The summit of ridge is 8 m above the level of water in the upper reservoir. Determine the minimum depth of the pipe below the summit of the ridge, if the absolute pressure head at the summit of siphon is not to fall below 3 m of water. Take f = 0.006 and atmospheric pressure head = 10.3 m of water. The length of siphon from the upper reservoir to the summit is 500 m. Find the discharge.

Data :

Dia, of siphon, d = 200 mm = 0.2 m.

Difference in level of 2 reservoirs, H = 40 m.

Length of pipe = L = 8000 m.

Height of ridge summit from water level in upper reservoir = 8 m.

Let the depth of the pipe below the summit of ridge = xm.

$\therefore$ height of siphon from the water surface in the upper reservoir = (8-x)m.

pressure heat at c, $\frac{pc}{\varrho g}$ = 3 m of water absolute.

Atmospheric pressure head, $\frac{pa}{\varrho g}$ = 10.3 m of water.

f = 0.006 (co-efficient of friction).

Length of siphon from upper reservoir to the summit, $L_1$ = 500 m.

Figure A.

Applying Bernoulli's Equation to point A and B, and taking datum line passing through B, we have, (shown in figure A).

$\frac{PA}{\varrho g} + \frac{VA^2}{2g} + Z_A = \frac{PB}{\varrho g} + \frac{VB^2}{2g} + Z_B + h_L$

0 + 0 + 40 = 0 + 0 + 0 + $\frac{4f \times L \times v^2}{d \times 2g}$

$40 = \frac{4 \times 0.006 \times 8000 \times v^2}{0.2 \times 2 \times 9.81}$

v = 0.904 m/s

Now applying Bernoulli's Equation to points A and C and assuming datum line passing through A, (Shown in Figure A)

$\frac{PA}{\varrho g} + \frac{v^2A}{2g} + Z_A = \frac{pc}{\varrho g} + \frac{vc^2}{2g} + Z_c + h_L$

$\because \frac{PA}{\varrho g} = 10.3m$ and $\frac{pc}{\varrho g} = 3m$

$\therefore 10.3 + 0 + 0 = 3 + \frac{v^2}{2g} + (8 - x) + \frac{4 \times f \times L_1 \times v^2}{d \times 2g}$

$\therefore 10.3 = 3 = \frac{0.904^2}{2 \times 9.81} + (8 - x) + \frac{4 \times 0.006 \times 500 \times (0.904)^2}{0.2 \times 2 \times 9.81}$

$\therefore$ x = 3.24 m.

$\therefore$ Discharge Q = Area x Velocity

= $\frac{\pi}{4} \times 0.2^2 \times 0.904$

Q = 0.0283 $m^3/s$

$\therefore$ The minimum depth of the pipe below the summit of the ridge is 3.24 m.