Assume in the figure below that the level of water in the tank remains the same,

Question

Assume in the figure below that the level of water in the tank remains the same, and that there is no friction loss in the pipe, entrance or nozzle.

(a) Determine the volumetric flow rate from the nozzle (i.e., the exit pipe with 2in diameter)

(b ) (b) Determine the pressure and velocity at points A, B, C, and D

Assume in the figure below that the level of water in the tank remains the same, and that there is no friction loss in the pipe, entrance or nozzle. Determine the volumetric flow rate from the nozzle (i.e., the exit pipe with 2in diameter) Determine the pressure and velocity at points A, B, C, and D

Explanation / Answer

Bernoulli's equation, in its simplfied form

p + q = po
po = total pressure (pressure in bottle)
p = static pressure (atmospheric pressure)
q = (density*v^2) / 2

Solve for velocity


Volumetric flow rate equation is as follows:
Q = A * v
where A = area of nozzle
v = velocity at nozzle (calculated above).

Q will give units of cubic metres / second.

Edit: I see your note saying that Bernoulli's equation is too complex - it really isn't, and really is the only way to solve your problem.

Edit 2 - Here is the solution, based on your numbers

120 psi internal pressure
3 cm^2 CSA at nozzle

For this solution I am assuming that 120 psi internal pressure is gauge pressure, as such:

po - p = 120 psi
po - p = 827.4 kPa

p + q = po
q = po - p
q = 827.4 kPa
q = 827,400 Pa
q = 0.5*(density*v^2)

density of water is 1000 kg / m^3

therefore:
q = 0.5*(1000*v^2) = 827,400 Pa
solve for v

v = 40.68 m/s

Q = A * v
where A = 3 cm ^2
or, A = 0.0003 m^2

Q = 0.0003 m^2 * 40.68 m/s
Q = 0.0122 m^3 / sec

1000 litres / cubic metre

Q = 12.2 litres / second

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