Pages From Power Generation Operation and Control by Allen J Wood & Bruce F Wollenberg-2

218 HYDROTHERMAL COORDINATION

Part 1

Let the hydroplant be limited to 10,OOO MWh of energy. Solve for T,*, the run time of the steam unit. The load is 90 x 168 = 15,120 MWh, requiring 5120 MWh to be generated by the steam plant.

The steam plant’s maximum efficiency is at J- = 50 MW. There- fore, the steam plant will need to run for 5120/50 or 102.4 h. The resulting schedule will require the steam plant to run at 50 MW and the hydroplant at 40 MW for the first 102.4 h of the week and the hydroplant at 90 MW for the remainder.

Part 2

Instead of specifying the energy limit on the steam plant, let the limit be on the volume of water that can be drawn from the hydroplants’ reservoir in 1 wk. Suppose the maximum drawdown is 250,000 acre-ft, how long should the steam unit run?

To solve this we must account for the plant’s q versus P characteristic. A different flow will take place when the hydroplant is operated at 40 MW than when it is operated at 90 MW. In this case,

q1 = [300 + 15(40)] x T, acre-ft

q2 = [300 + 15(90)] x (168 - T,) acre-ft

q 1 + q2 = 250,000 acre-ft and

Solving for T, we get 36.27 h.

7.4 THE SHORT-TERM HYDROTHERMAL SCHEDULING PROBLEM

A more general and basic short-term hydrothermal scheduling problem requires that a given amount of water be used in such a way as to minimize the cost of running the thermal units. We will use Figure 7.5 in setting up this problem.

The problem we wish to set up is the general, short-term hydrothermal scheduling problem where the thermal system is represented by an equivalent unit, P, , as was done in Chapter 6 . In this case, there is a single hydroelectric plant, PH. We assume that the hydroplant is not sufficient to supply all the load demands during the period and that there is a maximum total volume of water that may be discharged throughout the period of T,,, hours.

In setting up this problem and the examples that follow, we assume all spillages, s j , are zero. The only other hydraulic constraint we will impose initially is that the total volume of water discharged must be exactly as

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THE SHORT-TERM HYDROTHERMAL SCHEDULING PROBLEM 219

j = interval yj = inflowduringj V j = volume at end o f j qj = discharge during i s, = spillage discharge

during j

Equivalent steam unit

FIG. 7.5 Hydrothermal system with hydraulic constraints.

defined. Therefore, the mathematical scheduling problem may be set up as follows:

Problem: Min FT = njFj (7.21)

Subject to: C n j q j = qTOT total water discharge

j= 1

1m.x

j = 1

PHj - Psj = 0 load balance for j = 1 . . . j,,, 6 o a d j - where

nj = length of j‘” interval

nj = T,,, j = 1

and the loads are constant in each interval. Other constraints could be imposed, such as:

y t j = o = v, starting volume

yIj=jmax = VE ending volume

qmin 5 qj I qmax flow limits for j = 1 . . . j,,, fixed discharge for a particular hour 4. J J = Q .

Assume constant head operation and assume a q versus P characteristic is available, as shown in Figure 7.6, so that

= q(pH) (7.22)


COORDINATION 220 HYDROTHERMAL

5 6 *;

m P

P,(MW)

FIG. 7.6 Hydroelectric unit input-output characteristic for constant head.

We now have a similar problem to the take-or-pay fuel problem. The Lagrange function is

and for a specific interval j = k,

gives

and

gives

a 2 - = o apHk

(7.24)

(7.25)

This is sc.ied using the same techniques shown in Chapter Suppose we add the network losses to the problem. Then at each hour,

8 o a d j + 8 0 , s j - - psj = (7.26)

and the Lagrange function becomes

(7.27)


THE SHORT-TERM HYDROTHERMAL SCHEDULING PROBLEM 221

with resulting coordination equations (hour k):

(7.28)

(7.29)

This gives rise to a more complex scheduling solution requiring three loops, as shown in Figure 7.7. In this solution procedure, and c2 are the respective tolerances on the load balance and water balance relationships.

Note that this problem ignores volume and hourly discharge rate constraints.

h - y ITERATION WITH LOSSES

EQUATIONS

+ OUTPUT SCHEDULES

FIG. 7.7 A i.-y iteration scheme for hydrothermal scheduling.


222 HYDROTHERMAL COORDINATION

As a result, the value of y will be constant over the entire scheduling period as long as the units remain within their respective scheduling ranges. The value of y would change if a constraint (i.e., = V,,,, etc.) were encountered. This would require that the scheduling logic recognize such constraints and take appropriate steps to adjust y so that the constrained variable does not go beyond its limit. The appendix to this chapter gives a proof that y is constant when no storage constraints are encountered. As usual, in any gradient method, care must be exercised to allow constrained variables to move off their constraints if the solution so dictates.

EXAMPLE 7B

A load is to be supplied from a hydroplant and a steam system whose characteristics are given here.

Equivalent steam system: H = 500 + 8.OC + 0.0016P: (MBtu/h)

Fuel cost = 1.15 P/MBtu

150 MW I P, I 1500 MW

Hydroplant: q = 330 + 4.97PH acre-ft/h

OsP , s lOOOMW

q = 5300 + 12(P, - lO00) + 0.05(PH - 1000)2 acre-ft/h

lo00 < PH < 1 100 M W

The hydroplant is located a good distance from the load. The electrical losses are

~,,, = O.OOOOSP,Z M W

The load to be supplied is connected at the steam plant and has the following schedule:

2400-1200 = 1200 MW

1200-2400 = 1500 MW

The hydro-unit’s reservoir is limited to a drawdown of 100,000 acre-ft over the entire 24-h period. Inflow to the reservoir is to be neglected. The optimal schedule for this problem was found using a program written using Figure 7.7. The results are:

Time Period P steam P hydro Hydro-Discharge (acre-ft/h)

2400- 1200 567.4 668.3 3651.5 1200-2400 685.7 875.6 4681.7


Pages From Power Generation Operation and Control by Allen J Wood & Bruce F Wollenberg-2

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