1.7.

Optimal scheduling of Hydrothermal system:

Operation of a system having both hydro and thermal plants, far more
complex as hydro plants have negligible operating cost, but are required
to operate under constraints of water available for hydro generation in a
given period of time. The problem of minimizing the operating cost of a
hydrothermal system can be viewed as one of minimizing the fuel cost of
thermal plants under the constraint of water availability for hydro
generation over a given period of operation.
Water
inflow

Reservoir
PGT PGH

Thermal Hydro
plant plant
Pd Water
discharge
Figure (1.4) Fundamental hydro thermal system.

For a certain period of operation T, it is assumed that

(i) Storage of hydro reservoir at the beginning and the end of the period are
specified.
(ii) Water inflow to reservoir and load demand on the system.
The problem is to determine q(t), the water discharge as to minimize the
cost of thermal generation.
T

∫C
1
CT = (PGT ( t )dt
0
Under the following constraints:
(i) Meeting the load demand.
P GT(t) + P GH(t) – P L(t) – P D(t) = 0
(ii) Water availability
T T
X′(T) - X′(0) - ∫ J(t )dt + ∫ q( t )dt = 0
0 0
Where J(t) → is the water inflow.
X′(t) → water storage.
X′(0), X′(T) → are specified water storages at the beginning and at the end
of the optimization interval.
(iii) The hydro generation P GH(t) is a function of hydro discharge and water
storage.
P GH(t) = f[X′(t), q(t)].
 In the above problem, it is convenient to choose water discharges in all
subintervals except one as independent variables, while hydro
generations, thermal generations and water storges in all subintervals are
treated as dependent variables.

1.8. Hydro electric plant models:

Forebay

Generator
Gross
head
Trash racks and
Turbine
intake
After bay
Reservoir
Penstock
Draft tube

Figure (1.5) Hydro plant components

The diagram of a reservoir and hydroelectric plant shown in figure (1.5).
Let us consider some overall aspects of the falling water as it travels from
the reservoir through the penstock to the inlet gates through the
hydraulic turbine down the draft tube and out the tail race at the plant
exit. The power that the water can produce is equal to the rate of water
flow in cubic feet per second times a conversion coefficient that takes into
account the net head times the conversion efficiency of the turbine
generator. Conversion efficiences of turbine generators are typically in the
rage of 85 to 90%.
The hydroelectric project consists of a body of water impounded by a
dam, the hydroplant, and the exit channel (or) lower water body. The
head available to the turbine itself is slightly less than the gross head due
to the friction losses in the intake, penstock, and draft tube. This is
usually expressed as the net head and is equal to the gross head less the
flow losses. The flow losses can be very significant for low head plants.
The water level at the after bay is influenced by the flow out of the
reservoir including plant release and any spilling of water over the top of
the dam. The type of turbine used in a hydroelectric plant depends
primarily on the design head for the plant. The largest number of
hydroelectric projects use reaction type turbines. When three types of
reaction turbines are now is common use.
1) medium heads → Francis turbine is used
2) Low heads → propeller turbine is used.
3) High heads → impulse (or) pelton turbine is used.
Typically turbine performance results in an efficiency at full gate loading
of between 85% to 90%. The Francis turbine and the adjustable propeller
turbine may operate at 65 to 125%.
1.9 Scheduling problems:
PH PS
In the operation of a
hydro electric power q H S F
system, three general
categories of problem
arise. These depend on Hydro Steam
the balance between the
hydroelectric generation,
the thermal generation,
and the load. Fig. 1.6. Two unit hydrothermal system
The economic scheduling of these systems is really a problem is
scheduling water releases to satisfy all the hydraulic constraints and
meet the demand for electrical energy. Then the schedule is developed by
minimizing the production cost as in a conventional hydrothermal
system.
Figure (1.6), have two sources of electrical energy to supply a load, one is
hydro and another is steam. The hydro plant can supply the load by itself
for a limited time. For any time period j,
max
PHj > PLj j = 1 to …….jmax.
The energy available from the hydro plant is insufficient to meet the load.
jmax jmax
∑ PHj n j ≤ ∑ PLj n j
j=1 j=1
where n j = number of hours in period j
jmax
∑ n j = Tmax = Total int erval
j=1

Entire amount of energy from the hydroplant in such a way that the cost
of running the steam plant is minimized. The steam plant energy required
is
jmax jmax

∑P
j=1
Lj n j − ∑ PHj n j =
j=1
E
Steam energy
Load energy Hydro energy

Not require the steam unit to run for the entire interval of Tmax hours.
Ns
∑ PSj n j = E ; N S → number of periods the steam plant is run.
j=1
Ns
Then ∑ n j < Tmax The scheduling problem becomes
j=1
Ns
Min F t = ∑ F(PSj )n j
j=1
 Ns
Subject to ∑ PSj n j − E = 0 and the using Lagrange function is
j=1

NS  NS 
E − P n 
L= ∑ Sj j  ∑ Sj j 
F( P ) n + α
j=1  j=1 
∂L dF(PSj ) dF(PSi )
Then = − α = 0 for j = 1 to ……..NS ∴ =∝
∂PSj dPSj dPSj
This means that the steam plant should be run at constant incremental
cost for the entire period. Let the optimum value of system generated
power be P S* , which is the same for all time intervals the steam unit. The
total cost over interval is
Ns Ns
FT = ∑ F(PS* )n j = F(P S* ) ∑nj = F(P S*) T S
j=1 j=1
Ns
where TS = ∑nj = the total run time for the steam plant.
j=1
1.10. Short term hydro – thermal scheduling problem:

Sj j = internal
rj r j = inflow during j
V j = Volume at end of j
Qj = discharge during j
Vj
S j = spillage discharge during j

PHj PSj
Equivalent
H S steam unit

qj

Fig. 1.7. Hydrothermal system with hydraulic constraints

A basic short term hydrothermal scheduling problem requires that a given

amount of water be used in such away as to minimize the cost of running
the thermal units. We assume that the hydroplant is not sufficient to
supply to 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 Tmax hours. The mathematical scheduling problem may be set up
as follows:
 jmax
Min Ft = ∑ n j Fj
j=1
jmax
Subject to ∑ n j q j = q tot total water discharge.
j=1

PLj – P Hj – PSi = 0 Load balance for j = 1 to jmax

jmax
Where n j = length of jth interval ∑nj = Tmax.
j=1

Other constrains could be imposed,

V j| j = 0 = V s starting volume
V j| j=jmax = V E Ending volume
q min ≤ q I ≤ q max flow limits for j = 1……to jmax
q j = Qj fixed discharge for a particular hour.

PH (MW)
Figure (1.8) Hydroelectric unit input-output
characteristic for constant load

Assume constant head operation and assume a q versus p characteristic

is available as shown in figure.
The lagrange function is
jmax
 jmax 
L = ∑ [ n j F( PSj ) + λ j ( PLj − PHj − PSj )] + ν ∑ n j q j ( PHj ) − q tot 
j=1  j=1 
∂L
And for a specific interval j = k, =0
∂PSk
dFSk ∂L
gives n k = λ k and =0
dPSk ∂PHK
dqk
gives νn k = λk
dPHK
Suppose add the network losses to the problem. Then at each hour,
PLj + P loss j – P Hj – P Sj = 0
And the lagrange function becomes
jmax
 jmax 
L= ∑ [ n j F( Psj ) + λ j ( PLj + Ploss j − PHj − PSj )] + ν ∑ n j q j (PHj ) − q tot 
j=1  j=1 
With resulting coordination equations (hour k)
 dF( PSk ) ∂Ploss k dq (PHk ) ∂Ploss k
nk + λk = λ k; νn k + λk = λk
dPSk ∂PSk dPHk ∂PHk
This problem ignores volume and hourly discharge rate constraints. The
value of ν will be constant over the entire scheduling period as long as the
units remain with in their respective scheduling ranges.