Question

The nozzle bends the flow from vertically upward to 30 degrees with the horizontal as it discharges the water (at 20 degrees C) to the atmosphere at V = 125 ft/s.
The volume of water within the nozzle itself (above the flange) is 100 lb.

Find the horizontal and vertical forces that must be applied to the flange (by the pipe below it) to hold it in place.Area of flange = 1.0 ft^2Area of nozzle = 0.50 ft^2Volume of area above flange = 1.8 ft^3Vertical height from flange to nozzle = 2 ft

Answer 1

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer 2

Answer:

Fy = -11267.294 lbf

Explanation:

given data

nozzle flow =  30 degrees

discharges the water =  20 degrees C

volume of water =  100 lb

Area of flange = 1.0 ft²

Area of nozzle = 0.50 ft²

Volume of area flange = 1.8 ft³

Vertical height flange to nozzle = 2 ft

solution

we will apply here continuity equation that is

A1 × V1 = A2 × V2    .............1

put here value and we get volume V1 that is

V1 = $\frac{0.5\times 125}{1}$

V1 = 62.5 ft/s

and

now we will apply here Bernoulli equation that is

$\frac{p1}{\gamma 1} + \frac{V1^2}{2g} + z1 = \frac{p2}{\gamma 2} + \frac{V2^2}{2g} + z2$      .............................2

put here value and we will get

p1 = 0 + $\frac{62.4}{2\times 32.2}(125^2 - 62.5^2) + 62.4 (2)$

p1 = 11479.614 psf

so here moment in y will be

∑ Fy = m [  (Vo)y - (Vi)x ]

so here we get

p1 ×A1 + Fy - Wn - Ww = $\rho$ Q  [ V2 × sin30 - V1 ]

put here value and we get Fy

1147.614 × 1 + Fy - 100 - (62.4 × 1.8) = (1.94) × (0.5 ×130) × (125sin30 - 62.5)

solve it we get

Fy = -11267.294 lbf