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How to Derive Bernoulli’s Equation

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In this article, I will apply simple steps that will help you to understand how to derive Bernoulli’s equation.

Bernoulli’s Theorem: How to Derive Bernoulli’s Equation

Bernoulli’s theorem states that the sum of pressure, potential energy per unit volume, kinetic energy per unit volume of an ideal fluid that is flowing in a steady state is constant at each point along a stream line path in a pipe.

we can write the equation for the above theorem as

P_1+1/2\rho gv_1^2+\rho gh_1

The above expression is equal to

=P_2+1/2\rho gv_2^2+\rho gh_2

The above expression can further be broken down into

P_1+1/2\rho gv_1^2+\rho gh_1

which is equal to

=P_2+1/2\rho gv_2^2+\rho gh_2

and is equal to

=P_3+1/2\rho gv_3^2+\rho gh_3

and we can clearly see that it keeps going the same way which makes it to be constant. Thus, we can now rewrite the above equation as

P+1/2\rho gv^2+\rho gh=constant

How to Derive Bernoulli’s Equation

Here is how to derive Bernoulli’s equation, and i will break it into parts so that you will understand it better

Part 1

We all know that work is due to the change in energy and we can comfortably write it as

How to Derive Bernoulli’s Equation Part 1
\triangle W=\;\triangle E
Where\;\triangle E=\;\triangle K+\triangle P

Now we can write

\triangle W=\;\;\triangle K+\triangle P\dots\dots\dots\dots1
And\;\triangle K=Change\;in\;kinetic\;energy

Therefore, we can now say

W_{1\;}=\;F_1\;.\;\triangle X_1

In the presence of an angle, we can now write the above equation as

W_{1\;}=\;F_1\;.\;\triangle X_1\cos\theta

when we take θ as zero, we will then have

W_{1\;}=\;F_1\;.\;\triangle X_1\cos0

This implies that

W_{1\;}=\;F_1\;.\;\triangle X_1\cos\theta\;=\;F_1\;.\;\triangle X_1
W_1=\;F_1.\;\triangle X_{1.........}2

Since we know that pressure is equal to the force over area which is

P\;=\;\frac FA

we can now make Force, F subject of the formula by writing

<math xmlns=""><mi>F</mi><mo> </mo><mo>=</mo><mo> </mo><mi>P</mi><mi>A</mi></math>

Now we can substitute F into equation 2 and we will have

W_1=\;P_{1\;}A_1\triangle X_1.........3

Also, when θ = 1800 we will have Cosθ = Cos1800 = -1

We can now rewrite

W_2=F_2\triangle X_2Cos\theta


W_2=F_2\triangle X_2Cos180

which is equivalent to

W_2=-F_2\triangle X_2

The negative sign is because of Cos1800 = -1

Which is now

W_2=-\;P_{2\;}A_2\triangle X_2.........4

Part 2

How to Derive Bernoulli’s Equation Part 2

Now that we have understood how to relate work with pressure, area, and distance in equations 3 and 4. We will need to find out the relationship between work, pressure, mass and density


\triangle V=A\triangle X..........5


\triangle V=\frac m{\;\rho}......6


\;\rho=\frac m{\;V}

By putting equation 6 into 5, we will have

\frac m\rho=\;A\triangle X........7

Therefore, we can as well say

A_1\triangle X_1=\frac{m_1}\rho\;........8


A_2\triangle X_2=\frac{m_2}\rho\;........9

Now put equation 8 into 3, and equation 9 into 4 we will have




We now consider the summation of the total work done by the fluid

W = W1 + W2 …………..12

By adding substituting equations 10 and 11 into 12, we will now have


Part 3

Change in potential energy can be expressed as

\triangle P\;=\;mgh_2-mgh_1

Additionally, change in kinetic energy is equal to

\triangle K\;=\;\frac12mv_2^2-\frac12mv_1^2

Putting the above two equations into equation 1, we now have

\triangle W\;=

Part 4

Comparing equations 13 and 14, we now have


is equal to




Making m = m1 = m2, we can now say that


is equal to


By collecting like terms, we can now say that

P_1+\frac12\rho v_1^2+\rho gh_1\;=P_2+\frac12\rho v_2^2+\rho gh_2

and this implies that

P+\frac12\rho v^2+\rho gh\;=K\;=\;Cons\tan t

The above two equations are mathematical expression of Bernoulli’s equation.

Conclusion: How to Derive Bernoulli’s Equation

I hope after watching the video and going through the text, you have now learnt how to derive Bernoulli’s equation at ease

You may also like to read:

SS1 Lesson Note: Introduction to Physics For First Term

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