Summary Fluid Dynamics

Fluid Dynamics Summary 2020


1 Mass Balance (1a)
1.1 General Form of a Balance
ϕ = flow ϕ” = flux
dV X
= ϕv,in X|in − ϕv,out X|out + rx V
dt
X = volumetric concentration of quantity X

• x=ρ in [kg/m3 ]
• x = ρe in [J/m3 ]
• x = ρCp T in [J/m3 ]
• x = ρV in [N s/m3 ]

V = volume in [m3 ]
ϕv = volumetric flowrate in [m3 /s]
rx = volumetric production in [1/m3 s]


1.2 Mass and Component Balance
Total Mass Balance
dρ dM
V = = ϕm,in − ϕm,out
dt dt

• Assume that volume does not depend on time, V = constant.
• Mass is never produced, rx = 0

Mass balance for component A
dρA
V dt = ϕV ρA |in − ϕV ρA |out + rA V with ρ in kg/m3

Molar balance for component A
dcA
V dt = ϕV cA |in − ϕV cA |out + rA V with cA in mol/m3

Note: rA V does not have same dimensions for two formulas.

1.3 Recipe for Constructing a Balance
1. Make a drawing of the situation/system
2. Select parameter of interest (X)
3. Define the control volume (CV)
dX
4. General formulation: dt = in − out + production


1.4 Steady State Conditions
• Accumulation is zero, thus ϕm,in = ϕm,out
• Continuity Equation: A1 V1 = A2 V2




1

,1.5 Continuity Equation




dV X
dt = ϕv X|in − ϕv X|out + rx V with X = ρ (mass balance)

• Steady state, thus dV ρ
dt = 0, rx V = 0
– ϕv ρin = ϕv ρout
• ϕv = A ∗ Vlinear and Vlinear = ∆L2
∆t =V

– ρ1 A1 ∆L ∆L2
∆t = ρ2 A2 ∆t
1


– ρ1 A1 V1 = ρ2 A2 V2
• For in-compressible fluids: A1 V1 = A2 V2 and ρ = constant


2 Dimensional Analysis (1b)
Basic units: m, kg, s, K, mol
• Force: F = m ∗ a in [N ] = [ kgm
s2 ] ⇒
ML
T2

• Pressure: P = F
A = m∗a
A
kg
in [P a] = [ ms2] =
N
m2 ⇒ M
LT 2
2
M L2
• Power: P = F ∗ V = m ∗ a ∗ v in [W ] = [ kgm
s3 ] =
Nm
s ⇒ T3

Engineering Numbers

• Re = ρvD
η = inertial forces
viscous forces

• Nu = hD
λ = total heat transfer
heat transfer by conduction
√
• Ha = IDKr
k = mass transfer with chemical reaction
mass transfer without chemical reaction

• Le = λ
CpID = rate of heat transfer
rate of mass transfer


• Re > 400 → turbulent flow
• Re < 400 → laminar flow
• In between → transitional flow

π-theorem (Buckingham)
# dimensionless groups = # parameters - # basic units




2

, 3 Energy Balance (2)
dV X
= ϕv X|in − ϕv X|out + rx V
dt
with X = ρe M = ρV ϕm = ρϕv

this gives:
dρV e dM e
= = ϕm e|in − ϕm e|out + rx V
dt dt
with

p 1 2
e=u+ + v + gh
ρ 2
• u = thermal energy
• p
ρ = pressure energy

• 1 2
2v = kinetic energy
• gh = potential energy

dEt
= ϕm e|in − ϕm e|out + ϕw + ϕq
dt

• h=u+ p
ρ → enthalpy per unit mass
v2
• 2 → kinetic energy per unit mass
• ϕw → work per unit time applied to/by the system
• ϕq → heat exchange with the environment


3.1 Remarks Energy Balance
U pV U p
1. U = T S − pV u= m = TS − m u= m = Ts − ρ
3
• ρp = energy concentration per unit mass ⇒ m
N m Nm
2 ∗ kg = kg =
J
kg
• p in m2 = m3 → energy concentration per unit volume
N Nm


2. u1 − u2 = cv (T1 − T2 )
• For (almost in compressible) fluids → cv ≈ cp
→ u1 − u2 = Cp (T1 − T2 )


3.2 Mechanical Energy Balance
v2 v2
   
dEm p1 p2
= ϕm + 1 + gh1 − ϕm + 2 + gh2 + ϕw − ϕm ef r
dt ρ 2 ρ 2

ϕm ef r = production term that represents dissipation of mechanical energy per unit time into heat (i.e. into
internal energy).

Stationary Conditions
v2 v2
   
p1 p2
0 = ϕm + 1 + gh1 − ϕm + 2 + gh2 + ϕw − ϕm ef r
ρ 2 ρ 2
 
p2 − p1 1 2 2
0 = −ϕm + (v2 − v1 ) + g(h2 − h2 ) + ϕw − ϕm ef r
ρ 2

ϕm ef r always > 0 because mechanical energy decreases in the direction the flow.


3