Akram Ayache AUB MECH-314: Introduction to Fluid Mechanics Final Cheat Sheet Material Derivative:

D�Π = ∂�Π + (V��� ∙ ∇���)Π

Conservation of Mass:

0 = ∂�� ρ d�

��

+� ρ (V����/�� ∙ d��)

��

0 = ∂�ρ + ∇��� ∙ �ρV����

0 = ∂�ρ + ρ∇��� ∙ V���+ V��� ∙ �∇���ρ�Using ∂�ρ = 0 and Ma < 0.3 we get incompressible continuity: ∇��� ∙ V��� = 0

Conservation of Linear Momentum:

∑F�� = ∂�� ρV��� d�

��

+� ρV��� (V����/�� ∙ d��)

��

ρg�� − ∇���P + ∇��� ∙ τ�� = ρD�V��� ∑F�� − a����/�dm = ∂�� ρV��� d�

��

+� ρV��� (V����/�� ∙ d��)

��

∑F�� − ma����/� = m u − m �u�

ρ�g�

g�

g�

�− ∂�P

∂�P

∂�P

� + ∂�τ�� + ∂�τ�� + ∂�τ��

∂�τ�� + ∂�τ�� + ∂�τ��

∂�τ�� + ∂�τ�� + ∂�τ��

� = ρ ∂�V� + V� ∂�V� + V� ∂�V� + V� ∂�V�

∂�V� + V� ∂�V� + V� ∂�V� + V� ∂�V�

∂�V� + V� ∂�V� + V� ∂�V� + V� ∂�V�

�

For Newtonian Fluids τ�� = µ∇���V��� and τ� = τ � = µ(∂�V + ∂ V�) thus momentum balance yields the Navier-Stokes Equation:

ρg�� − ∇���P + µ∇�V��� = ρD�V��� ρ�g�

g�

g�

� − ∂�P

∂�P

∂�P

�+ µ ∂��� V� + ∂��

� V� + ∂��� V�

∂��� V� + ∂��

� V� + ∂��� V�

∂��� V� + ∂��

� V� + ∂��� V�

� = ρ ∂�V� + V� ∂�V� + V� ∂�V� + V� ∂�V�

∂�V� + V� ∂�V� + V� ∂�V� + V� ∂�V�

∂�V� + V� ∂�V� + V� ∂�V� + V� ∂�V�

�

For Inviscid Flow we set τ�� ≡ 0���� thus momentum balance yields the Euler Equation:

ρg�� − ∇���P = ρD�V��� ρ�g�

g�

g�

�− ∂�P

∂�P

∂�P

� = ρ ∂�V� + V� ∂�V� + V� ∂�V� + V� ∂�V�

∂�V� + V� ∂�V� + V� ∂�V� + V� ∂�V�

∂�V� + V� ∂�V� + V� ∂�V� + V� ∂�V�

�

Bernoulli’s Equation:

This equation applies for frictionless flow along a streamline:

∂�V ds +1

ρdp + VdV + gdz = 0

If we assume the flow steady and incompressible and integrate over a streamline we get:

p� − p�

ρ+

V�� − V�

�

2+ g(z� − z�) = 0

Conservation of Angular Momentum:

∑M���� = ∂�� ρ�r� × V����d�

��

+� ρ�r� × V����(V����/�� ∙ d��)

��

Euler’s Turbine Formula: T = ρQ�r�V�� − r�V��� ; Q=VnA

Sprinkler Formula: ω =��

�−

��

�

Conservation of Energy:

D�E = 0 = D�Q �+ D�W �= ∂�� ρed�

��

+� ρ�h + 0.5V� + gz��V����/�� ∙ d���

��

D�W� � = −� ρ�V����/�� ∙ d���

��

D�W� � = ��τ���� ∙ V����/��� ∙ d������

��

Under steady state the energy equation for an incompressible flow becomes similar to the Bernoulli’s Equation:

� p

ρg+αV2

2g+ z�

1inup

= � p

ρg+αV2

2g+ z�

2out

down

+ hfriction − hpump + hturbine ; h =Power

ρgQ

This motivates defining the Hydraulic Grade line which shows the energy head minus the velocity head.

Reynolds Number: It is the ratio of inertial to viscous forces thus gives a measure of the turbulence:

Re =m VτA

=ρV�A

μVA ÷ L=

ρV�μA

=ρVL

μ=

VL

ν Inertial

Viscous=

ρVr ∂rVr

µ ∂zz2 Vr

=0.5ρVr���2

0.5L�2µRVr��� = ρVr���Lµ

L

R= Re

L

R

Manometery and Surface Tension:

pA + ρ1gh − ρ2gh = pB

yields��� pA − pB =ρ2 − ρ1

gh

dF = �dL ; dE = �dA

How to use the Right Equation: In essence we have Linear Momentum: +Frictionless or Inviscid � Euler’s Equation +Integration over streamline � Bernoulli’s Equation

+Incompressibility � Head Equation +Irrotationality � Uniform Bernoulli Constant

The Nabla Operator in Cylindrical Coordinates:

∇���=

���

1

r+ ∂r

1

r∂θ

∂z �� =

���

1

r∂r(r ∙)

1

r∂θ

∂z �� ∇2=

���

1

r∂r(r ∂r ∙)

1

r2 ∂θθ2

∂zz2 �

� DtΠ = ∂tΠ + (Vr ∂r +1

rVθ ∂θ + Vz ∂z)Π

A Fluid becomes rotational if ζ� = 2ω��� = ∇��� × V��� ≠ 0 : *It is Viscous *Non-Inertial forces act upon it

*It feels Entropy gradients *It feels Density gradients

Laminar Flow Equations and the Friction Factor: Without pumps and turbines and applying the energy head equation in a pipe we get:

hf = z1 − z2 +p1 − p2

ρg= for laminar only

32µLV

ρgd2 =128µLQ

πρgd4 and note that Q =π∆pd4

128µL

Moreover the momentum equation equates the right hand side to

hf =4τL

ρgd hf = f

L

d

V2

2g where f = fcn�Re, ε ÷ d, shape� ∗ i

Equating at first then assuming laminar flow gives:

f = hfd

L

2g

V2 =π2

8

ghfd5

LQ2 =8τ

ρV2 ∗ ii f lam =8(8µV ÷ d)

ρV2 =64

Re ∗ iii

Moody Chart Formulas: Colebrook Formula:

1√f= −2 log10"ε ÷ d

3.7+

2.51

Re√f#

Haaland Formula: 1√f

= −1.8 log10�"ε ÷ d

3.7#1.11

+6.9

Re�

Dimensionless Head Loss Parameter:

ζ =gd3hf

Lυ2 =f Re2

2

Re= −$8ζ log10�ε ÷ d

3.7+

1.775$ζ � =ζ

32

Turbulent↑ Laminar↑

Pipe Flow and Design Problems: +Head Loss Problem(L,d,V): Get Re, then find f by formulas or charts then get hf. +Flow Rate Problem(L,d,hf): Get the dimensionless head loss parameter, then deduce Re by the second formula, finally get V from Re.OR: Get f by (i) which gives a relation of the type V=(C÷f)0.5 . Then guess f, get V and hence Re, then get a better f. +Pipe Diameter Problem(L,V,hf): Using ii to relate f and d (1), get Re in terms of d (2), get roughness in terms of d (3). Then guess f, get d from (1) get Re from (2), and the surface roughness from (3) then compute a better f. +Pipe Length Problem(V,d,hf): Get hp by dividing power by ρgQ, compute Re and the shape factor then get f by Colebrook or Haaland formula. Finally set hp and hf equal and get L.

Piping Systems:

∆h = hf + hminor =V2

2g"fL

D+ ΣK#

Dimensions:

Length L Mass flow MT-1 Kin Viscosity L2T-1 Density ML -3 Area L2 Pressure ML-1T-2 Surface Ten. MT-2 Temperature Θ

Volume L3 Strain rate T-1 Force MLT -2 Sp. Heat L2T-2Θ-1 Velocity LT-1 Angle 1 Moment ML2T-2 Sp. Weight ML-2T-2

Acceleration LT-2 Ang. speed T-1 Power ML2T-3 Conductivity MLT-3 Θ-1 Volume flow L3T-1 Viscosity ML-1T-1 Energy ML2T-2 Expansion Θ-1

Solving Multiple Pipe Systems:

In Series the first equation would be setting the flows equal. Then find the total head loss for the system as a CV by getting ∆z+∆p÷ρg and set it equal to the sum of individual head losses in every portion of the system. Estimate the individual friction factors, get one velocity, get the all Re of the system and figure a better estimate of the friction factors by Haaland’s relation.

In parallel again find the head loss for the system as one CV. Then set this head loss equal to individual head losses in every portion. The problem now is just as the Flow Rate Problem above where we solve for the individual friction factors and velocities one at a time.

In a junction we set the HGL height at the meeting point to be hJ (initial guess would be the intermediate value of zi)and so ∆hi=zi-hJ , for each member this yields a relation between the individual friction factors and velocities. Then we calculate the dimensionless head loss parameter ζ and deduce Re. Afterwards we get the friction factor and hence the velocity and we do that for every element. Fill out the following table and sum the flow rates: if the sum is positive increase the guess of hJ and if the sum is negative use a lower hJ . Repeat the iteration till the sum of flow rates converges to zero.

Reservoir hJ (guessed) zi - hJ fi Vi Qi

Turbulent Modeling: u

u∗=

1

0.41ln "ρR

u∗# + 5

Where u is the centerline velocity then to get other parameters:

τw = ρ�u∗�2 V = 0.85u ∆p =2τw∆L

R for horizontal pipe (see∗ i)

Extra Formulae:

β =ghfQ

3

Lυ5 σ =ευ

Q Re2.5 = −2"128β

π3 #0.5

log10�πσRe

14.8+

2.51Re1.5�128β ÷ π3�0.5�