Question

A horizontal water jet at 70 °F issues from a circular orifice in a large tank. The jet strikes a vertical plate that is normal to the axis of the jet. A force of 600 lbf is needed to hold the plate in place against the action of the jet.
If the pressure in the tank is 25 psig at point A parallel to the orifice in the tank, what is the diameter of the jet just down-stream of the orifice?

Answer 1

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

$P_{A} = 25 psig$

Force, F = $\frac{mv^{2} }{r}$

Mass, m = $\rho V$, where $\rho =$ density of water = 62.4 lbs/ft³

V = volume

F = $\frac{\rho Vv^{2} }{r}$

Area, A = V/r

$F = \rho A v^{2}$...........(1)

Applying Bernoulli's equation between point A and B on the orifice

$\frac{P_{A} }{\rho} + \frac{V_{A} ^{2} }{2}+ gh_{A} = \frac{P_{B} }{\rho} + \frac{V_{B} ^{2} }{2}+ gh_{B}$..................(2)

$h_{A} = h_{B}$

At point A, the initial rest position, $v_{A} = 0$

At the orifice, $P_{B} = 0$

Making the appropriate substitution, equation (2) becomes

$\frac{P_{A} }{\rho} = \frac{v_{B} ^{2} }{2}$

$P_{A} = 25 psig\\P_{A} = 25 * 144 psf\\P_{A} = 3600 psf$

$\rho = 1.94 slug/ft^{3}$

$\frac{3600 }{1.94} = \frac{v_{B} ^{2} }{2}\\v_{B} ^{2} = 3711.34\\v_{B} =\sqrt{3711.34} \\v_{B} = 60.92 ft/s$

Put the value of $v_{B}$

$600 = 1.94* A 60.92^{2}$

600 = 4320A

A = 600/7200

A = 0.0833 ft²

Area, $A = \frac{\pi d^{2}} {4}$

$0.0833 = \frac{\pi d^{2}} {4}\\d^{2} = (4*0.0833)/\pi \\d^{2} = 0.106\\d = 0.326 ft$