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Answers 1: Ua Flux Mass

1. The mass flux through two surfaces of a pipe must be the same, calculated as ρuA, where ρ is the density, u is the velocity component normal to the area, and A is the area. The mass flux is calculated to be 15.7 kg/s. 2. The force on a fluid within a control volume can be calculated using the steady-state momentum principle as F = ρU2A, where U is the uniform velocity. The force is calculated to be 503 N. 3. The decay of a toxin released in a room is modeled as an exponential decay process. The time for the concentration to reduce to 1 ppm is calculated to be 48 minutes

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Ruben Segar
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67 views4 pages

Answers 1: Ua Flux Mass

1. The mass flux through two surfaces of a pipe must be the same, calculated as ρuA, where ρ is the density, u is the velocity component normal to the area, and A is the area. The mass flux is calculated to be 15.7 kg/s. 2. The force on a fluid within a control volume can be calculated using the steady-state momentum principle as F = ρU2A, where U is the uniform velocity. The force is calculated to be 503 N. 3. The decay of a toxin released in a room is modeled as an exponential decay process. The time for the concentration to reduce to 1 ppm is calculated to be 48 minutes

Uploaded by

Ruben Segar
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Answers 1

Q1.
For S1:
π  0.12
mass flux  ρuA  1000  2   15.7 kg s 1
4
This must be the same as the mass flux through S2. (Otherwise, mass would accumulate
between these surfaces.)

To calculate the mass flux in general, use the component of velocity normal to the area (or,
equivalently, the projected area normal to the velocity):
mass flux  ρu n A  ρu  A

Answer: 15.7 kg s–1 (both surfaces).

CFD Answers 1 – 1
Q2.
Steady-state momentum principle:
force (on fluid) = (momentum flux)out – (momentum flux)in
where, for a uniform velocity:
momentum flux = mass flux × velocity

For the x-component of momentum and the fluid in a control volume encompassing the
region shown:
 F  0  (ρUA)U
Hence,
π  0.12
F  ρU 2 A  1000  8 2   503 N
4

Answer: 503 N.

CFD Answers 1 – 2
Q3.
(a) Volume of room:
V  30  8  5  1200 m 3
If  is the concentration (expressed here as mass of toxin per mass of fluid) then the initial
mass of gas in the room is
mass of fluid × concentration = mass of toxin
 (ρV ) 0  2 kg
2
 0   1.389  10 3
1.2  1200

Answer: 0 = 1390 ppm by mass.

(b)
change in amount of toxin = amount in – amount out
or, as a rate equation:
rate of change of amount of toxin = rate of entering – rate of leaving

d
(ρV)  0  (ρuA)
dt
d uA
  ( )  ,  = 0 at t = 0
dt V
d uA 0.5  6
  λ where λ   0.0025 s 1
dt V 1200

This is exponential decay, with solution


   0 e  λt

 e λt  0

1 
 t  ln 0
λ 

For the required concentration ( = 1 ppm):


1
t ln(1389)  2895 s  48 min
0.0025

Answer: 48 minutes.

CFD Answers 1 – 3
Q4.
The rate at which chemical enters the river (here, as mass of chemical per unit time) is
2.5
S  6.944  10 4 kg s 1
3600
At steady state, the flux of chemical through a downstream section equals the rate at which it
enters the river. If  is pollutant concentration (here, mass of chemical per volume of water):
(uA)  S
S 6.944  10 4
    2.31  10 4 kg m 3
uA 0.3  5  2

Answer: 2.3110–4 kg m–3.

CFD Answers 1 – 4

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