Bernoulli's Theorem - Practice Questions & MCQ

 For a point in a fluid flow, Bernoulli's Theorem relates between its pressure, its velocity and its height from a reference point.

Bernoulli's Theorem states that the total energy (Pressure energy, Potential energy, and Kinetic energy ) per unit volume or mass of an incompressible and non-viscous fluid in steady flow through a pipe remains constant throughout the flow. (Provided that  there is no source or sink of the fluid along the length of the pipe).

Mathematically for a liquid flowing through a pipe.

We can write Bernoulli's  equation as 

$
P+\rho g h+\frac{1}{2} \rho v^2=\text { constant }
$

$P \rightarrow$ Pressure energy per unit volume
$\rho g h \rightarrow$ Potential Energy per unit volume
$\frac{1}{2} \rho v^2 \rightarrow$ Kinetic Energy per unit volume

Bernoulli's Theorem can be proved with the help of work-energy theorem.
Bernoulli's equation also represents the conservation of mechanical energy in case of moving fluids.
- Bernoulli's theorem for the unit mass of liquid flowing through the pipe is given by

$
\frac{P}{\rho}+g h+\frac{1}{2} v^2=\text { constant }
$

If we divide the above equation by g we get

$
\frac{P}{\rho g}+h+\frac{v^2}{2 g}=\text { constant }
$

Where
$\mathrm{h}=$ gravitational head
$\frac{P}{\rho g} \rightarrow$ Pressure head
$\frac{v^2}{2 g} \rightarrow$ velocity head

• For the below figure

 With the help of Bernoulli's  equation 

We can write

$P_1+\rho g h_1+\frac{1}{2} \rho v_1^2=P_2+\rho g h_2+\frac{1}{2} \rho v_2^2=$ constant

The velocity of Efflux or Torricelli's Theorem-

If a liquid is filled in a vessel up to height H and a hole is made at a depth h below the free surface of the liquid as shown in fig.

Now take the level of the hole as reference level (i.e., zero point of potential energy)

And by applying Bernoulli's  equation we get

$v=\sqrt{2 g h}$

This v is called the Velocity of Efflux.

This formula is only valid when (Area of Hole) <<< (Area of the vessel)

Thus Torricelli's Theorem relates the speed of fluid flowing out of an orifice.

(Note-  The speed that an object would acquire in falling

from rest through a distance h is equal to $v=\sqrt{2 g h}$

And this is same as that of Velocity of Efflux.)

• The velocity of efflux is independent of the nature of liquid ($\rho$), the quantity of liquid in the vessel and the area of the orifice/hole.

• The velocity of efflux depends on h (i.e depth below the free surface)

            I.e Means Greater is the distance of the hole from the free surface of the liquid, greater will be the velocity of efflux

• As the distance of hole from the ground is (H-h) and its initial vertical velocity at hole is zero.

So Time taken by liquid  to reach the Ground =t is given by

$
T=\sqrt{\frac{2(H-h)}{g}}
$

Where
H - the height of the vessel
And $\mathrm{h}=$ depth below the free surface
- Range (x)-

During time t liquid is moving horizontally with constant velocity v ,
And it will hit the base level at a horizontal distance x as shown in the above figure.
This horizontal distance x is also called a Horizontal range.
Using $x=v t$
We get Range as

$
x=R=2 \sqrt{h(H-h)}
$

This range will be maximum when

$
h=\frac{H}{2}
$

And Maximum value of the range is H
Means $x_{\max }=R_{\max }=H$

REACTION FORCE  DUE TO EJECTION OF LIQUID

A cylindrical vessel has an opening of cross-sectional area ' a ' near the bottom. A disc is held against the opening to prevent liquid of density $\rho$ from coming out. If the height of the liquid above the opening is h. Let us analyse the force on the disc in this situation.

The disc experiences hydrostatic pressure from the liquid inside the vessel. Pressure at the level of the disc is

$
P_1=P_{\mathrm{um}}+\rho g h
$

The air pressure on the outside of disc is $P_2=P_{\text {atm... }}$
The net outward force $=\left(p_2-p_1\right) a=\rho g h a$.
Now the disc is moved a short distance away in the horizontal direction. The liquid comes out, strikes the disc inelastically and drops vertically downward. The water in this case will impart impulsive (impact) force on the disc.

When the disc is moved away, the liquid moves out with speed $v=\sqrt{2 g h}$.
The (mass per second), i.e, the rate of mass coming out of the opening is given by $\frac{d m}{d t}=\rho a v=\rho a \sqrt{2 g h}$.
Momentum per second imparted by water rightward is given by

$
\frac{d m}{d t} v=(\rho a v) v=\rho a v^2=2 \rho g h a
$

The change in momentum per second after striking the disc is

$
\Delta \vec{P}=\vec{P}_f-\vec{P}_i \Rightarrow 0-\left(\rho a v^2\right)=-\rho a v^2=-2 \rho g h a
$
 

Taking the rightward direction as positive, the force on the liquid is towards the left and its reaction with the disc is towards the right. The force obtained is twice the hydrostatic force. The force due to atmospheric pressure is cancelled out on both sides.

Note:

If the velocity of a liquid (or gas) of density $\rho$ coming out through an opening of the area of cross-section a is v, then there will be a thrust force due to liquid coming out the opening. The direction of the thrust force will be just opposite to the velocity direction.

Venturimeter-

• It is a device is used for measuring the rate of flow of liquid through pipes.

• This device based on application Bernoulli's theorem.

• The image of the Venturi Meter device is given below

For the above figure

a₁ and a₂ are an area of cross-section of tube A and B respectively

And $v_1$ and $v_2$ are the Velocities of the flow of liquid through $A$ and $B$ respectively
And $P_1$ and $P_2$ are the Liquid pressure at A and B respectively
Then $P_1-P_2=\rho g h$
Where $\rho=$ density of flowing liquid
And $\mathrm{h}=$ difference of fluid level between the vertical tube D and E
Now applying Bernoulli's equation for the horizontal flow of liquid we get

$
P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \cdots
$

If $\mathrm{V}=$ rate of the flow of liquid through a pipe
Then from the continuity equation, we can write

$
V=a_1 v_1=a_2 v_2
$

From equation (1) and (2) and (3)
We get
$\mathrm{V}=$ rate of the flow of liquid through pipe
As

$
V=a_1 a_2 \sqrt{\frac{2 g h}{a_1^2-a_2^2}}
$
 

• Aspirator pumps-

This works on the principle of Bernoulli's Theorem.

Example of Aspirator pumps is paint-gun, scent-spray or insect-sprayer, etc.

In such devices, high-speed air is passed over a tube T with the help of motion of a piston P in a cylinder C and this helps to spray the liquid L as shown in the above figure.

 The high-speed air creates low pressure in the tube and because of the low-pressure liquid rise in it. And thus liquid gets sprayed with expelled air.

• Change of plane of motion of spinning ball-

This can be with the help of  the principle of Bernoulli's Theorem

Magnus effect- When a spinning ball is thrown it deviates from its usual path in flight. This effect is called the Magnus effect. 

This effect plays a very important role in sports like cricket, tennis, and football, etc.

• Working of an aeroplane-

This is also based on  Bernoulli's principle.

• During  a tornado or hurricane, blowing off roofs by wind storms can be explained 

with the help of  the principle of Bernoulli's Theorem