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Example Questions (Exam Practice) - Fluid Dynamics
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- Rijksuniversiteit Groningen (RuG)
Example Questions (Exam Practice) - Fluid Dynamics
[Meer zien]Voorbeeld 3 van de 26 pagina's
Voorbeeld 3 van de 26 pagina's
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Voorbeeld van de inhoud
Fluid Dynamics Summary Examples and Tutorials 20201 Tutorial: Mass Balances (1)
1.1 Question 1
Consider two connected well-stirred containers with same volume V m3 . Water is pumped from v1 to v2 and
3
back at rate of ϕv ms . Both containers remain full and volume of tubes can be neglected. At t = 0, salt
concentration in volume 1 is c0 mol
m3 , volume 2 is filled with pure water.
a. Give relationship between salt concentration (c1 and c2 ) at any
time t.
Make a sketch. Salt is conserved, thus, at any time the total amount present
in V1 and V2 equals amount c0 initially present in V1 . V1 c1 + V2 c2 = V1 c0 , and
since V1 = V2 , we can write c1 + c2 = c0 . In contrast to mass, concentration is
not a conserved entity.
b. Give expression for c2 as function of time.
The mass or component balance for c2 in terms of concentration reads:
→ V dcdt = ϕv c1 − ϕv c2 → amount of salt entering V2 (per unit time) is amount
2
entering from V1 (> 0) minus amount leaving V2 (< 0).
To solve the differential equation above requires expressing c1 as function of c2 .
From the relation found in A we know c1 = c0 − c2 . We then obtain:
dc2
V = ϕv (c0 − 2c2 ) − ϕc2 = ϕv (c0 − 2c2 )
dt
Z
dc2
Z Figure 1: Question 1a
V dt = ϕv (c0 − 2c2 )dt
dt
Z Z
V dc2 = ϕ(c0 − 2c2 )dt
Z Z
dc2 ϕv
= dt
c0 − 2c2 V
Apply boundary conditions:
Z c2 Z t
dc2 ϕv
= dt
0 c0 − c2 0 V
1 c2 ϕv t
− ln(c0 − c2 ) = t
2 0 V 0
c0 − 2c2 2ϕv
= e− V t
c0
1 2ϕv
c2 (t) = c0 1 − e− V t
2
Figure 2: Question 1d
c. What is the final salt concentration in both containers?
→ check lim → ∞ → c2 = 21 c0 = c1
d. Draw schematically in one figure c1 and c2 as function of time.
See figure on the right.
1
,1.2 Question 2
Consider a factory that discharges a stream of 5 sl with a toxic
component, at concentration of 12mM . Governmental regu-
lations allow discharging this toxic component to the environ-
ment at concentrations of 2mM or below. To do so, the factory
dilutes the toxic stream into a effectively stirred tank of vol-
ume V . The natural water used for dilution has a background
concentration (cin ) of the same toxic component of 0.04mM .
The effluent is discharged to the nearby river. There are no
chemical reactions going on in the tank
a. Give the mass balance for the water and a molar
balance for the toxic component assuming steady-state
conditions. Figure 3: Question 2a
First, make a sketch (see figure on the right). We see that there are two types of balances:
1. Volumetric or solution (water + toxin) balance in sl . Has two incoming and 1 outgoing stream:
ϕv,in + 5 = ϕv,out
mmol l
2. Balance for only the toxin in s , which is the product of volumetric flow (in s and concentration (here
in mM , i.e. mmol
l ):
ϕv,in ∗ 0.04 + 5 ∗ 12 = ϕv,out ∗ 2
b. How much water (steady-state) is required for complying to governmental regulations?
Substitute the volumetric/solution balance into the toxin balance:
ϕv,in ∗ 0.04 + 5 ∗ 12 = (ϕv,in + 5) ∗ 2
0.04ϕv,in ∗ 60 = 2ϕv,in + 10
1.96ϕv,in = 50
l
ϕv,in = 25, 51
s
After a long time of operation the factory decides that the tank requires maintenance. The inflow of the toxic
component is stopped and they start flushing the tank with just natural water. The tank is still effectively
stirred and remains completely filled all the time. c. Give the concentration c as function of time.
The molar balance of the toxic:
dc
V = ϕv,in cin − ϕv,out cout
dt
Suppose flushing is with clean water, then there is no inflow of toxic, inflow term would be zero and the
integration is straightforward.
Suppose we use natural water with toxic component 0.04mM , then:
• cout = c due to effective stirring.
• ϕv,in = ϕv,out = ϕv due to steady state
This gives:
dc
V = ϕv cin − ϕv c = ϕv (cin − c)
dt
dc ϕv
=
dt(cin − c) V
2
, Integrate and use boundary conditions at t → c and at t0 → c0 :
Z c Z t
dc ϕc
dt = dt
c0 dt(cin − c) 0 V
Z c Z t
dc ϕc
= dt
c0 cin − c V 0
c ϕv t
−ln(cin − c) = t
c0 V 0
cin − c ϕv
= e− V t
cin − c0
ϕv
c(t) = cin − (cin − c0 )e− V t
2 Lecture: Dimensional Analysis (1b)
2.1 Example: Force on a Sphere in a Flow
Suppose we are given the following info for force on a sphere in a flow. We need to define an equation.
K = f (Du, V r, ρ, η) → K = Dua ∗ V rb ∗ ρc ∗ η d ∗ constant
First, define the different parameters in Mass/Length/Time (or put in basic units).
ML L M M
K = [N ] = Du = [L] Vr = ρ= η = Pa ∗ s =
T2 T L3 LT
a b c d
ML L M M
→ = L ∗ ∗ ∗
T2 T L3 LT
This gives: M = 1 = c + d L = 1 = a + b − 3c − d T = −2 = −b − 3c
Rewrite for d as function of the other parameters: a = 2 − d b=2−d c=1−d
Put the above back into the original equation:
K = Du2−d ∗ V r2−d ∗ ρ1−d ∗ η d ∗ constant
Rewrite into: d d
K η η
= ∗ constant = f = f (Re)
Du2 V r2 ρ DuV rρ DuV rρ
3
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