Chapter 6, Problem 2ICA

Final Assignment of this ICA: You have done several observation exercises. In these, you thought of observation as just a “stream of consciousness” with no regard to organization of your efforts. With your previous observation as a basis, generalize the search for observations into several (three to six or so) categories. The use of these categories should make the construction of a list of observations easier in the future.

Analyze the following problems using the SOLVEM method.

ICA 6-4

What diameter will produce a maximum discharge velocity of a liquid through an orifice on the side at the bottom of the cylindrical container? Consider diameters ranging from 0.2 to 2 meters.

To determine

Analyze the given problem using SOLVEM method and find out the diameter which will produce maximum discharge velocity of a liquid through the orifice.

Explanation of Solution

Discussion:

The problem has been solved using the SOLVEM method as follows.

Sketch:

The arrangement of cylindrical container is drawn as shown in Figure 1.

Objective:

Calculate the diameter to produce maximum velocity in a cylinder orifice.

Observations:

• The fluid is pushed out of the orifice because of the force of gravity acting on it.
• Fluid exits the tank horizontally.
• The speed of the discharge of the fluid increases with the increase in depth.
• Larger the container, more will the amount of liquid contained in it.
• Also, the liquid content in the container increase with the increase in depth.

List Variables:

A is area of the container,

D is diameter of the container $\left(\text{0}\text{.2 m to 2 m}\right)$,

g is Acceleration due to gravity $\left(9.8\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)$,

$\Delta H$ is height of the liquid above the orifice,

KE is kinetic energy,

m is mass of the fluid,

PE is potential energy,

V is velocity of the fluid, and

$\rho$ is density of the fluid.

Equations:

Consider the expression for density.

$\rho =\frac{m}{V}$ (1)

Here,

$\rho$ is the density,

$m$ is the mass, and

$V$ is the volume.

Consider the expression for area of a circle.

$A=\pi r^2$ (2)

Here,

$A$ is the area, and

$r$ is the radius.

Consider the expression for potential energy.

$PE=mg\Delta H$ (3)

Here,

$PE$ is the potential energy,

$m$ is the mass,

$g$ is the acceleration due to gravity, and

$\Delta H$ is the height.

Consider the expression for kinetic energy.

$KE=\frac{1}{2}m{V}^2$ (4)

Here,

$KE$ is the kinetic energy,

$m$ is the mass, and

$V$ is the speed.

The potential energy of the small volume of the fluid at surface will be equal to kinetic energy of the fluid which exits. Therefore,

Equate equation (3) and equation (4) as follows.

$mg\Delta H=\frac{1}{2}m{V}^2$ (5)

Manipulate:

The mass in the equation present on both the sides of the equation cancel out each other.

$g\Delta H=\frac{V^2}{2}$

Rearrange the equation above to solve for V.

$V=\sqrt{2g\Delta H}$

Therefore from the answer we see that the velocity is independent of the diameter of the container.

Conclusion:

Hence, the velocity is independent of the diameter of the container.