Question

part a-d

2 A paint sprayer pumps air through a constriction in a 0.025- m diameter pipe, as shown in the figure. The flow causes the pressure in the constricted area to drop and paint rises up the feed tube and enters the air stream. The speed of the air stream in the 0.025-m diameter sections is 5.00 m/s. The density of the air is 1.29 kg/m3, and the density of the paint is 1200 kg/m3. We can treat the air and paint as incompressible ideal fluids. .00 mis Air 2.5cm 250 cm Paint Hydrostatic equations can be used for the liquid paint in the container and the vertical feed tube dipped in the container, because the flow rate in the container and feed tube are relatively very slow. Hence determine the gauge pressure at the top of the vertical feed tube (which exposed to air), "just about" supported in the vertical tube. Hint: The gauge pressure needs to be negative in order to suck paint up, and support the column of paint in the vertical feed tube. paint liquid is Gauge pressure at top of vertical feed tube If the diameter of the constriction is d (meters), use the continuity equation to write an expression for the velocity of air (v)in the constriction in terms of d terms of d): (c) Use Bernouli's equation to write an expression for the gauge pressure in the constriction in terms of the velocity of air (v) in the constriction. The section of the horizontal pipe which is 0.025 m in diameter (where air comes out), is open to outside air, and so the pressure in the pipe is 1 atmosphere (1.01 x 105 Pa) Expression for gauge pressure in constriction (in terms of Vi): (d) Use above results to determine the maximum diameter (d) of the constriction that will allow the sprayer to operate?

Answer 1

a] In the Hydrostatic case, the Bernoulli's equation simplifies to:

$P_1-P_2 = [h_2-h_1]\rho g$

Since P₂ is the atmospheric pressure, P₁ - P₂ will therefore be the Gauge Pressure.

so, Gauge pressure = [0 - 0.125] x 1.29 x 9.8 = 1.58025 Pa.

b] Using Continuity equation,

A₁v₁ = A₂v₂

$v_2 = 5\frac{\pi (0.025/2)^2}{\pi (d/2)^2}=\frac{3.125\times 10^{-3}}{d^2}$

this is the expression for velocity of air.

c]

Use Bernoulli's equation,

$P_1 + h_1\rho g + \frac{1}{2}\rho v_1^2 = P_2 + h_2\rho g + \frac{1}{2}\rho v_2^2$

=> $P_g = [h_2-h_1]\rho g + \frac{1}{2}\rho [v_2^2-v_1^2]$

d] Use h₂ - h₁ = 0.125 m and substitute v₂ = 0.003125/d² to obtain the necessary diameter needed to pull the paint up the pipe.