3.2 LONG RANGE HYDRO THERMAL SCHEDULING

Let a power system consist of one large hydel plant and one thermal plant say. The two plants cater to the total system load P_D.

The hydro plant can generate power P_H more than power demand P_D for a limited time t. That is,

 

 

However, the energy available from the hydro plant is insufficient to meet the load because of the water constraints.

Let t_max be the number of hours required for an optimization solution (week/month/year) say.

The hydro energy produced is less than total energy required. The difference in energy shall be supplied from the thermal plant.

Neglecting losses,

Let the thermal plant be operating for t_Th hours, where t_Th < t_max. The energy produced by it in t_Th hours is:

Using Eqns. (3.2) and (3.3), the energy equivalent constraint can be written as:

It can be understood that the hydel plant is working for total t_max hours in the considered period. However, the thermal plant is working only for t_Th hours. The fuel cost of the thermal plant can be written as

 

In the span of t_Th hours, the total energy cost is:

Now, the problem is to minimize the value Eq.(3.6) subject to equality constraint Eq.(3.4) using Lagrangian technique, the constrained objective function can be written as unconstrained function using Lagrangian multiplier λ as:

Using the Kuhn-Tucker necessary condition for local minima of L, the condition for economic operation can be obtained. The condition can be obtained when the first derivative of L with respect to independent variable P_Th is set to zero.

i.e.,

For the period of t_Th, when both hydro and thermal generations are available, the economic operation in hydro thermal scheduling can be achieved when Eq.(3.7) is satisfied. The thermal plant is loaded according to Eq.(3.7) and the remaining part of load demand is met by the hydro plant.

3.2.1 Determination of Optimal Contribution of Thermal Plant

Let the total demand P_D exist for t_max hours. The P_D can be met from thermal and hydro plants. Let be the optimal generation value of the thermal plant.

Fig. 3.2 represents energy diagram for t_max hours.

The thermal energy is given by Eq.(3.3) as:

From Fig.3.2, the value of E_Th can be obtained by calculating the area of rectangle ABCD as:

From Eq.(3.10),

Using Eq.(3.5),

The total cost of thermal energy generation is:

For minimal thermal energy cost,

Using Eq.(3.13),

From Eq.(3.14)

Note: Eq.(3.15) represents optimal thermal generation, while Eq.(3.11) gives the time t_Th for which the thermal plant should be kept in service.

Example 3.1

A power system consists of one thermal and one hydro plant. The load is constant at 100 MW for a period of 45 days. The fuel cost curve of the thermal plant is represented by:

 

 

If the total energy produced by the hydel plant in the period is 80,000 MWH, calculate the optimal time for which the thermal plant should be kept in service and optimal thermal generation.

Solution:

 

Using Eq.(3.5), 50

Optimal thermal generation

Using Eq.(3.11)

Note: The total load 100 MW for a period of 1080 hours (45 days) shall be met as per the following conclusions.

Total load = 100 MW

Total duration = 1080 hours

In 531.263 hours,

 

 

In the remaining period (1080 – 531.263 = 548.737 hours), the total demand of 100 MW shall be met from the hydel plant only.

3.2.2 Short Term Hydro Thermal Scheduling

Let one thermal plant and one hydel plant together meet a total load demand of P_D MW. The duration of this run may range from a day to maximum of one week. Knowing the total volume of water available for discharge, the problem lies in the minimization of total cost subject to water discharge constraints.

Mathematically, the problem is solved as:

    Where F_c = total cost function (in Rs/hr)

                P_H = Hydro power generation (in MW)

                P_Th = Thermal power generation (in MW)

                    t = time interval in the concerned period (day/week)

                t_max = Maximum number of hours in the concerned period

                V_total = Total volume of water discharged in 0 to t_max hours

                q(t) = discharge in m³/sec

For the sake of simplifying the problem, the following assumptions are considered.

• The total power demand P_D during the period is constant.
• At the begining and in the end, the volume of water in the reservoir is specified.
• Due to technical problems, water cannot be discharged suddenly or stopped suddenly. i.e q_min ≤ q(t) ≤ q_max
• Hydro power generation is a function of discharge q(t).
• The spill-over of water from the dam is zero

  i.e S(t) = 0

The equality and inequality constraints and the statement of the problem are given in detail below.

The total cost function is a function of fuel costs of thermal and hydro plants. For a given P_D, the total cost function is:

where Fc(PTh, PH)	:	Total cost function, which is a function of thermal generation PTh and hydro generation PH (in Rupees)
FTh(t, PTh)	:	Cost of thermal generation. Computed at t, while generating PTh (in Rs/hr)
FH(t, PTh)	:	Cost of hydro generation. Computed at t, while generating PH (in Rs/hr)

The objective is to minimize F_C (P_Th, P_H)

In Eq.(3.19),

 

FC (PTh, PH)	= Total cost function
Fth (t, PTh)	= Fuel cost of thermal plant
γh	= Constant which converts incremental water into an equivalent incremental cost
W (t, PH)	= Water discharge to generate hydel power PH in the given time ‘t’.

It should be remembered that discharge is a function of hydro power generation.

The equality constraint is:

 

In the Eq.(3.20)

 

 

Assuming that there in no spillage, the inequality constraint is:

In the Eq.(3.21),

 

    W_Total = Total water discharge in the period from t = 1 to t = t_max hours.

      W(t) = Water discharge during the time ‘t’.

 

The constrained objective function Eq.(3.19) can be converted into an unconstrained function using Lagrangian technique as:

The optimal solution can be obtained by equating

Similarly optimal scheduling of hydro generation can be obtained by evaluating using Eq. (3.22),

Optimal hydro scheduling is possible when incremental cost of received power(λ) is equal to plant’s incremental cost of generation as:

Eqs. (3.24 and 3.25) are known as coordination equations and can be used for computation.

Algorithm

Step-1: Have an initial guess for λ_h

Step-2: Set t = 1

Step-3: Assume λ^deg

Step-4: The P_Th and P_H can be obtained using Eqs (3.24) and (3.25)

Step-5: Check if P_D – P_H – P_Th = 0. If this condition is satisfied GOTO Step-6. Otherwise take a new value of λ^deg and repeat Steps 3 to 5.

Step-6: t = t + 1 (goto next hour) If t > t_max GOTO Step-7. Else repeat Steps 3 to 5 for new load demand P_D.

Step-7: The water withdrawal from hydel plant for total hydel generation during the period should be computed.

Step-8: If the difference between the calculated water withdrawal and scheduled water withdrawal is not within the prescribed tolerance, then γ_n is adjusted and the procedure is repeated until the criterion is satisfied

Step-9: Print Results, END.