Name: Introduction to Fluid Mechanics
Uploaded: 2019-07-19
Duration: 1 min 36 s
Description: A fluid can be at rest like water in a glass or in motion like water flowing in a pipe. This video explains fluid mechanics dealing with the behaviour of the...

Question

State and prove Bernoulli&rsquo;s theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid.

Accepted Answer

Let us consider a flow of liquid through a pipe AB as shown in Figure. Let V be the volume of the liquid when it enters A in a time t which is equal to the volume of the liquid leaving B in the same time. Let $a_{A}$ , $v_{A}$ and $P_{A}$ be the area of cross-section of the tube, velocity of the liquid, and pressure exerted by the liquid at A respectively.
Let the force exerted by the liquid at A is $F_{A} = P_{A} a_{A}$
Distance travelled by the liquid in time t is $d = \frac{v_{A}}{t}$
Therefore, the work done is $W = F_{A} d = P_{A} a_{A} v_{A} t$
But $a_{A} v_{A} t = a_{A} d = V$ volume of the liquid entering at A. Thus, the work done is the pressure energy (at A), $W = F_{A} d = P_{A} V$

Pressure energy per unit volume at A
$\frac{P r e s s u r e e n e r g y}{v o l u m e} = \frac{P_{A} V}{V} = P_{A}$
Pressure energy per unit mass at A
Pressure energy
$\frac{P r e s s u r e e n e r g y}{m a s s} = \frac{P_{A} V}{m} = \frac{P_{A}}{\frac{m}{V}} = \frac{P_{A}}{ρ}$
Since m is the mass of the liquid entering at A in a given time, therefore, pressure energy of the liquid at A is
$E_{P A} = P_{A} V = P A V \times (\frac{m}{m}) = m \frac{P_{A}}{ρ}$
Potential energy of the liquid at A, $P E_{A} = m g h_{A}$
Due to the flow of liquid, the kinetic energy of the liquid at A, $K E_{A} = \frac{1}{2} m {v_{A}}^{2}$
Therefore, the total energy due to the flow of liquid at A, $E_{A} = E_{P A} + K E_{A} + P E_{A}$
$E_{A} = m \frac{P_{A}}{ρ} + \frac{1}{2} m {v_{A}}^{2} + m g h_{A}$
Similarly, let aB, vB, and PB be the area of cross section of the tube, velocity of the liquid, and pressure exerted by the liquid at B. Calculating the total energy at EB, we get
$E_{B} = m \frac{P_{B}}{ρ} + \frac{1}{2} m {v_{B}}^{2} + m g h_{B}$
From the law of conservation of energy, $E_{A} = E_{B}$
$m \frac{P_{A}}{ρ} + \frac{1}{2} m {v_{A}}^{2} + m g h_{A} = m \frac{P_{B}}{ρ} + \frac{1}{2} m {v_{B}}^{2} + m g h_{B} \frac{P_{A}}{ρ} + \frac{1}{2} {v_{A}}^{2} + g h_{A} = \frac{P_{B}}{ρ} + \frac{1}{2} {v_{B}}^{2} + g h_{B} = c o n s \tan t$
Thus, the above equation can be written as
$\frac{P}{ρ} + \frac{1}{2} v^{2} + g h = c o n s \tan t$

State and prove Bernoulli’s theorem for a flow of incompressible, non-viscous, and streamlined flow of fluid.

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