Question

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 17.1 m below the water level. If the rate of flow from the leak is 2.30 ? 10?3 m3/min, determine the following.

(a) Determine the speed at which the water leaves the hole. m/s

(b) Determine the diameter of the hole. mm

Answer 1

Let’s use Bernoulli’s equation.
P1 + ½ * Density * v1^2 + Density * g * y1 = P2 + ½ * Density * v2^2 + Density * g * y2
The left side is at the hole, and the right side is at the surface.
P1 and P2 are the atmospheric pressure. The atmospheric pressure is the same at the surface and at the hole. So, we can neglect the atmospheric pressure.

½ * Density * v1^2 + Density * g * y1 = ½ * Density * v2^2 + Density * g * y2
The density of the water is the same at the surface and at the hole. So we can divide both sides by Density.

½ * v1^2 + g * y1 = ½ * v2^2 + g * y2
Since the cross sectional area of tank is large and the cross sectional area hole is small, the downward velocity of water at the surface is much slower than the velocity of water through the hole. So we can neglect the velocity of water at the surface, v2.

½ * v1^2 + g * y1 = g * y2
Subtract g * y1 from both sides
½ * v1^2 = g * y2 – g * y1
½ * v1^2 = g * (y2 – y1)
Multiply both sides by 2
v1^2 = 2 * g * (y2 – y1)
v1 = ?[2 * g * (y2 – y1)]

(y2 – y1) is the distance from the surface to the hole = 17.1 m
v1 = ?(2 * 9.8 * 17.1) = ?335.16 m/s

Let’s imagine that a long horizontal pipe is attached to hole. The cross sectional area of the pipe is the same as the cross sectional area of the hole.
Each second the water moves ?335.16 meters.

The rate of flow from the leak is found to be 2.30 * 10-3 m^3/min.
Let’s convert to m^3/s by dividing by 60.
The rate of flow = 2.20 * 10-3 ÷ 60 = 3.833 * 10^-5 m^3/s
So each second 3.833 * 10^-5 m^3 of water flows out the hole and through the pipe. ..ANSWER

Volume per second = Cross sectional area * Distance per second
3.833 * 10^-5 = Cross sectional area * ?335.16
Cross sectional area = 3.667 * 10^-5 ÷ ?301.84
Cross sectional area = 2.09 * 10^-6 m^2
This is the cross sectional area of the pipe is approximately 2.09 * 10^-6 m^2.

So this is the cross sectional area of the hole is approximately 2.09 * 10^-6 m^2.

If the hole is circular, cross sectional area = ? * r^2
? * r^2 = 2.09 * 10^-6
r^2 = 2.09 * 10^-6 ÷ ?
r = ?(2.09 * 10^-6 ÷ ?)
Diameter = 2 * ?(2.09 * 10^-6 ÷ ?)
The diameter is approximately 1.63*10^-3 m = 1.63 mm .........ANSWER.
This is a very small hole. I use the pipe story, because it is the only that I can understand this problem. I hope this helps you understand how to solve this type of problem.

This morning I thought of a different way to solve this problem. The first equation in the fluid dynamics chapter in my college physics book is shown below.
Eq#1: ?i * A1 * v1 = ?2 * A2 * v2
? = density, A = cross sectional area, v = velocity
This is called the equation of continuity.
Density * Area * velocity = kg/m^3 * m^2 * m/s = kg/s = mass /second

As water flows through different size pipes, the mass of water which is flowing through each pipe in 1 second is the same. The water that flows out of the sink faucet comes from a large pipe in the alley. The water flows from that pipe to a ¾ inch pipe in our basement. Next, it flows through the ½ inch pipe which leads to the sink faucet. When I turn on the faucet, a specific mass of water flows into the sink each second. If 100 grams of water is flowing into the sink each second, a 100 grams is flowing through each of these pipes each second.

The density of the water flowing through each pipe is the same. So we can divide both sides of the equation of continuity by ?.

Eq#2: A1 * v1 = A2 * v2
Area * velocity = m^2 * m/s = m^3/s, m^3 = volume, s = seconds = time
Volume per second in ¾ inch pipe = Volume per second in ½ inch pipe
Eq#3: Area * velocity = Volume/time = m^3/s

Equation #2 is actually telling us that the volume of water which flows through each pipe in one second is the same for both pipes. . When I turn on the faucet, a specific volume of water flows into the sink each second. If 100ml of water is flowing into the sink each second, a 100 ml is flowing through each of these pipes each second.

Volume per second is the rate of flow!
In this problem, the rate of flow = 2.30 10-3 m3/min. = 3.833 * 10^-5 m^3/s and the velocity = ?335.16 m/s
Let’s put these numbers into Eq#3.
Area * ?335.16 = 3.833 * 10^-5
Area = 3.833 * 10^-5 ÷ ?335.16 = 2.09 * 10^-6 m^2
This is the same answer. So this a valid method for solving this problem.

Hope this explanation will help you.

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