Compute the discharge of a river from
the data given in table below. The current meter coefficients are; a = 0.04 and
b = 0.76.
Distance from bank (ft) |
Depth (ft) |
Meter Depth (ft) |
Revolutions |
Time (sec) |
2 |
1 |
0.6 |
10 |
30 |
4 |
1.5 |
0.9 |
22 |
30 |
6 |
4 |
0.8 |
35 |
30 |
|
|
3.2 |
28 |
30 |
8 |
3 |
0.6 |
45 |
30 |
|
|
2.4 |
28 |
30 |
10 |
2 |
1.2 |
33 |
30 |
12 |
1 |
0.6 |
22 |
30 |
Solution#2
As we have
V = a + bNs
Average width
W =(2+4-2/2)2 /2×2
(W = 2.25 ft
For rest of segments
W1 = (2÷2 + 2÷2)
(W1 = 2 ft
For depth less than 2ft,
velocity is measured at 60% of depth, hence it is the average velocity. For
depth more than 2ft, velocity is measured at 20% and 80% of depth,
Distance from bank (ft) |
Average width |
Depth (ft) |
Revolutions |
Time (sec) |
Ns Rev/sec |
Velocity a=bN |
Area |
discharge |
|
|
|
|
2 |
2.25 |
1 |
10 |
30 |
0.33333 |
0.29333 |
2.25 |
0.645326 |
|
|
|
|
4 |
2 |
1.5 |
22 |
30 |
0.73333 |
0.59735 |
3 |
1.79193 |
|
|
|
|
6 |
2 |
4 |
35 |
30 |
1.16666 |
0.926616 |
8 |
7.413328 |
|
|
|
|
|
2 |
|
28 |
30 |
0.93333 |
0.74933 |
0 |
0 |
|
|
|
|
8 |
2 |
3 |
45 |
30 |
1.526 |
1.182 |
6 |
7.08 |
|
|
|
|
|
2 |
|
28 |
30 |
0.93333 |
0.74933 |
0 |
0 |
|
|
|
|
10 |
2 |
2 |
33 |
30 |
1.112 |
0.876 |
4 |
3.504 |
|
|
|
|
12 |
2.25 |
1 |
22 |
30 |
0.73333 |
0.59733 |
2.25 |
1.34399 |
|
|
|
|
TOTAL= 21.778544
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