**Bernoulli’s Principle**

- For a streamline fluid flow, the sum of the pressure (P), the kinetic energy per unit volume (ρv
^{2}/2) and the potential energy per unit volume (ρgh) remain constant. - Mathematically:-
**P+ ρv**^{2}/2 + ρgh = constant- where P= pressure ,
- E./ Volume=1/2mv
^{2}/V = 1/2v^{2}(m/V) = 1/2ρv^{2} - E./Volume = mgh/V = (m/V)gh = ρgh

__Derive: Bernoulli’s equation__

**Assumptions**:

- Fluid flow through a pipe of varying width.
- Pipe is located at changing heights.
- Fluid is incompressible.
- Flow is laminar.
- No energy is lost due to friction:applicable only to non-viscous fluids.

- Mathematically: –
- Consider the fluid initially lying between B and D. In an infinitesimal timeinterval Δt, this fluid would have moved.
- Suppose v
_{1}= speed at B and v_{2}= speedat D, initial distance moved by fluid from to C=v_{1}Δt. - In the same interval Δtfluid distance moved by D to E = v
_{2}Δt. - P
_{1}= Pressureat A_{1,}P_{2}=Pressure at A_{2}. - Work done on the fluid atleft end (BC) W
_{1}= P_{1}A_{1}(v_{1}Δt). - Work done by the fluid at the other end (DE)W
_{2}= P_{2}A_{2}(v_{2}Δt)

- Suppose v
- Net work done on the fluid is W
_{1}– W_{2 }= (P_{1}A_{1}v_{1}Δt− P_{2}A_{2}v_{2}Δt) - By the Equation of continuity Av=constant.
- P
_{1}A_{1 }v_{1}Δt – P_{2}A_{2}v_{2}Δt where A_{1}v_{1}Δt =P_{1}ΔV and A_{2}v_{2}Δt = P_{2}ΔV.

- P
- Therefore Work done = (P
_{1}− P_{2}) ΔVequation (a)- Part of this work goes in changing Kinetic energy, ΔK = (½)m (v
_{2}^{2}– v_{1}^{2}) and part in gravitational potential energy,ΔU =mg (h_{2}− h_{1}).

- Part of this work goes in changing Kinetic energy, ΔK = (½)m (v
- The total change in energy ΔE= ΔK +ΔU = (½) m (v
_{2}^{2}– v_{1}^{2}) + mg (h_{2}− h_{1}). (i) - Density of the fluid ρ =m/V or m=ρV

- Therefore in small interval of time Δt, small change in mass Δm
- Δm=ρΔV (ii)

- Putting the value from equation (ii) to (i)

- ΔE = 1/2 ρΔV (v
_{2}^{2}– v_{1}^{2}) + ρgΔV (h_{2}− h_{1}) equation(b) - By using work-energy theorem: W = ΔE
- From (a) and (b)
- (P
_{1}-P_{2}) ΔV =(1/2) ρΔV (v_{2}^{2}– v_{1}^{2}) + ρgΔV (h_{2}− h_{1}) - P
_{1}-P_{2}= 1/2ρv_{2}^{2}– 1/2ρv_{1}^{2}+ρgh_{2}-ρgh_{1}(By cancelling ΔV from both the sides).

- After rearranging we get,
**P**_{1}+ (1/2)**ρ****v**_{1}^{2}+**ρg****h**_{1}= (1/2)**ρ****v**_{2}^{2}+**ρg****h**_{2} **P+(1/2)****ρv**^{2}+ρg**h = constant.****This is the Bernoulli’s equation**.

__Bernoulli’s equation: Special Cases__

- When a fluid is at rest. This means v
_{1}=v_{2}=0.

- From Bernoulli’s equation P
_{1}+ (1/2) ρ v_{1}^{2}+ ρg h_{1}= (1/2) ρ v_{2}^{2}+ ρg h_{2} - By puttingv
_{1}=v_{2}=0 in the above equation changes to- P
_{1}-P_{2}= ρg(h_{2}-h_{1}). This equation is same as when the fluids are at rest.

- P

- When the pipe is horizontal.h
_{1}=h_{2}.This means there is no Potential energy by the virtue of height.

- Therefore from Bernoulli’s equation(P
_{1}+ (1/2) ρ v_{1}^{2}+ ρg h_{1}= (1/2) ρ v_{2}^{2}+ ρg h_{2}) - By simplifying,P+(1/2) ρ v
^{2}= constant.

__Problem:-__

Water flows through a horizontal pipeline of varying cross-section.If the pressure of waterequals 6cm of mercury at a point where the velocity of flow is 30cm/s, what is the pressure at the another point where the velocity of flow is 50m/s?

** Answer**:-

At R_{1}:- v_{1} = 30cm/s =0.3m/s

P_{1}=ρg h=6×10^{-2}x13600x9.8=7997N/m^{2}

At R_{2}:- v_{2}=50cm/s=0.5m/s

From Bernoulli’s equation: – P+ (1/2) ρ v^{2}+ρg h=constant

P_{1}+ (1/2) ρ v_{1}^{2} = P_{2}(1/2) ρ v_{2}^{2}

7997+1/2x 1000x (0.3)^{2} = P_{2}+1/2x 1000x (0.5)^{2}

P_{2}=7917N/m^{2}

=ρg h_{2} = h_{2}x13600x9.8

h_{2} = 5.9cmHg.

**Torricelli’s law**

- Torricelli law states that the speed of flow of fluid from an orifice is equal to the speed that it would attain if falling freely for a distance equal to the height of the free surface of the liquid above the orifice.
- Consider any vessel which has an orifice (slit)filled with some fluid.
- The fluid will start flowing through the slit and according to Torricelli law the speed with which the fluid will flow is equal to the speed with which a freely falling bodyattains such that the height from which the body falls is equal to the height of the slit from the free surface of the fluid.
- Let the distance between the free surface and the slit = h
- Velocity with which the fluid flows is equal to the velocity with which a freely falling body attains if it is falling from a height h.

__Derivation of the Law:-__

- Let A
_{1}= area of the slit (it is very small), v_{1}= Velocity with which fluid is flowing out.- A
_{2}=Area of the free surface of the fluid,v_{2}=velocity of the fluid at the free surface.

- A
- From Equation of Continuity, Av=constant.Therefore A
_{1}v_{1}= A_{2}v_{2}.- From the figure, A
_{2}>>>A_{1}, This implies v_{2}<<v_{1}(This meansfluid is at rest on the free surface), Therefore v_{2}~ 0.

- From the figure, A
- Using Bernoulli’s equation,
- P+ (1/2) ρ v
^{2}+ρgh = constant.

- P+ (1/2) ρ v
- Applying Bernoulli’s equation at the slit:
- P
_{a}+(1/2) ρ v_{1}^{2}+ρgy_{1}(Equation 1) where P_{a}=atmospheric pressure,y_{1}=height of the slit from the base.

- P
- Applying Bernoulli’s equation at the surface:
- P+ρgy
_{2}(Equation 2) where as v_{2}=0 therefore (1/2) ρ v_{1}^{2}=0, y_{2}=height of the free surface from the base. - By equating(1) and (2),
- P
_{a}+ (1/2) ρ v_{1}^{2}+ ρgy_{1}= P+ρgy_{2} - (1/2) ρ v
_{1}^{2}= (P-P_{a}) + ρg(y_{2}-y_{1}) - =(P-P
_{a}) ρgh (where h=(y_{2}-y_{1})) - v
_{1}^{2}=2/ρ [(P-P_{a}) + ρgh] - Therefore
**v1=√2/ρ [(P-P**This is the velocity by which the fluid will come out of the small slit._{a}) + ρgh]. **v**_{1}is known as Speed of Efflux. This means the speed of the fluid outflow.

** Case1**:- The vessel is not closed it is open to atmosphere that means P=P

_{a}.

- Therefore
**v**.This is the speed of a freely falling body._{1}=√2gh - This is accordance to Torricelli’s law which states that the speed by which the fluid is flowing out of a small slit of a container is same as the velocity of a freely falling body.
Tank is not open to atmosphere but P>>P__Case2:-___{a}.- Therefore 2gh is ignored as it is very very large, hence
**v**_{1}= √2P/**ρ.** - The velocity with which the fluid will come out of the container is determined by the Pressure at the free surface of the fluid alone.

- Therefore 2gh is ignored as it is very very large, hence

** Problem: –**Calculate the velocity of emergence of a liquid from a hole in the side of a wide cell 15cm below the liquid surface?

__Answer:-__

By using Torricelli’slaw v1=√2gh

=√2×9.8x15x10^{-2} m/s

=1.7m/s

** Problem:- **Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984kgm

^{–3}. Determine the height of the wine column for normal atmospheric pressure.

** Answer**:-

Density of mercury, ρ_{1} = 13.6 × 10^{3}kg/m^{3}

Height of the mercury column, h_{1} = 0.76 m

Density of French wine, ρ_{2} = 984 kg/m^{3}

Height of the French wine column = h_{2}

Acceleration due to gravity, g = 9.8 m/s^{2}

The pressure in both the columns is equal, i.e.

Pressure in the mercury column= Pressure in the French wine column

ρ_{1}h_{1}g = ρ_{2}h_{2}g

h_{2}= ρ_{1}h_{1}/ ρ_{2}

= (13.6×10^{3}x0.76)/984

= 10.5 m

Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.