These two problems are handled by provdng a supplementary sgnal from the control center to each generaton unt on automatc generaton control (AGC).
The sgnal s derved from the Area Control Error (ACE) and, when receved at a generaton plant, actvates the speed changer motor to adjust the energy supply set pont to the generator. In these notes, we wll learn how the ACE s comted, and we wll see how t s used n correctng steady-state frequency error. 2. evew From our notes AGC, we defned the followng terms: Net Actual Interchange: AP j Net cheduled Interchange: P j Interchange Devaton (defned pg 396 of text): ΔP j =AP j -P j () Note that the above terms are each defned wth respect to two dstnct areas. 2
We also defned, n AGC, the followng terms: Actual Export: cheduled Export: n AP AP j (2) j n P P j (3) n P Net Devaton: P j (4) j Note that the above three terms are each defned wth respect to a sngle area. We also saw that the net devaton s related to the net actual and scheduled nterchanges, and to the actual and scheduled exports, by: P n j n j AP j P j n j n j P j ( AP j AP j P ) j P (5) Although your text does not defne net devaton, t does use t as the frst term n the ACE expresson of eq. (.3). 3
Fnally, we saw n our notes AGC2, when dscussng the mult-machne case, that P M f 6 f 6 P... K K (6) P... K K (7) If all unts have the same per-unt droop constant,.e., f = 2 = = K, then eqs. (6) and (7) become P M f 6 f P... K P... K (8) 6 (9) In eqs. (8) and (9), ΔP represents the change n total load so that t s postve f load ncreases, negatve f load decreases; postve f gen decreases, negatve f gen ncreases. 4
3. Area Control Error The Area Control Error (ACE) s composed of Net Devaton ΔP, from eq. (5), wrtten more compactly below: P AP P (5) When ΔP >, t means that the actual export exceeds the scheduled export, and so the generaton n area should be reduced. teady-state frequency devaton f f 6 () When Δf>, t means the generaton n the system exceeds the load and therefore we should reduce generaton n the area. From the above, our frst mlse may be to mmedately wrte down the ACE for area as: ACE P f (a) Alternatvely (and consstent wth the text) ACE P (b) But we note 2 problems wth eq. (). Frst, we are addng 2 quanttes that have dfferent unts. 5
Anytme you come across a relaton that adds 2 or more unts havng dfferent unts, beware. The second problem s that the magntudes of the two terms n eq. () may dffer dramatcally. If we are workng n MW and Hz (or rad/sec), then we may see ΔP n the s of MW whereas we wll see Δf (or Δω) n the hundredths or at most tenths of a Hz. The mplcaton s that the control sgnal, per eq. (), wll greatly favor the export devatons over the frequency devatons. Therefore we need to scale one of them. To do so, we defne area frequency characterstc as β. It has unts of MW/Hz. Your text shows (pg. 393) that f D f DLf (2a) where D D L f (2b) 6
D s the dampng coeffcent from the swng equaton (and represents the effect of synchronous generator wndage and frcton); D L s the load dampng coeffcent and represents the tendency of the load to decrease as frequency decreases (an effect manly attrbuted to nducton motors). The f or ω subscrpts ndcate whether we wll comte ACE usng f or ω (text uses ω). EPI [] provdes an nterestng fgure whch compares frequency senstvty for motor loads wth non-motor loads, shown below n Fg.. Fg. Fgure shows that motor loads reduce about 2% for every % drop n frequency. If we assume that non- 7
motor loads are unaffected by frequency, a reasonable composte characterstc mght be that total load reduces by % for every % drop n frequency, as ndcated by the total load characterstc n Fg.. To account for load senstvty to frequency devatons, we wll use parameter D L accordng to change n load D L change n frequency (a) from whch we may wrte: change n load D L (b) If our system has a % decrease n power for every % decrease n frequency, then D L =. The dampng terms of eq. (2) are usually sgnfcantly smaller than the regulaton term, so that a reasonable approxmaton s that f (2c) f 8
(2d) Then the ACE equaton becomes: ACE P Bff (2e) ACE P B (2f) where B f (or B ω ) s the frequency bas characterstc for area ; t s generally set equal to the area frequency characterstc, β f (or β ω ), as shown n your text on p. 396. 4. Example (smlar to Ex.5 n text) Consder the two-area nterconnecton shown n Fg. wth data gven as below. A MW A2 Fg. Area Area 2 Load=2, MW Load=4, MW Capacty=2, MW Capacty=42, MW Gen=9, MW Gen=4, MW =.5 =.5 The scheduled nterchange s MW flowng from A2 to A, so that scheduled exports are: 9
P =- MW P 2 = MW There are two parts to ths problem:.determne frequency and generaton of each area and the te lne flow after a MW loss of load n A, but before secondary control has taken effect (ths means we wll determne only the effect of prmary control). 2.epeat () after secondary control acton has taken effect. oluton: We assume that area capactes are the ratngs: =2 2 =42 We also note that the load decreases, therefore ΔP=- MW..From eq. (8), f P 6... Therefore K f.86*6. 484Hz so that f=6.484 Hz..5( ).86 [2 42]
Then we can comte the ncrease n generaton n each area per eq. (9). For Area : P P (2)( ) 322. M 6 2 42 Therefore For Area 2: P P M 9 322.6 8,677. 4MW 2P (42)( ) 677. M 2 4 2 42 Therefore P M 4 677.4 4,322. 6MW 2 MW MW It s of nterest now to obtan the actual exports, and we can do ths for each area by takng the dfference between load and generaton. The loads n areas and 2 were 2 and 4, but remember that we lost MW of load n area, so that ts value s now 9. Generaton levels were comted above. Therefore AP =8,677-9,=-322.6 MW AP 2 =4,322.6-4,=322.6 MW
o clearly the new te-lne flow s 322.6 MW from A2 to A. Note: For multple areas, calculatng te-lne flows requres a power flow soluton (DC power flow s sutable to use here). From eq. (5), net devaton s: P AP P 322.6 ( ) P AP P 322.6 () 2 2 2 677.4 677.4 2.To fnd the effect of supplementary control, we frst need to comte the frequency bas parameters. We neglect effects of dampng terms, so that we use eq. (2c) or (2d): f (2c) f (2d) But note that the droop constant n (2c) or (2d) s not n per-unt. To convert to per-unt, recall eqs. (3) and (4) from AGC2 notes: 2
f P M f / r f f P M P M (3a) (3b) / / P / M r f P M P M ubsttutng eq. (3a) or (3b) nto (4) gves: Therefore f P M / f / / P / M r r f P M f r r P f f r p u r M f f r r (4) (5a) (5b) (6a) (6b) We can use (6a) to calculate the frequency bas term of (2c), whch s used n the ACE equaton of (2d): ACE P B f (2d) f 3
Alternatvely, we can use (6b) to calculate the frequency bas terms of (2d), whch s used n the ACE equaton of (2e): ACE In the rest of ths example, we wll use (6a), (2c), and (2d). P B (2e) Applyng eq. (6a) to obtan f and f2, we get f.5 6 f r 2 f.5 6 f 2 r 2 42.5Hz/ MW.7429Hz/ MW Then the frequency bas terms are obtaned as B B f f 2 f f 2 6666.7 MW / Hz.5.7429 4MW / Now we can comte the ACE n each area: P B f ( AP P ) B f 677.4 322.68 Hz ACE f f ( -322.6- -) 6666.7(.484) MW 4
ACE 2 P 2 f 2 f ( AP2 P2 ) f 2f (322.6 -) 4(.484) 677.4 677.6 MW o what does ths mean? ACE ndcates that the generators n A receve a sgnal to decrease generaton by MW. ecall from page 9 that after prmary control acton, P M =8,677.4 MW. Wth a MW decrease, then P M =7,677.4 MW. ACE 2 ndcates that the generators n A2 do not change, so P M2 =4,322.6 MW. But now consder that prevous to the secondary control acton, we were at a steady state (wth steady-state frequency devaton of.484 so that f=6.484 Hz). Now the ACE sgnal has caused a decrease n A generaton by MW. Ths acton creates a load-generaton mbalance of ΔP=+ MW that wll cause unt prmary controllers to act n both areas. From eq. (8): 5
f P 6... Therefore K f.86*6. 484Hz.5( ).86 [2 42] so that f=6.484 -.484 = 6Hz. Then we comte the ncrease n generaton n each area per eq. (9). For Area : P P (2)( ) 322. M 6 2 42 Therefore For Area 2: P P M 7677.4 322.6 8, MW 2P (42)( ) 677. M 2 4 2 42 Therefore P M 4,322.6 677.4 4, MW 2 MW MW The exports for the two areas wll then be: AP =8,-9,=- MW AP 2 =4,-4,=+ MW A summary of the stuaton s gven below, where we see that the fnal control acton has resulted n A generaton entrely compensatng 6
for the A load decrease of MW (wth area exports unchanged). Area Area 2 Load=9, MW Load=4, MW Gen=8, MW Gen=4, MW [] Interconnected Power ystem Dynamcs Tutoral, Electrc Power esearch Insttute EPI T-7726, March 997. 7