Question

A reservoir has the following available data. Find the probable useful life of the reservoir.
Reservoir capacity = 16.8 Mm³ (Million cubic meter)
Average annual flood volume = 42 Mm³
Annual sediments = 13.33x 104 tones, Specific gravity of sediments = 2.2
Life of the reservoir terminates when 80% of initial capacity is filled up.
Capacity-inflow ratio vs trap efficiency in % age is given in table-2.
Assume 20 % of the capacity is filled up in first interval.

Table-2 Capacity-inflow ratio 0.1 0.2 Trap efficiency (%) 86 92 1 0.3 0.4 0.5 0.6 0.7| 94 95 96 96.5 97 0.8 97 0.9 97 97.5

Answer 1

Given:

Reservoir capacity = 16.8 Mm³

Avg. annual flood volume = 42 Mm³

Annual sediments = 13.33*104 tonnes

Sp. gravity of sediments =2.2

Sp. weight of sediments = 2.2*9.81 KN/m³

Life will terminate when 80% of initial capacity is filled up

Solution:

Volume of sediment = mass/density = (13.3*104*9.81*1000)/(2.2*9.81) =0.63 Mm³

Capacity inflow %	Capacity as volume (*106 m3) or Mm3	Capacity inflow ratio	Trap efficiency	Average trap efficiency (%) ($\eta$)	Annual sediment trapped (St) Mm3	Year to fill (Gap/St) yr
100	16.8	16.8/42 = 0.4	95
				(95+94.992)/2 = 94.996	(94.996*0.63)/100 =0.5985	3.36/0.5985 = 5.61
80	16.8*80/100 =13.44	13.44/42 =0.32	94.992
				(94.992+93.98)/2 = 94.486	(94.486*0.63)/100 =0.5953	3.36/0.5953 = 5.64
60	16.8*60/100 =10.08	10.08/42 = 0.24	93.98
				(93.98+91.976)/2 = 92.978	(92.978*0.63)/100 = 0.5858	3.36/0.5858 = 5.74
40	16.8*40/100 =6.72	6.72/42 = 0.16	91.976
				(91.978+85.828)/2 = 88.902	(88.902*0.63)/100 = 0.5601	3.36/0.5601 = 6
20	16.8*20/100 =3.36	3.36/42 = 0.08	85.828

Capcity inflow

Since 20% is filled up in the first interval, we take 20 as the interval in capacity inflow %

Capacity as volume  

= (Reservoir capacity * capcity inflow%)/100

Capacity inflow ratio

= Capacity as volume / Average annual flood volume

Trap efficiency

Trap efficiency is found out corresponding to the capacity inflow ratio woth the help of the table given in the question.

If direct values are not given, we have to interpolate them.

Average trap efficiency is the average value of 2 trap efficiencies.

For example:(95 + 94.992)/2 = 94.996

Annual sediment trapped

= (Avg trap efficiency * volume of sediment) /100

Here volume of sediment is already found out = 0.63 Mm³

year to fill

= interval between capacity as volume / Annual sediment trapped

Here the gap between capacity as volume = 3.36 Mm³

Total years of usefull life of reservoir = $\sum$ (yr to fill) = 5.61 + 5.64 + 5.74 + 6 = 22.99 yrs

23 years (appr.)