Answer to Question #171028 in General Chemistry for Md Asif Iqbal

Question

 A converging pipe bend with its centre line in a horizontal plane, changes the direction of pipe

line by 60° in clockwise direction and reduces the pipe line diameter from 30cm to 20cm in the

direction of flow. If the pressure indicated by Bourdan gauge at the centre line of 30cm diameter

entrance to the bend is 140 kN/m2 and the flow of water through the pipe line is 0.10m3/s,

determine the magnitude and direction of force on the bend due to moving water.

Answer 1

$\theta$ = 60^o

D₁ = 30 cm = 0.30m

A₁ = $\frac{\pi}{4} D_1^2$ = 0.0706m²

D₂ = 20 cm = 0.20m

A₂ = $\frac{\pi}{4} D_2^2$ = 0.0314m²

pressure = 140 kN/m² = 140,000 N/m²

Q = 0.10 m³/sec

elevation difference (Z₁ - Z₂)= 0.30-0.20 = 0.10m

V₁ = Q/A₁ = 0.10/0.0706 = 1.41 m/s

V₂ = Q/A₂ = 0.10/0.0314 = 3.18 m/s

applying bernoulli's equation

$\frac{p_1}{\rho g}+\frac{V_1^2}{2g} +{Z_1-Z_2}+ = \frac{p _2}{\rho g}+\frac{V_2^2}{2g}$

$\frac{140,000}{1000(9.81)}+\frac{1.41^2}{2(9.81)} + 0.10 =\frac{p_2}{1000(9.81)}+\frac{3.18^2}{2(9.81)}$

p₂ = 136,922.89 N/m²

forces on behind in x and y directions are given by

F_x = ρQ[V₁ - V₂ cos$\theta$] + p₁A₁ - p₂A₂cos$\theta$

= 1000(0.10)[1.41 - 3.18(cos60^o)] + 140,000(0.0706) - 136,922.89 (0.0314)cos60^o

= 7716.62N

F_y = ρQ(-V₂sin$\theta$ ) - p₂A₂sin$\theta$

= 1000(0.10) ( - 3.18(sin60^o) - 136,922.89(0.0314)sin60^o

= - 3998.76N

negative sign indicates the F_y is actinginthe downward direction

therefore resultant force ,

F_R = 2$\sqrt{F_x^2 + F_y^2}$ = 2 $\sqrt{7716.62^2 + (-3998.76)^2}$ = 8691.16 N

the angle made by resultant force with x-axis given by

tan$\theta$ = F_y/F_x = 3998.76/7716.62 =0.518

$\theta$ = 27.39