Question

x Your answer is incorrect. Try again. (Free jet) Water flows from a nozzle of triangular cross section as shown in the figure below. After it has fallen a distance of 2.7 ft, its cross section is circular (because of surface tension effects) with a diameter D = 0.11 ft. Determine the flowrate, Q. Equilateral triangle each side of length 0.19 ft 2.7 ft D=0.11 ft 1.2061 ft/s

Answer 1

Using the continuity equation:(Volumetric flow remains constant)

A1v1 = A2V2

Za[(z110)) = 1^((2610)

V2 = 1.645 vi

Now using Bernoulli's equation between points 1 and 2, we have:

Equilateral triangle each side of length 0.19 ft 2.7 ft D-0.11ft -

$\rm \frac{p_1}{\rho g} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\rho g} + \frac{v_2^2}{2g} + z_2$

We have p₁ = p₂ (Atmospheric pressure for the free jet)

+2.7 (164xy+ 。

((1.645) – 1)v *1 = 2.7

((1.645)2 – 1)v = 2.7 * 2 * 32.2

V1 = 10.1 ft/s

Then the volumetric flow rate is:

Q= Ajva = V3/0.192) x 10.1 = 0.1578 ft3/s. 4