6
HydroThermal Scheduling
OBJECTIVES
After reading this chapter, you should be able to:
 know the importance of hydrothermal coordination
 develop the mathematical modeling of longterm hydrothermal coordination
 study the Kirchmayer’s method for shortterm hydrothermal coordination
 study the advantages of hydrothermal plants combination
6.1 INTRODUCTION
No state or country is endowed with plenty of water sources or abundant coal and nuclear fuel. For minimum environmental pollution, thermal generation should be minimum. Hence, a mix of hydro and thermalpower generation is necessary. The states that have a large hydropotential can supply excess hydropower during periods of high water runoff to other states and can receive thermal power during periods of low water runoff from other states. The states, which have a low hydropotential and large coal reserves, can use the small hydropower for meeting peak load requirements. This makes the thermal stations to operate at high load factors and to have reduced installed capacity with the result economy. In states, which have adequate hydro as well as thermalpower generation capacities, power coordination to obtain a most economical operating state is essential. Maximum advantage of cheap hydropower should be taken so that the coal reserves can be conserved and environmental pollution can be minimized. The whole or a part of the base load can be supplied by the runoff river hydroplants, and the peak or the remaining load is then met by a proper mix of reservoirtype hydroplants and thermal plants. Determination of this by a proper mix is the determination of the most economical operating state of a hydrothermal system. The hydrothermal coordination is classified into longterm coordination and shortterm coordination.
6.2 HYDROTHERMAL COORDINATION
Initially, there were mostly thermal power plants to generate electrical power. There is a need for the development of hydropower plants due to the following reasons.
 Due to the increment of power in the load demand from all sides such as industrial, agricultural, commercial, and domestic.
 Due to the high cost of fuel (coal).
 Due to the limited range of fuel.
The hydroplants can be started easily and can be assigned a load in very short time. However, in the case of thermal plants, it requires several hours to make the boilers, super heater, and turbine system ready to take the load. For this reason, the hydroplants can handle fastchanging loads effectively. The thermal plants in contrast are slow in response. Hence, due to this, the thermal plants are more suitable to operate as base load plants, leaving hydroplants to operate as peak load plants.
FIG. 6.1 Fundamental hydrothermal system
The maximum advantage of cheap hydropower should be taken so that the coal reserves can be conserved and environmental pollution can be minimized. In a hydrothermal system, the whole or a part of the base load can be supplied by the runoff river hydroplants and the peak or the remaining load is then met by a proper coordination of reservoirtype hydroplants and thermal plants.
The operating cost of thermal plants is very high and at the same time its capital cost is low when compared with a hydroelectric plant. The operating cost of a hydroelectric plant is low and its capital cost is high such that it has become economical as well as convenient to run both thermal as well as hydroplants in the same grid.
In the case of thermal plants, the optimal scheduling problem can be completely solved at any desired instant without referring to the operation at other times. It is a static optimization problem.
The operation of a system having both hydro and thermal plants is more complex as hydroplants have a negligible operating cost but are required to run under the constraint of availability of water for hydrogeneration during a given period of time. This problem is the ‘dynamic optimization problem’ where the time factor is to be considered.
The optimal scheduling problem in a hydrothermal system can be stated as to minimize the fuel cost of thermal plants under the constraint of water availability for hydrogeneration over a given period of operation.
Consider a simple hydrothermal system, shown in Fig. 6.1, which consists of one hydro and one thermal plant supplying power to load connected at the center in between the plants and is referred to as the fundamental system.
To solve the optimization problem in this system, consider the real power generations of two plants P_{GThermal} and P_{GHydro} as control variables. The transmission power loss is expressed in terms of the B coefficient as
6.3 SCHEDULING OF HYDROUNITS IN A HYDROTHERMAL SYSTEM
 In case of hydrounits without thermal units in the system, the problem is simple. The economic scheduling consists of scheduling water release to satisfy the hydraulic constraints and to satisfy the electrical demand.
 Where hydrothermal systems are predominantly hydro, scheduling may be done by scheduling the system to produce minimum cost for the thermal systems.
 In systems where there is a close balance between hydro and thermal generation and in systems where the hydrocapacity is only a fraction of the total capacity, it is generally desired to schedule generation such that thermal generating costs are minimized.
6.4 COORDINATION OF RUNOFF RIVER PLANT AND STEAM PLANT
A runoff river hydroplant operates as the water is available in needed quantities. These plants are provided with a small pondage or reservoir, which makes it possible to meet the hourly variation of load.
The ratio of runoff during the rainy season to the runoff during the dry season may be as large as 100. As such the runoff river plants have very little from capacity. The usefulness of these runoff river plants can be considerably increased if such a plant is properly coordinated with a thermal plant. When such coordination exists, the hydroplant may carry the base load upto its installed capacity during the period of high stream flows and the thermal plant may carry the peak load. During the period of lean flow, the thermal plant supplies the base load and the hydroplant supplies the peak load. Thus, the load met by a thermal plant can be adjusted to conform to the available river flow. This type of coordination of a runoff river hydroplant with a thermal plant results in a greater utilization factor of the river flow and a saving in the amount of fuel consumed in the thermal plant.
6.5 LONGTERM COORDINATION
Typical longterm coordination may be extended from one week to one year or several years. The coordination of the operation of reservoir hydropower plants and steam plants involves the best utilization of available water in terms of the scheduling of water released. In other words, since the operating costs of hydroplants are very low, hydropower can be generated at very little incremental cost. In a combined operational system, the generation of thermal power should be displaced by available hydropower so that maximum decrement production costs will be realized at the steam plant. The longterm scheduling problem involves the longterm forecasting of water availability and the scheduling of reservoir water releases for an interval of time that depends on the reservoir capacities and the chronological load curve of the system. Based on these factors during different times of the year, the hydro and steam plants can be operated as base load plants and peak load plants and vice versa.
For the longterm drawdown schedule, a basic best policy selection must be made. The best policy is that should the water be used under the assumption that it will be replaced at a rate based on the statistically expected rate or should the water be released using a worstcase prediction?
Longterm scheduling is made based on an optimizing policy in view of statistically treated unknowns such as load, hydraulic inflows, and unit availability (i.e., steam and hydroplants).
The useful techniques employed for this type of scheduling problems include:
 the simulation of an entire longterm operational time period for a given set of operating conditions by using the dynamic programming method,
 composite hydraulic simulation models, and
 statistical production cost models.
For the longterm scheduling of a hydrothermal system, there should be required generation to meet the requirements of load demand and both hydro and thermal generations should be so scheduled so as to maintain the minimum fuel costs. This requires that the available water should be put to an optimum use.
6.6 SHORTTERM COORDINATION
The economic system operation of thermal units depends only on the conditions that exist from instant to instant. However, the economic scheduling of combined hydrothermal systems depends on the conditions existing over the entire operating period.
This type of hydrothermal scheduling is required for one day or one week, which involves the hourbyhour scheduling of all available generations on a system to get the minimum production cost for the given time. Such types of scheduling problems, the load, hydraulic inflows, and unit availabilities are assumed to be known.
Here also, the problem is how to supply load, as per the load cycle during the period of operation so that generation by thermal plants will be minimum. This condition will be satisfied when the value of hydropower generation rather than its amount is a maximum over a certain period. The basic problem is that determining the degree to which the minimized economy of operating the hydrounits at other than the maximum efficiency loading may be tolerated for an increased economy with an increased load or vice versa to result in the lowest total thermal power production costs over the specified operating period.
The factors on which the economic operation of a combined hydrothermal system depends are as follows:
 Load cycle.
 Incremental fuel costs of thermal power stations.
 Expected water inflow in hydropower stations.
 Water head that is a function of water storage in hydropower stations.
 Hydropower generation.
 Incremental transmission loss (ITL).
The following are the few important methods for shortterm hydrothermal coordination:
 Constant hydrogeneration method.
 Constant thermal generation method.
 Maximum hydroefficiency method.
 Kirchmayer’s method.
6.6.1 Constant hydrogeneration method
In this method, a scheduled amount of water at a constant head is used such that the hydropower generation is kept constant throughout the operating period.
6.6.2 Constant thermal generation method
Thermal power generation is kept constant throughout the operating period in such a way that the hydropower plants use a specified and scheduled amount of water and operate on varying power generation schedules during the operating period.
6.6.3 Maximum hydroefficiency method
In this method, during peak load periods, the hydropower plants are operated at their maximum efficiency; during offpeak load periods they operate at an efficiency nearer to their maximum–efficiency with the use of a specified amount of water for hydropower generation.
Kirchmayer’s method is explained in Section 6.8.
6.7 GENERAL MATHEMATICAL FORMULATION OF LONGTERM HYDROTHERMAL SCHEDULING
To mathematically formulate the optimal scheduling problem in a hydrothermal system, the following assumptions are to be made for a certain period of operation T (a day, a week, or a year):
 The storage of a hydroreservoir at the beginning and at the end of period of operation T are specified.
 After accounting for the irrigation purpose, water inflow to the reservoir and load demand on the system are known deterministically as functions of time with certainties.
The optimization problem here is to determine the water discharge rate q(t) so as to minimize the cost of thermal generation.
Objective function is
Subject to the following constraints:
(i) The real power balance equation
P_{GT}(t) + P_{GH}(t) = P_{L}(t) + P_{D}(t) + P_{D}(t)
i.e., P_{GT}(t) + P_{GH}(t) − P_{L}(t) − P_{D}(t) = 0 for t ∈ (0, T) (6.2)
where 
P_{GT}(t) is the real power thermal generation at time ‘t’, 

P_{GH}(t) the real power hydro generation at time ‘t’, 

P_{L}(t) real power loss at time ‘t’, and 

P_{D}(t) the real power demand at time ‘t’. 
(ii) Water availability equation:
where 
X′(t) is the water storage at time ‘t’, 

X′(0) the water storage at the beginning of operation time, T, 

X′(T) the water storage at the end of operation time, T, 

J(t) the water inflow rate, and 

q(t) the water discharge rate. 
(iii) Real power hydrogeneration
The real power hydrogeneration P_{GH}(t) is a function of water storage X′(t) and water discharge rate q(t)
6.7.1 Solution of problemdiscretization principle
By the discretization principle, the above problem can be conveniently solved. The optimization interval T is subdivided into N equal subintervals of Δt time length and over each subinterval, it is assumed that all the variables remain fixed in value.
The same problem can be reformulated as
subject to the following constraints:
(i) Power balance equation
where 
P^{K}_{GT} is the thermal generation in Kth interval, 

P^{K}_{GH} the hydro generation in Kth interval, 

P^{K}_{L} the transmission power loss in Kth interval and is expressed as 

, and 

P^{K}_{D} is the load demand in the Kth interval. 
(ii) Water availability equation:
where X′^{K} is the water storage at the end of interval K, j^{K} the water inflow rate in interval K, and q^{K} the water discharge rate in interval K.
Dividing Equation (6.7) by Δt, it becomes
X^{K} − X^{K − 1} − j^{K} + q^{K} = 0 for K = 1, 2 … N (6.8)
where is the water storage in discharge units.
x^{0} and x^{N} are specified as water storage rates at the beginning and at the end of the optimization interval, respectively.
(iii) The real power hydrogeneration in any subinterval can be written as
P^{K}_{GH} = h_{o} {1 + 0.5 e (X^{K} + X^{K − 1})} (q^{K} − ρ) (6.9)
where 
h_{o} = 9.81 × 10^{−3} h_{o}′; 

h_{o}′ is the basic water head which is corresponding to dead storage, 

e the water head correction factor to account for the variation in head with storage, and 

ρ the noneffective discharge (due to the need of which a hydro generation can run at noload condition). 
Equation (6.9) can be obtained as follows:
P^{K}_{GH} = 9.81 × 10^{−3} h^{K}_{av} (q_{K} − ρ) MW
where (q^{K} − ρ) is the effective discharge in m^{3}/s and h^{K}_{av} is the average head in the K^{th} interval and is given as
where A is the area of crosssection of the reservoir at the given storage
h^{K}_{av} = h′_{o} (1 + 0.5 e(X^{K} + X^{K−1}))
where , which is tabulated for various storage values
∴ P^{K}_{GH} = h_{o} {1 + 0.5e(X^{K} + X^{K − 1})} (q^{K − ρ}
where h_{o} = 9.81 × 10^{–3} h′_{o}.
The optimization problem is mathematically stated for any subinterval ‘K ’ by the objective function given by Equation (6.5), which is subjected to equation constraints given by Equations (6.6), (6.8), and (6.9).
In the above optimization problem, it is convenient to choose water discharges in all subintervals except one subinterval as independent variables and hydrogenerations, thermal generations, water storages in all subintervals and except water discharge as dependent variables; i.e., independent variables are represented by q^{K}, for K = 2, 3, …, N and for K ≠ 1. Dependent variables are represented by P^{K}_{GT}, P^{K}_{GH} X^{K}, and q^{1}, for K = 1, 2, …, N. [Since the water discharge in one subinterval is a dependent variable.]
Equation (6.8) can be written for all values of K = 1, 2, …, N:

i.e., X^{1} – X^{0} – j^{1} + q^{1} = 0 
for K = 1 

X^{2} – X^{1} – j^{2} + q^{2} = 0 
for K = 2 

X^{N} – X^{(N – 1)} – j^{N} + q^{N} = 0 
for k = N^{th} interval 
By adding the above set of equations, we get
Equation (6.10) is known as the water availability equation.
For K = 2, 3, …, N, there are (N – 1) number of water discharges (q’s), which can be specified as independent variables and the remaining one, i.e., q^{1}, is specified as a dependent variable and it can be determined from Equation (6.10) as
6.7.2 Solution technique
For obtaining a solution to the optimization problem in a hydrothermal system, a nonlinear programming technique in conjunction with the firstorder gradient method is used.
Define the Lagrangian function by augmenting the objective function (cost function) given by Equation (6.5) with equality constraints given by Equations (6.6), (6.8), and (6.9) through Lagrangian multipliers.
where λ_{1}^{K},λ_{2}^{K}, and λ_{3}^{K} are the Lagrangian multipliers that are dual variables. These are obtained by taking the partial derivatives of the Lagrangian function with respect to the dependent variables and equating them to zero.
Substituting Equation (6.8) in Equation (6.12) and differentiating the resultant equation with respect to q^{1}, we get
From the above equations, for any subinterval, the Lagrangian multipliers can be obtained as follows:
 λ_{1}^{K} can be obtained from Equation (6.13),
 λ_{2}^{K} can be obtained from Equation (6.14), and
 λ′_{2} can be obtained from Equation (6.16) and remaining λ_{2(K ≠ 1)}^{K} can be obtained from Equation (6.15).
The partial derivatives of the Lagrangian function with respect to independent variables give the gradient vector:
For optimality, the gradient vector should be zero , if there are no inequality constraints on the independent variables, i.e., on control variables (water discharges).
If not we have to find out the new values of control variables that will optimize the objective function, this can be achieved by moving in the negative direction of the gradient vector to a point, where the value of objective function is nearer to the optimal value.
It is an iterative process and this process is repeated till all the components of the gradient vector are closer to zero within a specified tolerance.
6.7.3 Algorithm
Step 1: 
Assume an initial set of independent variables, q^{K} for all subintervals except the first subinterval {i.e., q^{2}, q^{3} … q^{N}} 
Step 2: 
Obtain the values of dependent variables x^{K}, P^{K}_{GH},P^{K}_{GT} and q^{1} using Equations (6.8), (6.9), (6.6), and (6.11), respectively. 
Step 3: 
Obtain the Lagrangian multipliers λ^{K}_{1},λ^{K}_{3} λ_{2}, and λ^{K}_{2} using Equations (6.13), (6.14), (6.16), and (6.15), respectively. 

Obtain the gradient vector and check whether all its elements are close to zero within a specified tolerance, if so the optimal value is reached; if not, go to the next step. 
Step 5: 
Obtain new values of control variables using the firstorder gradient method, 
where α is a positive scalar, which defines the step length, and having a value depends on the problem on hand, then go to Step 2 and repeat the process.
The inequality constraints of the problem on dependent and independent variables can be handled in the case of an optimal power flow solution. Inequality constraints on independent variables check the Kuhn–Tucker condition (given in optimal power flow, Chapter V). The inequality constraints on dependent variables can be handled by augmenting the objective function through a penalty function.
The abovementioned solution method can be directly extended to a system having multihydro and multithermal plants.
Drawback: It requires large memory since the independent variables, dependent variables, and gradients need to be stored simultaneously.
A modified technique known as decomposition overcomes the above drawback. In the decomposition technique, optimization is carried out over each subinterval and a complete cycle of iteration is repeated, if the water availability equation does not check at the end of the cycle.
Example 6.1: A typical hydrothermal system is shown in Fig. 6.2. For a typical day, the load on the system varies in steps of eight hours each as 9, 12, and 8 MW, respectively. There is no water inflow into the reservoir of the hydroplant. The initial water storage in the reservoir is 120 m^{3}/s and the final water storage should be 75 m^{3}/s, i.e., the total water available for hydrogeneration during the day is 30 m^{3}/s.
FIG. 6.2 Fundamental hydrothermal system
Basic head is 30 m. Water head correction factor e is given to be 0.004. Assume for simplicity that the reservoir is rectangular so that e does not change with water storage. Let the noneffective water discharge be assumed as 3 m^{3}/s. The fuel costcurve characteristics of the thermal plant is C_{T} = 0.2 P_{GT}^{2 +}50 P_{GT +} 130 Rs./hr. Find the optimum generation schedule by assuming the transmission losses neglected.
Solution:
Given:
Fuel cost of the thermal plant, C_{T} = 0.2 P_{GT}^{2} + 50 P_{GT} + 130 Rs./hr
Incremental fuel cost,
Total time of operation, T 
= 24 hr 
No. of subintervals, N 
= 3 
Duration of each subinterval, Δt 
= 8 hr 
Initial water storage in reservoir, x′(0) 
= 120 m^{3}/s 
Final water storage, x′(3) 
= 75 m^{3}/s 
Basic water head, h′_{o} 
= 30 m 
Waterhead correction factor, e 
= 0.04 
Noneffective water discharge, ρ 
= 3 m^{3}/s 
Since there are three subintervals, (N−1), the number of water discharges of the corresponding subintervals can be specified as independent variables and the remaining one is specified as a dependent variable, i.e., the water discharges q^{2} and q^{3} are considered as independent variables and dependent variable q^{1}.
Let us assume the initial values to be
q^{2} = 15 m^{3}/s
q^{3} = 15 m^{3}/s
for the problem formulation P_{GH}, P_{GT}, x, and q^{1} are treated as independent variables.
The dependent variable q^{1} (water discharge in the first subinterval) can be obtained by Equation (6.11).
We have the water availability equation,
x^{K} – x^{k} ^{−1} – j^{K+} q^{K} = 0 for K=1, 2, … N
From the above equation, we have
x^{1} = x^{o} + j^{1} – q^{1} = 120 – 10 = 110 m^{3}/s
x^{2} = x^{1} + j^{2} – q^{2} = 110 – 15 = 95 m^{3}/s
We know the real power hydrogeneration at any interval K by Equation (6.9):

P^{K}_{GH} 
= 
h_{o} {1 + 0.5e(x^{K} + x^{K − 1})}(q^{k} − e) 


= 
9.81 × 10^{−3} h′_{o} {1 + 0.5e(x^{K} + x^{K − 1})}(q^{K} − ρ) 

P^{1}_{GH} 
= 
9.81 × 10^{− 3} × 30 {1 + 0.5 × 0.004(x^{1} + x^{o}) q^{1} − ρ} 


= 
9.81 × 10^{−3} × 30 {1 + 0.5 × 0.004 (110 + 120)} (10 − 3) 


= 
3.0077 MW 

P^{2}_{GH} 
= 
9.81 × 10^{−3} × 30{1 + 0.5 × 0.004(x^{2} + x^{1})q^{2} − ρ} 


= 
9.81 × 10^{−3} × 30 {1 + 0.5 × 0.004 (95 + 110)} (15 − 3) 


= 
4.9795 MW 

P^{3}_{GH} 
= 
9.81 × 10^{−3} × 30{1 + 0.5 × 0.004(x^{3} + x^{2})q^{2} − ρ} 


= 
9.81 × 10^{−3} × 30 {1 + 0.5 × 0.004 (75 + 95)} (20 − 3) 


= 
6.7041 MW 
The thermal power generations during the subintervals are
P^{1}_{GT} = P^{1}_{D} − P^{1}_{GH} = 9 − 3.0077 = 5.9923 MW
P^{2}_{GT} = P^{2}_{D} − P^{2}_{GH} = 12 − 4.9795 = 7.0205 MW
P^{3}_{GT} = P^{3}_{D} − P^{3}_{GH} = 8 − 6.7041 = 1.2959 MW
λ_{1}^{K} can be obtained from Equation (6.13):
i.e.,
By neglecting transmission losses, we have
⇒ λ_{1}^{1} = 0.4P^{1}_{GT} + 50 = 0.4 × 5.9923 + 50 = 52.3969 Rs./MWh
λ_{1}^{2} = 0.4P^{2}_{GT} + 50 = 0.4 × 7.0205 + 50 = 52.8082 Rs./MWh
λ_{1}^{3} = 0.4P^{3}_{GT} + 50 = 0.4 × 1.2959 + 50 = 50.5183 Rs./MWh
From Equation (6.14),
By neglecting transmission losses, we have
⇒ λ_{3}^{K} = λ_{1}^{K}
∴ λ_{3}^{1} = λ_{1}^{1} = 52.3969 Rs./MWh
λ_{3}^{2} = λ_{1}^{2} = 52.8082 Rs./MWh
λ_{3}^{3} = λ_{1}^{3} = 50.5183 Rs./MWh
From Equation (6.16), we have
⇒ λ_{2}^{1} = λ_{2}^{1}h_{o} {1 + 0.5e(2x^{o} + j^{1} − 2q^{1} + ρ)}
= 52.3969 × 9.81 × 10^{−3} × 30 {1 + 0.5 × 0.004 (2 × 120 − 2 × 10 + 3)
(since j = 0)
= 22.2979 Rs./MWh
From Equation (6.15), we have
For K = 1,
∴ λ_{2}^{2} 
= 
λ_{2}^{2} − λ_{3}^{2}0.5h_{o}e(q^{1} − ρ) − λ_{3}^{2}0.5h_{o}e(q^{2} − ρ) 

= 
22.2979 − {52.3969 × 0.5 × 9.81 × 10−3 × 30 × 0.004(10−3)} 


− {52.8082 × 0.5 × 9.81 × 10−3 × 30 × 0.004(15−3)} 

= 
22.2979 − 0.5889 

= 
21.709 Rs./MWh 
and for K = 2
∴ λ_{2}^{3} 
= 
λ_{2}^{2} − λ_{2}^{2} 0.5h_{o}e(q^{2} − ρ) − λ_{3}^{3}0.5h_{o}e(q^{3} − ρ) 

= 
21.709 − {52.8082 × 0.5 × 9.81 × 10−3 × 30 × 0.004(15 − 3)} 


−{50.5183 × 0.5 × 9.81 × 10−3 × 30 × 0.004(20 − 3)} 

= 
21.709 − 0.8784 

= 
20.8305 Rs./MWh 
i.e., λ_{2}^{1} 
= 
22.2979 Rs./MWh 
λ_{2}^{2} 
= 
21.709 Rs./MWh 
λ_{2}^{3} 
= 
20.8305 Rs./MWh 
From Equation (6.17), the gradient vector is
If the tolerance value for the gradient vector is 0.1, since for the above iteration, the gradient vector is not zero (≤ 0.1), i.e., the optimality is not satisfied here. Then, for the second iteration, obtain the new values of control variables (q^{K}_{new}, for K ≠ 1) by using the firstorder gradient method as follows:
(∵ α is a positive scalar)
Let us consider α = 0.5,
∴ q_{new}^{2} = (q^{2})^{1} = 15 − 0.5(0.1685) = 14.9157 m^{3}/s
Similarly, q_{new}^{3} = (q^{3})^{1} = 15 − 0.5(1.4134) = 19.2933 m^{3}/s
and from Equation (6.11),

⇒ q^{1} 
= x^{o} – x^{3} – (q^{2 +} q^{3}) (since j^{K} = 0) 

q^{1} 
= 120 – 75 – (14.9157 + 19.2933) 


= 10.791 m^{3}/s 
To obtain the optimal generation schedule in hydrothermal coordination, the procedure is repeated for the next iteration and checked for a gradient vector. If the gradient vector becomes zero within a specified tolerance, then that will be the optimum generation schedule, otherwise the iterations are to be carried out.
6.8 SOLUTION OF SHORTTERM HYDROTHERMAL SCHEDULING PROBLEMS—KIRCHMAYER’S METHOD
In this method, the coordination equations are derived in terms of penalty factors of both plants for obtaining the optimum scheduling of a hydrothermal system and hence it is also known as the penalty factor method of solution of shortterm hydrothermal scheduling problems.
Let 
P_{GTi} be the power generation of i^{th} thermal plant in MW, 

P_{GHj} be the power generation of j^{th} hydroplant in MW, 

be the incremental fuel cost of i^{th} thermal plant in Rs./MWh, 

w_{j} be the quantity of water used for power generation at j^{th} hydroplant in m^{3}/s, 

be the incremental water rate of j^{th} hydroplant in m^{3}/s/MW, 

be the incremental transmission loss of i^{th} thermal plant, 

be the incremental transmission loss of j^{th} hydel plant, 

λ be the Lagrangian multiplier, 

γ_{j} be the constant which converts the incremental water rate of hydel plant j into an incremental cost, 

n be the total number of plants, 

α be the number of thermal plants, 

n−α be the number of hydroplants, and 

T be the time interval during which the plant operation is considered. 
Here, the objective is to find the generation of individual plants, both thermal as well as hydel that the generation cost (cost of fuel in thermal) is optimum and at the same time total demand (P_{D}) and losses (P_{L}) are continuously met.
As it is a shortrange problem, there will not be any appreciable change in the level of water in the reservoirs during the interval (i.e., the effects of rainfall and evaporation are neglected) and hence the head of water in the reservoir will be assumed to be constant.
Let K_{j} be the specified quantity of water, which must be utilized within the interval T at each hydrostation j.
Problem formulation
The objective function is to minimize the cost of generation:
i.e.,
subject to the equality constraints
and
where w_{j} is the turbine discharge in the j^{th} plant in m^{3}/s and K_{j} the amount of water in m^{3} utilized during the time period T in the j^{th} hydroplant.
The coefficient γ must be selected so as to use the specified amount of water during the operating period.
Now, the objective function becomes
Substituting K_{j} from Equation (6.21) in the above equation, we get
For a particular load demand P_{D}, Equation (6.20) results as
For a particular hydroplant x, Equation (6.23) can be rewritten as
By rearranging the above equation, we get
From Equation (6.22), the condition for minimization is
The above equation can be written as
For hydroplant x,
Multiplying the above equation by ,
Substitute for from Equation (6.24) in Equation (6.27), we get
Rewriting the above equation as
∴ ΔP_{GTi} ≠ 0 and ΔP_{GHj} ≠ 0, Equation (6.28) becomes
and
Equations (6.29) and (6.30) can be written in the form:
and
From Equations (6.31) and (6.32), we have
where (I_{C})_{i} is the incremental fuel cost of the i^{th} thermal plant and (I_{W})_{j} the incremental water rate of the j^{th} hydroplant.
Equations (6.34) and (6.35) may be expressed approximately as
where and are the approximate penalty factors of the i^{th} thermal plant and the j^{th} hydroplant, respectively.
Equations (6.34) and (6.35) are the coordinate equations, which are used to obtain the optimal scheduling of the hydrothermal system when considering the transmission losses.
In the above equations, the transmission loss P_{L} is expressed as
The power generation of a hydroplant P_{GHj} is directly proportional to its head and discharge rate w_{j}.
When neglecting the transmission losses, the coordination equations become
Example 6.2: A twoplant system having a steam plant near the load center and a hydroplant at a remote location is shown in Fig. 6.3. The load is 500 MW for 16 hr a day and 350–MW, for 8 hr a day.
The characteristics of the units are
C_{1} = 120 + 45 P_{GT} + 0.075 P^{2}_{GT}
w_{2} = 0.6 P_{GH} + 0.00283 P_{2}_{GH} m^{3}/s
Loss coefficient, B_{22} = 0.001 MW^{−1}
Find the generation schedule, daily water used by the hydroplant, and daily operating cost of the thermal plant for γ_{j} = 85.5 Rs./m^{3}hr.
Solution:
Given: C_{1} = 120 + 45 P_{GT} + 0.075 P^{2}_{GT}
Coordination equation for thermal unit is
45 + 0.15 P_{GT} + 0.075 P^{2}_{GT}
FIG. 6.3 A typical twoplant hydrothermal system
For the hydrounit, the coordination equation is
Since the load is nearer to the thermal plant, the transmission loss is only due to the hydroplant and therefore B_{TT} = B_{TH} = B_{HT} = 0:
Power balance equation, P_{GT} + P_{GH} = P_{D} + P_{L} and the condition for optimal scheduling is
When P_{D} = 500 MW
0.15 P_{GT} + 45 = 85.5(0.6 + 5.66 × 10^{−3}P_{GH})
(0.15 P_{GT} + 45) (1 − 0.002 P_{GH}) = 85.5 (0.6 + 5.66 × 10−3P_{GH})
0.15 P_{GT} + 45 − 3 × 10^{−4} P_{GT}P_{GH} − 0.09 P_{GT} = 51.3 + 0.48393 P_{GH}
0.57393 P_{GH} − 0.15 P_{GT} + 3 × 10^{−4} P_{GT} P_{GH} + 6.3 = 0 (6.39)
and
P_{GT} + P_{GH} = 400 + 0.001 P^{2}_{GT}
P_{GT} = 400 + 0.001 P^{2}_{GT} − P_{GH} (6.40)
Substituting Equation (6.40) in Equation (6.39), we get
0.57393 P_{GH} − 0.15(400 + 0.001 P^{2}_{GH} − P_{GH}) + 3 × 10^{−4}
P_{GH} (400 + 0.001 P^{2}_{GH} − P_{GH} ) + 6.3 = 0
By solving the above equation, we get
P_{GH} = 81.876 MW
By substituting the P_{GH} value in Equation (6.40), we get
P_{GT} = 424.8 MW
P_{L} = 6.70367 MW
When P_{D} = 350 MW
Equation (6.40) can be modified as
P_{GT} = 350 + 0.001 P^{2}_{GH} − P_{GH} (6.41)
Substituting Equation (6.41) in Equation (6.39), we get
0.57393 P_{GH} − 0.15(350 + 0.001 P^{2}_{GH} − P_{GH}) + 3 × 10^{−4}
P_{GH} (350 + 0.001 P^{2}_{GH} − P_{GH}) + 6.3 = 0
By solving the above equation, we get
P_{GH} = 58.5851 MW
By substituting the P_{GH} value in Equation (6.41), we get
P_{GT} = 294.847 MW
P_{L} = 3.43221 MW
Daily water used by the hydroplant

w 

0.6 P_{GH} + 0.00283 P^{2}_{GH} m^{3}/s 


= 
Daily water quantity used for a 500 MW load for 16 hr + daily water quantity used for a 350 MW load for 8 hr 


= 
{[0.6 × 81.876 + 0.00283 × (81.876)^{2}] × 14 + [0.6 × 58.586 + 0.00283 × (58.586)^{2}] × 8} × 3600 


= 
5.21449 × 10^{6} m^{3} 
Daily operating cost of the thermal plant is:

C_{1} 
= 
(120 + 45 P_{GT} + 0.075 P^{2}_{GT}) 


= 
Operating cost of the thermal plant for meeting the 500 MW load for 16 hr + operating cost of the thermal plant for meeting the 350 MW load for 8 hr 


= 
[120 + 45 × 424.8 + 0.075(424.8)^{2}] × 16 + [120 + 45 × 294.85 + 0.075(424.8)^{2}] × 8 


= 
Rs. 6,83,589.96 per day 
Example 6.3: A twoplant system that has a hydroplant near the load center and a steam plant at a remote location is shown in Fig. 6.4. The load is 400 MW for 14 hr a day and 200 MW, for 10 hr a day.
The characteristics of the units are
C_{1} = 150 + 60 P_{GT} + 0.1 P^{2}_{GT} Rs/hr
w_{2} = 0.8 P_{GH} + 0.000333 P^{2}_{GH} m3/s
FIG. 6.4 A typical twoplant hydrothermal system
Loss coefficient, B_{22} = 0.001 MW^{−1}
Find the generation schedule, daily water used by the hydroplant, and the daily operating cost of a thermal plant for γ_{j} = 77.5 Rs./m^{3}hr.
Solution:
Equations for thermal and hydroplants are
Since the load is nearer to the hydroplant, the transmission loss is only due to the thermal plant and therefore B_{HH} = B_{TH} = B_{HT} = 0:
When P_{D} = 400 MW
The power balance equation is

P_{GT} + P_{GH} 
= 
P_{D} + P_{L} 


= 
400 + 0.001 P^{2}_{GT} 

P_{GH} 
= 
400 + 0.001 P^{2}_{GT} − P_{GT} (6.42) 
The condition for optimal scheduling problem is
0.2 P_{GT} + 60 = 77.5 (0.8 + 6.6 × 10^{−4} P_{GH}) (1 − 0.002 P_{GT})
0.2 P_{GT} + 60 = 62 + 0.051615 P_{GH} − 0.124 P_{GT} − 1.032 × 10^{−4} P_{GH} P_{GT}
0.2 P_{GT} + 0.124 P_{GT} + 1.032 × 10^{−4} P_{GH}P_{GT} − 0.0516 P_{GH} − 2 = 0 (6.43)
Substituting P_{GH} from Equation (6.42) in Equation (6.43), we get
0.2 P_{GT} + 0. 124 P_{GT} + 1.032 × 10^{−4} P_{GT} (400 + 0.001 P^{2}_{GT} − P_{GT})
− 0.0516(400 + 0.001 P^{2}_{GT} − P_{GT}) − 2 = 0
By solving the above equation, we get
P_{GT} = 55.4 MW
By substituting the P_{GT} value in Equation (6.42), we get
P_{GH} = 347.66 MW
P_{L} = 3.069 MW
When P_{D} = 200 MW
From Equation (6.42), the power balance equation becomes
P_{GH} = 200 0.001 P^{2}_{GT} − P_{GT} (6.44)
Substituting P_{GH} from Equation (6.44) in Equation (6.43), we get
0.2 P_{GT} + 0.124 P_{GT} + 1.032 × 10^{−4} P_{GT} (200 + 0.001 P^{2}_{GH} − P_{GT})
− 0.0516 (200 + 0.001 P^{2}_{GT} − P_{GT}) − 2 = 0
By solving the above equation, we get
P_{GT} = 31.575 MW
By substituting the P_{GH} value in Equation (6.44), we get
P_{GH} = 169.421 MW
P_{L} = 0.9969 MW
Daily operating cost of the thermal plant

C_{1} 
= 
150 + 60 P_{GT} + 0.1 P^{2}_{GT} 


= 
Daily operating cost of the thermal plant for meeting a 400 MW load for 14 hr + daily operating cost of the thermal plant for meeting a 200 MW load for 10 hr 


= 
[150 + 60 × 55.4 + 0.1 × (55.4)^{2}] × 14 + [150 + 60 × 31.575 + 0.1 × (31.575)^{2}] × 10 


= 
Rs. 74,374.80 
Daily operating cost of the hydroplant

w 
= 
0.8 P_{GH} + 0.000333 P^{2}_{GH} m^{3}/s 


= 
Daily water quantity used for the 400 MW load for 14 hr + daily water quantity used for the 200 MW load for 10 hr 


= 
{[0.8 × 347.66 + 0.000333 × (347.66)^{2}] × 14 + [0.8 × 169.421 + 0.000333 × (169.421)^{2}] × 10 × 3600 


= 
21.2696 × 10^{6} m^{3} 
Example 6.4: A twoplant system that has a thermal station near the load center and a hydropower station at a remote location is shown in Fig. 6.5.
The characteristics of both stations are
C_{1} = (26 + 0.045P_{GT})P_{GT} Rs./hr
w_{2} = (7 + 0.004P_{GH})P_{GH} m^{3}/s
and γ_{2} = Rs. 4 × 10^{−4}/m^{3}
The transmission loss coefficient, B_{22} = 0.0025 MW^{−1}.
Determine the power generation at each station and the power received by the load
when λ = 65 Rs./MWh.
Solution:
Here, n = 2
Transmission loss,
Since the load is near the thermal station, the power flow is from the hydrostation only; therefore, B_{12} = B_{11} = 0:
For the thermal power station, the coordination equation is
FIG. 6.5 Twoplant system
For a hydropower station, the coordination equation is
By solving the above equation, we get
P_{GH} = 199.99 MW
Transmission loss, P_{L} = B_{22} P^{2}_{GH} = 0.0025(199.99)^{2} = 99.993 MW
Therefore, the power received by the load, P_{D} = P_{GT} + P_{GH} − P_{L} = 433.33 + 622.38 − 193.68 = 533.327 MW.
Example 6.5: For the system of Example 6.4, if the load is 750 MW for 14 hr a day and 500 MW for 10 hr on the same day, find the generation schedule, daily water used by the hydroplant, and the daily operating cost of thermal power.
Solution:
When load, P_{D} = 750 MW
The power balance equation, P_{GT} + P_{GH} = P_{D} + P_{L}
= 750 + 0.0025 P^{2}_{GH}
P_{GT} = 750 + 0.0025 P^{2}_{GH} − P_{GH} (6.45)
The condition for optimality is
(26 + 0.09 P_{GT}) (1 − 5 × 10^{− 3}P_{GH}) = 28 × 10^{−4} + 32 × 10^{−7} P_{GH} (6.46)
Substituting P_{GT} from Equation (6.45) in Equation (6.46), we get
[26 + 0.09 (750 + 0.0025 P^{2}_{GH} − P_{GH})](1 − 5 × 10^{− 3} P_{GH}) = 28 × 10^{−4} + 32 × 10^{−7} P_{GH}
− 1.125 × 10^{−6} P^{3}_{GH} + 6.75 × 10^{−4}P^{2}_{GH} − 0.5574 P_{GH} + 25.9922 = 0
By solving the above equation, we get
P_{GH} = 200 MW
Substituting the P_{GH} value in Equation (6.45), we get
P_{GT} = 650 MW and P_{L} = 100 MW
When load, P_{D} = 400 MW
Equation (6.45) can be modified as
P_{GT} = 400 + 0.0022 P^{2}_{GH} − P_{GH} (6.47)
Substituting the above equation in Equation (6.46), we get
[26 + 0.09 (400 + 0.0025 P^{2}_{GH} − P_{GH})](1 − 5 × 10^{− 3} P_{GH}) = 28 × 10^{−4} + 32 × 10^{−7} P_{GH}
− 1.125 × 10^{−6} P^{3}_{GH} + 6.75 × 10^{−4}P^{2}_{GH} − 0.3999 P_{GH} + 61.9972 = 0
By solving the above equation, we get
P_{GH} = 200 MW
Substituting the P_{GH} value in Equation (6.47), we get
P_{GT} = 300 MW and P_{L} = 100 MW
Daily operating cost of the hydroplant

w_{2} 
= 
(7 + 0.004P_{GH}) P_{GH} m^{3}/s 


= 
Daily water quantity used for a 750 MW load for 14 hr + daily water quantity used for a 400 MW load for 10 hr 


= 
{[7 × 200 + 0.004 × (200)^{2}] × 14 + [7 × 200 + 0.004 × (200)^{2}] × 10} × 3600 


= 
134.784 × 10^{6} m^{3} 
Daily operating cost of the thermal plant

C_{1} 
= 
(26 + 0.045P_{GT}) P_{GT} Rs./hr 


= 
Daily operating cost of a thermal plant for meeting a 750 MW load for 14 hr + daily operating cost of thermal plant for meeting a 400 MW load for 10 hr 


= 
[26 × 650 + 0.045 × (650)^{2}] × 14 + [26 × 300 + 0.045 × (300)^{2}] × 10 


= 
Rs. 6,21,275 
Example 6.6: A load is feeded by two plants, one is thermal and the other is a hydroplant. The load is located near the thermal power plant as shown in Fig. 6.6. The characteristics of the two plants are as follows:
C_{T} = 0.04P_{GT}^{2} + 30P_{GT} + 20 Rs./hr
w_{H} = 0.0012 P_{GH}^{2} + 7.5P_{GH} m^{3}/s
γ_{H} = 2.5 × 10^{–5} Rs./m^{3}
FIG. 6.6 Twoplant system
The transmission loss coefficient is B_{22} = 0.0015 MW^{−1}. Determine the power generation of both thermal and hydroplants, the load connected when λ = 45 Rs./MWh.
Solution:
Given:
Transmission loss,
The load is located near the thermal plants; hence, the power flow to the load is only from the hydroplant:
i.e., B_{11} = B_{12} = 0
∴ P_{L} = B_{22} P_{G2}^{2} = B_{22} P_{GH}^{2} = 0.0015 P_{GH}^{2}
The incremental transmission loss of the thermal plant is
Penalty factor of the thermal plant,
The incremental transmission loss of the hydroplant is
Penalty factor of the hydroplant,
The condition for hydrothermal coordination is
and
or (0.000216 P_{GH} + 0.675) 
= 
(1 − 0.003 P_{GH})45 
0.135216 P_{GH} 
= 
44.325 
∴ P_{GH} 
= 
327.809 MW 
Transmission loss, P_{L} = B_{22} P_{GH}^{2} = 0.0015(327.809)^{2} = 161.188 MW
The load connected, P_{D} = P_{GT} + P_{GH} − P_{L} = 187.5 + 327.809 − 161.188 = 354.121 MW
Example 6.7: For Example 6.6, determine the daily water used by the hydroplant and the daily operating cost of the thermal plant with the load connected for totally 24 hr.
Solution:
From Example 6.6,
The load connected, P_{D} 
= 354.121 MW 
Generation of the thermal plant, P_{GT} 
= 187.5 MW 
Generation of the hydroplant, P_{GH} 
= 327.809 MW 
The daily water used is

w_{H} 
= 
0.0012 P_{GH}^{2} + 7.5 P_{GH} m^{3}/s 


= 
[0.0012 P_{GH}^{2} + 7.5 P_{GH}] × 3,600 m^{3}/hr 


= 
[0.0012 P_{GH}^{2} + 7.5 P_{GH}] × 3,600 × 24 m^{3}/day 
Substituting the value of P_{GH} = 327.809 MW in the above equation, we have

w_{H} 
= 
[0.0012(327.809)^{2} + 7.5 × 327.809] × 3,600 × 24 


= 
223.56 × 10^{6} m^{3} 
Daily operating cost of the thermal plant 
= 
(0.04 P_{GT}^{2} + 30 P_{GT} + 20) Rs./h 

= 
Rs. [0.04(187.5)^{2} + 30 (187.5) + 20] × 24 

= 
Rs.1,69,230 
Example 6.8: In a twoplant operation system, the hydroplant operates for 8 hr during each day and the steam plant operates throughout the day. The characteristics of the steam and hydroplants are
C_{T} = 0.025P_{GT}^{2} + 14P_{GT} + 12 Rs./hr
w_{H} = 0.002P_{GH}^{2} + 28P_{GH} m^{3}/s
When both plants are running, the power flow from the steam plant to the load is 190 MW and the total quantity of water used for the hydroplant operation during 8 hr is 220 × 10^{6} m^{3}.
Determine the generation of a hydroplant and cost of water used. Neglect the transmission losses.
Solution:
The cost of the thermal plant is
C_{T} = (0.025 P_{GT}^{2} + 14P_{GT} + 12) Rs./hr
The incremental fuel cost of the thermal plant is
and for the hydroplant, w_{H} = (0.002P_{GH}^{2} + 28P_{GH}) m^{3}/s
The incremental water flow is
For hydrothermal scheduling, the optimal condition is
(since losses are neglected, L_{T} = 1)
Power flow to the load from the thermal plant, P_{GT} = 190 MW (given). By substituting the value of P_{GT} = 190 MW in the above equation, we get
λ = 0.05(190) + 14 = 23.5 Rs./MWh
The total quantity of water used during a onehour operation is

w_{H} 
= 0.0012 P_{GH}^{2} + 7.5P_{GH} m^{3}/s 


= [0.0012 P_{GH}^{2} + 7.5P_{GH}] × 3,600 m^{3}/hr 
For an 8hr operation, the quantity of water used is
Let the cost of water be γ_{H} Rs./hr/m^{3}/s.
From Equation (6.48)
Example 6.9: A twoplant system with no transmission loss shown in Fig. 6.7(a) is to supply a load shown in Fig. 6.7(b).
The data of the system are as follows:
C_{1} = (32 + 0.03P_{GT})P_{GT}
w_{2} = (8 + 0.004P_{GH})P_{GH} m^{3}/s
The maximum capacity of the hydroplant and the steam plant are 450 and 250 MW, respectively. Determine the generating schedule of the system so that 150.3478 million m^{3} water is used during the 24hr period.
FIG. 6.7 (a) Twoplant system; (b) daily load curve
Solution:
(i) Constant hydrogeneration
If P_{GH} is the hydropower in MW generated in 24 hr, then we have

(8 + 0.004 P_{GH})P_{GH} × 24 × 60 × 60 
= 
150.3478 × 10^{6} 

8 P_{GH} + 0.004 P^{2}_{GH} 
= 
1,740.136 

0.004 P^{2}_{GH} + 8 P_{GH} − 1,740.136 
= 
0 
By solving the above equation, we get
P_{GH} = 197.929 MW
During the peak load of 600 MW
Hydrogeneration, P_{GH} = 197.929 MW
Thermal generation, P_{GT} = 600  P_{GH} = 600  197.929 = 402.071 MW
During offpeak load of 400 MW
Hydrogeneration, P_{GH} = 197.929 MW
Thermal generation, P_{GT} = 400  P_{GH} = 400  197.929 = 402.071 MW
The running cost of a steam plant for 24 hr is

C_{1} 
= 
(32 + 0.03P_{GT})P_{GT} × 12/at _{600 MW} + (32 + 0.03P_{GT})P_{GT} × 12/at _{400 MW} 


= 
(32 + 0.03 × 402.071) 402.071 × 12 + (32 + 0.03 × 202.071) 202.071 × 12 


= 
Rs. 3,04,888.288 
(ii) Constant thermal generation
If P_{GH} is the hydropower during the peak load period
(P_{GH} – 200) is the hydropower during the offpeak load period
Given w_{2} = (8 + 0.004P_{GH})P_{GH} m^{3}/s
{(8 + 0.004P_{GH})P_{GH} + [8 + 0.004(P_{GH} − 200)](P_{GH} − 200)} × 12 × 3,600 = 150.3478 × 10^{6}
After simplification, we get
8 × 10 ^{− 3} P_{GH}^{2} + 14.4P_{GH} – 4,920.273 = 0
∴ P_{GH} = 93.74793 MW
The generation scheduling is given as follows:
Hydro  Thermal (P_{D} – P_{GH})  

Peak (600) 
293.75 MW 
306.25 MW 
Offpeak (400) 
93.75 MW 
306.25 MW 
The steam plant operating cost for 24 hr is

C_{1} 
= 
(32 + 0.03P_{GT})P_{GT} 


= 
(32 + 0.03 × 306.25) 306.25 × 12 + (32 + 0.03 × 306.25) 306.25 × 12 


= 
Rs. 3,02,728.125 
(iii) Equal incremental plant costs
Let P′_{GT} and P′_{GH} be the steam generation and hydrogeneration during peak loads, P″_{GT} and P″_{GH} the steam generation and hydrogeneration during offpeak loads, respectively.
For peak load conditions:
The value of λ′ should be so chosen as to make
For offpeak periods:
The value of λ″ should be chosen so as to make
For the whole operating period, γ_{2} should be chosen so as to use the same value of water, i.e., 150.3478 million m^{3} during the 24hr period.
{(8 + 0.004 P′_{GH} )P′_{GH} +(8 + 0.004 P″_{GH} )P″_{GH} } × 12 × 3,600 = 150.3478 × 10^{6} (6.55)
All the above equations can be solved by a hitandtrail or an iterative method:

P′_{GT} 
= 
276.362 MW, P′_{GH} = 323.638MW 

λ′ 
= 
48.58172 Rs./MWh 



− (8 + 0.004 × 323.638) × 323.638 + 3,480.273= (8 + 0004 P‴_{GH} )P″_{GH} 

8 P′_{GH} 
= 
+ 0.004 P″^{2}_{GH} = 472.2 
By solving the above equation, we get

P″_{GH} + 57.38 MW P″_{GT} 
= 
342.62 MW 

λ″ 
= 
52.5572 Rs. /MWh 

γ_{2} 
= 
6.2131 Rs./hr/m3/s 
The thermal operating cost

C_{1} 
= 
(32 + 0.03 P_{GT})P_{GT} 


= 
(32 + 0.03 × 276.362) × 276.362 × 12 + (32 + 0.03 × 342.62) × 342.62 × 12 


= 
Rs. 3,07,444.279 
(iv) Maximum hydroefficiency method
Let it be assumed that the maximum efficiency of a hydrounit occurs at 275 MW.
Therefore, the hydropower plant supply is 275 MW during the peak load. The amount of water used during peak load hours:

w_{2}  = (8 + 0.004P_{GH})P_{GH} m^{3}/s 
= (8 + 0.004 × 275) × 275 × 12 × 3,600 = 108.108 × 10^{6} 
Water available for offpeak hydrogeneration:
= total water available – water available at peak load
= 150.34 × 10^{6} − 108.108 × 10^{6} = 42.2398 × 10^{6} m^{3}
The real power generation of a hydroplant P_{GH} during offpeak hours is found by using
(8 + 0.004P_{GH}) P_{GH} × 12 × 3600 = 4,22,39,800
0.004P_{GH}^{2} + 8P_{GH} – 977.77 = 0
P_{GH(Offpeak load)} = 115.5461 MW
The generation scheduling is given as follows:
Hydro  Thermal (P_{D} – P_{GH})  

Peak (600) 
273 MW 
325 MW 
Offpeak (400) 
115.546 MW 
284.4539 MW 
The daily operating cost of a thermal plant

C_{1} 
= 
(32 + 0.03P_{GT})P_{GT} 


= 
(32 + 0.03 × 325) 325 × 12 + (32 + 0.03 × 284.4539) 284.4539 × 12 


= 
Rs. 12,43,023.55 
Example 6.10: A thermal station and a hydrostation supply an area jointly. The hydrostation is run 16 hr daily and the thermal station is run through 24 hr. The incremental fuel cost characteristics of the thermal plant are
C_{T} = 6 + 12 P_{GT} + 0.04P_{GT}^{2} Rs./hr
If the load on the thermal station, when both plants are in operation, is 350 MW, the incremental water rate of a hydropower plant . The total quantity of water utilized during a 16hr operation of the hydroplant is 450 million m^{3}. Find the generation of the hydroplant and cost of water use. Assume that the total load on the hydroplant is constant for the 16hr period.
Solution:
Given: C_{T} = 6 + 12P_{GT} + 0.04P_{GT}^{2}
P_{GT} = 350 MW (given)
∴ 12 + 0.08 × 350 = λ
λ = 40Rs./MWh
The total quantity of water used during 16 hr of operation of a hydroplant is
(28 + 0.03P_{GH})P_{GH} × 16 × 3, 600 = 450 × 10^{6}
0.03P_{GH}^{2} + 28P_{GH} = 7,812.5
0.03P_{GH}^{2} + 28P_{GH} – 7,812.5 = 0
By solving the above equation, we get
P_{GH} = 224.849 MW
If the cost of water used is γ, then we have
γ (28 + 0.03P_{GH}) = λ
γ (28 + 0.03 × 224.849) = 40
∴ γ = 1.15122 Rs./hr/m^{3}/s
6.9 ADVANTAGES OF OPERATION OF HYDROTHERMAL COMBINATIONS
The following advantages are obtained by operation combination of hydrothermal power plants.
6.9.1 Flexibility
The power system reliability and security can be obtained by the combined operation of hydro and thermal units. It provides the reserve capacity to meet the random phenomena of forced outage of units and unexpected load implied on a system.
Thermal plants require an appreciable time for starting and for being put into service. Hydroplants can be started and put into operation very quickly with lower operating costs. Hence, it is required to operate hydroplants economically as baseload plants as well as peak load plants. Hydroplants are most preferable to operate as peak load plants such that their operation improves the flexibility of the system operation and makes the thermal plant operation easier.
6.9.2 Greater economy
The runoff river hydroplants would generally meet the entire or part of the base loads, and thermal plants should be set up to increase the firm capacity of the system. The remaining power demand can be met by a combination of reservoirtype hydroplants, thermal plants, and nuclear plants. In every power system, a certain ratio of hydropower to total power demand will result in a minimum overall cost of supply.
6.9.3 Security of supply
Water availability must depend on the season. It is high during the rainy season and may be reduced due to the occurrence of draught during longer plants. Problems arise in the thermal power plant operation due to transportation of coal, unavailability of labor, etc. It is found that the forced outages of hydroplants are few compared to those in thermal plants.
The above facts suggested the operation of hydrothermal systems to maintain the reliability and security of supply to the consumers.
6.9.4 Better energy conservation
During heavy runoff periods, the generation of hydropower is more, which results in the conservation of fossil fuels. During draught periods, more steam power has to be generated such that the availability of water needs the minimum needs like drinking and agricultural events.
6.9.5 Reserve capacity maintenance
For the operation of a power system, it is necessary that every system has some certain reserve capacity to meet the forced outages and unexpected load demands. By the combined operation of hydro and thermal plants, the reserve capacity maintenance is reduced.
Example 6.11: MATLAB program on hydrothermal scheduling without inflow and without loss. Find the optimum generation for a hydrothermal system for a typical day, wherein load varies in three steps of 8 hr each as 15, 25, and 8 MW, respectively. There is no water inflow into the reservoir of the hydroplant. The initial water storage in the reservoir is 180 m^{3}/s and the final water storage should be 100 m^{3}/s. The basic head is 35 m and the waterhead correction factor e is 0.005. Assume for simplicity that the reservoir is rectangular so that ρ does not change with water storage. Let the noneffective water discharge be assumed as 4 m^{3}/s. The incremental fuel cost (IFC) of the thermal power plant is . Further transmission losses may be neglected.
PROGRAM IS UNDER THE FILE NAME hydrothermal.m
clc;
clear;
load =[15;25;8]; %load in MW
time =[8;8;8]; time of each load
ni=length(load(:,1);
x(1)=180;
x(ni+1)=100;
h0=35;
e=0.005;
ro=4;
ifc=[2.0 25.0];
delt=time(1);
alpha=0.5;
%assuming in initial discharges
q(2)=20;
q(3)=22;
maxgrad=1;
iter=0;
while(maxgrad>0.1)
iter=iter+1;
sumq=0;
for I=2:ni
sumq=sumq+q(i);
end
q(1)=x(1)−x(ni+1)−sumq;
for I=2:ni
x(i)=x(i−1)−q(i−1); %water storage before ith interval
end
for i=1:ni
pd(i)=load(i,1);
pgh(i)= 9.81*0.001*h0*(1+0.5*e*(x(i)+x(i+1)))*(q(i)−ro);
if pgh(i)>pd(i)
pgh(i)=pd(i);
pgt(i)=0;
else
pgt(i)=pd(i)−pgh(i);
end
lambda1(i)=ifc(1,1)*pgt(i)+ifc(1,2);
lambda3(i)=lambda1(i);%since losses are neglected
if i==1
lambda2(i)=lambda3(i)*9.81*0.001*h0*(1+0.5*e*(2*x(i)−
2*q(i)+ro));
else
lambda2(i)=lambda2(i−1)−0.5*9.81*0.001*h0*e*(lambda3
(i−1)*(q(i−1)−ro)+lambda3(i)*(q(i)−ro));
end
if i~=1
grad(i)=lambda2(i)−lambda3(i)*9.81*0.001*h0*(1+0.5*e*(2*x
(i)−2*q(i)+ro));
end
end
maxgrad=(max(abs(grad)));
for i=2:ni
q(i)=q(i)−alpha*(grad(i));
end
end
pgh
pgt
iter
RESULTS:
pgh = 12.4706 
21.8178 
5.4672 
pgt = 2.5294 
3.1822 
2.5328 
netPG = 15 
25 
8 
iter = 15 


Example 6.12: MATLAB program on hydrothermal scheduling with inflow and without losses. Find the optimum generation for a hydrothermal system for a typical day, wherein load varies in three steps of 8 hr each as 15, 25, and 8 MW, respectively. There is water inflow into the reservoir of the hydroplant in three intervals of 2, 4, and 3 m^{3}/s. The initial water storage in the reservoir is 180 m^{3}/s and the final water storage should be 100 m^{3}/s. The basic head is 35–m and the waterhead correction factor e is 0.005. Assume for simplicity that the reservoir is rectangular so that ρ does not change with water storage. Let the noneffective water discharge be assumed as 4 m^{3}/s. The IFC of the thermal power plant is Further transmission losses may be neglected.
PROGRAM IS UNDER THE FILE NAME hydrothermalinflow.m
clc;
clear;
load=[15;25;8]; %load in MW
time=[8;8;8]; %time of each load
ni=length(load(:,1));
x(1)=180;
x(ni + 1)=100;
h0=35;
e=0.005;
ro=4;
ifc=[2.0 25.0];
delt=time(1);
alpha=0.5;
J=[2;4;3];
%assuming initial discharges
q(2)=20;
q(3)=22;
maxgrad=1;
iter=0;
while(maxgrad>0.1)
iter=iter + 1;
sumq=0;
sumj=0;
for i=2:ni
sumq=sumq + q(i);
end
for i=1:ni
sumj=sumj + J(i,1);
end
q(1)=x(1)−x(ni+1)−sumq + sumj;
for i=2:ni
x(i)=x(i−1)−q(i−1) + J(i,1); %water storage before ith
interval
end
for i=1:ni
pd(i)=load(i,1);
pgh(i)=9.81*0.001*h0*(1 + 0.5*e*(x(i) + x(i + 1)))*(q(i)−ro);
if pgh(i)>pd(i)
pgh(i)=pd(i);
pgt(i)=0;
else
pgt(i)=pd(i)−pgh(i);
end
lambda1(i)=ifc(1,1)*pgt(i) + ifc(1,2);
lambda3(i)=lambda1(i); %since losses are neglected
if i=1
lambda2(i)=lambda3(i)*9.81*0.001*h0*(1 + 0.5*e*(2*x(i)−
2*q(i) + J(i,1) + ro));
else
lambda2(i)=lambda2(i−1)−0.5*9.81*0.001*h0*e*(lambda3
(i−1)*(q(i−1)−ro) + lambda3(i)*(q(i)−ro));
end
if i~=1
grad(i)=lambda2(i)−
lambda3(i)*9.81*0.001*h0*(1 + 0.5*e*(2*x(i)−
2*q(i) + J(i,1) + ro));
end
end
maxgrad=max(abs(grad));
for i=2:ni
q(i)=q(i)−alpha*(grad(i));
end
end
pgh
pgt
iter
RESULTS:
pgh = 14.0553 
23.6463 
7.3583 
pgt = 0.9447 
1.3537 
0.6417 
netPG = 15 
25 
8 
iter = 15 


Example 6.13: MATLAB program on hydrothermal scheduling without inflow and with losses. Find the optimum generation for a hydrothermal system for a typical day, wherein load varies in three steps of 8 hr each as 15, 25, and 8 MW, respectively. There is no water inflow into the reservoir of the hydroplant. The initial water storage in the reservoir is 180 m^{3}/s and the final water storage should be 100 m^{3}/s. The basic head is 35 m and the waterhead correction factor e is 0.005. Assume for simplicity that the reservoir is rectangular so that ρ does not change with water storage. Let the noneffective water discharge be assumed as 4 m^{3}/s. The IFC of the thermal power plant is . Further transmission losses are considered and are taken as follows: .
PROGRAM IS UNDER THE FILE NAME hydrothermalloss.m
clc;
clear;
load=[15;25;8]; %load in MW
time=[8;8;8]; %time of each load
ni=length(load(:,1));
x(1)=180;
x(ni + 1)=100;
h0=35;
e=0.005;
ro=4;
ifc=[2.0 25.0];
delt=time(1);
alpha=0.5;
dpl=0.5;
%assuming initial discharges
q(2)=20;
q(3)=22;
maxgrad=1;
iter==0;
while(maxgrad>0.1)
iter=iter + 1;
sumq=0;
for i=2:ni
sumq=sumq + q(i);
end
q(1)=x(1)−x(ni + 1)−sumq;
for i=2:ni
x(i)=x(i−1)−q(i−1); %water storage before ith interval
end
for i=1:ni
pd(i)=load(i,1);
pgh(i)=9.81*0.001*h0*(1 + 0.5*e*(x(i) + x(i + 1)))*(q(i)−ro);
if pgh(i)>pd(i)
pgh(i)=pd(i);
pgt(i)=0;
else
pgt(i)=pd(i)−pgh(i);
end
lambda1(i)=(ifc(1,1)*pgt(i) + ifc(1,2))/(1−dpl);
lambda3(i)=lambda1(i)*(1−dpl);
if i==1
lambda2(i)=lambda3(i)*9.81*0.001*h0*(1 + 0.5*e*(2*x(i)−
2*q(i) + ro));
else
lambda2(i)=lambda2(i−1)−0.5*9.81*0.001*h0*e*(lambda3
(i−1)*(q(i−1)−ro) + lambda3(i)*(q(i)−ro));
end
if i~=1
grad(i)=lambda2(i)−
lambda3(i)*9.81*0.001*h0*(1 + 0.5*e*(2*x(i)−2*q(i) + ro));
end
end
maxgrad=(max(abs(grad)));
for i=2:ni
q(i)=q(i)−alpha*(grad(i));
end
end
pgh
pgt
iter
grad
RESULTS:
pgh = 12.4706 
21.8178 
5.4672 
pgt = 2.5294 
3.1822 
2.5328 
grad = 0 
−0.0765 
0.0826 
netPG = 15 
25 
8 
iter = 15 


Example 6.14: MATLAB program on hydrothermal scheduling with inflow and with losses. Find the optimum generation for a hydrothermal system for a typical day, wherein load varies in three steps of 8 hr each as 15, 25 and 8 MW, respectively. There is water inflow into the reservoir of the hydroplant in three intervals of 2, 4, 3 m^{3}/s. The initial water storage in the reservoir is 180 m^{3}/s and the final water storage should be 100 m^{3}/s. The basic head is 35 m and the waterhead correction factor e is 0.005. Assume for simplicity that the reservoir is rectangular so that ρ does not change with water storage. Let the noneffective water discharge be assumed as 4 m^{3}/s. The IFC of the thermal power plant is . Further transmission losses are considered and are taken as follows:
PROGRAM IS UNDER THE FILE NAME hydrothermalinflowloss.m
clc;
clear;
load=[15;25;8]; %load in MW
time=[8;8;8]; %time of each load
ni=length(load(:,1));
x(1)=180;
x(ni + 1)=100;
h0=35;
e=0.005;
ro=4;
ifc=[2.0 25.0];
delt=time(1);
alpha=0.5;
J=[2;4;3];
dpl=0.15;
%assuming initial discharges
q(2)=20;
q(3)=22;
maxgrad=1;
iter=0;
while(maxgrad>0.1)
iter=iter + 1;
sumq=0;
sumj=0;
for i=2:ni
sumq=sumq + q(i);
end
for i=1:ni
sumj=sumj + J(i,1);
end
q(1)=x(1)−x(ni + 1)−sumq + sumj;
for i=2:ni
x(i)=x(i−1)−q(i−1) + J(i,1);  %water storage before ith 
interval 
end
for i=1:ni
pd(i)=load(i,1);
pgh(i)=9.81*0.001*h0*(1 + 0.5*e*(x(i) + x(i + 1)))*(q(i)−ro);
if pgh(i)>pd(i)
pgh(i)=pd(i);
pgt(i)=0;
else
pgt(i)=pd(i)−pgh(i);
end
lambda1(i)=(ifc(1,1)*pgt(i) + ifc(1,2))/(1−dpl);
lambda3(i)=lambda1(i)*(1−dpl);
if i=1
lambda2(i)=lambda3(i)*9.81*0.001*h0*(1 + 0.5*e*(2*x(i)−
2*q(i) + J(i,1) + ro));
else
lambda2(i)=lambda2(i−1)−0.5*9.81*0.001*h0*e*(lambda3
(i−1)*(q(i−1)−ro) + lambda3(i)*(q(i)−ro));
end
if i~=1
grad(i)=lambda2(i)−
lambda3(i)*9.81*0.001*h0*(1 + 0.5*e*(2*x(i)−
2*q(i) + J(i,1) + ro));
end
end
maxgrad=max(abs(grad));
for i=2:ni
q(i)=q(i)−alpha*(grad(i));
end
end
pgh
pgt
iter
RESULTS:
pgh = 14.0553 
23.6463 
7.3583 
pgt = 0.9447 
1.3537 
0.6417 
netPG = 15 
25 
8 
iter = 15 


KEY NOTES
 The optimal scheduling problem in the case of thermal plants can be completely solved at any desired instant without referring to the operation at other times. It is a static optimization problem.
 The optimal scheduling problem in the hydrothermal system is a dynamic optimization problem where the time factor is to be considered.
 The optimal scheduling problem in a hydrothermal system can be stated as minimizing the fuel cost of thermal plants under the constraint of water availability for hydrogeneration over a given period of operation.
 The methods of hydrothermal coordination are:
 Constant hydrogeneration method.
 Constant thermal generation method.
 Maximum hydroefficiency method.
 Kirchmayer’s method.
 Constant hydrogeneration method—A scheduled amount of water at a constant head is used such that the hydropower generation is kept constant throughout the operating period.
 Constant thermal generation method—Thermal power generation is kept constant throughout the operating period in such a way that the hydropower plants use a specified and scheduled amount of water and operate on varying power generation schedule during the operating period.
 Maximum hydroefficiency method—During peakload periods, the hydropower plants are operated at their maximum efficiency; during offpeak load periods, they operate at an efficiency nearer to their maximum efficiency with the use of a specified amount of water for hydropower generation.
 Kirchmayer’s method—The coordination equations are derived in terms of penalty factors of both plants for obtaining the optimum scheduling of the hydrothermal system and hence it is also known as the penalty factor method of solution of shortterm hydrothermal scheduling problems.
 Longterm hydrothermal scheduling problems can be solved by the discretization principle.
 In the longterm hydrothermal scheduling problem, it is convenient to choose water discharges in all subintervals except one subinterval as independent variables and hydrogenerations, thermal generations, water storages in all subintervals, and excepted water discharge as dependent variables,
i.e., Independent variables are represented by q^{K}, for K = 2, 3,…,N
≠ 1
Dependent variables are represented by P^{K}_{GH}, P^{K}_{GT}, X^{K,} and q^{1}, for K = 1,2,…,N. [Since the water discharge in one subinterval is a dependent variable.]
 For optimality of longterm hydrothermal scheduling, the gradient vector should be zero, i.e., .
SHORT QUESTIONS AND ANSWERS
 Why is the optimal scheduling problem in the case of thermal plants referred to as a static optimization problem?
Optimal scheduling problem can be completely solved at any desired instant without referring to the operation at other times.
 The optimization problem in the case of a hydrothermal system is referred to as a dynamic problem. Why is it so?
The operation of the system having hydro and thermal plants have negligible operation costs but is required under the constraint of water availability for hydrogeneration over a given period of time.
 What is the statement of optimization problem of hydrothermal system?
Minimize the fuel cost of thermal plants under the constraint of water availability for hydrogeneration over a given period of time.
 In the optimal scheduling problem of a hydrothermal system, which variables are considered as control variables?
Thermal and hydropower generations (P_{GT} and P_{GH}).
 Fastchanging loads can be effectively met by which type of plants?
Hydroplants.
 Generally, which type of plants are more suitable to operate as baseload and peak load plants?
Thermal plants are suited for baseload plants and hydroplants are suited for peak load plants.
 Whole or part of the base load can be supplied by which type of hydroplants?
Runoff river type.
 The peak load or remaining base load is met by which type of plants?
A proper coordination of reservoirtype hydroplants and thermal plants.
 In the optimal scheduling problem of a hydrothermal system, what parameters are assumed to be known as the function of time with certainty?
Water inflow to the reservoir and load demand.
 What is the mathematical statement of the optimization problem in the hydrothermal system?
Determine the water discharge rate q(t) so as to minimize the cost of thermal generation.
 Write the objective function expression of hydrothermal scheduling problem.
 Write the constraint equations of the hydrothermal scheduling problem.
P_{GT}(t) + P_{GH}(t) − P_{L}(t) − P_{D}(t) = 0
for t ∈ (O,T)—Real power balance equation
—P_{GH}(t) = (X‴ (t), q(t))
 By which principle can the optimal scheduling problem of a hydrothermal system be solved?
Discretization principle.
 Write the expression for real power hydrogeneration in any subinterval ‘K’?
P^{K}_{GH} = h_{o} {1 + 0.5e (X^{K} + X ^{K − 1})} (q^{K} − ρ)
 Define the terms of the above real power hydrogeneration.
P^{K}_{GH} = h_{o} {1 + 0.5e (X^{K} + X ^{K − 1})} (q^{K} − ρ)
where h_{0} = 9.81 × 10^{–3} h_{0}^{1}, h_{o}^{1} is the basic water head that corresponds to dead storage, e the waterhead correction factor to account for the variation in head with water storage, X^{k} the water storage at interval k, q^{k} the water discharge at interval k, and ρ the noneffective discharge.
 In the optimal scheduling problem of a hydrothermal system, which variables are used to choose as independent variables?
Water discharges in all subintervals except one subinterval:
i.e., e^{k}_{k ≠1}, for q^{2}, q^{3} ⋯ q^{N}
where k = 2, 3 … N (k is subinterval).
 Which parameters are used as dependent variables?
Thermal, hydrogenerations, water storages at all subintervals, and water discharge at excepted subintervals are used as dependent variables,
i.e., P_{GT}^{k}, P_{GH}^{k} X^{k}, and q^{1}
 In solving the optimal scheduling problem of a hydrothermal system, for ‘N’ subintervals (i.e., k = 1, 2, …, N), N−1 number of water discharges q’s can be specified as independent variables except one subinterval. Write the expression for water discharge in the excepted subinterval, which is taken as a dependant variable.
 Which technique is used to obtain the solution to the optimization problem of the hydrothermal system?
A nonlinear programming technique in conjunction with a firstorder negative gradient method is used to obtain the solution to the optimization problem.
 Write the expression for a Lagrangian function obtained by augmenting the objective function with constraint equations in the case of a hydrothermal scheduling problem.
 What is the gradient vector?
The partial derivatives of the Lagrangian function with respect to independent variables are
 What is the condition for optimality in the case of a hydrothermal scheduling problem?
The gradient vector should be zero:
 The condition for optimality in a hydrothermal scheduling problem is that the gradient vector should be zero. If this condition is violated, how will we obtain the optimal solution?
Find the new values of control variables, which will optimize the objective function. This can be achieved by moving in the negative direction of the gradient vector to a point where the value of the objective function is nearer to an optimal value.
 For a system with a multihydro and a multithermal plant, the nonlinear programming technique in conjunction with the firstorder gradient method is also directly applied. However, what is the drawback?
It requires large memory since the independent and dependant variables, and gradients need to be stored simultaneously.
 By which method can the drawback of the nonlinear programming technique be overcome when applied to a multihydro and multithermal plant system and what is its procedure?
By the method of decomposition technique. In this technique, the optimization is carried out over each subinterval and a complete cycle of iteration is replaced, if the water availability equation does not check at the end of the cycle.
 For shortrange scheduling of a hydrothermal plant, which method is useful?
Kirchmayer’s method or the penalty factor method is useful for shortrange scheduling.
 What is Kirchmayer’s method of obtaining the optimum scheduling of a hydrothermal system?
In Kirchmayer’s method or the penalty factor method, the coordination equations are derived in terms of penalty factors of both hydro and thermal plants.
 What is the condition for optimality in a hydrothermal scheduling problem when considering transmission losses?
where i represents the thermal plant and j represents the hydroplant.
 What is the meaning of the terms and ?
is the incremental cost of the i th thermal plant and
is the incremental water rate of the jth hydroplant.
 What is shortterm hydrothermal coordination?
Shortterm hydrothermal coordination is done for a fixed quantity of water to be used in a certain period (i.e., 24 hr).
 What are the scheduling methods for shortterm hydrothermal coordination?
 Constant hydrogeneration method.
 Constant steam generation method.
 Maximum hydroefficiency method.
 Equal incremental production costs and solution of coordination equations (Kirchmayer’s method).
 What is the significance of the coefficient γ_{j}?
γ_{j} represents the incremental water rates into incremental costs which must be so selected as to use the desired amount of water during the operating period.
 Write the condition for optimality in the problem of a short range hydrothermal system according to Kirchmayer’s method when neglecting transmission losses.
 What is the significance of terms and ?
is the penalty factor of the i^{th} thermal plant and
is the penalty factor of the j^{th} hydroplant
These terms are very much useful in getting the optimality in a hydrothermal scheduling problem, which is solved by Kirchmayer’s method.
 Write the condition for optimality in an optimal scheduling problem of a short range hydrothermal system with approximate penalty factors.
MULTIPLECHOICE QUESTIONS
 When compared to a hydroelectric plant, the operating cost of the thermal plant is very _____ and its capital cost is _____.
 High, low.
 High, high.
 Low, low.
 Low, high.
 When compared to a thermal plant, the operating cost and capital cost of a hydroelectric plant are:
 High and low.
 Low and high.
 Both high.
 Both low.
 The optimal scheduling problem in the case of thermal plants is:
 Static optimization problem.
 Dynamic optimization problem.
 Static as well as dynamic optimization problem.
 Either static or dynamic optimization problem.
 The operation of the system having hydro and thermal plants is more complex. In this case, the optimal scheduling problem is:
 Static optimization problem.
 Dynamic optimization problem.
 Static as well as dynamic optimization problem.
 Either static or dynamic optimization problem.
 The optimal scheduling problem in the case of a thermal plant can be completely solved at any desired instant:
 With reference to the operation at other times.
 Without reference to the operation at other times.
 Case (a) or case (b) that depends on the size of the plant.
 None of these.
 The time factor is considered in solving the optimization problem of _____.
 Hydro plants.
 Thermal plants.
 Hydrothermal plants.
 None of these.
 The objective function to the optimization problem in a hydrothermal system becomes:
 Minimize the fuel cost of thermal plants.
 Minimize the time of operation.
 Maximize the water availability for hydrogeneration.
 All of these.
 The optimal scheduling problem of a hydrothermal system is solved under the constraint of:
 Fuel cost of thermal plants for thermal generation.
 Time of operation of the entire system.
 Water availability for hydrogeneration over a given period.
 Availability of coal for thermal generation over a given period.
 To solve the optimization problem in a hydrothermal system, which of the following variables are considered as control variables?
 P_{D thermal} and P_{G hydro}.
 Q_{D thermal} and Q_{D hydro}.
 P_{G thermal} and P_{D hydro}.
 P_{G thermal} and P_{G hydro}.
 In which system is the generation scheduled generally such that the operating costs of thermal generation are minimized?
 Systems where there is a close balance between hydro and thermal generation.
 Systems where the hydrocapacity is only a fraction of the total capacity.
 Both (a) and (b).
 None of these.
 Thermal plants are more suitable to operate as _____ plants leaving hydroplants to operate as _____ plants.
 Base load, base load.
 Peak load, peak load.
 Peak load, base load.
 Base load, peak load.
 In hydrothermal systems, the whole are part of the base load that can be supplied by:
 Runoff rivertype hydroplants.
 Reservoirtype hydroplants.
 Thermal plants.
 Reservoirtype hydroplants and thermal plants with proper coordination.
 In a hydrothermal system, the peak load can be met by:
 Runoff rivertype hydroplants.
 Reservoirtype hydroplants.
 Thermal plants.
 Reservoirtype hydroplants and thermal plants with proper coordination.
 For an optimal scheduling problem, it is assumed, which parameter is known deterministically as a function of time?
 Water inflow to the reservoir.
 Power generation.
 Load demand.
 Both (a) and (c).
 In a hydrothermal system, the optimization problem is stated as determining _____ so as to minimize the cost of thermal generation.
 Load demand (P_{D}).
 Water storage (X).
 Water discharge rate (q(t)).
 Water inflow rate (J(t)).
 Which of the following equations is considered as a constraint to the optimization problem of a hydrothermal system?
 Real power balance equation.
 Water availability equation.
 Real power hydrogeneration as a function of water storage.
 All of these.
 The water availability equation is:
 P_{GH}(t) + P_{GH}(t) − P_{L}(t) − P_{D}(t) = 0, t∈(0,T).
 P_{GH}(t) = f(X′(t), q(t)).
 None of these.
 In the optimization problem of a hydrothermal system, the constraint real power hydrogeneration is a function of:
 Water inflow rate (J(t)).
 Water storage (X).
 Water discharge rate (q(t))
 (i) and (ii).
 (ii) and (iii).
 (i) and (iii).
 None of these.
 The optimization scheduling problem of a hydrothermal system can be conveniently solved by _____ principle.
 Dependence.
 Discretization.
 Dividing.
 None of these.
 In the discretization principle, the real power hydrogeneration at any subinterval ‘k’ can be expressed as:
 P^{K}_{GH} = h_{o} {1 + 0.5e (X^{k − 1} + X^{k})} (q^{k} − ρ)
 P^{K}_{GH} = h_{o} {1 − 0.5e (X^{k − 1} + X^{k})} (q^{k} − ρ)
 P^{K}_{GH} = h_{o} {1 + 0.5e (X^{k − 1} + X^{k})} (q^{k} − ρ)
 P^{K}_{GH} = h_{o} {1 + 0.5e (X^{k − 1} + X^{k − 1})} (q^{k} − ρ)
 In the optimization problem of a hydrothermal system, which of the following are closed as independent variables?
 Water storages in all subintervals except one subinterval.
 Water inflows in all subintervals except one subinterval.
 Water discharges in all subintervals except one subinterval.
 Hydro and thermal generations, water storages at all subintervals, and water discharge at one subinterval.
 In the optimal scheduling problem of a hydrothermal system, which of the following are closed as dependent variables?
 Water storages in all subintervals except one subinterval.
 Water inflows in all subintervals except one subinterval.
 Water discharges in all subintervals except one subinterval.
 Hydro and thermal generations, water storages at all subintervals, and water discharge at one subinterval.
 To obtain the solution to the optimization problem of a hydrothermal system, which of the following technique is used?
 Nonlinear programming technique in conjunction with the firstorder gradient method.
 Linear programming technique in conjunction with the firstorder gradient method.
 Nonlinear programming technique in conjunction with the multipleorder gradient method.
 Linear programming technique in conjunction with the multipleorder gradient method.
 In a hydrothermal system for optimality, the condition is:
 Gradient vector should be zero.
 Gradient vector should be positive.
 Gradient vector should be negative.
 None of these.
 For multihydro and multithermal plants, the optimization problem can be solved by a modified technique, which is known as:
 Discretization technique.
 Decomposition technique.
 Decoupled technique.
 None of these.
 In Kirchmayer’s method of solution of optimization problem in a hydrothermal system, the coordination equations are derived in terms of _____ of both plants.
 Power generations.
 Power demands.
 Penalty factors.
 All of these.
 γ_{j} is used as a constant, in an optimization problem of a hydrothermal system, which converts:
 Fuel cost of a thermal plant into an incremental fuel cost.
 Incremental water rate of a hydroplant into an incremental cost.
 Incremental water inflow rate into an incremental discharge rate.
 None of these.
 The power generation of a hydroplant P_{GH} is directly proportional to:
 Plant head.
 Water discharge ω_{j}.
 Both (a) and (b).
 None of these.
 The main advantages of the operation of a hydrothermal system are:
 Greater economy.
 Security of supply and flexibility.
 Better energy conservation.
 Reduction in reserve capacity maintenance.
Regarding the above statement, which is correct?
 (i) and (ii).
 (ii) and (iii).
 all except (iii).
 All of these.
 The coordination equations used to obtain the optimal scheduling of a hydrothermal system when considering transmission losses are:
 None of these.
 As far as possible, hydroplants are used for baseload operation since:
 Their capital cost is high.
 Their operation is easy.
 Their capital cost is low.
 Their efficiency is low.
 A thermal plant gives minimum cost per unit of energy generated when used as a _____ plant.
 Peak load.
 Baseload plant.
 Simultaneously as baseload and peak load plant.
 None of these.
 In the combined operation of steam plant and runoff river plants, the sites of hydro and steam plants can be found with the help of _____.
 Demand curve.
 Input–output curve.
 Load curve.
 Chronological load curve.
 Longterm hydrothermal coordination can be done by:
 Plotting the basic rule curve.
 Plotting no spillrule curve.
 Plotting the full reservoir storage curve.
 All of these.
 _____ hydrothermal coordination is done for the available water and is to be used in a given period (24 hr).
 Longterm.
 Shortterm.
 Both (a) and (b).
 None of these.
 Hydrothermal coordination is necessary only in countries with:
 Ample coal resources.
 Ample water resources.
 Both (a) and (b).
 None of these.
 In shortterm hydrothermal coordination,
 No spillrule curve is used.
 Spillrule curve is used.
 Here no rule curve is used due to constraints.
 None of these.
 The units of incremental water rate are:
 m^{3}/sMW.
 m^{3}s/MW.
 ms/MW.
 m^{2}s/MW.
 Hydrogeneration is a function of:
 Water head.
 Water discharge.
 Water inflow.
 Both (a) and (b).
 In the longterm hydrothermal coordination,
 Basic rule curve is plotted.
 No spill curve.
 No full reservoir storage curve.
 All of these.
 In the combined operation of a steam and a runoff river plant, the sizes of hydro and steam plants can be obtained with the help of:
 Load curve.
 Demand curve.
 Chronological load curve.
 None of these.
REVIEW QUESTIONS
 Explain the hydrothermal coordination and its importance.
 Describe the types of hydrothermal coordination.
 What are the factors on which economic operation of a combined hydrothermal system depends?
 What are the important methods of hydrothermal coordination? Explain them in brief.
 Explain the mathematical formulation of longterm hydrothermal scheduling.
 Explain the solution method of longterm hydrothermal scheduling by discretization principle.
 Develop an algorithm for the solution of longterm hydrothermal scheduling problem.
 Derive the condition for optimality of shortterm hydrothermal scheduling problem.
 What are the advantages of hydrothermal plants combinations?
PROBLEMS
 The system shown in Fig. 6.8(a) is to supply a load shown in Fig. 6.8(b). The data of the system are as follows:
C_{T} = (16 + 0.01P_{GT})P_{GT}Rs./hr
w_{2} = (4 + 0.0035P_{GH})P_{GH}m^{3}/s
The maximum capacity of a hydroplant and a steam plant are 400 and 270 MW, respectively. Determine the generating schedule of the system so that 130.426 million m^{3} water is used during the 24hr period.
 A thermal station and a hydrostation supply an area jointly. The hydrostation is run 12 hr daily and the thermal station is run throughout 24 hr. The incremental fuel cost characteristic of the thermal plant is
C_{T} = 3 + 5P_{GT} + 0.02P_{GT} Rs/hr
If the load on the thermal station, when both plants are in operation, is 250 MW, the incremental water rate of the hydropower plant is
The total quantity of water utilized during the 12hr operation of a hydroplant is 450 million m^{3}. Find the generation of the hydroplant and the cost of water used. Assume that the total load on the hydroplant is constant for the 12hr period.
 A twoplant system that has a thermal station near the load center and a hydropower station at a remote location is shown in Fig. 6.9.
The characteristics of both stations are:
C_{T} = (20 + 0.03P_{GT})P_{GT}Rs./hr
w_{2} = (8 + 0.002P_{GH})P_{GH}m^{3}/s
and γ_{2} = Rs. 5 × 104/m^{3}
The transmission loss coefficient, B_{22} = 0.0005.
 if the load is 700 MW for 15hr a day and 500 MW for 9 hr on the same day, find the generation schedule, daily water used by hydroplant, and the daily operating cost of the thermal power.
 Determine the power generation at each station and the power received by the load when λ = 50 Rs./MWh.
 A twoplant system that has a hydrostation near the load center and a thermal power station at a remote location is shown in Fig. 6.10.
The characteristics of both stations are
C_{T} = (20 + 0.03P_{GT})P_{GT}Rs./hr
w_{2} = (8 + 0.002P_{GH})P_{GH}m^{3}/s
and γ_{2} = Rs. 5.5/m^{3}
The transmission loss coefficient, B_{22} = 0.0005.
 If the load is 700 MW for 15 hr a day and 500 MW for 9 hr on the same day, find the generation schedule, daily water used by the hydroplant, and the daily operating cost of thermal power.
 Determine the power generation at each station and the power received by the load when λ = 50 Rs. /MWh.
FIG. 6.8 Illustration for Problem 1; (a) twoplant system; (b) daily load curve
FIG. 6.9 Twoplant system
FIG. 6.10 Twoplant system