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

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

G L Mittal and TARUN MITTAL Solutions for Chapter: Motion in Fluids : Viscosity, Exercise 1: QUESTIONS

G L Mittal Physics Solutions for Exercise - G L Mittal and TARUN MITTAL Solutions for Chapter: Motion in Fluids : Viscosity, Exercise 1: QUESTIONS

Questions from G L Mittal and TARUN MITTAL Solutions for Chapter: Motion in Fluids : Viscosity, Exercise 1: QUESTIONS with Hints & Solutions

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