Chapter 10.8 - Streamline Flow

In the previous section, we saw the details about buoyancy and flotation. In this section, we will see streamline flow

First we will see some basics. It can be written in 4 steps:

1. In fig.10.30(a) below, a fluid is flowing through a pipe

• The pipe is indicated by the two grey curves

    ♦ The pipe has a curved shape

    ♦ The area of cross section of the pipe increases as we move towards the right

• The fluid is indicated in orange color

    ♦ It flows from left to right

Fig.10.30

2. A blue curve is shown inside the pipe. How is this blue curve obtained?

• The answer can be written in 4 steps

(i) Pick any one particle in the fluid

(ii) Draw the path followed by that particle

(iii) We will get a smooth curve 

(iv) The blue curve in fig.a is obtained by drawing the path followed by a particle

3. But what is the significance of the path followed by a single particle?

• The answer can be written in 4 steps:

(i) We drew the blue curve by following a single particle

(ii) But in fact, there are numerous particles which flow along the same path

(iii) The particle that we chose has

    ♦ Numerous predecessors which flowed along the same path

    ♦ Numerous successors which will flow along the same path

(iv) Since the path is used by so many particles, it has much significance

■ A 3D view of the pipe, the fluid and the path inside is shown in fig.10.30(c)

4. There will be a large number of such paths inside the pipe

• It is like:

    ♦ A large number of small pipes within the large pipe

    ♦ All the small pipes will be having the same uniform cross sectional area

    ♦ This uniform cross sectional area will be equal to the size of the particle

    ♦ In later sections, we will see how particles are able to move in such a disciplined manner

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We have seen some basics. Let us now write the three important properties of the blue curve

1. The first property is related to velocity of particles. It can be written in 4 steps

(i) In fig.10.30(b) above, three points P, Q and R are marked on the curve

(ii) We said that, we chose a single particle to draw the curve

• When that particle reaches P, it’s velocity will be v_P

• This v_P will be tangential to the curve at P

    ♦ We know how to draw the tangent to a circle at any point on that circle

    ♦ In the same way, we can draw a tangent at any given point on a curve

    ♦ We will learn it in math classes

• So, to show the ‘direction of movement’ of the particle at P, all we need to do is:

    ♦ Draw the tangent at P

• When that particle reaches Q, it’s velocity will be v_Q

    ♦ This v_Q will be tangential to the curve at Q

• When that particle reaches R, it’s velocity will be v_R

    ♦ This v_R will be tangential to the curve at R

(iii) Now comes the interesting part

• Every predecessor of our particle

    ♦ Had the same velocity v_P when it crossed P

    ♦ Had the same velocity v_Q when it crossed Q

    ♦ Had the same velocity v_R when it crossed R

• Every successor of our particle

    ♦ Will have the same velocity v_P when it crosses P

    ♦ Will have the same velocity v_Q when it crosses Q

    ♦ Will have the same velocity v_R when it crosses R

(iv) We can write about the relation between various velocities

• We know that, velocity has both magnitude and direction

    ♦ Obviously, in our present case, v_P ≠ v_Q ≠ v_R

    ♦ Because, their directions are different

• But their magnitudes may be equal

    ♦ It will depend on the area of cross section of the pipe. We will see those details later

2. The second property is that:

The curves do not cross each other

• This can be explained in 4 steps:

(i) Imagine two curves crossing each other

(ii) A particle coming through one curve will reach a ‘point of intersection’

(iii) From that point onwards, the particle can choose to go in any of the two directions

• This is not allowed. A particle must follow only a single path

(iv) So the curves do not cross each other 

3. The third property is related to the 'permanent nature' of the curve

• This can be explained in 2 steps:

(i) The flow is continuously taking place inside the pipe

    ♦ We begin to observe the flow when the reading in the stop-watch is t₁

    ♦ We stop the observation when the reading in the stop-watch is t₂

• So we observe the flow for a time duration of t = (t₂-t₁)

(ii) During this t, the shape of the curve will not change

• It is like a road marked on a map. That road does not change in shape

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■ If the above three conditions are satisfied, the flow is called: steady flow

■ Another name for steady flow is: streamline flow

■ Each path in the flow is called a streamline

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• Now we will see equation of continuity

• It can be explained in 14 steps

1. The pipe in fig.10.30 is shown again in fig.10.31(a) below:

Fig.10.31

• At the points A, B and C, imaginary planes cut through the pipe

    ♦ These planes are perpendicular to the pipe at the respective points

• A 3D view is shown in fig.b
2. When flow takes place through the pipe, streamlines penetrate through these planes
• Consider the plane at A
    ♦ Numerous streamlines will penetrate through this plane
• Each of those streamlines carry particles
• So we can write:
Numerous particles continuously pass through the imaginary plane at A
3. We want the total mass of the particles which pass the plane at A, in a small interval of time Δt
    ♦ Let us begin the observation when the reading in the stop-watch is t₁

    ♦ Let us stop the observation when the reading in the stop-watch is t₂

• Let the difference between t₁ and t₂ be very small, say Δt

    ♦ That means, (t₂-t₁) = Δt

    ♦ That means, we observe the flow for a very short time duration of Δt
4. Numerous particles will be passing the plane at instant t₁

• Pick any one of those particles. We will call it: particle 1 or p₁
    ♦ At the instant t₁, let it's velocity be v_(p1)

• We know that: distance = velocity × time

• So in a time duration of Δt, that particle will travel a distance of: v_(p1) × Δt

5. This is the distance traveled by the particle p₁

• Different particles will be having different velocities when they cross the plane:

    ♦ v_(p1), v_(p2), v_(p3), v_(p4) so on . . . 

■ But it is possible to find an average velocity which is applicable to all particles
• Let this average velocity be v_A

• That is:
    ♦ Take any point which cross the plane at A
    ♦ When it just crosses the plane, it’s velocity is assumed to be equal to v_A

■ So all particles will travel a distance of (v_A × Δt) in the duration Δt

■ Then volume of all the particles which cross the plane in the duration of t will be equal to (A_A × v_A × Δt)

    ♦ ∵ Volume of a prism = Base area × height

    ♦ A_A is the cross sectional area of the pipe at A

• This volume is indicated by green color in fig.10.32(a) below

    ♦ Fig.10.32(b) shows a 3D view

    ♦ The pipe is cut length wise. The volume shown in green color, has a near cylindrical shape

Fig.10.32

6. Here we must recognize two important assumptions. This can be explained in two steps:

(i) The velocity v_A is that velocity when the particles just cross the plane at A

• When they move away from the plane (that is., with passage of time), the velocity changes
• But in our present case, Δt is very small

• So we can assume that, the particles travel with the same velocity v_A for the entire duration of Δt

(ii) Similar is the case with area

• A_A is the cross sectional area of the pipe at A

• When the particles move away from the plane, they will reach new positions, where the cross sectional area is different
• But in our present case, Δt is very small

• In this small duration, the particles will travel only a very small distance
• So we can assume that, the area remains the same A_A for the entire length of (v_A × Δt)

7. Based on the two assumptions, we can write:
• In the duration of Δt, numerous particles will cross the plane at A

    ♦ Total volume of all those particles will be equal to (A_A × v_A × Δt) 

8. We can write it in terms of mass also:

• In the duration of Δt, numerous particles will cross the plane at A

    ♦ Total mass of all those particles will be equal to (A_A × v_A × Δt × ρ)

    ♦ Where ρ is the density of the fluid

9. Now we will consider the masses at B and C

Since we have discussed the mass at A in detail, the masses at B and C can be written in just 3 steps:

(i) We started the observation at t₁

• We did the observation for a duration of Δt

    ♦ In that duration of Δt starting from the same t₁, numerous particles will have crossed the plane at B

    ♦ In that duration of Δt starting from the same t₁, numerous particles will have crossed the plane at C

(ii) We can apply the same steps that we applied to the plane at A. We will get:

• In the duration of Δt, numerous particles will have crossed the plane at B

    ♦ Total volume of all those particles will be equal to (A_B × v_B × Δt) 

    ♦ A_B is the cross sectional area of the pipe at B

    ♦ v_B is the velocity of the particles at B

    ♦ This total volume is shown in red color in fig.10.32

• In the duration of Δt, numerous particles will have crossed the plane at C

    ♦ Total volume of all those particles will be equal to (A_C × v_C × Δt) 

    ♦ A_C is the cross sectional area of the pipe at C

    ♦ v_C is the velocity of the particles at C

    ♦ This total volume is shown in magenta color in fig.10.32

(iii) Now we can write about the masses. We will get:

• In the duration of Δt, numerous particles will have crossed the plane at B

    ♦ Total mass of all those particles will be equal to (A_B × v_B × Δt × ρ)

    ♦ Where ρ is the density of the fluid

• In the duration of Δt, numerous particles will have crossed the plane at C

    ♦ Total mass of all those particles will be equal to (A_C × v_C × Δt × ρ) 

10. If the fluid is in-compressible,

    ♦ The mass flowing out
    ♦ must be equal to
    ♦ The mass flowing in
• So we get:

(A_A × v_A × Δt × ρ) = (A_B × v_B × Δt × ρ) = (A_C × v_C × Δt × ρ)

From this we get:

Eq.10.5: (A_A × v_A) = (A_B × v_B) = (A_C × v_C)
■ This equation is called the equation of continuity
11. It is clear that, the product (A×v) will be the same at all sectional planes along the pipe

• So we can write: A×v = constant

■ The product (A×v) is called volume flux

• Another name for volume flux is flow rate

12. Since Av is a constant, we can write: $\mathbf\small{\rm{v \propto \frac{1}{A}}}$ 

• That means, when area increases, velocity decreases and vice versa

• We will see it's applications in later sections

13. 'Flow rate' or 'Rate of flow' implies:

    ♦ How many liters flow in unit time

    ♦ or

    ♦ How many cubic meters flow in unit time

• So in general, 'flow rate' is 'volume per unit time'

14. Let us see how this 'volume per unit time' is related to 'Av'

• We can use dimensional analysis to establish the relation

• It can be written in 3 steps:

(i) Dimensions of 'cubic meters per second' is obtained as:

$\mathbf\small{\rm{\frac{[L^3]}{[T]}=[L^3 T^{-1}]}}$

(ii) Dimensions of 'Av' is obtained as:

$\mathbf\small{\rm{[L^2] \times \frac{[L]}{[T]}=[L^3 T^{-1}]}}$

(iii) We see that both dimensions are the same

• So 'Av' is equivalent to 'volume per unit time' 

An example:

A fluid flows through a pipe of uniform cross sectional area. If the rate fo flow is 3 m_³s_^-1  and velocity of flow is 0.5 ms_^-1, what is the cross sectional area of the pipe?

Solution:

1. We have: Flow rate = Av

2. Substituting the given values, we get: 3 (m_³s_^-1) = A 0.5 (ms_^-1)

3. Thus we get: $\mathbf\small{\rm{A= \frac{3(m^3 s^{-1})}{0.5(m \; s^{-1})}=6\;m^2}}$

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• An interesting point can be written in 6 steps:
(i) In step (10) above, we have seen that, each of the total masses at A, B and C in a duration of Δt are the same

(ii) If masses are the same, the volumes at A, B and C will also be the same
• Those volumes are the three cylindrical shapes in fig.10.32(b)
(iii) We can write:
Volume of green cylinder = Volume of red cylinder = Volume of magenta cylinder
(iv) We see that the base areas of the cylinders (A_green, A_red, A_magenta) are different
• Even then, we get the same volume because, the lengths (l_green, l_red, l_magenta) are different
(v) We have:
    ♦ A_green < A_red < A_magenta

    ♦ l_green > l_red > l_magenta

(vi) When area is greater, the length becomes lesser because:

    ♦ The particles have a lower velocity at sections of greater cross sectional areas
    ♦ So they can travel only a short distance in the duration of Δt

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• We have discussed steady flow and equation of continuity

• Steady flow is possible only when the speed of flow is below a certain value

■ This limiting speed is called critical speed

• If the speed of flow is above critical speed, the streamlines will cross each other

• We will not be able to identify the streamlines any more. Such a flow is called turbulent flow

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In the next section, we will see Bernoulli's principle

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Chapter 10.5 - Pascal's law

In the previous section, we saw the details about mercury barometer. In this section, we will see Pascal's law

Pascal's law can be explained in 7 steps:

1. Consider the arrangement shown in fig.10.22 below:

Fig.10.22

• A horizontal cylinder is shown in pink color

• It has a piston on it’s left end

2. Three vertical cylinders having different diameters are taken out from the horizontal cylinder

• We know that, though the vertical cylinders have different diameters, the liquid levels will be the same (in our present case, it is h₁)

• The pressure in the horizontal cylinder will be equal to 𝛒gh₁

    ♦ Where 𝛒 is the density of the liquid in the cylinder

• So initial pressure at the bottom of each vertical cylinder will be the same: 𝛒gh₁

3. Now, push the piston towards the right

• The pressure in the horizontal cylinder will increase

• The liquid will rise to a new ‘common level’ in all the vertical cylinders 

4. Note that: A new ‘common level’ (h₂) is attained

■ That means, the ‘increase in pressure’ in the horizontal cylinder is distributed equally among the vertical cylinders

    ♦ Each cylinder gets the same 'extra pressure'

    ♦ So the final pressure in each cylinder will be the same: 𝛒gh₂

■ Pascal’s law of transmission of fluid pressure states that:

Whenever external pressure is applied on any part of a fluid in a container, it is transmitted undiminished and equally in all directions

5. A simple demonstration of the law can be given using a filled balloon

• When we apply a force on any one portion of the balloon, the air inside will apply an outward force on every point of the balloon

• This is shown in fig.4.23 below:

Fig.10.23

6. The word 'undiminished' is used to indicate that, the 'applied external pressure' is completely available to act on every part of the fluid

    ♦ The distance between 'the point of application' to 'a far end of the fluid body' has no significance    

    ♦ If a pressure P is applied, both 'far ends' and 'near ends' will experience the same P

7. 'Equally in all directions' indicate that, the shape of the container has no significance

    ♦ The fluid may be contained inside a vessel of complicated shape

    ♦ Even then, if a pressure P is applied, all points of the fluid will experience the same P

The Hydraulic lift

• Working of the hydraulic lift is based on Pascal’s law. The details can be written in 5 steps:

1. In fig.10.24, we see two pistons separated by a liquid

    ♦ The smaller piston has an area of A₁

    ♦ The larger piston has an area of A₂

Fig.10.24

2. A downward force F₁ is applied on the smaller piston

• Then the pressure experienced by the liquid will be $\mathbf\small{\rm{\frac{F_1}{A_1}}}$

3. According to Pascal’s law, this pressure will be experienced on the whole liquid body

• The liquid body will apply this pressure on the sides and roof of the container

■ So the larger piston will experience the same upward pressure $\mathbf\small{\rm{\frac{F_1}{A_1}}}$

4. Let F₂ be the upward force experienced by the larger piston

• Then the pressure experienced by the larger piston is equal to $\mathbf\small{\rm{\frac{F_2}{A_2}}}$

• This pressure is equal to $\mathbf\small{\rm{\frac{F_1}{A_1}}}$

5. So we can write: $\mathbf\small{\rm{\frac{F_1}{A_1}=\frac{F_2}{A_2}}}$

• From this we get: $\mathbf\small{\rm{F_2=\left(\frac{A_2}{A_1} \right )F_1}}$

• We see that: F₁ is multiplied by a factor $\mathbf\small{\rm{\frac{A_2}{A_1}}}$

    ♦ 'A₂ in the numerator' is larger than 'A₁ in the denominator'

    ♦ So the factor $\mathbf\small{\rm{\frac{A_2}{A_1}}}$ will be larger than 1

    ♦ So F₂ will be greater than F₁

■ That means, we can:

    ♦ lift a large load placed on the larger piston

    ♦ by applying a small force at the smaller piston

■ The factor $\mathbf\small{\rm{\frac{A_2}{A_1}}}$ is the mechanical advantage of the hydraulic lift

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Solved example 10.12

Two syringes of different cross sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?

Solution:

Part (a):

1. The arrangement is shown in fig.10.25(a) below:

Fig.10.25

• We have: $\mathbf\small{\rm{F_2=\left(\frac{A_2}{A_1} \right )F_1}}$

2. Substituting the known values, we get:

$\mathbf\small{\rm{F_2=\left(\frac{0.25 \times \pi \times 3^2}{0.25 \times \pi \times 1^2} \right )\times 10=90\;N}}$

Part (b):

1. Water is an incompressible liquid

• So all the volume which is pushed out from the left syringe will enter the right syringe

2.Volume pushed out from the left syringe will be (A₁ × 6)

• Let x be the distance moved by the piston of the right syringe. This is shown in fig.b

• So the volume entering the right syringe will be (A₂ × x)

3. Equating the two volumes, we get:

A₁ × 6 = A₂ × x

⇒ $\mathbf\small{\rm{x=\left(\frac{A_1}{A_2} \right )6}}$

4. Substituting the known values, we get: $\mathbf\small{\rm{x=\left(\frac{0.25 \times 1^2}{0.25 \times 3^2} \right )\times 6=0.67\;cm}}$

 

Solved example 10.13

In a car lift compressed air exerts a force F₁ on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm (Fig 10.7). If the mass of the car to be lifted is 1350 kg, calculate F₁ . What is the pressure necessary to accomplish this task? (g = 9.8 ms^-2 ).

Solution:

Part (a):

1. We have: $\mathbf\small{\rm{F_1=\left(\frac{A_1}{A_2} \right )F_2}}$

2. Substituting the known values, we get:

$\mathbf\small{\rm{F_1=\left(\frac{0.25 \times 5^2}{0.25 \times 15^2} \right )\times 1350 \times 9.8=1470\;N}}$

Part (b):

1. We are applying a force of 1470 N on the small piston

2. The small piston has an area of (𝝅 × 5²) cm²

• So we are applying a pressure of:

$\mathbf\small{\rm{\frac{1470}{\pi \times 5^2}}}$ = 18.73 N cm^-2 = 1.873 × 10⁵ N m^-2

3. Normal atmospheric pressure at sea level is 1.013 × 10⁵ N m^-2

• So 1.873 × 10⁵ N m^-2 is nearly twice the atmospheric pressure    

Solved example 10.14

A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². What maximum pressure would the smaller piston have to bear ?

Solution:

1. We have: $\mathbf\small{\rm{\frac{F_1}{A_1}=\frac{F_2}{A_2}}}$

2. Substituting the known values, we get: $\mathbf\small{\rm{\frac{F_1}{A_1}=\frac{3000 \times 9.8}{425 \times 10^{-4}}}}$ = 6.92 × 10⁵ Nm^-2

Hydraulic brake

• The working of the hydraulic brake system is also based on Pascal’s law. This can be explained in 4 steps:

1. When the driver apply a force on the brake pedal, he is actually pushing the piston of a ‘master cylinder’

2. The fluid inside this cylinder will move out and apply pressure on four other pistons

    ♦ These pistons are situated at the four wheels of the automobile

    ♦ These pistons have a larger area than the piston in the master cylinder

          ✰ So a small force is sufficient at the brake pedal

3. The large pistons press against the brake lining of the wheels

• Thus the wheels slow down and come to a stop

• A schematic diagram of the system can be seen here 

4. The advantage of this system can be explained in three steps:

(i) We know that:

• The pressure will be transmitted undiminished and equally in all directions

(ii) So all wheels will experience the same braking force

• Also all wheels will experience the braking force at the same time

(iii) This will enhance safety while applying brakes

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In the next section, we will see buoyancy and flotation

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