SME
Transport Phenomena
Transport analogies
Use this for “Turbulent air flows in a smooth tube at Re = 30,000, Pr = 0.71. Estimate Nu from the friction factor and compare it with Dittus–Boelter.”
Diagram
δT = δ·Pr⁻¹ᐟ³ = 1.126δ.
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
,
jH
=
f
2
Nu
=
jH
×
Re
×
Pr
⅓
NuDB
=
0.023
×
Re
0.8
×
Pr
0.4
Formulas
f
=
0.079
R
e
0.25
0.005283
=
0.079
50
,
000
0.25
jH
=
f
2
0.002642
=
0.005283
2
Nu = jH × Re × Pr1 ÷ 3
117.3
=
0.002642
×
50,000
×
0.7
1 ÷ 3
NuDB = 0.023 × Re0.8 × Pr0.4
114.5
=
0.023
×
50,
000
0.8
×
0.7
0.4
Assumptions
•
Turbulent flow in a smooth tube, 4000 < Re < 10⁵: Blasius gives the Fanning f.
•
The analogy holds for 0.6 < Pr < 60; Dittus–Boelter (Pr0.4, fluid heated) needs Re above about 10,000.
•
Only skin friction carries over: no form drag, as behind a bluff body.
Step-by-step
WE KNOW
Re = 50,000, Pr = 0.7
FIND
Fanning friction factor (f), Colburn j-factor (jH), Nusselt number (Nu), Dittus–Boelter Nu (NuDB)
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 Colburn j-factor (jH)
Use jH = f ÷ 2
Chilton–Colburn: the heat j-factor equals half the Fanning friction factor.
jH
=
0.005283
2
jH = 0.002642
Step 3 · Find Nusselt number (Nu)
Use Nu = jH × Re × Pr1 ÷ 3
The
j-factor
is
Nu
÷
(RePr^(
1
3
));
the
cube
root
of
Pr
sets
the
thermal
layer
apart.
Nu
=
0.002642
×
50,000
×
0.7
1 ÷ 3
Nu = 0.002642 × 50,000 × 0.8879
Nu = 132.1 × 0.8879
Nu = 117.3
Step 4 · Find Dittus–Boelter Nu (NuDB)
Use NuDB = 0.023 × Re0.8 × Pr0.4
Dittus–Boelter for a heated fluid, a check on the analogy: the two should agree within a few percent.
NuDB
=
0.023
×
50,
000
0.8
×
0.7
0.4
NuDB = 0.023 × 5,743.5 × 0.86704
NuDB = 132.1005 × 0.86704
NuDB = 114.5
Check
Put the numbers back in. Both sides should match.
0.005283
=
0.079
50
,
000
0.25
✓
0.002642
=
0.005283
2
✓
117.3
=
0.002642
×
50,000
×
0.7
1 ÷ 3
✓
114.5
=
0.023
×
50,
000
0.8
×
0.7
0.4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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.”
Prandtl and Schmidt numbers
Use this for “Engine oil at 80 °C has cₚ = 2131 J/(kg·K), μ = 0.032 Pa·s and k = 0.138 W/(m·K). Find Pr and say what it compares.”