This document contains information about calculating the storage volume of two reservoirs using different methods. For the first reservoir: - The storage volume is calculated as 2.5 Mha-m using the cone, prismoidal, and trapezoidal methods based on area-elevation data ranging from 200-300m in 20m intervals. For the second reservoir: - The storage volume is calculated as 1.5 Mha-m using the cone, trapezoidal, and prismoidal methods based on area-elevation data including an interpolated value for 270m elevation.

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
(km2
)
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
(km2
)
Equation Volume in km2
-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 km2
-m
By considering the last two areas the volume (V2) will be
V2= (A5+A6)
V2= (320+400)
V2= 7200 km2
-m
Total Volume (VT) will be;
VT= V1+ V2
VT = (17800+7200) km2
-m
VT =25000 km2
-m
VT =25000Mm2
-m = 2.5Mha-m
Total storage volume of reservoir = 2.5Mha-m
(c) By trapezoidal formula
∆v = (A1+2A2+2A3+…………….2An-1+ An)
∆v = (150 + 400 + 2(175 + 210 + 270 + 320)
∆v = 25,000 km2
-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)km2
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 (2402
) (i)
270 = a0 + a1 (260) + a2 (2602
) (ii)
320 = a0 + a1 (280) + a2 (2802
) (iii)
By means of calculator:
a0 = -1290
a1 = 9.25
a2 = -0.0125
Then A (270) = -1290 + 9.25*(270) – 0.0125*(2702
)
A (270) = 296.25 km2

4. 4
The table that include the Full Reservoir Level of 270 m
Elevation(m) 200 220 240 260 270
Areas (km2
) 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
(km2
)
Equation Volume in km2
-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(km2
) Equation
V = *(A1 + A2)
Area in km2
-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 km2
-m
By considering the last two areas the volume (V2) will be
V2= (A3+A4)
V2= (210+270)
V2= 4800 km2
-m
For the elevation of the interval of 10 m
V3= *(A3+A4)
V3 = *(270 + 296.25)
V3= 2831.25 km2
-m
VT=V1 + V2 + V3
VT=7066.67 + 4800 + 2831.25
VT=14697.92 km2
-m
VT=1.5Mha-m
Total storage volume of reservoir = 1.5Mha-m