Physics Lesson 9.6.3 - Bernoulli Equation

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Welcome to our Physics lesson on Bernoulli Equation, this is the third lesson of our suite of physics lessons covering the topic of Bernoulli Equation, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

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 × Δ_x1 on the liquid, while atmosphere does the work W₂ = F₂ × Δ_x2 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

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

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

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

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 × Δ_x1 = A₂ × Δ_x2.

Thus, we can write

Simplifying volume V from both sides, we obtain

Rearranging the terms of the above equation, we obtain

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

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 (v_A = ?)
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 P_B and P_A from both sides of Bernoulli equation as both point are under the influence of atmospheric pressure P_atm (both of them are in contact with the atmosphere). Thus, the Bernoulli equation becomes

ρ × v²_B/2 + ρ × g × h_b
= ρ × v²_A/2 + ρ × g × h_a

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

0 = v²_A/2 + g × h_a
v²_A = -2 × g × h_a
v_a = r = √-2 × g × h_a
= 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 × v²_A/2 + ρ × g × h_a
= ρ × v²_C/2 + ρ × g × h_c

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

v²_A/2 = v²_C/2 + g × h_c

Hence,

v²_C/2 = v²_A/2 - g × h_c

Multiplying both sides by 2, we obtain

v²_C = v²_A - 2 × g × h_c
v_c = √v²_A - 2 × g × h_c
= √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 h_C = 1.1 m - 0.4 m - 0.2 m = 0.5 m, we can determine the falling time t using the kinematic equation

h_c = g × t²/2

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

t² = 2 × h_c/g ⇒ t
= √2 × h_c/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 = v_c × 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 P_atm = 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 = F_tap/A_hose + P_atm
= 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

You have reached the end of Physics lesson 9.6.3 Bernoulli Equation. There are 3 lessons in this physics tutorial covering Bernoulli Equation, you can access all the lessons from this tutorial below.

More Bernoulli Equation Lessons and Learning Resources

Density and Pressure Learning Material

Tutorial ID	Physics Tutorial Title	Tutorial	Video Tutorial	Revision Notes	Revision Questions
9.6	Bernoulli Equation
Lesson ID	Physics Lesson Title			Lesson	Video Lesson
9.6.1	What is an Ideal Fluid?
9.6.2	The Equation of Continuity
9.6.3	Bernoulli Equation

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