Mumbai University > Civil Engineering > Sem 5 > Applied Hydraulics 1

Marks: 10M

Year: May 2016

Given:-

Dia of bend D = $D_1 = D_2$ = 30cm = 0.3m

$\therefore \text{Area A} = A_1 = A_2 = \frac{\pi}{4} \times D^2 = \frac{\pi}{4} \times (0.3)^2 = 0.07068m^2$

velocity $v = V_1 = V_2 = 4 m/s\\ \theta = 45^0$

Discharge Q = A x V

= 0.07068 x 4

= 0.2827$m^3/s$

pressure head = 15m of water or $\frac{p}{\rho g}$ = 15 m of water

$P = 15 \times \rho g = 15 \times 1000 \times 9.81 = 147150N/m^2$

pressure intensity P = $P_1 = P_2 = 147150N/m^2$

$V_{1x} = 4m/s \hspace{1cm} V_2 cos 45^0 = 4 \times 0.7071\\ V_{1y} = 0 \hspace{1.8cm} V_{2y} = V_2 sin 45^0 = 4 \times 0.7071$

$(P_1 A_1)_x = P_1A_1 = 147150 \times 0.07068 \\ (P_1 A_1)_y = 0\\ (P_2 A_2)_x = -P_2 A_2 cos 45^0\\ (P_2 A_2)_y = -P_2 A_2 sin45^0$

$f_x = \rho Q[V_{1x} - V_{2x}] + (P_1 A_1)_x + (P_2 A_2)_x \\ \hspace{0.5cm}= 1000 \times 0.2827 [4 - 4 \times 0.7071] + 147150 \times 0.07068 - P_2 A_2 cos 45^0 \\ \hspace{0.5cm}= 331.21 + 147150 \times 0.07068 - 147150 \times 0.07068 \times 0.7071 \\ \hspace{0.5cm}= 331.21 + 10400.56 - 7354.23 \\ \hspace{0.5cm}= 3377.54N$

Force along y - axis

$F_y = \rho g [V_1 y - V_2 y] + (P_1A_1)_y + (P_2 A_2)_y \\ \hspace{0.5cm}= 1000 \times 0.2827 [0 - 4 \times 0.7071] + 0 + [-P_2A_2 sin 45^0] \\ \hspace{0.5cm}= 282.7 [0 - 2.8284] + 0 + [-147150 \times 0.07068 \times 0.7071] \\ \hspace{0.5cm}= -799.58 - 7354.23 \\ \hspace{0.5cm}= - 8153.81N$

Resultant force

$F_R = \sqrt{F_x^2 + F_y^2} \\ \hspace{0.5cm}= \sqrt{(3377.54)^2 + (8153.81)^2}\\ \hspace{0.5cm}= 8825.66N$

The angle made by $F_R$ with x-axis

$tan \theta = \frac{F_y}{F_x} = \frac{8153.81}{3377.54} = 2.414\\ \theta = tan^{-1} (2.414) = 67^0 29$