Name: Surface Tension and Its Causes
Uploaded: 2019-07-19
Duration: 3 min 11 s
Description: Surface tension is caused by attraction of particles in the surface layer by the bulk of liquid .Here we wiss study about the causes and variation in surface tension. Let's watch.

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.

Important Questions on Properties of Matter

Learn and Prepare for any exam you want !