Physics, 01.04.2020 20:31 kfull6027

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 from: shay7298

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.

Answer from: Quest

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the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia

Answer from: Quest

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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 reduci...

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