SME
Transport Phenomena · Problem type
Sherwood number by the same analogy
Use this for “Water vapour diffuses into air flowing at Re = 20,000 in a wet smooth tube, Sc = 0.61. Estimate Sh from the friction factor.”
Diagram
δc = δ·Sc⁻¹ᐟ³ = 1.186δ.
The thinner the layer, the steeper its gradient at the wall and the faster the transfer.
This is an example. Type your own values to start a new problem.
f
=
0.079
Re
0.25
Sh
=
f
2
×
Re
×
Sc
⅓
Formulas
f
=
0.079
R
e
0.25
0.005283
=
0.079
50
,
000
0.25
Sh
=
f
2
×
Re
×
Sc^(1
÷
3)
111.4
=
0.005283
2
×
50,000
×
0.6
1 ÷ 3
Assumptions
•
Turbulent flow in a smooth tube, 4000 < Re < 10⁵: Blasius gives the Fanning f.
•
Chilton–Colburn for mass holds for 0.6 < Sc < 2500, with low mass-transfer rates.
•
Only skin friction carries over: no form drag, as behind a bluff body.
Step-by-step
WE KNOW
Re = 50,000, Sc = 0.6
FIND
Fanning friction factor (f), Sherwood number (Sh)
Step 1 · Find Fanning friction factor (f)
Use f = 0.079 ÷ Re0.25
The Blasius fit for the Fanning friction factor in a smooth tube.
f
=
0.079
50
,
000
0.25
f
=
0.079
14.953
f
=
0.005283
Step 2 · Find Sherwood number (Sh)
Use Sh = f ÷ 2 × Re × Sc1 ÷ 3
The
mass
j-factor
Sh
÷
(ReSc^(
1
3
))
equals
f
2,
as
the
heat
one
does.
Sh
=
0.005283
2
×
50,000
×
0.6
1 ÷ 3
Sh
=
0.005283
2
×
50,000
×
0.8434
Sh = 0.002642 × 50,000 × 0.8434
Sh = 132.1 × 0.8434
Sh = 111.4
Check
Put the numbers back in. Both sides should match.
0.005283
=
0.079
50
,
000
0.25
✓
111.4
=
0.005283
2
×
50,000
×
0.6
1 ÷ 3
✓
Topic check
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