In the previous section, we saw the details about streamline flow and equation of continuity. In this section, we will see Bernoulli's equation
Bernoulli's equation can be derived in 12 steps:
1. In fig.10.33(a) below, a fluid (shown in orange color) is flowing through a pipe
• The pipe has a varying cross sectional area
• The pipe has a varying cross sectional area
 |
| Fig.10.33 |
♦ If velocity changes, it means that, there is an acceleration
♦ If there is an acceleration, it means that, there is a force
♦ If there is a force, work will be done by that force
• In fig.10.33(b), a portion of the fluid is marked as AB
♦ It is shown in green color
■ We want to find the work done on AB
3. The portion AB is shown separately in fig.10.34(a) below:
• The fluid in the lower portion of AB will exert a pressure P1 on AB
♦ This is indicated by the lower red arrow
• The fluid in the upper portion of AB will exert a pressure P2 on AB
♦ This is indicated by the upper red arrow
■ We want to find the work done on AB
3. The portion AB is shown separately in fig.10.34(a) below:
• The fluid in the lower portion of AB will exert a pressure P1 on AB
♦ This is indicated by the lower red arrow
• The fluid in the upper portion of AB will exert a pressure P2 on AB
♦ This is indicated by the upper red arrow
 |
| Fig.10.34 |
• If A1 is the area of cross section at A, the force will be equal to (P1 × A1)
• If A2 is the area of cross section at B, the force will be equal to (P2 × A2)
■ So the portion AB is acted upon by:
♦ a force (P1 × A1) at A
■ So the portion AB is acted upon by:
♦ a force (P1 × A1) at A
♦ a force (P2 × A2) at B
5. To obtain work, we multiply force by displacement
So next, we want the displacement. It can be written in 6 steps:
(i) The flow is taking place continuously
• We want the displacement suffered by AB is a very small duration of Δt
• We want the displacement suffered by AB is a very small duration of Δt
(ii) Let in the small duration Δt, the portion AB move forward in such a way that:
♦ The bottom surface which was at A, reaches a new position A’
♦ This is shown in fig.10.34(b) above
(iii) So a certain volume is vacated. This vacated volume is shown in yellow color
• Based on our discussion in the previous section, we can write:
♦ Since Δt is small, the yellow volume can be considered as a cylinder
• The length of this yellow cylinder will be equal to (v1 × Δt)
♦ Where v1 is the velocity at A
• So volume of this yellow cylinder will be equal to (A1 × v1 × Δt)
♦ Where A1 is the cross sectional area at A
(iv) So AB vacated certain volume at the bottom
• Obviously, some volume at the top will be newly occupied. This newly occupied volume is shown in violet color
♦ The top surface which was at B, has now moved to B’
(v) Similar to what we wrote in (iii), we can write:
♦ Since Δt is small, the violet volume can be considered as a cylinder
• The length of this violet cylinder will be equal to (v2 × Δt)
♦ Where v2 is the velocity at B
• So volume of this yellow cylinder will be equal to (A2 × v2 × Δt)
♦ Where A2 is the cross sectional area at B
(vi) Thus we get the displacements:
• The displacement at A
= length of the yellow cylinder
= (v1 × Δt)
• The displacement at B
= length of the violet cylinder
= (v2 × Δt)
6. Now we can find the works:
(i) Work done at A
= Force at A × Displacement at A
= (P1 × A1) × (v1 × Δt)
• Rearranging this, we get:
Work done at A = P1 × [A1 × (v1 × Δt)]
⇒ Work done at A = P1 × Vyellow
♦ Where Vyellow is the volume of the yellow cylinder
(ii) Work done at B
= Force at B × Displacement at B
= (P2 × A2) × (v2 × Δt)
• Rearranging this, we get:
Work done at B = P2 × [A2 × (v2 × Δt)]
⇒ Work done at B = P2 × Vviolet
♦ Where Vviolet is the volume of the violet cylinder
7. So net work done
= Work done at A - Work done at B
= (P1 × Vyellow) - (P2 × Vviolet)
• But from equation of continuity, we have: Vyellow = Vviolet
♦ We can denote both of them commonly as ΔV
♦ 'Δ' is used because, it is the small change in volume taking place in the duration Δt
■ So net work done = [(P1 × ΔV) - (P2 × ΔV)] = [(P1 - P2)ΔV]
8. Here we can apply the law of conservation of energy:
♦ A portion of this work is used to change the kinetic energy
♦ The remaining is used to change the potential energy
(See section 6.8 in chapter 6)
9. First we will see the change in kinetic energy
• It can be written in 4 steps:
(i) In the duration Δt, a certain mass has moved from A to B
• Obviously, this mass is equal to
♦ the mass of fluid in Vyellow
♦ which is same as
♦ the mass of fluid in Vviolet
(ii) We have:
♦ Mass of fluid in Vyellow = Vyellow × ρ
♦ Mass of fluid in Vyellow = Vviolet × ρ
(iii) But from (7), we have: Vyellow = Vviolet = ΔV
■ So from (i) we get:
During Δt, a mass (ΔV×ρ) moves from A to B
(iv) Now we can write the kinetic energies:
• When the mass is at A, it's velocity is v1
♦ So the kinetic energy at A = $\mathbf\small{\rm{\frac{1}{2}m v^2=\frac{1}{2}(\Delta V \times \rho) v_1^2}}$
• When the mass is at B, it's velocity is v2
♦ So the kinetic energy at B = $\mathbf\small{\rm{\frac{1}{2}m v^2=\frac{1}{2}(\Delta V \times \rho) v_2^2}}$
■ Thus we get:
♦ Change in kinetic energy = $\mathbf\small{\rm{\left[\frac{1}{2}(\Delta V \times \rho) v_2^2-\frac{1}{2}(\Delta V \times \rho) v_1^2\right]=\frac{1}{2}(\Delta V \times \rho)\left[ v_2^2- v_1^2\right]}}$
10. Next we will see the change in potential energy
• It can be written in 2 steps:
(i) During Δt, a mass (ΔV×ρ) moves from A to B
Based on this, we can write the potential energies:
• When the mass is at A, it's potential energy is mgh = (ΔV×ρ)gh1
♦ Where h1 is the height of A from datum
♦ See fig.10.34(b) above
• When the mass is at B, it's potential energy is mgh = (ΔV×ρ)gh2
♦ Where h2 is the height of B from datum
(ii) So change in potential energy = (ΔV×ρ)gh2 - (ΔV×ρ)gh1
= (ΔV×ρ)g(h2 - h1)
11. So we obtained three items:
♦ Work done [step (7)]
♦ Change in kinetic energy [step (9)]
♦ Change in potential energy [step (10)]
• We can apply the relation that we wrote in (8). We get:
$\mathbf\small{\rm{(P_1-P_2)\Delta V=\frac{1}{2}(\Delta V \times \rho)\left[ v_2^2- v_1^2\right]+(\Delta V \times \rho)g(h_2-h_1)}}$
• Dividing both sides by ΔV, we get:
$\mathbf\small{\rm{(P_1-P_2)=\frac{1}{2}\rho\left[ v_2^2- v_1^2\right]+\rho g(h_2-h_1)}}$
• Rearranging this, we get:
Eq.10.6: $\mathbf\small{\rm{P_1+\frac{1}{2}\rho v_1^2+ \rho g h_1=P_2 + \frac{1}{2}\rho v_2^2+\rho gh_2}}$
• This is known as Bernoulli's equation
■ A and B can be any two points along the length of the pipe. So we can write:
Eq.10.7: $\mathbf\small{\rm{P+\frac{1}{2}\rho v^2+ \rho g h= constant}}$
12. We see that, there are three terms in the equation
Let us analyse them separately. The analysis can be written in steps:
(i) The first term is P, which is pressure
• It’s dimensions can be obtained as:
$\mathbf\small{\rm{\frac{Force}{Area}=\frac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]}}$
(ii) The second term is $\mathbf\small{\rm{\frac{1}{2}\rho v^2}}$
• It’s dimensions can be obtained as:
$\mathbf\small{\rm{Density \times velocity^2 = [ML^{-3}]\times [LT^{-1}]^2=[ML^{-1}T^{-2}]}}$
(iii) The third term is $\mathbf\small{\rm{\rho g h}}$
• It’s dimensions can be obtained as:
$\mathbf\small{\rm{Density \times acceleration \times height = [ML^{-3}]\times [LT^{-2}]\times [L]=[ML^{-1}T^{-2}]}}$
(iv) We see that, all three terms have the same dimensions
• Indeed, we can add them only if they have the same dimensions
(v) Now consider the quantity: Energy per unit volume
• It's dimensions can be obtained as:
$\mathbf\small{\rm{\frac{Energy}{Volume}=\frac{Force \times Distance}{Volume}=\frac{[MLT^{-2}]\times [L]}{[L^3]}=[ML^{-1}T^{-2}]}}$
• This is the same dimensions of the three terms
■ So each of the three terms in the Bernoulli's equation is equivalent to:
Energy per unit volume
■ We can write:
The total energy per unit volume will be a constant at all points along a pipe
• So we have seen the derivation of Bernoulli's equation. Next we will see some more details related to the equation:
• In the previous sections, we have seen that, the pressure at a point is given by: ρgz
♦ Where z is the depth of that point below the surface of the fluid
• Now, in Bernoulli’s equation, we see a similar term: ρgh
♦ Where h is the height of the point above datum
■ We want to know the relation between the two items: ρgz and ρgh
• It can be explained using an example. We will write it in 19 steps:
1. Fig.10.35 below shows a simple water supply system
• Water is taken out from the yellow tank
♦ First the water flows through the red pipe of larger diameter
♦ Then it flows through the magenta pipe of smaller diameter
♦ A tap is taken out from the magenta pipe
♦ The yellow pipe of varying diameter is connected to the tap
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| Fig.10.35 |
2. In the fig.10.35, the tap is closed. So there is no water in the yellow pipe
• But in fig.10.36(a) below, the tap is open
♦ So water will enter the yellow pipe and will flow out
• But the flow is suppressed by connecting a ‘manometer containing a heavy liquid’
 |
| Fig.10.36 |
• Four points A, B, C and D are marked along the path of flow
• A and D are on the same horizontal level
♦ They are at a depth of z1 from the top surface of water
♦ They are at a height of h1 from the datum
• B and C are on the same horizontal level
♦ They are at a depth of z2 from the top surface of water
♦ They are at a height of h2 from the datum
4. Pressures at various points:
• The pressures at A and D can be obtained as:
♦ PA = PD = Pa + ρgz1
✰ Where Pa is the atmospheric pressure
• The pressures at B and C can be obtained as:
♦ PB = PC = Pa + ρgz2
5. Pressure differences:
• Pressure difference between A and B = (PB - PA) = [Pa + ρgz2 - (Pa + ρgz1)] = ρg(z2 - z1)
• Pressure difference between D and C = (PC - PD) = [Pa + ρgz2 - (Pa + ρgz1)] = ρg(z2 - z1)
6. Next we will apply Bernoulli’s equation at the four points
(The fluid is stationary. So velocity is zero at all points. We need not write the second term in the equation)
(i) At point A:
♦ Pressure energy per unit volume = Pa + ρgz1
✰ Where Pa is the atmospheric pressure
♦ Potential energy per unit volume = ρgh1
♦ Total energy per unit volume = (Pa + ρgz1 + ρgh1) = Pa + ρg(z1 + h1)
(ii) At point B:
♦ Pressure energy per unit volume = Pa + ρgz2
♦ Potential energy per unit volume = ρgh2
♦ Total energy per unit volume = (Pa + ρgz2 + ρgh2) = Pa + ρg(z2 + h2)
(iii) At point C:
♦ Pressure energy per unit volume = Pa + ρgz2
♦ Potential energy per unit volume = ρgh2
♦ Total energy per unit volume = (Pa + ρgz2 + ρgh2) = Pa + ρg(z2 + h2)
(iv) At point D:
♦ Pressure energy per unit volume = Pa + ρgz1
♦ Potential energy per unit volume = ρgh1
♦ Total energy per unit volume = (Pa + ρgz1 + ρgh1) = Pa + ρg(z1 + h1)
7. According to Bernoulli's equation, the total energies must be same at all points
• So we can write:
Pa + ρg(z1 + h1) = Pa + ρg(z2 + h2) = Pa + ρg(z2 + h2) = Pa + ρg(z1 + h1)
♦ The first and fourth items are the same. We need not write them twice
♦ The second and third items are the same. We need not write them twice
• So in effect, we get: Pa + ρg(z1 + h1) = Pa + ρg(z2 + h2)
• Rearranging this, we get:
(ρgz2 - ρgz1) = (ρgh1 - ρgh2)
⇒ ρg(z2 - z1) = ρg(h1 - h2)
8. But from (5), we have:
ρg(z2 - z1) = (PB - PA)
9. So the result in (7) can be modified as:
(PB - PA) = ρg(z2 - z1) = ρg(h1 - h2)
• From fig.10.36(a), it is clear that:
(z2 - z1) = (h1 - h2) = Level difference between A and B
10. So we can write:
• When the fluid is stationary,
♦ The pressure difference between any two points
♦ is equal to
♦ ρg × level difference between the two points
11. Next, let us compare points which are on the same horizontal level:
• First we will compare A and D:
• From (7) we get: Pa + ρg(z1 + h1) = Pa + ρg(z1 + h1)
• It is clear that:• When the fluid is stationary,
♦ The pressure difference between any two points on the same horizontal level
♦ is equal to
♦ zero
• This result can be obtained from (10) also:
♦ The level difference between the two points
♦ on the same horizontal level
♦ is equal to
♦ zero
• So pressure difference = (ρg × 0) = 0
12. In fig.10.36(b), the manometer is removed. So the water begins to flow out through the yellow pipe
♦ We will apply Bernoulli's equation at the four points again
♦ This time we have to include the kinetic energy term also
(i) At point A:
♦ Pressure energy per unit volume = Pa + PA
♦ Kinetic energy per unit volume = 1⁄2 ×ρ×vA2
♦ Potential energy per unit volume = ρgh1
♦ Total energy per unit volume = (Pa + PA + 1⁄2 ×ρ×vA2 + ρgh1)
(ii) At point B:
♦ Pressure energy per unit volume = Pa + PB
♦ Kinetic energy per unit volume = 1⁄2 ×ρ×vB2
♦ Potential energy per unit volume = ρgh2
♦ Total energy per unit volume = (Pa + PB + 1⁄2 ×ρ×vB2 + ρgh2)
(iii) At point C:
♦ Pressure energy per unit volume = Pa + PC
♦ Kinetic energy per unit volume = 1⁄2 ×ρ×vC2
♦ Potential energy per unit volume = ρgh2
♦ Total energy per unit volume = (Pa + PC + 1⁄2 ×ρ×vC2 + ρgh2)
(iv) At point D:
♦ Pressure energy per unit volume = Pa
✰ Point D is open to atmosphere. So only atmospheric pressure is present at D
♦ Kinetic energy per unit volume = 1⁄2 ×ρ×vD2
♦ Potential energy per unit volume = ρgh1
♦ Total energy per unit volume = (Pa + 1⁄2 ×ρ×vD2 + ρgh1)
13. The total energies at all points should be the same. So we get:
(Pa + PA + 1⁄2 ×ρ×vA2 + ρgh1) = (Pa + PB + 1⁄2 ×ρ×vB2 + ρgh2)
= (Pa + PC + 1⁄2 ×ρ×vC2 + ρgh2) = (Pa + 1⁄2 ×ρ×vD2 + ρgh1)
14. Considering points A and D, we get:
(Pa + PA + 1⁄2 ×ρ×vA2 + ρgh1) = (Pa + 1⁄2 ×ρ×vD2 + ρgh1)
⇒ (PA + 1⁄2 ×ρ×vA2) = (1⁄2 ×ρ×vD2)
⇒ PA = 1⁄2 ×ρ×(vA2 - vD2)
15. So we can write:
• When the water is flowing:
♦ The pressure energy at A
♦ is equal to
♦ The change in kinetic energy between points A and D
■ In other words, the pressure energy at A is utilized for obtaining the ‘kinetic energy change’ between A and D
16. Considering points B and C, we get:
(Pa + PB + 1⁄2 ×ρ×vB2 + ρgh2) = (Pa + PC + 1⁄2 ×ρ×vC2 + ρgh2)
⇒ (PB + 1⁄2 ×ρ×vB2) = (PC + 1⁄2 ×ρ×vC2)
♦ The velocity at B will not be equal to the velocity at C
✰ This is because of the difference in areas of the pipes
■ So we can write:
Even though B and C are on the same horizontal level, PB will not be equal to PC
17. The result in (16) can be rearranged as:
(PB - PC) = (1⁄2 ×ρ×vC2 - 1⁄2 ×ρ×vB2)
■ So we can write:
The difference in pressure energy (PB - PC) is utilized to change the kinetic energy between B and C
18. Considering points A and C, we get:
(Pa + PA + 1⁄2 ×ρ×vA2 + ρgh1) = (Pa + PC + 1⁄2 ×ρ×vC2 + ρgh2)
⇒ (PA + 1⁄2 ×ρ×vA2 + ρgh1) = (PC + 1⁄2 ×ρ×vC2 + ρgh2)
♦ The velocity at A will not be equal to the velocity at C
✰ This is because of the difference in areas of the pipes
♦ There is difference in heights also
19. The result in (18) can be rearranged as:
(PA - PC) = (1⁄2 ×ρ×vC2 - 1⁄2 ×ρ×vA2) + (ρgh2 - ρgh1)
■ So we can write:
The difference in pressure energy (PA - PC) is utilized for two items:
♦ to change the kinetic energy between A and C
♦ to change the potential energy between A and C
(Pa + PB + 1⁄2 ×ρ×vB2 + ρgh2) = (Pa + 1⁄2 ×ρ×vD2 + ρgh1)
⇒ (PB + 1⁄2 ×ρ×vB2 + ρgh2) = (1⁄2 ×ρ×vD2 + ρgh1)
♦ The velocity at B will not be equal to the velocity at D
✰ This is because of the difference in areas of the pipes
♦ There is difference in heights also
19. The result in (20) can be rearranged as:
PB = (1⁄2 ×ρ×vD2 - 1⁄2 ×ρ×vB2) + (ρgh1 - ρgh2)
■ So we can write:
The pressure energy PB is utilized for two items:
♦ to change the kinetic energy between B and D
♦ to change the potential energy between B and D
• We must always remember that, there are situations where Bernoulli’s equation cannot be applied as such. This can be explained in 4 steps:
1. We applied the law of conservation of energy
• We assumed that, all work is utilized for changing two items:
♦ Kinetic energy and potential energy
2. But all the energy is not available in this way. Some energy will be lost to overcome friction
• The friction related to fluids is called viscosity. We will learn about it in a later section
• Bernoulli’s equation is applicable only to non-viscous fluids
3. Also, the fluid must be in-compressible
• If the fluid is compressible, it’s volume and density will change
• In such a situation, the equation is not valid
4. Bernoulli’s equation is applicable only for streamline flow
• If the flow is turbulent, velocity and pressure fluctuate constantly
• In such a situation, the equation is not valid
In the next section, we will see a few more applications of Bernoulli's equation
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