Name: Fluid Mechanics
Uploaded: 2019-07-19
Duration: 6 min 9 s
Description: This video enables students to understand that fluid mechanics comprises fluid statics and fluid dynamics. Students also learn about steady and turbulent flo...

Question

Using Bernoulli’s theorem, prove the following formula for an ideal fluid, $p_{1} - p_{2} = \frac{ρ}{2} ({v_{2}}^{2} - {v_{1}}^{2})$ , where $p_{1}$ and $p_{2}$ are the pressures and $v_{1}$ and $v_{2}$ are the velocities of flow of a liquid of density $ρ$ at the ends of a horizontal tube. There is no friction in the tube.

Accepted Answer

When an incompressible and non-viscous liquid (or gas) flows in streamlined motion from one place to another, then at every point of its path the total energy per unit volume (Pressure energy $+$ Kinetic energy $+$ Potential energy) is constant. That is,

$P + \frac{1}{2} ρ v^{2} + ρ g h = c o n s \tan t$ .

Thus, Bernoulli’s theorem is in one way the principle of conservation of energy for a flowing liquid (or gas).

Suppose an incompressible and non-viscous liquid is flowing in streamlined motion through a tube $X Y$ of non-uniform cross-section. Let $A_{1}$ and $A_{2}$ be the areas of cross-section of the tube at the ends $X$ and $Y$ respectively, which are at heights $h_{1}$ and $h_{2}$ from the surface of the earth. Let $p_{1}$ be the pressure and $v_{1}$ the velocity of flow of the liquid at $X$ and $p_{2}$ and $v_{2}$ the respective quantities at $Y$ . Since the area $A_{2}$ is smaller than $A_{1}$ , the velocity $v_{2}$ is greater than $v_{1}$ .
Work done per second on the liquid entering the tube at $X$ is,

Force $\times$ Distance $= p_{1} \times A_{1} \times v_{1}$

Similarly, work done per second by the liquid leaving the tube at $Y$ is, $p_{2} \times A_{2} \times v_{2}$

Thus, net work done on the liquid is,

$p_{1} \times A_{1} \times v_{1} = p_{2} \times A_{2} \times v_{2} . . . . . (i)$

But $A_{1} \times v_{1}$ and $A_{2} \times v_{2}$ are respectively the volumes of the liquid entering at $X$ and leaving at $Y$ per second. These volumes must be equal and so,

$A_{1} \times v_{1} = A_{2} \times v_{2} = \frac{m}{ρ}$ , where $m$ is the mass of the liquid entering at $X$ or leaving at $Y$ in $1$ second and $ρ$ is the density of the liquid. Thus, from the above equation net work done on the liquid is, $(p_{1} - p_{2}) \frac{m}{ρ} . . . . . (i i)$

Now, increase in kinetic energy of the liquid is, $\frac{1}{2} m ({v_{2}}^{2} - {v_{1}}^{2})$ .

Decrease in potential energy of the liquid is, $m g (h_{1} - h_{2})$

This increase in energy is due to the net work done on the liquid, that is,

Net work done $=$ Net increase in energy.

Hence, $(p_{1} - p_{2}) \frac{m}{ρ} = \frac{1}{2} m ({v_{2}}^{2} - {v_{1}}^{2}) - m g (h_{1} - h_{2})$ .

If the two ends of the tube are at the same height from the reference level then, $h_{1} = h_{2}$ or

$(p_{1} - p_{2}) \frac{m}{ρ} = \frac{1}{2} m ({v_{2}}^{2} - {v_{1}}^{2})$ or

$p_{1} - p_{2} = \frac{ρ}{2} ({v_{2}}^{2} - {v_{1}}^{2})$

Important Questions on Motion in Fluids : Viscosity

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Using Bernoulli’s theorem, prove the following formula for an ideal fluid, p1-p2=ρ2(v22-v12), where p1 and p2 are the pressures and v1 and v2 are the velocities of flow of a liquid of density ρ at the ends of a horizontal tube. There is no friction in the tube.

Important Questions on Motion in Fluids : Viscosity

Learn and Prepare for any exam you want !