FE Exam Review Course: Fluid Mechanics

Fluid Properties

1. Density ρ, kg/m3 or slugs/ft3

2. γ = ρg = specific weight; units N/m3

3. v = 1/ρ = specific volume; m3/kg or ft3/slugs4. µ = dynamic viscosity; Pa.s = kg/(m.s) or lb.sec/ft2

5. ν = µ/ρ = kinematic viscosity; m2/s or ft2/s6. s = ρ/ρwater is specific gravity.

Equation of State for Ideal Gas

P = ρRT ; where P and T must be absolute values.

Equation of State for Incompressible Fluid

ρ = constant

Physical Constants

1. γwater = 62.4 lb/ft3 = 9800N/m3

2. ρwater = 1.94 slugs/ft3 = 1000kg/m3

3. γair = 0.076 lb/ft3 at standard atmospheric temperature and pressure4. swater = 1.05. smercury = 13.66. Runiversal is universal gas constant = 8.31434 N.m/mole−K = 1545 ft.lb/lb.mole.0R7. Rair = 287 N−m/kg.K = 53.35 ft.lb/lbm−0 R8. Standard atmospheric pressurePatm = 14.7 psi = 101.35 kPa = 33.9 ft of water = 29.9 in of Hg = 760 mm Hg.

Commonly Used Terms

1. Force N = kg.m/s2

2. Stress N/m2

3. Pressure is a normal stress; N/m2

4. Shear Stress τ ; N/m2

5. Strain rate du/dy; 1/s6. Dynamic viscosity µ = Shear stress / Strain rate : τ = µdu

dy

7. Reynolds numberRe = ρUL

µ= UL

ν

ρ is density, U is some reference velocity, µ is dynamic viscosity, and L is some reference lengthscale.

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Hydrostatics

1. Pressure due to hydrostatics changes with height. Pressure increases in the direction ofgravity.

2.∂p

∂z= −ρg = −γ (1)

3. For an incompressible fluid:

P2 − P1 = ρg(h2 − h1) = γ(h2 − h1) (2)

4. Pressure head: h = Pρg

= Pγ

5. Pressure force: Fp =∫PdA

6. Force on an immersed surface (curved or plane:)

(a) Horizontal component of the force equals force on a vertical projection of the area.

– Horizontal component = ρghcA, where hc is the height of the centroid of the pro-jected area from the water level, A is the projected area.

– The horizontal force acts at the centroid of the pressure prism.

(b) Vertical component equals weight of fluid (or displaced fluid) above the area.

– Vertical component = ρgVdisplaced

– The vertical component acts at the centroid of the displaced fluid region.

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Conservation Laws and Flow Problems

1. Conservation of Mass or Continuity

For a single inlet, single outlet problem with steady flow:

ρ1A1V1 = ρ2A2V2 (3)

where 1 and 2 are the two regions through which fluid crosses the control volume, A1 andA2 are the areas of the two sections, V1 and V2 are velocity components normal to theareas, and ρ1 and ρ2 are fluid densities.

2. Conservation of Linear Momentum

For a single inlet, single outlet problem with steady flow:

m(−→V 2 −

−→V 1) =

∑−→F (4)

where m = ρAV is the mass flux. The above is a vector equation.

The x− component:m(V2x − V1x) =

∑Fx (5)

The y − component:m(V2y − V1y) =

∑Fy (6)

3. Conservation of Energy (Mechanical Energy):For a single inlet, single outlet problem with steady, adiabatic flow, constantfluid density

P1

ρg+V 2

1

2g+ z1 =

P2

ρg+V 2

2

2g+ z2 +H +

∑hL (7)

where 1 is inlet, and 2 is outlet, hL indicates head loss due to friction, curves, bends etc.,H is the head due to mechanical work (positive for turbine work (net work output) andnegative for pump work (net work input).

Bernoulli equation: (conservation of energy with no heat input/output, no work input/output,steady, isothermal flow)

P

ρg+V 2

2g+ z = constant; along a streamline (8)

4. PowerP = ρgQH; Q is volume flow rate (9)

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Pipe Flow

1. Reynolds Number Re = ρV Dµ

= V Dν

where D is the hydraulic diameter of the pipe.

2. The flow inside a pipe is typically laminar when Re < 2100

3. The flow inside a pipe is typically transitional (laminar to turbulent 2100 < Re < 4000

4. The flow inside a pipe is turbulent when Re > 4000

5. Along an energy grade line:

EGL = z +P

ρg+V 2

2g= constant (10)

6. Along a hydraulic grade line:

HGL = z +P

ρg= constant (11)

7. Frictional head loss (Major head loss) in a pipe:

hL = fL

D

V 2

2g(12)

where L is the length, D is the diameter of the pipe, V is the average velocity, and f is theDarcy friction factor.

8. For laminar flows: f = 64Re

9. For turbulent and transitional flows: f = fcn(Re, εD, duct shape), whereε is the wall rough-

ness.

10. Plot of the Darcy friction factor f versus the Re for various εD

is called the Moody chart. IfRe and ε

Dare known, friction factor can be readily obtained by reading the values from the

chart. For a given moderate Reynolds number, the frictional head loss in a pipe is largerfor higher surface roughness. For very large Reynolds numbers (fully turbulent flows) thefriction factor is a weak function of the surface roughness.

11. Minor losses in pipes and pipe systems can occur due to sudden expansions/contractions,turning, bends, joints, sharp corners, at the inlet/exit etc.

hL,minor = KLV 2

2g; KL is a constant factor. (13)

12. Hydraulic diameter for non-circular pipes:

Dh =4A

P(14)

where A is the cross-sectional area and P is the wetted perimeter.

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Some More Advanced Topics

The following pages are NOT included on your FE exam, however, are given here just for com-pleteness.

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1 Mathematical Definitions

1.1 Gradient of a Scalar, 5η• Cartesian (i, j, k are the unit vectors in x, y and z directions)

5η =∂η

∂xi+

∂η

∂yj +

∂η

∂zk (15)

• Cylindrical (er, eθ, ez are the unit vectors in r, θ and z directions)

5η =∂η

∂rer +

1

r

∂η

∂θeθ +

∂η

∂zez (16)

Note that the gradient of a scalar (η) is a vector.

1.2 Divergence of a Vector, 5 · ~V• Cartesian (~V = ui+ vj + wk)

5 · ~V =∂u

∂x+∂v

∂y+∂w

∂z(17)

• Cylindrical (~V = vrer + vθeθ + vz ez)

5 · ~V =1

r

∂

∂r(rvr) +

1

r

∂vθ∂θ

+∂vz∂z

(18)

Note that divergence of a vector (~V ) is dot product of the gradient vector and ~V . This resultsin a scalar quantity (e.g. RHS of the above equations are scalars and not vectors).

1.3 Curl of a Vector, 5× ~V

• Cartesian (~V = ui+ vj + wk)

5× ~V =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

u v w

∣∣∣∣∣∣ (19)

• Cylindrical (~V = vrer + vθeθ + vz ez)

5× ~V =

∣∣∣∣∣∣1rer eθ

1rez

∂∂r

∂∂θ

∂∂z

vr rvθ vz

∣∣∣∣∣∣ (20)

1.4 Vorticity

ξ = 5× ~V (21)

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2 Lagrangian-Eulerian Formulations

2.1 Material or Total Rate of Change

Dβ

Dt=∂β

∂t+(~V · 5

)β (22)

When β is a scalar Φ

• CartesianDΦ

Dt=∂Φ

∂t+ u

∂Φ

∂x+ v

∂Φ

∂y+ w

∂Φ

∂z(23)

• CylindricalDΦ

Dt=∂Φ

∂t+ vr

∂Φ

∂r+vθr

∂Φ

∂θ+ vz

∂Φ

∂z(24)

When β is a vector ~A = ai+ bj + ck

• CartesianD ~A

Dt=∂ ~A

∂t+ u

∂ ~A

∂x+ v

∂ ~A

∂y+ w

∂ ~A

∂z(25)

• CylindricalD ~A

Dt=∂ ~A

∂t+ vr

∂ ~A

∂r+vθr

∂ ~A

∂θ+ vz

∂ ~A

∂z(26)

Note that if β is a vector, the above relation has three components (one for each component of~A. In other words, the above equations are in reality three equations (one each for a, b and c).

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5 Drag and Lift on Immersed Objects

1. Drag and lift forces in immersed objects cannot be readily obtained using analytical toolsexcept for very simple cases (flat plate boundary layer) under laminar conditions.

2. Drag and lift forces are typically obtained from experiments or numerical analysis anddocumented in terms of drag coefficient (CD) and lift coefficient (CL).

CD =DragForce

1/2ρV 2A; CL =

LiftForce

1/2ρV 2A(57)

where V is the average free-stream velocity of the fluid upstream, A is the relevant area ofthe object, and ρ is the density of the fluid. The area A usually is based on:

• Frontal area: the projected area of the body as seen from the stream suitable for thickimmersed objects such as cylinders, spheres, cars, trucks, missiles, projectiles, andtorpedoes.

• Planform area: the body area as seen from the above (or in the plan view), suitablefor wide, flat bodies such as airfoil wings, hydrofoils etc.

• Wetted area: customary for surface ships, and barges.

3. Drag force on an immersed object consists of contributions from two main forces: pressureforces on the surface of the object, and frictional forces due to viscosity of fluid. These arecalled pressure drag and friction drag, respectively. Then

CD = CD,press + CD,fric (58)

The drag and lift coefficients are typically functions of the Reynolds number, and thesurface roughness

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6 Dimensional Analysis and Similitude

1. Common non-dimensional numbers encountered in fluid mechanics:

Reynolds Number, Re =ρVL

µ=

Inertia Forces

Viscous Forces(59)

Mach Number, Ma =U

a=

Flow Speed

Sound Speed(60)

Froude Number, Fr =U2

gL=

Inertia Force

Gravitational Force(61)

Drag Coefficeint, CD =FD

12ρU2A

=Drag Force

Dynamic Force(62)

Lift Coefficeint, CL =FL

12ρU2A

=Lift Force

Dynamic Force(63)

Frictional fractor f =hf

V2

2gL/D

=Frictional Head Loss

Velocity Head(64)

Skin Friction Coefficient cf =τwall

ρV2

2

=Wall Shear Stress

Dynamic Pressure(65)

Pressure Coefficient Cp =p− p∞ρV2

2

=Static Pressure

Dynamic Pressure(66)

(67)

2. Similitude and Modeling: Flow conditions for a model test are completely similar if allrelevant dimensionless parameters have the same corresponding values for the model andthe prototype

3. Geometric Similarity: A model and prototype are geometrically similar if and only if allbody dimensions in all three co-ordinates have the same linear scale ratio. All angles arepreserved in geometric similarity. All flow directions are also preserved.

4. Kinematic Similarity: The motions of two systems are kinematically similar if homologousparticles lie at homologous points at homologous times.

5. Dynamic Similarity: Dynamic similarity exists when the model and the prototype have thesame length scale ratio, time scale ratio, and force scale (or mass scale) ratio. Dynamicsimilarity requires geometric similarity between the prototype and the model. In addition,the model and prototype non-dimensional force and pressure coefficients must be identical.

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