# Physics Tutorial: Bernoulli Equation

In this Physics tutorial, you will learn:

• What is an ideal fluid?
• How does the flowing speed relate to the thickness of the tube?
• What does the Bernoulli equation say about flowing fluids?
• What are some special cases of Bernoulli Equation use?
• How to calculate the flowing speed of a liquid from a hole?

## Introduction

Suppose you want to wet the garden but the hose is not long enough to reach all parts of the garden. What do you do in such a case?

Do you think water velocity is the same if you open two holes in the lateral sides of a container filled with water if holes are at different heights? Why?

These questions belong to Hydrodynamics - a part of fluid dynamics that studies liquids in motion. In this tutorial, we will explain the basics of this theory accompanying it with examples.

## What is an Ideal Fluid?

Since the mathematical apparatus used to study the behaviour of real fluid is too complicated, we use a simplified concept that helps us study fluids with a satisfactory approximation. This concept is known as "Ideal Fluid" and obviously, it contains some restrictions compared to real fluids. These restrictions are:

It is a kind of flow, in which the velocity of the fluid at a particular fixed point does not change with time. Thus, the flow of ideal fluids is steady.

For example, all liquid particles passing through the point A will have the same speed vA if the flow of liquid is steady.

### 2- Incompressible flow

Real fluids are slightly compressible. However, since this compression is very small, we neglect it when dealing with ideal gases and thus, we consider them as incompressible. This helps us a lot because we can take the volume of ideal fluids as constant when the number of molecules doesn't change.

### 3- Non-viscous flow

In simple words, viscosity represents the resistance of a fluid to the flow. Thus, ideal fluids are considered as non-viscous as they do not make any resistance to the flow.

### 4- Non-rotational flow

This means fluid particles do not rotate during the flow but they move only forward. In simple words, this means the fluid streams do not interfere with each other even when they have to pass through an obstacle as shown in the figure.

## The Equation of Continuity

Let us consider again the first scenario discussed in the "Introduction" paragraph. Thus, from experience we know that if we want to increase the flowing speed of water in order to reach the farthest parts of the garden, we must partially close the hose opening using the thumb. In this way, we decrease the flowing area in the output. This means the cross sectional flowing area and the flowing speed are somehow inversely proportional to each other (smaller cross-sectional area → higher flowing speed and larger cross sectional area → lower flowing speed.

But, how can we prove mathematically the above assumption? For this, let's consider a kind of bottle opened in both ends as shown below.

Let's consider a liquid sample flowing in the two ends of the tube (the coloured sections); one at the beginning and the other at the end of the tube. Since there is the same amount of liquid in both sections (m1 = m2 = m and ρ1 = ρ2 = ρ), we have

m1/ρ1 = m2/ρ2

This means that

V1 = V2

Since the volume of a cylinder is calculated by the formula V = A × Δx (where Δx here stand for height of cylinder), we obtain for volume in both sections

A1 × ∆x1 = A2 × ∆x2

Here the cylinder height Δx acts as a kind of particles displacement. From uniform motion we know that

∆x = v × ∆t

where v here represents the flowing speed and Δt the time interval during which this flow occurs.

Since there is the same fluid in both coloured part and given that ideal liquids are incompressible, we have for the flowing rate in both sections

V1/∆t1 = V2/∆t2

Therefore, Δt1 = Δt2 = Δt. Hence, substituting ∆x = v × ∆t in the equation

A1 × ∆x1 = A2 × ∆x2

for both sections, we obtain

A1 × v1 × ∆t = A2 × v2 × ∆t

Simplifying Δt from both sides, we obtain

A1 × v1 = A2 × v2

Thus, we proved the above assumption, i.e. the cross sectional area of a flowing fluid and its flowing speed are inversely proportional to each other. The above equation is known as the "equation of continuity for ideal fluids".

### Example 1

A 4 cm thick hose is connected to a water tap from which water comes out at 0.8 m/s. The hose is 12 m long if stretched. How much area in the output part of the hose must be block with thumb so that water reaches a point, which is 15 m away from the tap? The hose is taken in the horizontal position at 1 m above the ground. Take g = 10 m/s2 if needed.

### Solution 1

First, we have to determine by what speed the water must come out from the hose, i.e. to determine v2 in the equation of continuity. This speed is calculated by kinematic methods. Thus, since water comes out horizontally from the hose (v0y = 0) and the height from the ground is h = 1 m, we obtain for the falling time t:

h = g × t2/2 ⇒ t = √2h/g
= √2 × 1/10
= 0.45 s

Hence, given that the horizontal distance of water is Δs = 15 m - 12 m = 3 m, we obtain for the output speed v2 (which corresponds to v0x in kinematics):

v2 = ∆x/t
= 3 m/0.45 s
= 6.67 m/s

Now, let's calculate the flowing area A2 allowed at the output by blocking the rest with thumb. Given that A1 = 4 cm2, v1 = 0.8 m/s and v2 = 6.67 m/s, we obtain from the equation of continuity

A1 × v1 = A2 × v2
A2 = A1 × v1/v2
= 4 × 0.8/6.67
≈ 0.48 cm2

Thus, the area we must block by thumb is

Ablocked = A1 - A2
= 4 cm2 - 0.48 cm2
= 3.52 cm2

## Bernoulli Equation

Consider the situation described in the figure below, in which a liquid is flowing from left to right through a hose with different heights (h1 and h2) and different thicknesses (A1 and A2).

Obviously, we don't expect the liquid have the same flowing speed in both sides, so we write these speeds as v1 and v2 respectively.

Let's consider a water sample of volume V (V1 = V2 = V). It is clear that if we want to send this sample from the left to the right end of the hose, we must use an input force F1 which when multiplied by the input area A1 gives the input pressure P1 (F1 = P1 × A1), otherwise, the liquid will not raise on the right part, because of the gravity. However, since the right end of the hose is open and therefore it is in contact with air, we have an opposing force F2 caused by the atmospheric pressure P2 on the output area A2 (F2 = P2 × A2).

Both the above forces do some work on the liquid. Thus, the source does the work W1 = F1 × Δx1 on the liquid, while atmosphere does the work W2 = F2 × Δx2 on the liquid. Since the input work is positive and the output work is negative, we obtain for the resultant work done on the system

∆W = W1-W2
= F1 × ∆x1 - F2 × ∆x2

This change in work contributes in the change of the mechanical energy ME of the system. Thus, since mechanical energy is the sum of kinetic and gravitational potential energy, we can write

∆W = ∆ME
= ∆KE + ∆GPE

Given that KE = m × v2 / 2 and GPE = m × g × h, we obtain

∆W = ∆KE + ∆GPE
∆W = KE2-KE1 + GPE2-GPE1
F1 × ∆x1-F2 × ∆x2
= m × v22/2 - m × v12/2 + m × g × h2 - m × g × h1

Substituting forces F1 and F2 by P1 × A1 and P2 × A2 respectively, and also expressing mass as a product of density and volume (m = ρ × V), we obtain

P1 × A1 × ∆x1 - P2 × A2 × ∆x2
= ρ × V × v22/2 - ρ × V × v12/2 + ρ × V × g × h2-ρ × V × g × h1

Also, since there is the same liquid in both sections considered, we have the same volume as stated earlier, i.e. V1 = V2 = V = A1 × Δx1 = A2 × Δx2.

Thus, we can write

P1 × V - P2 × V
= ρ × V × v22/2 - ρ × V × v12/2 + ρ × V × g × h2 - ρ × V × g × h1

Simplifying volume V from both sides, we obtain

P1 - P2
= ρ × v22/2 - ρ × v12/2 + ρ × g × h2 - ρ × g × h1

Rearranging the terms of the above equation, we obtain

P1 + ρ × v12/2 + ρ × g × h1
= P2 + ρ × v22/2 + ρ × g × h2

The above formula gives the Bernoulli Equation for a fluid in two states, 1 and 2. It is named after the Swiss scientists Daniel Bernoulli, who analysed the flowing properties in fluids during the 18th century. Generalizing this equation for all possible states, we obtain

P + ρ × v2/2 + ρ × g × h = constant

### Example 2

A cylinder completely filled with water is 1.1 m high. If we open a small hole in the lateral side of cylinder (at point A), 40 cm below the water surface, calculate:

1. The emerging speed of water from the hole (vA = ?)
2. If we open another identical hole at 20 cm below the first one (at point C), how far from the lower base will the water emerging from the second hole be, when it falls on the ground? (Δx = ?)

Take the upper surface of water as still. Also, take g = 10 m/s2 and ρwater = 1000 kg/m3.

### Solution 2

a Let's try to cancel out some of the terms from the Bernoulli equation

P1 + ρ × v12/2 + ρ × g × h1
= P2 + ρ × v22/2 + ρ × g × h2

This can be achieved by trying to choose an appropriate point of reference for the initial state. Thus, if we choose the point B at the water surface as a reference point, we write the index B instead of 1 and A instead of 2. In this case, we can cancel PB and PA from both sides of Bernoulli equation as both point are under the influence of atmospheric pressure Patm (both of them are in contact with the atmosphere). Thus, the Bernoulli equation becomes

ρ × v2B/2 + ρ × g × hb
= ρ × v2A/2 + ρ × g × ha

Since we have chosen the point B as a reference, we have hB = 0 and hA = - 40 cm = - 0.40 m. Also, we can cancel out the water density ρ from both sides of equation. Finally, we have vB = 0 as water in the upper surface is still. Hence, we obtain

0 = v2A/2 + g × ha
v2A = -2 × g × ha
va = r = √-2 × g × ha
= r = √-2 × 10 × (-0.40)
= r = √8.00
= 2.83 m/s

b The same reasoning (and procedure) can be used to find the emerging speed of water from the point C. Then, since water trajectory is parabolic, we can use the equations of projectile motion to determine the horizontal displacement Δx.

This time, the Bernoulli equation is

ρ × v2A/2 + ρ × g × ha
= ρ × v2C/2 + ρ × g × hc

Thus, choosing now the position A as reference point, we obtain hA = 0 and hC = - 20 cm = - 0.20 m. Also, we cancel out again the density of water ρ and pressures PA and PC as both points are in contact with the atmosphere, i.e. PA = PC = Patm. Thus, the Bernoulli equations for this situation becomes

v2A/2 = v2C/2 + g × hc

Hence,

v2C/2 = v2A/2 - g × hc

Multiplying both sides by 2, we obtain

v2C = v2A - 2 × g × hc
vc = √v2A - 2 × g × hc
= √2.832 - 2 × 10 × (-0.20)
= √8 + 4
= √12
= 3.46 m/s

Given that the height of the point C from the ground is hC = 1.1 m - 0.4 m - 0.2 m = 0.5 m, we can determine the falling time t using the kinematic equation

hc = g × t2/2

as in the vertical direction, we consider water as falling freely. Thus,

t2 = 2 × hc/g ⇒ t
= √2 × hc/g
= √2 × 0.5/10
= √0.1
= 0.32 s

Therefore, the horizontal distance from the cylinder lower base in which the water coming out from the point C falls on the ground is

∆x = vc × t
= 3.46 m/s × 0.32 s
= 1.11 m

Thus, water coming out from the point C falls 1.11 m on the right of the cylinder.

### The Simplified Version of Bernoulli Equation

In situations involving the Bernoulli Equation

P1 + ρ × v21/2 + ρ × g × h1
= P2 + ρ × v22/2 + ρ × g × h2

the terms ρ × g × h1 and ρ × g × h2 are often cancelled out because the changes in flowing height are small, i.e. h1 ≈ h2. Therefore, we obtain a simplified version of Bernoulli equation:

P1 + ρ × v21/2
= P2 + ρ × v22/2

It is obvious that the terms ρ × v21/2 and ρ × v22/2 represent pressure as we cannot add two quantities that are not of the same type. More precisely, they represent the dynamic pressure as they involve the flowing speed of liquid v. On the other hand, the terms P1 and P2 represent the static pressure we have discussed earlier. Therefore, we can obtain a simplified version of Bernoulli equation, which states that:

Total Pressure = Static Pressure + Dynamic Pressure = Constant

This statement is similar to the definition of mechanical energy we have discussed in Section 5 (ME = KE + PE = constant).

### Special Cases of Bernoulli Equation Applications

Not all terms of the Bernoulli Equation

P1 + ρ × v21/2 + ρ × g × h1
= P2 + ρ × v22/2 + ρ × g × h2

are always present in a given situation. Thus, for example, in static liquids (when liquids are not flowing), the terms related to flowing velocities are not considered. Therefore, the Bernoulli Equation becomes

P1 + ρ × g × h1 = P2 + ρ × g × h2

Furthermore, if we take the reference point at one of the given positions (for example at the position 1, the term related to the height h1 is cancelled out as h1 = 0. Therefore, the Bernoulli Equation is further simplified and becomes

P1 = P2 + ρ × g × h2

On the other hand, when height difference is negligible as discussed in the previous section, the height-related terms cancels out from Bernoulli Equation, so it becomes

P1 + ρ × v21/2
= P2 + ρ × v22/2

Finally, if there is the same pressure in both parts examined (for example, if both positions are in contact with the atmosphere as in the previous example, the static pressure-related terms P1 and P2 cancel out from the Bernoulli equation, and we therefore obtain

ρ × v21/2 + ρ × g × h1
= ρ × v22/2 + ρ × g × h2

Since in most cases there is the same liquid flowing throughout the tube, the density cancels out. Also, if we take one of the positions as a reference point (for example the position 1, so h1 = 0), the Bernoulli Equation becomes

v21/2 = v22/2 + g × h2

### Example 3

If the pushing force of the tap in the figure below is 12 N and the hose is 4 cm2 thick, what is the water speed at the output if we block three quarters of the hose's end by placing the thumb on it? Water comes out from the tap at 2 m/s and the entire hose is in the horizontal position. Take g = 10 m/s2 and ρwater = 1000 kg/m3. Also take the atmospheric pressure Patm = 100 000 Pa.

### Solution 3

Let's convert the input and output areas into m2 first. Thus, A1 = 4 cm2 = 0.0004 m2 and A2 = A1 - 3/4 A1 = 1/4 A1 = 1/4 × 4 cm2 = 1 cm2 = 0.0001 m2.

Since there is the same height in both ends of the hose, the height-related term in the Bernoulli Equation

P1 + ρ × v21/2 + ρ × g × h1 = P2 + ρ × v22/2 + ρ × g × h2

cancel out. Thus, we have

P1 + ρ × v21/2 = P2 + ρ × v22/2

The input pressure is obtained by the equation

P1 = Ftap/Ahose + Patm
= 12 N/0.0004 m2 + 100 000 Pa
= 30 000 Pa + 100 000 Pa
= 130 000 Pa

(Here we must consider the total pressure as input, not only water pressure, because water is in contact with the atmosphere at the source. If it were not so, water would turn back to the source when we close the tap due to the pushing force exerted by the atmosphere at the output.)

Now let's substitute the values in the (reduced) Bernoulli Equation. Thus,

130 000 + 1000 × 22/2 = 1000 × v22/2 + 100 000
130 000 + 2000 = 500 × v22 + 100 000
32 000 = 500 × v22
v22 = 32000/500 = 64
v2 = √64
= 8 m/s

## Summary

The mathematical apparatus used to study the behaviour of real fluid is too complicated. Therefore we use a simplified concept that helps us study fluids with a satisfactory approximation. This concept is known as "Ideal Fluid" and obviously, it contains some restrictions compared to real fluids. They are:

1. 1- Steady flow. It is a kind of flow, in which the velocity of the fluid at a particular fixed point does not change with time.
2. 2- Incompressible flow. Real fluids are slightly compressible. However, since this compression is very small, we neglect it when dealing with ideal gases and thus, we consider them as incompressible.
3. 3- Non-viscous flow. Ideal fluids are considered as non-viscous as they do not make any resistance to the flow.
4. 4- Non-rotational flow. This means fluid particles do not rotate during the flow but they move only forward.

The equation of continuity for ideal fluids states "the cross sectional area of a flowing fluid and its flowing speed are inversely proportional to each other." The equation of continuity for two situations 1 and 2, is

A1 × v1 = A2 × v2

where A stands for cross sectional area of the tube and v for the flowing speed.

Bernoulli Equation gives the relationship between static and dynamic pressure exerted in a flowing fluid. Its states that

Total Pressure = Static Pressure + Dynamic Pressure = Constant

or

P + ρ × v2/2 + ρ × g × h = constant

If we consider two situations 1 and 2, the Bernoulli equation is written as

P1 + ρ × v21/2 + ρ × g × h1 = P2 + ρ × v22/2 + ρ × g × h2

It is not for sure that all terms of the Bernoulli equation are to be used in a given situation. Some terms may cancel out according to the situation involved. Therefore, usually we obtain simplified versions of Bernoulli equations when solving problems related to hydrodynamics.

## Bernoulli Equation Revision Questions

1. What is the thickness (in cm2) at the right end of the open tube shown in the figure?

1. 1.20 cm2
2. 1.25 cm2
3. 1.62 cm2
4. 12.5 cm2

2. What is the water speed at the lowest end of the pipe shown in the figure if water at the highest end of the pipe is at rest? Both ends of the pipe are open. Take g = 10 m/s2, ρwater = 1000 kg/m3 and Patm = 100 000 Pa if needed. The larger area of the pipe is much greater than the smallest one.

1. 63.2 m/s
2. 44.7 m/s
3. 6.3 m/s
4. 4.5 m/s

3. Compare the flowing speeds of water from the points A, B and C shown in the figure below if the holes are identical and the upper part of the barrel is open.

1. vA > vB > vC
2. vA = vB = vC
3. vA = vB < vC
4. vA < vB < vC