Question 3.18: Turbulent Thermal Pipe Flow Consider fully-developed hot air......
Turbulent Thermal Pipe Flow
Consider fully-developed hot air flow in a metal tube (\rm\dot m = 0.05 \;kg/s; L = 5 m; D = 0.15 m ). Because of the cold ambient (T_\infty =0^\circ \rm C,~h_{ambient}=6\frac{W}{m^2K} ), the air inlet temperature T_1=103^\circ\rm C drops to T_2=77^\circ\rm C at the tube outlet. Find the heat loss and surface temperature at x = L.
\rm Nu_D=0.023\,Re_D^{4/5}Pr^{1/3} (3.80)
| Concepts | Assumptions | Sketch |
| • Energy balance \rm\dot Q=\dot mc_p(T_1-T_2) | • Steady-state |  |
| • Radial heat loss resistance in series | • Air \hat = Ideal gas | |
| • \rm Nu_D-correlation Eq. (3.80) | • Constant properties and parameters | |
| • Tube wall effects negligible |
• Properties (App. B) for T_{\rm ref} ≈ 360 K , Pr = 0.7, \rm c_p = 1,010J/kg·K, k = 0.03 W/m·K, μ = 208.2 × 10^{−7} N·s/m²
• Reynolds number \rm Re_D=\frac{4\dot m}{\mu\pi D} := 20,384 (i.e., turbulent)
• Heat loss over entire tube: \rm\dot Q=\dot mc_p(T_1-T_2)~~\dot Q=1313\,J/s
• Tube surface temperature \rm T_s at x = L (see thermal resistance network):
Hence,
\rm q_s(x=L)=\frac{T_2-T_\infty }{\frac{1}{h_i} +\frac{1}{h_{amb.}} } (E.3.18.1)
where \rm h_i=h(x=L)=Nu_D\frac{k}{D} and \rm Nu_D=0.023\,Re_D^{4/5}Pr^{1/3}.
Finally, with \rm Nu_D=57.22 and hence \rm h_i=11.44\frac{W}{m^2\cdot K} , so that:
\rm q_s(x=L)=305\,J/(m^2\cdot s) (E.3.18.2)
Now, applying again
\rm q_s=\frac{T_2-T_s}{1/h_i} (E.3.18.3)
we obtain
T_s=77^\circ \rm C-\frac{305}{11.44} =50.34^\circ C (E.3.18.4)
