Class 11 Physics
Mechanical Properties of Fluids
Bernoulli’s Principle
- For a streamline fluid flow, the sum of the pressure (P), the kinetic energy per unit volume (ρv2/2) and the potential energy per unit volume (ρgh) remain constant.
- Mathematically:- P+ ρv2/2 + ρgh = constant
- where P= pressure ,
- E./ Volume=1/2mv2/V = 1/2v2(m/V) = 1/2ρv2
- 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 v1= speed at B and v2= speedat D, initial distance moved by fluid from to C=v1Δt.
- In the same interval Δtfluid distance moved by D to E = v2Δt.
- P1= Pressureat A1, P2=Pressure at A2.
- Work done on the fluid atleft end (BC) W1 = P1A1(v1Δt).
- Work done by the fluid at the other end (DE)W2 = P2A2(v2Δt)
- Net work done on the fluid is W1 – W2 = (P1A1v1Δt− P2A2v2Δt)
- By the Equation of continuity Av=constant.
- P1A1 v1Δt - P2A2v2Δt where A1v1Δt =P1ΔV and A2v2Δt = P2ΔV.
- Therefore Work done = (P1− P2) ΔVequation (a)
- Part of this work goes in changing Kinetic energy, ΔK = (½)m (v22 – v12) and part in gravitational potential energy,ΔU =mg (h2 − h1).
- The total change in energy ΔE= ΔK +ΔU = (½) m (v22 – v12) + mg (h2 − h1). (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 (v22 – v12) + ρgΔV (h2 − h1) equation(b)
- By using work-energy theorem: W = ΔE
- From (a) and (b)
- (P1-P2) ΔV =(1/2) ρΔV (v22 – v12) + ρgΔV (h2 − h1)
- P1-P2 = 1/2ρv22 - 1/2ρv12+ρgh2 -ρgh1(By cancelling ΔV from both the sides).
- After rearranging we get,P1 + (1/2) ρ v12 + ρg h1 = (1/2) ρ v22 + ρg h2
- P+(1/2) ρv2+ρg h = constant.
- This is the Bernoulli’s equation.
The flow of an ideal fluid in a pipe ofvarying cross section. The fluid in asection of length v1Δt moves to the sectionof length v2Δt in time Δt.
Bernoulli’s equation: Special Cases
- When a fluid is at rest. This means v1=v2=0.
- From Bernoulli’s equation P1 + (1/2) ρ v12 + ρg h1 = (1/2) ρ v22 + ρg h2
- By puttingv1=v2=0 in the above equation changes to
- P1-P2= ρg(h2-h1). This equation is same as when the fluids are at rest.
- When the pipe is horizontal.h1=h2.This means there is no Potential energy by the virtue of height.
- Therefore from Bernoulli’s equation(P1 + (1/2) ρ v12 + ρg h1 = (1/2) ρ v22 + ρg h2)
- By simplifying,P+(1/2) ρ v2 = 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 R1:- v1 = 30cm/s =0.3m/s
P1=ρg h=6x10-2x13600x9.8=7997N/m2
At R2:- v2=50cm/s=0.5m/s
From Bernoulli’s equation: - P+ (1/2) ρ v2+ρg h=constant
P1+ (1/2) ρ v12 = P2(1/2) ρ v22
7997+1/2x 1000x (0.3)2 = P2+1/2x 1000x (0.5)2
P2=7917N/m2
=ρg h2 = h2x13600x9.8
h2 = 5.9cmHg.
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