Answer to Question #260731 in Mechanical Engineering for Pearl

Question

1. A piston of 46 mm diam moves concentrically in a cylinder of 50 mm diam. The cylinder is filled with water and when the piston moves into the cylinder the water flows through the gap surrounding the piston. If the velocity of the piston is 75 mm/s relative to the cylinder, what is the velocity of the flow through the gap.

2. A 100 mm diam pipe carrying 1.8 m² / min of water suddenly enlarges to 150 mm diam. Find (a) the loss of head due to the sudden enlargement, (b) the difference in pressure in kN / m² in the two pipes taken between points just outside the area of disturbance due to change of section. The axis of the pipe is horizontal.

3.Water flows down a straight inclined pipe. At a point A, 45 m above datum, the section of the pipe suddenly doubles in area. The pressure in the smaller pipe at A is 860 kN / m² and the velocity of flow in the larger pipe is 2.4 m / s. Find the pressure in kN / m² at a point B which is 2.5 m above datum.

Answer 1

Solution:

(2)

(a)

$h_L=\frac{(v_1-v_2)^2}{2g}$

We know;

$Q=A_1v_1=A_2v_2$

For the smaller Pipe;

$\frac{1.8}{60}=π\frac{0.1^2}{4}v_1$

$v_1=3.820m/s$

For the larger pipe;

$\frac{1.8}{60}=π\frac{0.15^2}{4}v_2$

$v_2=1.698m/s$

Now;

$h_L=\frac{(3.820-1.698)^2}{2×9.81}=0.2295m$

(b)

Since ;

$z_1=z_2$,Then;

$\frac{p_1}{\rho g}+\frac{v_1^2}{2g}=\frac{p_2}{\rho g}+\frac{v_2^2}{2g}+h_L$

Resolve as;

$\frac{p_1-p_2}{\rho g}=\frac{v_2^2-v_1^2}{2g}+h_L$

By substitution;

$\frac{p_1-p_2}{\rho g}=\frac{1.698^2-3.820^2}{2×9.81}+0.2295$

$p_1-p_2=-0.3672×1000×9.81$

$p_1-p_2=-3.603kN/m^2$

(Negative sign shows a decrease in pressure)