Physics 106 Lecture 13 Fluid Mechanics SJ 7th Ed.: Chap 14.1 to 14.5

• What is a fluid? • Pressure • Pressure varies with depth • Pascal’s principle • Methods for measuring pressure • Buoyant forces • Archimedes principle • Fluid dyypnamics assumptions • An ideal fluid • Continuity Equation • Bernoulli’s Equation

Goals

Ideal fl uid

Equation of continuity: Mass conservation

Bernoulli’s principle: Work-Energy theorem

1 Ideal Fluids

• non-viscous – Negligible internal friction (not sticky) • laminar (steady, streamline flow) – all particles passing through the same point have the same velocity • incompressible – the density remains constant • irrotational – no local rotation

Flow rate Mass flow rate = amount of mass crossing area A per unit time = a “current” sometimes called a “mass flux”

cross-section area A

velocity v volume of fluid in cylinder = dV = Adx dM ρdV I === mass flow rate mass dt dt dx == ρρ A Av dt length dx I ≡ mass flow rate = ρAv during time dt mass Volume flow rate = volume of fluid crossing area A per unit time

Ivol ≡ volume flow rate = Av Mass flow rate per area

Jmass ≡ mass flow/unit area = ρv

2 Equation of Continuity: conservation of mass • An ideal fluid is moving through a pipe of nonuniform diameter Æ The rate of mass entering one end equals the mass leaving at the other end

mass flow rate at "1" = ρ111A vAv==mass flow rate at "2" ρ 2 22

• Ideal fluid is incompressible so: ρ2

ρ1 = ρ2 = a cons tan t

∴ A1v1 = A2v2

Æ Equation of continuity ρ1

Where the pipe narrows, the fluid moves faster, and vice versa

Example:

Water flows through a fire hose of diameter 6.0 cm at a rate of 0.012 m^3/s. The fire hose ends in a nozzle of inner diameter 4.0 cm. What is the speed with which the water exits the nozzle?

3 Bernoulli’s Equation – Work-Energy theorem for ideal fluid

p2 pressure pressure p1 y2

y1

Question: If height and area change, how do the pressure and velocity of fluid change?

Derivation of Bernoulli’s Equation from Work-Energy theorem The fluid is incompressible. The continuity equation (no leaks) requires:

ΔV = A1Δx1 = A2Δx2 m1 = m2 ≡ m = ρΔV The net work done on fluid volume by external pressures:

ΔW = F1dx1 − F2dx2 = P1A1dx1 − P2A2dx2 = [P1 − P2 ] ΔV The change in the mechanical energy: ΔE = E − E = 1 mv2 − 1 mv2 + mgy − mgy 2 1 2 2 2 1 2 1 Equate to the net work and divide by the volume ΔV: 1 2 1 2 P − P = ρv − ρv + ρg(y − y ) 1 2 2 2 2 1 2 1 Rearrange: 1 2 1 2 P + ρv + ρgy = P + ρv + ρgy 1 2 1 1 2 2 2 2

Pressure P 1 Equal volumes flow in equal times Pressure P2 speed v 1 diiidue to continuity equation speed v2 height y 1 height y2 Volume Volume ΔV = A1Δx1 ΔV = A2Δx2

time t time t + Δt

4 Bernoulli’s Equation from analogy with Work-Energy theorem for solids

For solids,

net net ⎛⎞⎛⎞1122 Wnon−− conservative=Δ=+−+ F non conservative x⎜⎟⎜⎟ mv f mgh f mv i mgh i ⎝⎠⎝⎠22

For ideal fluids

⎛⎞⎛⎞1122 pp1222−=⎜⎟⎜⎟ρρ v + gh − ρρ v 11 + gh ⎝⎠⎝⎠22

that is, 11 p++ρρ v22 gh= p ++ ρρ v gh 11122222 1 or, pvgh++ρρ2 is conserved. 2

Bernoulli’s Equation – Work-energy theorem

1 2 1 2 P + ρv + ρgy = P + ρv + ρgy 1 2 1 1 2 2 2 2 1 2 That is, P + ρv + ρgy = Constant 2

For a fluid at rest Bernoulli’s Equation reduces to the earlier result for variation of pressure with depth:

P1 = P2 + ρg[y2 - y1]

For flow at constant altitude (y1 = y2), the pressure is lower where the fluid flows faster: 1 2 1 2 P + ρv = P + ρv 1 2 1 2 2 2 (Gases are compressible and are non-ideal fluid, but the above pressure / speed relation holds true.)

5 Bernoulli’s Equation : Principle behind airplane wing

Bernoulli’s Equation & venturi meter

6 Bernoulli’s Equation : Principle behind curve ball

Water flows in a tube that changes cross-sectional area.

A1= 2.0 m^2

A2=1.0 m^2

V1=3.0m/s Same height V2=?

P1 P2

iClicker Q

What is v2? a) 1.5 m/s b) 3.0 m/s c) 6.0 m/s

7 Water flows in a tube that changes cross-sectional area.

A1= 2.0 m^2

A2=1.0 m^2

V1=3.0m/s Same height V2=? P1 P2

Example: Find P1 - P2. Water density is 1000 kg/m^3

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