Respuesta :
Answer:
The net resultant force is 23.64 kN with an angle of 12.9Ā° with positive x-axis.
Explanation:
The mass flow rate of the fluid at inlet is equal to:
[tex]m=pA_{1} V_{1} =p\frac{\pi }{4} d_{1}^{2} V_{1}[/tex]
Where
p = density = 1000 kg/mĀ³
dā = inlet diameter = 30 cm = 0.3 m
Vā = velocity of fluid = 5 m/s
[tex]m=1000*\frac{\pi }{4} *0.3^{2} *5=353.43kg/s[/tex]
This mass flow is constant through the duct, thus:
minlet = moulet
[tex]pA_{1} V_{1} =pA_{2} V_{2} \\353.43=p\frac{\pi }{4} d_{2}^{2} V_{2}[/tex]
Where
dā = outlet diameter = 15 cm = 0.15 m
Vā = ?
Clearing Vā:
[tex]353.43=1000\frac{\pi }{4} *0.15^{2} *V_{2} \\V_{2} =20m/s[/tex]
Bernoulli's equation will be used to calculate the pressure at the outlet:
[tex]\frac{P_{1} }{pg} +\frac{V_{1}^{2} }{2g} +z_{1} =\frac{P_{2} }{pg} +\frac{V_{2}^{2} }{2g} +z_{2}[/tex]
Where:
Pā = pressure at inlet = 300x10Ā³ Pa
Pā = pressure at exit = ?
zā and zā = datum heads = zā = 0.5 m zā = 0
[tex]\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[/tex]
The force in x-direction using the moment expression is equal to:
[tex]F_{x} =-\beta mV_{1} -P_{1} A_{1}[/tex]
Where
Ī² = momentum flux = 1.04
[tex]F_{x} =-1.04*353.43*5-300x10^{3} *(\frac{\pi }{4} )*0.3^{2} =-23043.6N=-23.04kN[/tex]
The force in y-direction using the moment expression is:
[tex]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[/tex]
The net resultant force is equal to:
[tex]F_{net} =\sqrt{F_{x}^{2}+F_{y}^{2} } =\sqrt{(-23.04^{2})+(-5.28^{2}) } =23.64kN[/tex]
The angle of resultant force is equal to:
Īø = tanā»Ā¹(Fy/Fx) = tanā»Ā¹(-5.28/-23.04) = 12.9Ā° with positive x-axis.
The net resultant force exerted on the reducer by the water is :
- 23.64 kN Ā ā 12.9Ā° Ā in the positive x-axis
Given data :
p = 1000 kg/mĀ³
dā = 0.3 m
Vā = 5 m/s
dā = 0.15 m
First step : Calculate the mass flow rate of the fluid at inlet
m = pAāVā Ā = p*[tex]\frac{\pi }{4}[/tex] *dĀ²ā * Vā Ā ----- ( 1 )
insert values into equation ( 1 )
m = 353.43 kg/s
Note : mass flow rate at inlet = mass flow rate at outlet
ā“ 353.43 kg/s = p * [tex]\frac{\pi }{4}[/tex] * dĀ²ā * Vā ----- ( 2 )
Vā = 353.43 / ( 1000 * [tex]\frac{\pi }{4}[/tex] * 0.15Ā² )
Ā Ā = 20 m/s
Next step : Determine the outlet pressure ( Pā )
To calculate the outlet pressure we will apply Bernoulli's equation
Bernoulli's equation : [tex]\frac{P_{1} }{pg} + \frac{V_{1} ^{2} }{2g} + Z_{1} = \frac{P_{2} }{pg} + \frac{V^{2} _{2} }{2g} + Z_{2}[/tex] Ā ------- ( 3 )
where : Pā = 300 kPa
Ā Ā Ā Ā Ā Ā Ā Pā = ?
Ā Ā Ā Ā Ā Ā Ā Zā = 0.5 , Ā Zā = 0
Insert values into equation ( 3 )
Pā = 117425.7 kPa
Next step : Determine force in the x - direction
Fx = -Ī²mVā - PāAā Ā ------ ( 4 )
where : Ī² ( momentum flux correction factor ) = 1.04
Insert values into equation ( 4 )
Fx = - 23.04 kN
Next : Ā Determine Force in the y - direction
Fy = Ā -Ī²mVā + PāAā Ā --- ( 5 )
where : Ī² ( momentum flux correction factor ) = 1.04
insert values into equation ( 5 )
Fy = - 5.28 kN
Final step : Determine the net resultant force
Fnet = [tex]\sqrt{fy^{2} + fx^{2} }[/tex] = 23.64 kN
Angle of net resultant force = Tanā»Ā¹ ( Fy / Fx ) = Tanā»Ā¹ ( -5.28 / -23.04 )
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā = 12.9Ā°
Hence we can conclude that The net resultant force exerted on the reducer by the water is : 23.64 kN ā 12.9Ā° Ā in the positive x-axis
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