Question

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#Some_Question_From_Fluids (Class 11th)

◘ Prove Bernoulli's Theorem!

♣ Prove Equation Of Continuity!

♠ What is Stream Line Flow?


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Answer 1

$\bold{ BERNOULLI'S\: THEOREM$
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It states that when a non - Viscous liquid and incompressible liquid flows through a tube of non - uniform cross section area, then at each point, the total energy of the liquid ( Kinetic + Potential + Pressure energy ) per unit volume will remain constant.

P +$\frac{1}{2}$ ρ v² + ρgh = constant

$\bold{Proof}$ :

Consider an ideal liquid of density ρ,is flowing through the tube in streamline flow, then the force experienced by the liquid enters at the end in time t,

F₁ = P₁ A₂

Work done on the liquid at end x,

W₁ = F₁ × v ₁ ∆t

W₁ = P₁ A₁ ( v ₁ ∆t )

W₁ = P₁ ∆v₁

Similarly, work done by the liquid at end Y,

W₂ = P₂ ∆v₂

Net work done on the liquid,

W = ( P₁ ∆v₁ ) - ( P₂ ∆v₂ )

From equation of continuity,

A₁ V₁ = A₂ V₂

A₁ V₁ ∆t = A₂ V₂ ∆t

∆v₁ = ∆v₂

W = ( P₁ - P₂ ) ∆v

W = ( P₁ - P₂ )$\frac{m} {ρ}$ ---> ( i )

Kinetic Energy,

∆K =$\frac{1}{2}$ mv₂ ² - $\frac{1}{2}$ mv₁²

Potential Energy,

∆P = mgh₂ - mgh₁

Total energy of the liquid,

∆E = ∆K + ∆P

∆E = ( $\frac{1}{2}$ mv₂ ² - $\frac{1}{2}$ mv₁² ) + ( mgh₂ - mgh₁ )

From the law of conservation of energy,

W = ∆E

( P₁ - P₂ )$\frac{m} {ρ}$ = ($\ frac{1}{2}$ mv₂² - $\frac{1}{2}$ mv₁² ) + ( mgh₂ - mgh₁ )

( P₁ - P₂ ) = $\frac{1}{2}$ ρv₂² - frac{1}{2} ρv₁² + ρgh₂ -ρgh₁

P₁ +$\frac{1}{2}$ ρv₁² +ρgh₁ = P₂ +$\frac{1}{2}$ ρv₂² + ρgh₂

P + $\frac{1}{2}$ ρv² +ρgh = constant.
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$\bold{ EQUATION \:OF \:CONTINUITY$

It states that during the streamlined flow of an ideal liquid ( incompressible, no-Viscous) through a pipe of varying Cross section, the product of area of cross - section and velocity of flow remains constant.

AV = constant

$\bold {PROOF}$ :

Consider a liquid of density ρ flows through the tube XY, then volume of liquid enters at end v₁ = A₁v₁∆t and mass of the liquid enters at end X₁,

m₁ = A₁v₁ ∆t ρ

Similarly, mass of liquid going out through end y,

m₂ = A₂ v₂ ∆t ρ

According to law of conservation of mass,

m₁ = m₂

A₁v₁ ∆t ρ = A₂ v₂ ∆t ρ

A₁v₁ = A₂ v₂

AV = constant.

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$\bold{ STREAMLINE \:FLOW}$

It is the flow of liquid in which each particle of the liquid passing through a point travels along the same path and with same velocity as the preceding particle passing through the same point.

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Answer 2

Consider a fluid of negligible viscosity moving with laminar flow
Let the velocity, pressure and area of the fluid column be v1, P1 and A1 at Q and v2, P2 and A2 at R. Let the volume bounded by Q and R move to S and T where QS = L1, and RT = L2. If the fluid is incompressible:
A1L1 = A2L2
The work done by the pressure difference per unit volume = gain in k.e. per unit volume + gain in p.e. per unit volume. Now:
Work done = force x distance = p x volume 
Net work done per unit volume = P1 - P2 
k.e. per unit volume = ½ mv2 = ½ Vρ v2 = ½ρv2 (V = 1 for unit volume)

Therefore:

k.e. gained per unit volume = ½ ρ(v22 - v12)

p.e. gained per unit volume = ρg(h2 – h1)

where h1 and h2 are the heights of Q and R above some reference level. Therefore:

P1 - P2 = ½ ρ(v12 – v22) + ρg(h2 - h1) 
P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + rgh2

Therefore: 

P + ½ ρv2 + ρgh is a constant

For a horizontal tube h1 = h2 and so we have: 

P + ½ ρv2 = a constant

This is Bernoulli's theorem You can see that if there is a increase in velocity there must be a decrease of pressure and vice versa.

No fluid is totally incompressible but in practice the general qualitative assumptions still hold for real fluids.

Consider a fluid of negligible viscosity moving with laminar flow, as shown in Figure 1. 

Let the velocity, pressure and area of the fluid column be v1, P1 and A1 at Q and v2, P2 and A2 at R. Let the volume bounded by Q and R move to S and T where QS = L1, and RT = L2. If the fluid is incompressible:

A1L1 = A2L2

The work done by the pressure difference per unit volume = gain in k.e. per unit volume + gain in p.e. per unit volume. Now:

Work done = force x distance = p x volume 
Net work done per unit volume = P1 - P2 
k.e. per unit volume = ½ mv2 = ½ Vρ v2 = ½ρv2 (V = 1 for unit volume)

Therefore:

k.e. gained per unit volume = ½ ρ(v22 - v12)

p.e. gained per unit volume = ρg(h2 – h1)

where h1 and h2 are the heights of Q and R above some reference level. Therefore:

P1 - P2 = ½ ρ(v12 – v22) + ρg(h2 - h1) 
P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + rgh2

Therefore: 

P + ½ ρv2 + ρgh is a constant

For a horizontal tube h1 = h2 and so we have: 
P + ½ ρv2 = a constant
This is Bernoulli's theorem You can see that if there is a increase in velocity there must be a decrease of pressure and vice versa.

No fluid is totally incompressible but in practice the general qualitative assumptions still hold for real fluids.
R = A v = constant

Where,

R= volume flow rate,

A= flow area, and

v= flow velocity

Assumption:

To derive the continuity equation, it is important to assume a tube having a single entry and a single exit.

Derivation:

Consider the diagram:

Now, consider the fluid flows for a short interval of time in the tube. So, assume that short interval of time as Δt. In this time, the fluid will cover a distance of Δx1 with a velocity v1 at the lower end of the pipe.

At this time, the distance covered by the fluid will be->

Δx1 = v1Δt

Now, at the lower end of the pipe, the volume of the fluid that will flow into the pipe will be->

V = A1 Δx1 = A1 v1 Δt

It is known that mass (m) = Density (ρ) × Volume (V). So, the mass of the fluid in Δx1 region will be->

Δm1= Density × Volume

=> Δm1 = ρ1A1v1Δt ——–(Equation 1)

Now, the mass flux has to be calculated at the lower end. Mass flux is simply defined as the mass of the fluid per unit time passing through any cross-sectional area. For the lower end with cross-sectional area A1, mass flux will be->

Δm1/Δt = ρ1A1v1 ——–(Equation 2)

Similarly, the mass flux at the upper end will be->

Δm2/Δt = ρ2A2v2 ——–(Equation 3)

Here, v2 is the velocity of the fluid through the upper end of the pipe i.e. through Δx2 , inΔttime andA2, is the cross-sectional area of the upper end.

In this, the density of the fluid between the lower end of the pipe and the upper end of the pipe remains same with time as the flow is steady. So, the mass flux at the lower end of the pipe is equal to the mass flux at the upper end of the pipe i.e. Equation 2 = Equation 3.

Thus,

ρ1A1v1 = ρ2A2v2 ——–(Equation 4)

This can be written in a more general form as->

ρ A v = constant

With equation proves the law of conservation of mass in fluid dynamics. Also, if the fluid is incompressible, the density will remain constant for steady flow. So,ρ1 =ρ2.
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