Determination of reservoir storage capacity

Methods of of determining storage capacity of a reservoir

1. QUESTION 01

A reservoir has the following areas enclosed by contours at various elevation. Determine the

Elevation
(m)
200 220 240 260 280 300
Area of
contours
(km
2
)
150 175 210 270 320 400


Solution.

(a) By Cone method.



The total reservoir storage volume is 2.49 2.5 Mha-m












Elevation
Interval (m)
Area
(km
2
)
Equation Volume in km
2
-m
200 150

√
3246.79
220 175
220 175

√
3844.69
240 210
240 210

√
4787.45
260 270
260 270

√
5892.93
280 320
280 320

√
7185.14
300 400
TOTAL STORAGE VOLUME 24957

2. 2

(b)By Prismoidal method

∆v=


((A1+An) + 4(Sum Even Areas) + 2(sum Odd areas))

V1=


((150+320) + 4(175+270) + 2(210))
V1= 17,800 km
2
-m
By considering the last two areas the volume (V2) will be
V2=


(A5+A6)
V2=


(320+400)
V2= 7200 km
2
-m

Total Volume (VT) will be;
VT= V1+ V2
VT = (17800+7200) km
2
-m
VT =25000 km
2
-m
VT =25000Mm
2
-m = 2.5Mha-m
Total storage volume of reservoir = 2.5Mha-m
(c) By trapezoidal formula
∆v =


(A1+2A2+2A3+…………….2A n-1+ An)

∆v =


(150 + 400 + 2(175 + 210 + 270 + 320)

∆v = 25,000 km
2
-m

Total storage volume of reservoir = 2.5 Mha-m

3. 3

QUESTION 02

By considering the Full Reservoir Level of 270 m, its corresponding area can be calculated by
using the quadratic interpolation as follows


Elevation(E)m Areas(A)km
2
240 210
260 270
280 320

From the formulae of quadratic interpolation

A (E) = a0 + a1 (E) + a2 (E)
2

210 = a0 + a1 (240) + a2 (240
2
) (i)
270 = a0 + a1 (260) + a2 (260
2
) (ii)
320 = a0 + a1 (280) + a2 (280
2
) (iii)


By means of calculator:
a0 = -1290
a1 = 9.25
a2 = -0.0125

Then A (270) = -1290 + 9.25*(270) – 0.0125*(270
2
)

A (270) = 296.25 km
2

4. 4

The table that include the Full Reservoir Level of 270 m

Elevation(m) 200 220 240 260 270
Areas (km
2
) 150 175 210 270 296.25


(a) To determine the capacity of the reservoir by Cone formula

The total reservoir storage volume is 1.47 1.5 Mha-m

(b) By Trapezoidal formula


The total reservoir storage volume is 1.47 1.5 Mha-m


Elevation
Interval (m)
Area
(km
2
)
Equation Volume in km
2
-m
200 150

√
3246.79
220 175
220 175

√
3844.69
240 210
240 210

√
4787.45
260 270
260 270

√
2830.24
270 296.25
TOTAL STORAGE VOLUME 14709.17
Elevation Interval
(m)
Area(km
2
) Equation
V =


*(A1 + A2)
Area in km
2
-m
200 150


3250
220 175
220 175


3850
240 210
240 210


4800
260 270
260 270


2832.5
270 296.25
TOTAL STORAGE VOLUME 14732.5

5. 5

(c) By Prismoidal method

∆v=


((A1+An) + 4(Sum Even Areas) + 2(sum Odd areas))

V1=


((150+210) + 4(175) + 2(0))
V1= 7066.67 km
2
-m
By considering the last two areas the volume (V2) will be
V2=


(A3+A4)
V2=


(210+270)
V2= 4800 km
2
-m
For the elevation of the interval of 10 m
V3=


*(A3+A4)
V3 =


*(270 + 296.25)
V3= 2831.25 km
2
-m
VT=V1 + V2 + V3
VT=7066.67 + 4800 + 2831.25
VT=14697.92 km
2
-m
VT=1.5Mha-m
Total storage volume of reservoir = 1.5Mha-m