Question

Ql. A homogeneous earth dam 30 m high has a free board of 1.5 m. A flownet was constructed and the following results are noted
no. of potential drop is 12
no. of flow channel is 3
The dam has a 18m long horizontal filter and it's downstream end. Calculate the seepage across the dam per day. If the width of the dam be 200m and it's co. efficient of permeability of the soil be 3.55 *10^-4 cm/sec.​

Answer 1

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3

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Explanation:

Answer 2

Answer:

So, the seepage across the dam per day is 25 x $10^{-5}$ $\frac{m^{3} }{sec}$.

Explanation:

Given data:

K = 3.55 x $10^{-4}$$\frac{cm^{3} }{sec}$.

(0.000355/100) = 0.00000355 $\frac{m^{3} }{sec}$, OR.

3 x $10^{-8}$$\frac{m^{3} }{sec}$

No. of potential drop, Nd = 12

no. of flow channel, Nf = 3

We know discharge (Q) = kH x number of flow channel/ number of Drops.

where, k = coefficient of permeability and H = water head.

H = (30 -1.5) =  28.5 m.

$Nf$ = 3,

$Nd$ = 12,

Then formula;

Q = kh x ($\frac{Nf}{Nd}$).

Q =0.00000355 x 28.5 x ($\frac{3}{12}$)

Q = 0.00002529 OR 25 x $10^{-5}$ $\frac{m^{3} }{sec}$.

Hence, the seepage across the dam per day is 25 x $10^{-5}$ $\frac{m^{3} }{sec}$.

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