Question 9.9: Calculate the critical heat flux for the conditions of Examp...
Calculate the critical heat flux for the conditions of Example 9.8 assuming that the tube is horizontal rather than vertical.
The factor \gamma_{H} is calculated from Equation (9.100) using fluid properties from Example 9.1.
\gamma_{H}=\left(\frac{G D}{\mu_{L}}\right)\left(\frac{\mu_{L}^{2}}{g_{c} \sigma D \rho_{L}}\right)^{-1.58}\left[\frac{\left(\rho_{L}-\rho_{V}\right) g D^{2}}{g_{c} \sigma}\right]^{-1.05}\left(\frac{\mu_{L}}{\mu_{V}}\right)^{6.41}
=\left(\frac{300 \times 0.0212}{156 \times 10^{-6}}\right)\left[\frac{\left(156 \times 10^{-6}\right)^{2}}{1.0 \times 8.2 \times 10^{-3} \times 0.0212 \times 567}\right]^{-1.58}
\times\left[\frac{(567 – 18.09) \times 9.81(0.0212)^{2}}{1.0 \times 8.2 \times 10^{-3}}\right]^{-1.05}\left(\frac{156 \times 10^{-6}}{7.11 \times 10^{-6}}\right)^{6.41}
\gamma_{H}=1.1326 \times 10^{21}
Equation (9.99) is now used to calculate the critical heat flux:
\frac{\hat{q}_{c}}{G \lambda}=575 \gamma_{H}^{-0.34}(L / D)^{-0.511}\left(\frac{\rho_{L}-\rho_{V}}{\rho_{V}}\right)^{1.27}\left(1+\Delta H_{\text {in }} / \lambda\right)^{1.64}
\frac{\hat{q}_{c}}{G \lambda}=2.750 \times 10^{-4}
\hat{q}_{c}=2.750 \times 10^{-4} \times 300 \times 272,000 \cong 22,440 W / m ^{2}
This result is much lower than the critical heat flux for vertical flow calculated in Example 9.8, demonstrating that the effect of tube orientation on critical heat flux can be very significant