Respuesta :
Answer:
Fnet=(3666.47i-8487.33j) lbf
Explanation:
The velocity in the nozzle V1 is:
[tex]v_{1}=\frac{Q}{A}=\frac{"v_{j}A_{2} }{A} =2v_{j}(\frac{\frac{\pi }{4}d^{2} }{\frac{\pi }{4}D^{2} } )[/tex]
where Q is the discharge, vj is the velocity in the jet, d is the diameter of the jet and D is the diameter of the nozzle. Replacing values:
[tex]v_{1}=2*80.2(\frac{\frac{\pi }{4}1^{2} }{\frac{\pi }{4}4^{2} } )=10.02 fps[/tex]
In x-direction the momentum is:
∑Fx=∑m-miVix
[tex]Fx_{}+p_{1}(\frac{\pi }{4}D^{2})=p(\frac{\pi }{4}d^{2})v_{j}^{2}-p(\frac{\pi }{4}d^{2})v_{j}^{2}sin\alpha -p(\frac{\pi }{4}d^{2})v_{1}^{2}[/tex]
where
p1=pressure=43 psig
D=4 in=0.33 ft
d=1=0.083 ft
α=30°
p=density=1.94 slug/ft^3
Replacing values and clearing Fx:
Fx=3666.47 lbf
In x-direction the momentum is:
∑Fy=∑m-miViy
[tex]F_{y}=p(\frac{\pi }{4}d^{2} )v_{j}(-v_{j}cos\alpha )[/tex]
Replacing previous values, we have:
Fy = -8487.33 lbf
The net force is equal to:
Fnet=Fx+Fy
Fnet=(3666.47i-8487.33j) lbf
Answer:
the force required are
- Fx = - 523.5 Ibf ( x axis)
- Fy = - 58.94 Ibf ( y axis)
Explanation:
Data given
V2 = V3 = 80.2 ft/s
P1 = 43 psig = 6192 Ib/ft^2
note: P2 = P3 = Patm = 0 Ib/ ft^2
D2 = D3 = 1 inch
D1 = 4 inches
p = 62.4 Ibm/ft3 = 1.94 slug/ft3
calculate the cross sectional Areas of the water jets
A 1 = [tex]\frac{\pi*D1^{2} }{4} = \frac{\pi }{4 }* ( \frac{4}{12} )^{2}[/tex] = 0.0873 ft2
A2 = [tex]\frac{\pi }{4} * ( \frac{1}{12})^{2}[/tex] = 0.005454 ft2
calculate the mass flow rate
M2 = M3 = P *A2*V2 = 1.94 * 0.005454 * 80.2 = 0.8486 slug/s
calculate the mass balance
M1 = M2 + M2 = 0.8486 + 0.8486 = 1.6972 Ibm/ s
next calculate the velocity at each section
velocity at section 1 of pipe
= [tex]\frac{M1}{pA1}[/tex]
V1 = [tex]\frac{1.6972}{1.94 * 0.0873}[/tex] = 10.02 ft/s
calculate the linear momentum
F = m ( Vout - Vin )
along the x - direction
Fx + (P1 A1) - (P2 A2) + (P3 A3) sin 30 = (M2 V2) + M3 ( -V3 sin 30) - (M1 V1)
Fx + 540.56 = 68.06 -34.03 - 17
Fx = -523.5 Ibf
Along the y - direction
Fy + P3 A3 COS 30 = M3 ( -V3 COS 30 ) - 0
Fy + 0 = 0.8486 ( -80.2 cos 30 )
Fy = - 58.94 Ibf
