Answer :
Answer:
The diameter of the jet just down the orifice is 0.326 ft
Explanation:
Temperature of the horizontal water = 70⁰C
Force, F = 600 lbf
[tex]P_{A} = 25 psig[/tex]
Force, F = [tex]\frac{mv^{2} }{r}[/tex]
Mass, m = [tex]\rho V[/tex], where [tex]\rho =[/tex] density of water = 62.4 lbs/ft³
V = volume
F = [tex]\frac{\rho Vv^{2} }{r}[/tex]
Area, A = V/r
[tex]F = \rho A v^{2}[/tex]...........(1)
Applying Bernoulli's equation between point A and B on the orifice
[tex]\frac{P_{A} }{\rho} + \frac{V_{A} ^{2} }{2}+ gh_{A} = \frac{P_{B} }{\rho} + \frac{V_{B} ^{2} }{2}+ gh_{B}[/tex]..................(2)
[tex]h_{A} = h_{B}[/tex]
At point A, the initial rest position, [tex]v_{A} = 0[/tex]
At the orifice, [tex]P_{B} = 0[/tex]
Making the appropriate substitution, equation (2) becomes
[tex]\frac{P_{A} }{\rho} = \frac{v_{B} ^{2} }{2}[/tex]
[tex]P_{A} = 25 psig\\P_{A} = 25 * 144 psf\\P_{A} = 3600 psf[/tex]
[tex]\rho = 1.94 slug/ft^{3}[/tex]
[tex]\frac{3600 }{1.94} = \frac{v_{B} ^{2} }{2}\\v_{B} ^{2} = 3711.34\\v_{B} =\sqrt{3711.34} \\v_{B} = 60.92 ft/s[/tex]
Put the value of [tex]v_{B}[/tex]
[tex]600 = 1.94* A 60.92^{2}[/tex]
600 = 4320A
A = 600/7200
A = 0.0833 ft²
Area, [tex]A = \frac{\pi d^{2}} {4}[/tex]
[tex]0.0833 = \frac{\pi d^{2}} {4}\\d^{2} = (4*0.0833)/\pi \\d^{2} = 0.106\\d = 0.326 ft[/tex]