Three Level Rule Curve for Optimum Operation of a Multipurpose Reservoir using Genetic Algorithms

Abstract

Finding optimal policies for real-life reservoir systems operation (RSO) is a challenging task as the available analytical methods cannot handle the arbitrary functions of the problem. Most of the methods employed are numerical or iterative type and are computer dependent. Since the computer resources in terms of memory and CPU time are limited efficient algorithms are necessary to deal with the RSO problems. In this paper we present a Genetic Algorithms (GA) optimized rule curve (RC) model for monthly operation of a multipurpose reservoir which maximizes hydropower produced while meeting the irrigation demands with a given reliability. Instead of the usual single target storage for each period the proposed model considers three sets of target storages, namely dry, normal, and wet storages, based on the beginning of the period storage level. The reservoir considered is Bhadra Multipurpose Reservoir, in the state of Karnataka, India, which supplies water to irrigation fields through two canals while generating hydropower with turbines installed at each of the canal heads and at the river bed. Optimization ability and robustness of GA-RC approach are ascertained through simulation with a different inflow sequence for which global optimum is computed using Dynamic Programming. Further, a 15 year real-time simulation of the reservoir using historical inflows and demands showed significant improvement in the benefit, i.e. power produced, without compromising on the irrigation demands throughout the operation period.

ANNEXURE

Table 4 Energy production details of turbines

Where H_g is gross head (elevation of water level above river bed) in meters and is given as:

$$ Hg=20.983+0.04877S-3.780\times {10}^{-5}{S}^2+1.720\times {10}^{-8}{S}^3-3.035\times {10}^{-12}{S}^4 $$

And S is reservoir storage in M.cu.m

a (Energy expressions)

Power produced, P by a flow Q with a net head, H is given as

$$ P= wQH=9.81\ QH\kern0.75em kW $$

Where w is unit weight of water (=9810 N); Q is flow through turbines in cu.m/s; H is net head (= actual reservoir water level - tailwater level) in m.

Energy produced, E, with a constant flow of Q cu.m/s in a standard month, having 30.4375 days which is the average number of days for a month in a four year period, is

$$ E=P\times 30.4375\times 24=9.81\ QH\times 730.5\ kWh\ \left( kilowatt- hour\right) $$

If Q is expressed in M.cu.m/month (million cubic meters per month),

$$ E=9.81\times Q\left(\frac{10^6}{30.4375\times 24\times 3600}\right)\times H\times 730.5\ \mathrm{kWh}=2725\ QH\ kWh $$

Using plant efficiency as 0.85 and losses due to friction in penstocks and valves as 2%,

$$ E=0.85\times 0.98\times 2725\ QH\kern0.5em kWh\kern0.5em =0.0022699\ QH\kern1em M. kWh\ \left( million\ kilowatt- hour\right) $$