Neglecting minor losses, find the increase in discharge if 4f = 0.04. The head at inlet is 300 mm.

Data :

Dia. of pipe, D = 0.6m.

Length of pipe L = 1.5 km = 1.5 x 1000 = 1500m. 4 f = 0.04

Head at inlet, h = 300 mm = 0.3m.

Head at outlet, which is an atmospheric head = 0

$\therefore$ Head Loss = $H_f$ = Head at inlet - Head at outlet.

$h_f$ = 0.3 - 0

$h_f$ = 0.3m.

Length of another parallel pipe, $L_1 = \frac{1500}{2} = 750m$

Dia of another parallel pipe, $d_1$ = 0.6m.

Case 1. Discharge for a single pipe of length 1500 m and d = 0.6m

$h_f = \frac{4fLv^2}{d \times 2g}$

Where V = Velocity of flow for single pipe.

$\therefore$ 0.3 = $\frac{4 \times 0.1 \times 1500 \times v^2}{0.6 \times 2 \times 9.81}$

V = 0.2426 m/s

and Discharge $Q = V \times A = 0.2426 \times \frac {\pi}{4}\times 0.6^2$

Q = 0.0685 $m^3/s$

Case 2.

When an addition pipe of length 750m and diameter 0.6m is connected in parallel with the last half of the pipe.

Let Q1, Q2 be discharge in 1st and 2nd parallel pipe.

$\therefore Q = Q_1 + Q_2$

Q is discharge in main pipe, when pipes are parallel.

But length and diameter of each parallel pipe is same.

$\therefore Q_1 = Q_2 = Q/2$

Now, Consider the flow through ABC

= Head lost through AB + Head lost through BC. ---- (1)

But head lost due to friction through ABC = 0.3m (given)

$h f_(AB)$ = $\frac{4f \times 750 \times v^2}{0.6 \times 2 \times 9.81}$

(V = Velocity of flow through AB)

But $V = \frac{Q}{A} = \frac{Q}{\pi \times (0.6)^2} = \frac{4Q}{\pi \times 0.6^2}$

$\therefore h f_(AB) = \frac{4 \times 0.01 \times 750}{0.6 \times 2 \times 9.81} \times (\frac{4Q}{\pi \times 0.36})^2 = 31.87 Q^2$ --------( 2 )

$hf_(BC)$ = $\frac{4 \times 0.1 \times 750}{0.6 \times 2 \times 9.81} \times [\frac{Q/2}{\pi \times 0.6^2}]^2$

( $\therefore V_1 = \frac{Q}{\pi \times 0.6^2}$)

$\therefore hf (BC) = 7.969Q_2$ ---------- (3)

Now, Substituting (2) and (3) in (1)

$\therefore 0.3 = 31.87 Q^2 + 7.969 Q^2$

$\therefore Q = 0.0867 m^3/s$

$\therefore$ Increase in discharge = Q - Q

= 0.0867 - 0.0685

= 0.0182 $m^3/s$