Question 13.10: Using the same data as in the previous example, draw the mas......
Using the same data as in the previous example, draw the mass flow curve and (a) determine the storage required to meet the given demand; (b) investigate the maximum possible demand that could be obtained, and any additional storage requirement. For convenience, assume each month has 30 days.
The cumulative mass inflow and outflow curves must be drawn. Assuming a uniform demand throughout the year, the cumulative outflow line (ΣO) will be straight with a value of zero initially and 12 × 5.81 × 10^{6} = 69.72 \times 10^{6} m³ at the end of the year. With units of 10^{6} m³, the monthly inflow (I) and cumulative inflow (ΣI ) is:
The cumulative inflow and outflow curves are shown in Fig. 13.19. The reservoir is initially full. From April to the end of September, ΣI is less than ΣO so the reservoir empties, the difference AB representing the volume of reservoir storage required if the demand is to be met. From October onwards, inflow exceeds outflow and the reservoir refills. The reservoir is full again where the two lines (ΣO and ΣI) cross. In February and March, ΣI exceeds ΣO so the difference CD represents the total amount of water that passes over the dam’s spillway. This could be prevented by providing additional reservoir capacity equivalent to CD, i.e. by raising the height of the dam.(a) For the specified total demand of 5.81 × 10^{6} m³/month, the required storage is given by the ordinate AB = 21.90 × 10^{6} m³.
(b) As indicated above, a volume CD = 5.38 × 10^{6} m³ passes over the spillway and is ‘wasted’. Thus the reservoir could support a slightly higher demand as indicated by the dashed line OD. This represents a total demand of 6.26 \times 10^6 m ^3 / month \text {, or } 5.45 \times 10^6 m ^3 / month with the compensation flow deducted. However, to achieve this yield the reservoir storage must be increased to AE = 24.60 × 10^{6} m³. The reservoir would still refill at D, at the end of the 12 month period.
