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Stator ventilation duct

The air passes through the stator ventilation duct.

Problem Type:
Axisymmetric problem of heat transfer.

Geometry:

D = 30 mm, L = 4000 mm.

Given:
Speed of air in the channel v = 10 m/s,
Duct surface convection coefficient α = 50 W/K·m2,
Thermal load q = 300 W/m2.

Problem:

Define the air temperature rise.

Solution:

The channel is divided into n sections. The heat flux q from each section is calculated. The consumed energy dQ can be found as:

dQ = q·t, where
q - thermal flux (W),
t - time of air passage through the section (sec.): t = (L / n) / v.

Then the air temperature rise is calculated as:

dT = dQ / (dV·C·ρ), where
dQ - consumed energy (J),
dV - volume (m3),
C - thermal capacity of air (J / kg·K),
ρ - density of air (kg/m3).

Results:

0-iteration
 Section number Convection temperature, °C Thermal flux, W Consumed energy, J 1 0 18.85 0.754 2 0 18.85 0.754 3 0 18.85 0.754 4 0 18.85 0.754 5 0 18.85 0.754 6 0 18.85 0.754 7 0 18.85 0.754 8 0 18.85 0.754 9 0 18.85 0.754 10 0 18.85 0.754

1 iteration
 Section number Convection temperature, °C Thermal flux, W Consumed energy, J 1 0 19.95 0.798 2 2.41 18.90 0.756 3 4.82 18.85 0.754 4 7.24 18.85 0.754 5 9.65 18.85 0.754 6 12.06 18.85 0.754 7 14.47 18.86 0.754 8 16.89 18.84 0.754 9 19.30 18.81 0.752 10 21.71 17.74 0.709

2 iteration
 Section number Convection temperature, °C Thermal flux, W Consumed energy, J 1 0 19.97 0.799 2 2.55 18.88 0.755 3 4.97 18.85 0.754 4 7.38 18.85 0.754 5 9.80 18.85 0.754 6 12.21 18.85 0.754 7 14.62 18.85 0.754 8 17.03 18.82 0.753 9 19.44 18.47 0.739 10 21.85 18.10 0.724

3 iteration
 Section number Convection temperature, °C Thermal flux, W Consumed energy, J 1 0 19.97 0.799 2 2.56 18.88 0.755 3 4.97 18.85 0.754 4 7.38 18.85 0.754 5 9.80 18.84 0.754 6 12.21 18.85 0.754 7 14.62 18.85 0.754 8 17.03 18.82 0.753 9 19.44 18.47 0.739 10 21.81 18.10 0.724

Air temperature rise is dT = 21.81 °C

* Reference: John H. Lienhard IV, John H. Lienhard V, A heat transfer textbook.