The diameter of a pipe bend is 30 cm at inlet and 15 cm at outlet and the flow is turned

through $120^0$(angle measured in clockwise direction between direction of fluid flow at

inlet and outlet) in a vertical plane. The axis at inlet is horizontal and the centre of the

outlet section 1.5 m below the centre of the inlet section. Total volume of water in the

bend is $0.9 m^3$. Neglecting friction, calculate the magnitude and direction of the force

exerted on the bend by water flowing through it at 250 litr/s and when inlet pressure is

$0.15 N/mm^2$

Given: $D_1$=30 cm (Inlet diameter of pipe bend)

$D_2$=15 cm (Outlet diameter of pipe bend)

$θ=120°$ (Obtuse angle between inlet and outlet axis of pipe bend in a vertical plane)

$Z_1-Z_2$=1.5 m (Elevation difference between inlet and outlet)

$V=0.9 m^3$ (Volume of water in the bend)

Q ̇=250 lts/s (Discharge through the pipe)

$P_1=0.15 N/mm^2=0.15∙10^6 N/m^2$ (Inlet Pressure)

Neglect friction⇒$h_l$=0 (Head loss due to friction in the pipe bend)

To Find:

R(Net force on the bend in magnitude and direction)

Sol:

Consider the Control Volume (CV) consisting of the fluid in the pipe bend,

and

$v_2=\frac{Q }{A_2} =\frac{0.250}{0.01767}$=14.1483 m/s

Now, Pressure at outlet section is unknown. We apply Bernoulli’s Principle between the inlet and outlet section

$P_1/ρg+(v_1^2)/2g+Z_1=P_2/ρg+(v_2^2)/2g+Z_2+h_l$

Substituting in the above equation,

$\frac{0.15∙10^{6}}{9810}+\frac{3.5366^2}{2g}+Z_1-Z_2=\frac{P_2}{ρg}+\frac{14.1483^{2}}{2g}$+0

Since, ρ=$1000 kg/m^3$ (Density of water) and $g=9.81 m/s^2$ (acceleration due to gravity)

15.2905+0.6375+1.5=$\frac{P_2}{9810}$+10.2026+0

$\frac{P_2}{9810}$=7.2254

$P_2=70.882 kN/m^2$

Assuming direction of force on the fluid CV

Substituting in the above equation, ∴$0.15∙10^6 (0.07069)+70.882∙10^3 (0.01767)cos⁡60^{\circ}+R_x=1000(0.25)[-14.1483 cos⁡60^{\circ}-3.5366]$

∴$11229.74+R_x=-2653.6875$

∴$R_x=-13882.43 N (→)$

∴$R_x=13.882 kN (←)$(force on the fluid)

∴$R_x=13.882 kN (→)$(force on the pipe bend)

Along the Y axis, the momentum equation is given by, $(↑)$direction taken as positive

$∑F_y =(m ̇v)_outflow-(m ̇v)_inflow$

∴$∑F_y =m ̇_2 v_2y-m ̇_1 v_1y$

∴$∑F_x =m ̇[v_2x-v_1x ]$

∴$0+P_2 A_2 sin⁡60^{\circ}+ρVg+R_y=ρQ ̇[-v_2 sin⁡60^{\circ}-0]$

where, $ρVg$=Weight of fluid in the pipe bend acting vertically downwards Substituting in the above equation,

∴$70.882∙10^3 (0.01767) sin⁡60^{\circ}-9810(0.9)+R_y=1000(0.25)[-14.1483 sin⁡〖60^{\corc}]$

∴-7744.32+R_y=-3063.2

∴$R_y=4681.12 N (↑)$

∴$R_y=4.681 kN (↑)$

∴$R_y=4.681 kN (↓)$(force on the pipe bend)

Hence, Resultant force on the pipe bend=R=$\sqrt{(R_x^2+R_y^2)}$

∴R=$\sqrt{13.882^{2}+4.681^{2}}$=14.65 kN

and direction

$ϕ=tan^{-1}⁡(\frac{R_y}{R_x })=18.634°$ (↘clockwise from horizontal)