Question 9.4: The nucleate boiling of Example 9.1 takes place on a tube bu...

The critical heat flux for the tube bundle is given by Equation (9.24):

\hat{q}_{c, b u n d l e}=\hat{q}_{c, t u b e} \phi_{b}

The critical heat flux for a single tube was calculated in Example 9.3, and any of the three results obtained there can be used here. Choosing the result calculated by the Mostinski correlation, we have:

\hat{q}_{c, \text { tube }}=398,000  W / m ^{2}

To determine the bundle correction factor, first calculate the bundle geometry parameter, \psi_{b}, which for plain tubes is:

\psi_{b}=\frac{D_{b}}{n_{ t } D_{o}}=\frac{34}{520  \times  1.0}=0.0654

Since this value is less than 0.323, the bundle correction factor is:

\phi_{b}=3.1 \psi_{b}=3.1 \times 0.0654 \cong 0.203

Therefore,

\hat{q}_{c, b u n d l e}=398,000 \times 0.203 \cong 81,000  W / m ^{2}

The heat-transfer coefficient for a tube bundle is given by Equation (9.19):

h_{b}=h_{n b} F_{b}+h_{n c}

We will use the value of h_{nb} calculated in Example 9.1 by the modified Mostinski correlation, i.e., h_{nb} = 1396 W/m²· K. Also, from Example 9.1 ,  ΔT_{e} = 16.2 K. Since this value is much greater than 4 K, a rough estimate is sufficient for the natural convection coefficient, h_{nc}. For an organic compound, Palen’s recommendation for hydrocarbons is adequate, so we take h_{nc} ≅ 250 W/m²· K . The value of F_{b} is calculated using Equation (9.20):

                            F_{b}=1.0+0.1\left[\frac{0.785 D_{b}}{C_{1}\left(P_{T} / D_{o}\right)^{2} D_{o}}-1.0\right]^{0.75}

                             =1.0+0.1\left[\frac{0.785  \times  34}{1.0(1.25 / 1.0)^{2}  \times  1.0}-1.0\right]^{0.75}

                               F_{b} \cong 1.803

Therefore,

                              h_{b}=1396 \times 1.803+250=2767 \cong 2770 W / m ^{2} \cdot K