Calculator™ © - Free Online Calculators

Online Calculators since 2009

- Biquad Filter Coefficient Calculator
- Magnetic Field Inside A Toroid Calculator
- Root Mean Square Speed Calculator
- Final Temperature Of Mixture Calculator
- Roche Limit Calculator
- Coulomb’s Law Practice Questions
- Electric Charges. Conductors and Insulators Practice Questions
- Angular Frequency Of Oscillations In Rlc Circuit Calculator
- Radiative Heat Transfer Calculator
- Radio Line of Sight Calculator
- Angle Of Refraction Calculator

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?

Tutorial ID | Title | Tutorial | Video Tutorial | Revision Notes | Revision Questions | |
---|---|---|---|---|---|---|

9.6 | Bernoulli Equation |

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.

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 v_{A} if the flow of liquid is steady.

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.

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.

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.

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 (m_{1} = m_{2} = m and ρ_{1} = ρ_{2} = ρ), we have

This means that

V_{1} = V_{2}

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

A_{1} × ∆x_{1} = A_{2} × ∆x_{2}

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

Therefore, Δt_{1} = Δt_{2} = Δt. Hence, substituting ∆x = v × ∆t in the equation

A_{1} × ∆x_{1} = A_{2} × ∆x_{2}

for both sections, we obtain

A_{1} × v^{1} × ∆t = A_{2} × v^{2} × ∆t

Simplifying Δt from both sides, we obtain

A_{1} × v^{1} = A_{2} × v^{2}

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"**.

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/s^{2} if needed.

First, we have to determine by what speed the water must come out from the hose, i.e. to determine v_{2} in the equation of continuity. This speed is calculated by kinematic methods. Thus, since water comes out horizontally from the hose (v_{0y} = 0) and the height from the ground is h = 1 m, we obtain for the falling time t:

h = *g × t*^{2}*/**2** ⇒ t = √***2***h**/**g*

= √*2 × 1**/**10*

= 0.45 s

= √

= 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 v_{2} (which corresponds to v_{0x }in kinematics):

v_{2} = *∆x**/**t*

=*3 m**/**0.45 s*

= 6.67 m/s

=

= 6.67 m/s

Now, let's calculate the flowing area A_{2} allowed at the output by blocking the rest with thumb. Given that A_{1} = 4 cm^{2}, v_{1} = 0.8 m/s and v_{2} = 6.67 m/s, we obtain from the equation of continuity

A_{1} × v_{1} = A_{2} × v_{2}

A_{2} = *A*_{1} × v_{1}*/**v*_{2}

=*4 × 0.8**/**6.67*

≈ 0.48 cm^{2}

A

=

≈ 0.48 cm

Thus, the area we must block by thumb is

A_{blocked} = A_{1} - A_{2}

= 4 cm^{2} - 0.48 cm^{2}

= 3.52 cm^{2}

= 4 cm

= 3.52 cm

Consider the situation described in the figure below, in which a liquid is flowing from left to right through a hose with different heights (h_{1} and h_{2}) and different thicknesses (A_{1} and A_{2}).

Obviously, we don't expect the liquid have the same flowing speed in both sides, so we write these speeds as v_{1} and v_{2} respectively.

Let's consider a water sample of volume V (V_{1} = V_{2} = 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 F_{1} which when multiplied by the input area A_{1} gives the input pressure P_{1} (F_{1} = P_{1} × A_{1}), 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 F_{2} caused by the atmospheric pressure P_{2} on the output area A_{2} (F_{2} = P_{2} × A_{2}).

Both the above forces do some work on the liquid. Thus, the source does the work W_{1} = F_{1} × Δ_{x1} on the liquid, while atmosphere does the work W_{2} = F_{2} × 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

∆W = W_{1}-W_{2}

= F^{1} × ∆x_{1} - F^{2} × ∆x_{2}

= F

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

= ∆KE + ∆GPE

Given that KE = m × v^{2} / 2 and GPE = m × g × h, we obtain

∆W = ∆KE + ∆GPE

∆W = KE_{2}-KE_{1} + GPE_{2}-GPE_{1}

F^{1} × ∆x_{1}-F^{2} × ∆x_{2}

=*m × v*^{2}_{2}*/**2** - **m × v*^{1}^{2}*/**2** + m × g × h*_{2} - m × g × h_{1}

∆W = KE

F

=

Substituting forces F_{1} and F_{2} by P_{1} × A_{1} and P_{2} × A_{2} respectively, and also expressing mass as a product of density and volume (m = ρ × V), we obtain

P^{1} × A_{1} × ∆x_{1} - P^{2} × A_{2} × ∆x_{2}

=*ρ × V × v*^{2}_{2}*/**2** - **ρ × V × v*^{1}^{2}*/**2** + ρ × V × g × h*_{2}-ρ × V × g × h_{1}

=

Also, since there is the same liquid in both sections considered, we have the same volume as stated earlier, i.e. V_{1} = V_{2} = V = A_{1} × Δ_{x1} = A_{2} × Δ_{x2}.

Thus, we can write

P^{1} × V - P^{2} × V

=*ρ × V × v*^{2}_{2}*/**2** - **ρ × V × v*^{1}^{2}*/**2** + ρ × V × g × h*_{2} - ρ × V × g × h_{1}

=

Simplifying volume V from both sides, we obtain

P^{1} - P^{2}

=*ρ × v*^{2}_{2}*/**2** - **ρ × v*^{1}^{2}*/**2** + ρ × g × h*_{2} - ρ × g × h_{1}

=

Rearranging the terms of the above equation, we obtain

P^{1} + *ρ × v*^{1}^{2}*/**2** + ρ × g × h*_{1}

= P^{2} + *ρ × v*^{2}_{2}*/**2** + ρ × g × h*_{2}

= P

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 × v*^{2}*/**2** + ρ × g × h = constant*

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:

- The emerging speed of water from the hole (v
_{A}= ?) - 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/s^{2} and ρwater = 1000 ** kg/m^{3}**.

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

P^{1} + *ρ × v*^{1}^{2}*/**2** + ρ × g × h*_{1}

= P^{2} + *ρ × v*^{2}_{2}*/**2** + ρ × g × h*_{2}

= P

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

=

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 = *v*^{2}_{A}*/**2** + g × h*_{a}

v^{2}_{A} = -2 × g × h_{a}

v_{a} = r = √**-2 × g × h**_{a}

= r = √**-2 × 10 × (-0.40) **

= r = √**8.00**

= 2.83 m/s

v

v

= r = √

= r = √

= 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

=

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

Hence,

Multiplying both sides by 2, we obtain

v^{2}_{C} = v^{2}_{A} - 2 × g × h_{c}

v_{c} = √**v**^{2}_{A} - 2 × g × h_{c}

= √**2.83**^{2} - 2 × 10 × (-0.20)

= √**8 + 4**

= √**12**

= 3.46 m/s

v

= √

= √

= √

= 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 × t*^{2}*/**2*

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

t^{2} = *2 × h*_{c}*/**g** ⇒ t*

= √*2 × h*_{c}*/**g*

= √*2 × 0.5**/**10*

= √**0.1**

= 0.32 s

= √

= √

= √

= 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

= 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.

In situations involving the Bernoulli Equation

P^{1} + *ρ × v*^{2}_{1}*/**2** + ρ × g × h*_{1}

= P^{2} + *ρ × v*^{2}_{2}*/**2** + ρ × g × h*_{2}

= P

the terms ρ × g × h_{1} and ρ × g × h_{2} are often cancelled out because the changes in flowing height are small, i.e. h_{1} ≈ h_{2}. Therefore, we obtain a simplified version of Bernoulli equation:

P^{1} + *ρ × v*^{2}_{1}*/**2*

= P^{2} + *ρ × v*^{2}_{2}*/**2*

= P

It is obvious that the terms ** ρ × v^{2}_{1}/2** and

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).

Not all terms of the Bernoulli Equation

P^{1} + *ρ × v*^{2}_{1}*/**2** + ρ × g × h*_{1}

= P^{2} + *ρ × v*^{2}_{2}*/**2** + ρ × g × h*_{2}

= P

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

P^{1} + ρ × g × h_{1} = P^{2} + ρ × g × h_{2}

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 h_{1} is cancelled out as h_{1} = 0. Therefore, the Bernoulli Equation is further simplified and becomes

P^{1} = P^{2} + ρ × g × h_{2}

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

P^{1} + *ρ × v*^{2}_{1}*/**2*

= P^{2} + *ρ × v*^{2}_{2}*/**2*

= P

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 P_{1} and P_{2} cancel out from the Bernoulli equation, and we therefore obtain

=

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 h_{1} = 0), the Bernoulli Equation becomes

If the pushing force of the tap in the figure below is 12 N and the hose is 4 cm^{2} 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/s^{2} and ρwater = 1000 ** kg/m^{3}**. Also take the atmospheric pressure P

Let's convert the input and output areas into m^{2} first. Thus, A_{1} = 4 cm^{2} = 0.0004 m^{2} and A_{2} = A_{1} - 3/4 A_{1} = 1/4 A_{1} = 1/4 × 4 cm^{2} = 1 cm^{2} = 0.0001 m^{2}.

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

P^{1} + *ρ × v*^{2}_{1}*/**2** + ρ × g × h*_{1} = P^{2} + *ρ × v*^{2}_{2}*/**2** + ρ × g × h*_{2}

cancel out. Thus, we have

P^{1} + *ρ × v*^{2}_{1}*/**2** = P*^{2} + *ρ × v*^{2}_{2}*/**2*

The input pressure is obtained by the equation

P^{1} = *F*_{tap}*/**A*_{hose}* + P*_{atm}

=*12 N**/**0.0004 m*^{2}* + 100 000 Pa*

= 30 000 Pa + 100 000 Pa

= 130 000 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 × 2*^{2}*/**2** = **1000 × v*^{2}_{2}*/**2** + 100 000*

130 000 + 2000 = 500 × v^{2}_{2} + 100 000

32 000 = 500 × v^{2}_{2}

v^{2}_{2} = *32000**/**500** = 64*

v^{2} = √**64**

= 8 m/s

130 000 + 2000 = 500 × v

32 000 = 500 × v

v

v

= 8 m/s

Enjoy the "Bernoulli Equation" physics tutorial? People who liked the "Bernoulli Equation" tutorial found the following resources useful:

- Physics tutorial Feedback. Helps other - Leave a rating for this tutorial (see below)
- Density and Pressure Revision Notes: Bernoulli Equation. Print the notes so you can revise the key points covered in the physics tutorial for Bernoulli Equation
- Density and Pressure Practice Questions: Bernoulli Equation. Test and improve your knowledge of Bernoulli Equation with example questins and answers
- Check your calculations for Density and Pressure questions with our excellent Density and Pressure calculators which contain full equations and calculations clearly displayed line by line. See the Density and Pressure Calculators by iCalculator™ below.
- Continuing learning density and pressure - read our next physics tutorial: Adhesive and Cohesive Forces. Surface Tension and Capillarity

The following Physics Calculators are provided in support of the Density and Pressure tutorials.

- Bernoulli Principle Calculator
- Buoyancy Calculator
- Fluid Density Calculator
- Height Of Liquid Rise In Capillary Tubes Calculator
- Liquid And Total Pressure Calculator
- Solid Pressure Calculator
- Surface Tension Force Calculator

You may also find the following Physics calculators useful.

- Conical Pendulum Calculator
- Lc Resonance Calculator
- Galaxies Receding Speed Calculator
- Angle Of Refraction Calculator
- Mutual Inductance Calculator
- Energy Of Photons Calculator
- Amount Of Substance Obtained Through Electrolysis Calculator
- Fresnel Reflectance Of S Polarized Light Calculator
- Intensity And Loudness Of Sound Waves Calculator
- Magnetic Field Inside A Solenoid Calculator
- Electrostatic Energy Of A Uniformly Charged Sphere Calculator
- Total Magnetic Moment Of An Electron Calculator
- Lorentz Force Calculator
- Total Energy Of Hydrogen Like Atoms Calculator
- Cosmic Time At Redshift Calculator
- Conservation Of Momentum In 1 D Calculator
- Prism Minimum Angle Of Deviation Calculator
- Capacitance Of A Sphere Calculator
- Circuit Parallel Inductance Calculator
- Distance Of Planet From The Sun Calculator