A fluid flows through a horizontal 0.1 in diameter pipe

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Water flowing in a horizontal 30-cm-diameter pipe at 5 m/s and 300 kPa gage enters a 90° bend reducing section, which connects to a 15-cm-diameter vertical pipe. The inlet of the bend is 50 cm above the exit. Neglecting any frictional and gravitational effects, determine the net resultant force exerted on the reducer by the water

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Answer:

The net resultant force is 23.64 kN with an angle of 12.9° with positive x-axis.

Explanation:

The mass flow rate of the fluid at inlet is equal to:

m=pA_{1} V_{1} =p\frac{\pi }{4} d_{1}^{2} V_{1}

Where

p = density = 1000 kg/m³

d₁ = inlet diameter = 30 cm = 0.3 m

V₁ = velocity of fluid = 5 m/s

m=1000*\frac{\pi }{4} *0.3^{2} *5=353.43kg/s

This mass flow is constant through the duct, thus:

minlet = moulet

pA_{1} V_{1} =pA_{2} V_{2} \\353.43=p\frac{\pi }{4} d_{2}^{2} V_{2}

Where

d₂ = outlet diameter = 15 cm = 0.15 m

V₂ = ?

Clearing V₂:

353.43=1000\frac{\pi }{4} *0.15^{2} *V_{2} \\V_{2} =20m/s

Bernoulli's equation will be used to calculate the pressure at the outlet:

\frac{P_{1} }{pg} +\frac{V_{1}^{2} }{2g} +z_{1} =\frac{P_{2} }{pg} +\frac{V_{2}^{2} }{2g} +z_{2}

Where:

P₁ = pressure at inlet = 300x10³ Pa

P₂ = pressure at exit = ?

z₁ and z₂ = datum heads = z₁ = 0.5 m z₂ = 0

\frac{300x10^{3} }{1000*9.8} +\frac{5^{2} }{2*9.81} +0.5=\frac{P_{2} }{1000*9.8} +\frac{20^{2} }{2*9.81} +0\\32.36=\frac{P_{2} }{1000*9.8} +20.39\\P_{2} =117425.7kPa

The force in x-direction using the moment expression is equal to:

F_{x} =-\beta mV_{1} -P_{1} A_{1}

Where

β = momentum flux = 1.04

F_{x} =-1.04*353.43*5-300x10^{3} *(\frac{\pi }{4} )*0.3^{2} =-23043.6N=-23.04kN

The force in y-direction using the moment expression is:

F_{y} =-\beta mV_{2} +P_{2} A_{2} =-1.04*300x10^{3} *20+117.43*1000*\frac{\pi }{4} *0.15^{2} =-5276.18N=-5.28kN

The net resultant force is equal to:

F_{net} =\sqrt{F_{x}^{2}+F_{y}^{2} } =\sqrt{(-23.04^{2})+(-5.28^{2}) } =23.64kN

The angle of resultant force is equal to:

θ = tan⁻¹(Fy/Fx) = tan⁻¹(-5.28/-23.04) = 12.9° with positive x-axis.


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