Cheatsheet 2122 - Summary Fluid Mechanics PDF

Preview

Summary

CN2122 FLUID MECHCANICS Chapter 2: Dimensional Analysis Quantity Measure Formula (M L θ) (SI) Mass M Length L Time θ Force M L θ-2 Velocity L θ-1 Acceleration L θ-2 Pressure M L-1 θ-2 Pressure gradient M L-2 θ-2 Viscosity M L-1 θ-1 Density M L-3 Work M L2 θ-2 Power M L2 θ-3 Momentum M L θ-1 Energy M...

Description

CN2122 FLUID MECHCANICS Chapter 2: Dimensional Analysis Quantity Measure Formula (M L θ) (SI) Mass M Length L Time θ Force M L θ-2 Velocity L θ-1 Acceleration L θ-2 M L-1 θ-2 Pressure Pressure gradient M L-2 θ-2 M L-1 θ-1 Viscosity Density M L-3 Work M L2 θ-2 M L2 θ-3 Power Momentum M L θ-1 Energy M L2 θ-2 M L-1 θ-2 Stress Surface tension, σ M θ-2

Chapter 6 – Conservation of Momentum

Pressure Gauge

List all parameters involved Choose 3 primary dimensions N = P (linear momentum), η = Rank = 3 (x, y, z), Isolate target groups (E, F), For CS, inflow -ve, Outflow +ve Set-up Eqn, F = SR to CV is +ve π1= xaybzcE • Reducing elbow  force of fluid, Rx = - Kx (M0L0θ0 = _____), Solve a,b,c. V2, P2

Find all πi , Then equate π1 = f (π2, π3,…) Rx

V1, P1

Euler number

Eu

gc (- ΔP) / (ρ v2)

Pressure force/Inertial force

Froude number Mach number Reynolds number Weber number

Fr Ma Re We

v / (g L)1/2 v/c ρDv/μ ρ v2 / (σ / L)

Inertial force/gravitational F

Ry

Loss of Head at Abrupt Enlargements

 Upstream Head loss at Abrupt Contractions K

Chapter 9: Analysis of Fluid in Laminar Flow Flow entry-1dim, after entry length (Le)  2dim Re2100, Le/D ~ 4.4 ReD^1.6 (Turbulent)

Moment of Momentum Inertial force/Viscous force Inertial force/Surface tension F

Chapter 3: Fluid Static -- Basic Eqn of fluid static, max rate of ∆ of P in g direction. Constant P is ┴ to gravitational line. @ Constant ρ & g at -ve z direction,  + @ fluid surface  Barometer 

Squared Edge Well Rounded Radius >0.14d Etched out < d/2

K = 0.5 K=0

K=1 To solve, use Newton Raphson method. Knowing f(v), find f’(v) then iterate. vf(v)f’(v)f(v)/f’(v) v-f(v)/f’(v)v How to solve qns (start with energy eqn) (CHAPTER 11): • Pressure drop problem, find (-ΔP)/ ρ 1) Simplify:Adiabatic Q=0 | No pump/turbine Ws=0 | no KE change Δv=0 Chapter 8: Shear Stress in Laminar Flow | Isothermal Δu=0 | incompressible flow ρ=cont | Steady flow, d/dtcv=0 Shear stress of 2) Equate viscous term with major/minor loss unidirectional flow  3) Find Re eqn 4) Solve iteratively for friction factor etc. K = blotzmann cons Viscosity of gases  Blasius: φ = 0.0792Re --0.25 for initial guessColebrook or chen to find φ Viscosity of Liquid 

• Reducing elbow  External force

 Downstream

x

• Velocity/flow rate 1) Simplify Energy 2) Equate major loss into energy eqn

3) Get qn for φf Q2= __ and using Re = 4ρQ / (π μ D) for iteration φ  Q  Re  φ (from M.D.) Q  Re  φ (from M.D.)  Q OR

Rem = Re rt φf

Choose z momentum as w&r in xy  w x r  ez Vapour pressure

Manometer  (General eqn)

A

Hagen-Poiseulle Eqn  Laminar flow

y A

 Velocity Profile

x y

x B

B

Torque applied to shaft = Avg velocity used, include momentum correction factor, β •Laminar, β=4/3 •Turbulent, β=1.01~1.04

Turbulent Flow in pipe

•Pipe sizing 3) Get Eqn for D in φ and Re in D. 4) Using Re = 4ρQ / (π μ D) for iteration btw Re and D5 (in terms of φ) D  Re –Moody dia φf  D (repeat)

Chapter 7: Conservation of Energy N=E, η = e = E/m

Uniform Rectilinear Acceleration  Variable Density Fluid 

Viscous Work

Chapter 11: Flow in close conduits Adiabatic flow

Q = ṁ/ρ = A v

Newton Raphson - f(D) = __D5 + __D – φ - find f’(D) = - Dnew = Dold – f(D)/f’(D)

Constant Temp Adiabatic conditions

Isothermal flow Incompressible flow, ρ = constant

Chapter 4 – Fluid in Motion

Fanning Friction factor (φ): Eu = 2 φf L/D Eu =



Pump: P2, v2, A2

 Colebrook Eqn N = Extensive property. η = N/m = Intensive property

P1, v1, a1

z2 – z1

 Chen Eqn

Chapter 5 – Conservation of Mass  Moody Diagram N = M, η = 1 Fixed CV  Inflow/influx = -ve, Outflow/efflux = +ve

Math: sin2α = (1- cos 2α)/2 | cos2α = (1+cos 2α) /2 sin2α = 2sin α cos α ʃsin x dx = - cos x + C | ʃcos x dx = sin x + C

Incompressible flow, drop this

Bernoulli Eqn Hg = 13530 kg/m3 H20 = 1000kg/m3

General: f avg =

1 𝑏−𝑎

𝑏 ∫𝑎 𝑓(𝑥)𝑑𝑥

1N = 1 kg m s-2 1dyn = 10-5 N = 1g cm s-1 1Pa = 1 N m-2 = 1kg m-1 s-2

Q (flow rate) = vA G (mass flow rate)

Non-circular conduits, replace D in e/D with Deq = 4 rH (hydraulic radius) rH = A/Lp. A = Cross sectional area, Lp= wetted perimeter • Pressure drop due to fittings

Condition: No Heat flow, No shaft work, No energy loss, Steady Flow, Isothermal flow

 Major Loss

 Minor Loss (fittings)

Boundary Conditions btw 2 surfaces: (for part 2) Solid-fluid: Vf = Vs Liq-gas: Momentum flux = v grad in liquid≈0 [ 𝜏liq = 𝜏air = 0, ρair=0] Liq-liq: Momentum flux⊥interface & v continuous across interface ie equal

Single Particle system (Eldin) Part 2, Chapt 1 Equations of Fluid Flow Stokes’ Law, Drag force, FD=3πxμU, U= relative velocity Continuity Eqn (mass conservation) Single Particle Reynolds number, Rep = xUρf / μ R’ = force per unit  Drag Coefficient, CD = R’/(1/2 ρf U2), R’ = FD/(πx2/4) projected area  x = particle diameter, ρf = fluid density, μ = fluid viscosity CD = 24/Rep (stokes law) Navier-Stokes Eqn

Inertial force

 Know Rep to get CD

Stokes’

General

Newton’s

UT single particle terminal velocity For intermediate Rep, use CDRe2 ratio, sub Rep = 10,100, 1000, find CD then plot onto drag coeff graph. Plot cuts @ sphericity  Rep

Use this to find x

= xvUTρf / μ Multiple Particle Systems Voidage, ε = vol frac occupied by fluid = Vfluid/Vtotal Effective viscosity, με = μ/f(ε) Mbed= AH(1-ξ)ρparticle Ave suspension density, ρave = ε ρf + (1 – ε) ρp For stokes law Urel = relative velocity of particle to fluid UrelT = UT ε f(ε) = hindered settling velocity Actual fluid velocity Uf = Ufs/ε  (Ufs =superficial fluid vel = Qf/A) Actual particle velocity Up = Ups/(1- ε) (Ups=superficial parti vel = Qp/A) Batch Settling When settling, no net flow Qp+Qf=0  Uf = -Up (1-ε)/ε Stokes’ Newton’s Eqn General Use this Rep

X = particle diam D= vessel diam Ar =

Fluid Flow Through Packed Bed D = tube diam μ = fluid viscosity Laminar flow: Hagen-Poiseuille eqn Using equivalent diam, length and actual fluid vel through packed bed

or

Outside Euler’s eqn Blasius Solution for Laminar Boundary Layer uniform flow across a flat plate, pressure is constant BC, @y=0, Vx=Vy=0 @y large enough, Vx=V∞

gravity Pressure Viscous effect

To Solve problems, 1) Choose Coordinate system 2) Identify Driving Force & Flow features 1. Pressure gradient, body force Continuity eqn 2. Moving surface a. Steady State, v ≠ f(time) b. Incompressible fluid, ρ=const Can Use NS c. Newtonian fluid, μ= const d. Infinitely large plate, ignore end effects, no depth changes e. Fully develop region, V? = f(?) only. 3) Simplify Navier Stoke Eqn -- For pressure term, look at a. Direction of flow, same direction, keep it b. If direction is supported or not 4) Identify Boundary Conditions V? = ? @ ? . e.g Vx = 0 @ y = 0 5) Solve the Eqn (from N.S) by using boundary eqn Viscous Stress Tensor

𝜏 𝑥𝑦 𝑥𝑦, x = direction of normal vector to surface, y = direction of force - force per unit area (Nm-2) - linear r/s with fluid velocity (see table) Torque = length x force = length x (stress x area) Liquid air interface -consider force btw liq air interface, 𝜏yz liq = 𝜏yz air, μ air~0 Flow rate Q = ʃ ʃ V? dA  Q per width = Q/dx = ʃ Vz dy -circular, Q = ʃ ʃ Vz r dr dθ Po

Cauchy’s Momentum Eqn

Governing Eqn Inside Boundary 

>

PL

Reynold number: ratio of inertial to viscous force Small Re1  Inviscid flow, high flow rate Creeping flow, small Re Stokes Eqn Inviscid flow, Large Re Euler’s Eqn,

 converting n & f,f’,f’’ -select % of boundary layer thickness 𝛿, n=5, f’=0.99  Vx = 0.99V∞. n=5= 𝛿 /g(x)  𝛿 = 5 rt(v x / v∞) - calculate shear stress (object to fluid) from table

,

, tube diam=

Wetted peri = SBA SB = part surface area/volume of bed

Carman-Kozeny eqn:

CDx – local drag coefficient. CDL = overall mean drag coefficent Von Karman Momentum object other than Flat-plate, P not constant @object surface (Vx=Vy=0), NS becomes  dP/dx = 0, use Blasius’ approach dP/dx > 0, adverse pressure gradient dP/dx < 0, favourable pressure gradient Boundary Layer separation (dP/dx > 0)

General Eqn 

Conditions:1) steady, inviscid, incompressible. 2) valid along a streamline Von Karman Momentum Integral Eqn

Turbulent

Packed bed Re  Fluidization

If we dk εmf use εmf = 0.4 Wen and Yu, valid for 0.01 0 | Radial: β2 = 90 Actual H smaller due energy loss H = H0 – aQ2 Efficiency of fluid machine Pump:

KE1 + PE1 + P1 = KE2 + PE2 + P2  P1=P2 + ρ/2(V22 – V12)

Potential Theory ϕ -- velocity potential

BC4 @y=0, d2 Vx/dy2 = 0

Sludging occur when 2 of these conditions are satisfied

Velocity Diagram Ideal: 1) Negligible torque due to surface force 2) Inlet&outlet flows (rel to blade) tangent to blade to reduce energy loss Vz=0 R1=entrance radius 3) Uniform flow at inlet and outlet R2=exit radius Hence, Vt1=0, Vt2=U2-Vn2 cot β2 W=width ω=rpm Q = m/ρ H = U2Vt2/g m= Vrb = V – U Liner H-Q relationship U1 = r1 ω

@separation point, dVx/dy=0, 𝜏yx = 0 No more shear stress

-no fluid rotation, voracity of velocity, Von Karman Momentum Integral Eqn (Flate plate) - Using ϕ , find Vx,Vy,Vz  V = rt(Vx2 +Vy2 + Vz2) BC1 @y=0, Vx=0 Stream function Ψ – at any Re, incompressible,2D flow (not for turbulent) BC2 @y=𝛿, Vx=V∞ BC3 @y=𝛿, dVx/dy=0

F = fraction of powder < 45 μm

Theoretical Head added to flow:

Conditions: Incompressible, Newtonian fluid, inviscid flow, object of any shape

Turbulent 

Minimum bubbling velocity, Umb

Pump and Turbines Pumps, do work on fluid +ve torque. Turbine, extract work -ve torque

Rate of work done on(turbine)/by(pump) rotor (Mechanical Power) = ω = angular velocity

(1st order ODE)

laminar ,K3=5 

Laminar

L= eddy size

• Angular Momentum conservation -- Assume: steady state frictionless axisymmetric flow, no pressure effect, uniform flow at inflow and outflow +ve = pump, -ve = turbine

Bernoulli’s Eqn for,



Reynolds Stress = stress from velocity fluctuations (turbulent) Overall stress (exclude P) = Time ave stress + Re stress. - given Vx, Vy, sub into Vi eqn, (period = 2π/ω = t1) Re Stress - Find V’i = Vx – Vi,  Find Re stress Eddy Viscosity

Stream line, dΨ = 0 Tangent of stream line  dy/dx = Vy/Vx Stream function Ψ and Flow rate Q Flow rate for any cross-section btw 2 streamlines=difference in stream fn

Sign not impt, Looking at Magnitude

Momentum boundary Layer - thin layer where viscous effect is impt - governed by NS eqn Drag Coefficient F= drag force (N=kg m s-2) A= surface area of object projection ┴ to flow (m2) Ρ= density (kg m-3) V∞= external velocity (m s-1)

Pump efficiency 

Turbine:

Don’t use 1 and 4 tgt

- Using given Vx, add in boundary condi, Use BC4 if have 4 variable Keep highest order constant non-zero - Simplify Vx = ____, find - Then sub into VKeqn, - Solve for 𝛿 – then get - Sub into CDx and find CDL Turbine Efficiency  - find drag force F using this formula  Turbulent Flow F = 2 x CDL x ρv2/2 x A Fluid Velocity = mean value + fluctuation value

i = x,y or z

Determining Operating point (tanks, so v=o) Where system curve (Hp = ~~) intersects with pump curve.

Set t1 =Period = 2π/ω

Average velocity fluctuation = 0  Prevent cavitation, NPSHA>NPSHR Mean kinetic energy

Intensity of turbulence

(A:Available, R:Required)...