This process is repeated for the number of days for which flow values of Q IN is
available. At the end of the simulation, it is possible to obtain the number of days that
water cannot be supplied, demand flow rate, capacity and total spill volume.
Notes: (1) In the simulation here, if N J ¼ 10 days, it means that in theory, water
would not be available for supply during 10 days. In practice, the water supplier
would probably carry out some water rationing. The simulation shows however that
either the reservoir capacity is too small, or the intended drawoff is too much. (2) In
the formulation of Eq. (13.1), evaporation (daily or otherwise) has been omitted for
simplification purposes. The volume of evaporation may be estimated either as an
average daily volume or worked out – rather more accurately, but more complicated
also – by using the curve of volume v/s surface area of the reservoir to obtain surface
areas which can be multiplied by the daily evaporation rate, when the volume V 1 or
V 2 change daily. In both cases, the volume evaporated can be added to the drawoff
volume V OUT .
Thirty years ago, a Fortran program running on an IBM PC/AT computer (using
a 80,286 processor) for 20 years of data (7305 days, including leap years) took about
30 minutes for a run. With current desktop computers, and Excel, the same calculations take barely seconds, not to say occur instantaneously. This implies that as
soon as a parameter (say Q OUT ) is changed by the analyst, he can see the results for
N J , V SPILL , etc. immediately on the screen. Thus, if the analyst imposes the flow at
Q OUT ¼ 0.5 m
3 /s, he can vary V MAX , until he reaches this flow with N J ¼ 0. If need
be, he can refine his assumptions of V MAX to quite exact values through judicious
interpolation/bisection of other results, as he gets them on his screen. He can thus
obtain a series of values corresponding to the capacity v/s flow curve (lowest curve
on Fig. 13.3). Conversely, he can obtain the same curve, by imposing N J ¼ 0, and
then, for different integer values of V MAX , determine the flow available (again with
interpolation/bisection of previous screen results).
The other curves are obtained in a similar way. Thus, Fig. 13.3 shows also the
upper line for N J ¼ 40, i.e. for 40 days during the 20 year period during which there
would be no water supplied or rationing (of course over longer periods). One
immediate (and mistaken) conclusion would be to say that 40 days over 7305 days
merely represent a mere 0.5%, not worth much bother. Unfortunately, when going
through the list of these 40 days, one finds that they are clustered over a narrow range
of about only some 60 days.
A better conclusion would be to stick to N J ¼ 0 during the actual operation of the
reservoir, and that drawing off higher flows might lead to disaster.
There are many other possible uses of this simulation model. Assume a “what-if”
scenario of climate change which has reduced all flows by X %. This can be easily
accommodated in the algorithm or in the initial set of river flows. X may be varied
between 0 (no change) up to some 20% (or more) and either way (increase or
decrease of flows).
Figure 13.4 shows the results of these calculations, obtained for different reservoir capacities, while imposing N J ¼ 0.
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13 Modelling in Infrastructure Planning
available. At the end of the simulation, it is possible to obtain the number of days that
water cannot be supplied, demand flow rate, capacity and total spill volume.
Notes: (1) In the simulation here, if N J ¼ 10 days, it means that in theory, water
would not be available for supply during 10 days. In practice, the water supplier
would probably carry out some water rationing. The simulation shows however that
either the reservoir capacity is too small, or the intended drawoff is too much. (2) In
the formulation of Eq. (13.1), evaporation (daily or otherwise) has been omitted for
simplification purposes. The volume of evaporation may be estimated either as an
average daily volume or worked out – rather more accurately, but more complicated
also – by using the curve of volume v/s surface area of the reservoir to obtain surface
areas which can be multiplied by the daily evaporation rate, when the volume V 1 or
V 2 change daily. In both cases, the volume evaporated can be added to the drawoff
volume V OUT .
Thirty years ago, a Fortran program running on an IBM PC/AT computer (using
a 80,286 processor) for 20 years of data (7305 days, including leap years) took about
30 minutes for a run. With current desktop computers, and Excel, the same calculations take barely seconds, not to say occur instantaneously. This implies that as
soon as a parameter (say Q OUT ) is changed by the analyst, he can see the results for
N J , V SPILL , etc. immediately on the screen. Thus, if the analyst imposes the flow at
Q OUT ¼ 0.5 m
3 /s, he can vary V MAX , until he reaches this flow with N J ¼ 0. If need
be, he can refine his assumptions of V MAX to quite exact values through judicious
interpolation/bisection of other results, as he gets them on his screen. He can thus
obtain a series of values corresponding to the capacity v/s flow curve (lowest curve
on Fig. 13.3). Conversely, he can obtain the same curve, by imposing N J ¼ 0, and
then, for different integer values of V MAX , determine the flow available (again with
interpolation/bisection of previous screen results).
The other curves are obtained in a similar way. Thus, Fig. 13.3 shows also the
upper line for N J ¼ 40, i.e. for 40 days during the 20 year period during which there
would be no water supplied or rationing (of course over longer periods). One
immediate (and mistaken) conclusion would be to say that 40 days over 7305 days
merely represent a mere 0.5%, not worth much bother. Unfortunately, when going
through the list of these 40 days, one finds that they are clustered over a narrow range
of about only some 60 days.
A better conclusion would be to stick to N J ¼ 0 during the actual operation of the
reservoir, and that drawing off higher flows might lead to disaster.
There are many other possible uses of this simulation model. Assume a “what-if”
scenario of climate change which has reduced all flows by X %. This can be easily
accommodated in the algorithm or in the initial set of river flows. X may be varied
between 0 (no change) up to some 20% (or more) and either way (increase or
decrease of flows).
Figure 13.4 shows the results of these calculations, obtained for different reservoir capacities, while imposing N J ¼ 0.
378
13 Modelling in Infrastructure Planning