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
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| 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
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 vP
• This vP 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 vQ
♦ This vQ will be tangential to the curve at Q
• When that particle reaches R, it’s velocity will be vR
♦ This vR will be tangential to the curve at R
(iii) Now comes the interesting part
• Every predecessor of our particle
♦ Had the same velocity vP when it crossed P
♦ Had the same velocity vQ when it crossed Q
♦ Had the same velocity vR when it crossed R
• Every successor of our particle
♦ Will have the same velocity vP when it crosses P
♦ Will have the same velocity vQ when it crosses Q
♦ Will have the same velocity vR 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, vP ≠ vQ ≠ vR
♦ 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
■ 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
• 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:
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| Fig.10.31 |
♦ These planes are perpendicular to the pipe at the respective points
• 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
• 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
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
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)
♦ 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 vA
• Let this average velocity be vA
• 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 vA
♦ Take any point which cross the plane at A
♦ When it just crosses the plane, it’s velocity is assumed to be equal to vA
■ So all particles will travel a distance of (vA × Δt) in the duration Δt
■ Then volume of all the particles which cross the plane in the duration of t will be equal to (AA × vA × Δt)
♦ ∵ Volume of a prism = Base area × height
♦ AA 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
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| Fig.10.32 |
6. Here we must recognize two important assumptions. This can be explained in two steps:
(i) The velocity vA 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
• But in our present case, Δt is very small
• So we can assume that, the particles travel with the same velocity vA for the entire duration of Δt
(ii) Similar is the case with area
• AA 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
• 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 AA for the entire length of (vA × Δt)
• So we can assume that, the area remains the same AA for the entire length of (vA × Δt)
7. Based on the two assumptions, we can write:
• In the duration of Δt, numerous particles will cross the plane at A
• In the duration of Δt, numerous particles will cross the plane at A
♦ Total volume of all those particles will be equal to (AA × vA × 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 (AA × vA × 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 (AB × vB × Δt) ♦ AB is the cross sectional area of the pipe at B
♦ vB 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 (AC × vC × Δt) ♦ AC is the cross sectional area of the pipe at C
♦ vC 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 (AB × vB × 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 (AC × vC × Δt × ρ)
10. If the fluid is in-compressible,
♦ The mass flowing out
♦ must be equal to
♦ The mass flowing in
• So we get:
♦ must be equal to
♦ The mass flowing in
• So we get:
(AA × vA × Δt × ρ) = (AB × vB × Δt × ρ) = (AC × vC × Δt × ρ)
From this we get:
Eq.10.5: (AA × vA) = (AB × vB) = (AC × vC)
■ 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
■ 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 m3s-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 (m3s-1) = A 0.5 (ms-1)
3. Thus we get: $\mathbf\small{\rm{A= \frac{3(m^3 s^{-1})}{0.5(m \; s^{-1})}=6\;m^2}}$
(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 (Agreen, Ared, Amagenta) are different
• Even then, we get the same volume because, the lengths (lgreen, lred, lmagenta) are different
(v) We have:
♦ Agreen < Ared < Amagenta
• 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 (Agreen, Ared, Amagenta) are different
• Even then, we get the same volume because, the lengths (lgreen, lred, lmagenta) are different
(v) We have:
♦ Agreen < Ared < Amagenta
♦ lgreen > lred > lmagenta
(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
♦ So they can travel only a short distance in the duration of Δt
• 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
In the next section, we will see Bernoulli's principle
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