What lies between Δx E, which represents the steam valve, and ΔP M, which is the mechanical power into the synchronous machine?

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A 2.0 Introduction In the lat et of note, we developed a model of the peed governing mechanim, which i given below: xˆ K ( Pˆ ˆ) E () In thee note, we want to extend thi model o that it relate the actual mechanical power into the machine (intead of Δx E ), o that we can then examine the relation between the mechanical power into the machine and frequency deviation. What lie between Δx E, which repreent the team valve, and ΔP, which i the mechanical power into the ynchronou machine? 2.0 Extended model Your text (p. 38) doe not go into great detail in regard to the turbine model but rather argue that it repond much like the peed governing ytem, which i a ingle time-contant ytem.

Being a ingle time-contant ytem implie that there i only one pole (the characteritic equation ha only one root). hinking in term of invere LaPlace tranform, thi mean that the repone to a tep change in valve opening will be exponential (a oppoed to ocillatory). In thee note, we imply confirm that thi i the cae. Analytically, thi mean that the relation between the change in valve opening Δx E and the change in mechanical power into the generator ΔP i given by: P K xˆ Subtituting () into (2) reult in which i K K P ( Pˆ ˆ ) P E (2) (3) K K Pˆ ˆ (4) 2

Let aume that K and K are choen o that K K =, then eq. (4) become: P Pˆ ˆ (5) A block diagram repreenting eq. (5) i given in Fig. (Fig..4 in text). Δω ΔP + Σ - ΔP Fig. 3.0 echanical power and frequency Let expand (5) o that Pˆ Pˆ ˆ (6) onider a tep-change in power of ΔP and in frequency of Δω, which in the LaPlace domain i: 3

ˆ (7a) P P ˆ (7b) Subtitution of (7a) and (7b) into (6) reult in: Pˆ P (8) One eay way to examine eq. (8) i to conider ΔP (t) for very large value of t, i.e., for the teady-tate. o do thi, recall that the variable ΔP in eq. (8) i a LaPlace variable. o conider the correponding time-domain variable under the teady-tate, we may employ the final value theorem, which i: lim t f ( t) lim fˆ( ) (9) 0 Applying eq. (9) to eq. (8), we get: 4

5 P P P P t P P t lim lim ˆ lim ) ( lim 0 0 0 (0) herefore, when conidering the relation between teady-tate change, P P () hi i eq. (.6) in your text. ake ure that you undertand that in eq. (), ΔP, ΔP, and Δω in eq. () are ime-domain variable (not LaPlace variable) Steady-tate value of the time-domain variable (the value after you wait along time) Becaue we developed eq. () auming a tepchange in frequency, you might be milead into thinking that the frequency change i the

initiating change that caue the change in mechanical power ΔP. However, recall Fig. 3 of A note, repeated below for convenience a Fig. 2 in thee note. Fig. 2 he frequency change expreed by Δω in eq. () i the frequency deviation at the end of the imulation. he ΔP in eq. (), aociated with Fig. 2, i not the amount of generation that wa outaged, 6

but rather the amount of generation increaed at a certain generator in repone to the generation outage. So ΔP and Δω are the condition that can be oberved at the end of a tranient initiated by a load-generation imbalance. hey are condition that reult from the action of the primary governing control. In other word, the primary governing control will operate (in repone to ome frequency deviation caued by a load-generation imbalance) to change the generation level by ΔP and leave a teady-tate frequency deviation of Δω. Although we have not developed relation for ω, P, and P (but rather Δω, ΔP, and ΔP ), let aume we have at our dipoal a plot of P v. ω for a certain etting of P =P. Such a plot appear in Fig. 3. (he text make the following aumption on pg 383 (I added the italicized text): the local behavior (a characterized by 7

eq. ()) can be extrapolated to a larger domain. ) P Δω P Slope=-/ ΔP P Fig. 3 An important aumption behind Fig. 3 i that the adjutment to the generator et point, deignated by P =P, i done by a control ytem (a yet untudied), called the econdary or upplementary control ytem, which reult in ω=ω 0. he plot, therefore, provide an indication of what happen to the mechanical power P, and the frequency ω, following a diturbance from thi pre-diturbance condition for which P =P and ω= ω 0. ω 0 ω 8

It i clear from Fig. 3 that the local behavior i P. characterized by If we were to change the generation et point to P =P 2, under the aumption that the econdary control that actuate uch a change maintain ω 0, then the entire characteritic move to the right, a hown in Fig. 4. P P 2 P Fig. 4 ω 0 ω We may invert Fig. 3, o that the power axi i on the vertical and the frequency axi i on the horizontal, a hown in Fig. 5. 9

ω ω 0 ΔP Δω P Slope=- P Fig. 5 P Fig. 6 illutrate what happen when we change the generation et point from P =P to P =P 2, ω ω 0 P P 2 Fig. 6 P 0

It i conventional to illutrate the relationhip of frequency ω and mechanical power P a in Fig. 5 and 6, rather than Fig. 3 and 4. (egardle, however, be careful not to fall into the trap that it i howing P a the caue and ω a the effect. A repeated now in different way, they are both effect of the primary control ytem repone to a frequency deviation caued by a load-generation imbalance). From uch a picture a Fig. 5 and 6, we obtain the terminology droop, in that the primary control ytem act in uch a way o that the reulting frequency droop with increaing mechanical power. he contant, previouly called the regulation contant, i alo referred to a the droop etting.

4.0 Unit ecall eq. (), repeated here for convenience. P P () With no change to the generation et point, i.e., ΔP =0, then where we ee that P (2) P (3) We ee then that the unit of mut be (rad/ec)/w. A more common way of pecifying i in perunit, where we per-unitize top and bottom of eq. (3), o that: pu / 0 P / S r P pu pu (4) where ω 0 =377 and S r i the three-phae VA rating of the machine. When pecified thi way, relate fractional change in ω to fractional change in P. 2

It i alo ueful to note that 2f f pu f pu 0 2f 0 f0 (5) hu, we ee that per-unit frequency i the ame independent of whether it i computed uing rad/ec or Hz, a long a the proper bae i ued. herefore, eq. (4) can be expreed a pu P pu pu f P pu pu (6) 5.0 Example onider a 2-unit ytem, with data a follow: en A: S A =00 VA, pua =0.05 en B: S B =200 VA, pub =0.05 he load increae, with appropriate primary peed control (but no econdary control) o that the teady-tate frequency deviation i 0.0 Hz. What are ΔP A and ΔP B? 3

Solution: 0.0 f pu 0. 00067 pu 60 f pu ( 0.00067) PpuA 0. 0033pu 0.05 pua f pu ( 0.00067) PpuB 0. 0033pu 0.05 pub Note!!! Since pua = pub (and ince the teadytate frequency i the ame everywhere in the ytem), we get ΔP pua =ΔP pub, i.e., the generator pick up the ame amount of perunit power (given on their own bae). But let look at it in W: PA PpuASA 0.0033(00) 0. 33W PB PpuBSB 0.0033(200) 0. 66W oncluion: When two generator have the ame per-unit droop, they pick-up (compenate for load-gen imbalance) in proportion to their VA rating. 4

In North America, droop contant for mot unit are et at about 0.05 (5%). 6.0 ultimachine cae Now let conider a general multimachine ytem having K generator. From eq. (6), for a load change of ΔP W, the i th generator will repond according to: pui f P i / 60 Si f Pi / S 60 i pui (7) he total change in generation will equal ΔP, o: S P pu... Solving for Δf reult in f 60 S S K f P... Kpu Kpu 60 (8) K (9) pu Subtitute eq. (9) back into eq. (7) to get: S 5

P i S i pui f 60 S i pui S P... Kpu K (20) pu If all unit have the ame per-unit droop contant, i.e., pui = pu = = Kpu, then eq. (20) become: P i S i pui f 60 S S SiP... S K (2) which generalize our earlier concluion for the two-machine ytem that unit pick up in proportion to their VA rating. hi concluion hould drive the way an engineer perform contingency analyi of generator outage, i.e., one hould reditribute the lot generation to the remaining generator in proportion to their VA rating, a given by eq. (2). 6