Question 9.1: Pure component nucleate boiling takes place on the surface o...

(a) For the Forster–Zuber correlation, first calculate ΔT_{e} and ΔP_{sat}·

ΔT_{e} = T_{w}  –  T_{sat} = 453.7  –  437.5 = 16.2 K

ΔP_{sat} = P_{sat}( T_{w})  –  P_{sat}( T_{sat})

= 416.6 – 310.3

ΔP_{sat} =  106.3 kPa = 106,300 Pa

All the physical properties appearing in the correlation are given in proper SI units except the surface tension, which is:

\sigma  =  8.2  dyne / cm =8.2  \times  10^{-3} N / m

Substituting the appropriate values into Equation (9.1) yields:

h_{n b}=0.00122 \frac{k_{L}^{0.79}  C_{P, L}^{0.45} \rho_{L}^{0.49} g_{c}^{0.25} \Delta T_{e}^{0.24} \Delta P_{s a t}^{0.75}}{\sigma^{0.5} \mu_{L}^{0.29} \lambda^{0.24} \rho_{V}^{0.24}}

=\frac{0.00122(0.086)^{0.79}(2730)^{0.45}(567)^{0.49}(1.0)^{0.25}(16.2)^{0.24}(106,300)^{0.75}}{\left(8.2  \times  10^{-3}\right)^{0.5}\left(156  \times  10^{-6}\right)^{0.29}(272,000)^{0.24}(18.09)^{0.24}}

h_{n b}=5512  W / m ^{2} \cdot K

The heat flux at the pipe wall is given by:

\hat{q}=h_{n b} \Delta T_{e}=5512  \times  16.2=89,294 \cong 89,300  W / m ^{2}

(b) For the Mostinski correlation, first calculate the pressure correction factor using Equation (9.4).

P_{ r }=P / P_{c}=\frac{310.3}{2550}=0.1217

F_{P}=1.8 P_{r}^{0.17}+4 P_{r}^{1.2}+10 P_{r}^{10}

F_{P}=1.8(0.1217)^{0.17}+4(0.1217)^{1.2}+10(0.1217)^{10}=1.5777

Since \Delta T_{e} =  16.2 K is known, use Equation (9.3b) to calculate the heat-transfer coefficient.

h_{n b}=1.167 \times 10^{-8} P_{c}^{2.3} \Delta T_{e}^{2.333} F_{P}^{3.333}

=1.167 \times 10^{-8}(2550)^{2.3}(16.2)^{2.333}(1.5777)^{3.333}

h_{n b}=2421  W / m ^{2} \cdot K

The heat flux is calculated as before:

\hat{q}=h_{n b} \Delta T_{e}=2421 \times 16.2=39,220  W / m ^{2}

(c) The calculation is the same as in part (b) except that Equation (9.5) is used to calculate the pressure correction factor.

F_{P}=2.1 P_{r}^{0.27}+\left[9+\left(1-P_{r}^{2}\right)^{-1}\right] P_{r}^{2}

=2.1(0.1217)^{0.27}+\left\{9+\left[1-(0.1217)^{2}\right]^{-1}\right\}(0.1217)^{2}

F_{P}=1.3375

h_{n b}=1.167  \times  10^{-8}(2550)^{2.3}(16.2)^{2.333}(1.3375)^{3.333}

h_{n b}=1396  W / m ^{2} \cdot K

\hat{q}=h_{n b} \Delta T_{e}=1396  \times  16.2=22,615  W / m ^{2}

(d) To use the Cooper correlation, substitute \hat{q}=h_{n b} \Delta T_{e} in Equation (9.6b) to obtain:

h_{n b}=55  \hat{q}^{0.67}  P_{r}^{0.12}\left(-\log _{10} P_{ r }\right)^{-0.55} M^{-0.5}            (9.6b)

h_{n b}=55\left(h_{n b} \Delta T_{e}\right)^{0.67} P_{r}^{0.12}\left(-\log _{10} P_{ r }\right)^{-0.55} M^{-0.5}

Solving for h_{n b} gives:

The heat flux is then:

\hat{q}=h_{n b} \Delta T_{e}=23,214 \times 16.2 \cong 376,070  W / m ^{2}

(e) For the Stephan–Abdelsalam correlation, the theoretical bubble diameter is calculated using Equation (9.12). A contact angle of 35º is assumed for a non-cryogenic organic compound

d_{B}=0.0146 \theta_{c}\left[\frac{2 g_{c} \sigma}{g\left(\rho_{L}  –  \rho_{V}\right)}\right]^{0.5}=0.0146  \times  35\left[\frac{2  \times  1.0  \times  8.2  \times  10^{-3}}{9.81(567  –  18.09)}\right]^{0.5}

d_{B}=8.918 \times 10^{-4} m

The five dimensionless parameters are calculated using Equations (9.7) to (9.11). Since the heat flux is unknown, it is retained in Z_{1}.

Z_{1}=\frac{\hat{q} d_{B}}{k_{L} T_{s a t}}=\frac{\hat{q}  \times  8.918  \times  10^{-4}}{0.086  \times  437.5}=2.370 \times 10^{-5} \hat{q}

Z_{2}=\frac{\alpha_{L}^{2} \rho_{L}}{g_{c} \sigma d_{B}}=\frac{\left(5.556  \times  10^{-8}\right)^{2}  \times  567}{1.0  \times  8.2  \times  10^{-3}  \times  8.918  \times  10^{-4}}=2.393 \times 10^{-7}

Z_{3}=\frac{g_{c} \lambda d_{B}^{2}}{\alpha_{L}^{2}}=\frac{1.0  \times  272,000\left(8.918  \times  10^{-4}\right)^{2}}{\left(5.556  \times  10^{-8}\right)^{2}}=7.008 \times 10^{13}

Z_{4}=\rho_{ V } / \rho_{ L }=18.09 / 567=0.031905

Z_{5}=\left(\rho_{ L }-\rho_{ V }\right) / \rho_{ L }=(567-18.09) / 567=0.9681

Substituting these values into Equation (9.13) yields:

Substituting \hat{q}=h_{n b} \Delta T_{e}= h_{n b} × 16.2 gives:

h_{n b}=\left(k_{L} / d_{B}\right) \times 0.04402\left(16.2  h_{n b}\right)^{0.674}

=\left(0.086 / 8.918 \times 10^{-4}\right) \times 0.04402(16.2)^{0.674}  h_{n b}^{0.674}

h_{n b}=27.739  h_{n b}^{0.674}

h_{n b}=26,709  W / m ^{2} \cdot K

Finally, the heat flux is given by:

\hat{q}=h_{n b} \Delta T_{e}=26,709 \times 16.2=432,686 W / m ^{2}

The calculated values of the heat-transfer coefficient are summarized in the following table:

The predictions of the various correlations differ by a huge amount, the ratio of largest to smallest value being nearly 20. One reason for the wide variation is that nucleate boiling is very sensitive to the precise condition of the surface on which boiling occurs. Although the factors that govern the nucleation process are reasonably well understood (7–10), it is not practical to specify the detailed surface characteristics required to rigorously model the phenomenon. Even if this could be done, it would be of dubious utility in equipment design because the surface characteristics change in an uncontrollable manner over time due to aging, corrosion, fouling, cleaning procedures, etc. As a result, the variability exhibited by the calculated values actually reflects the variability observed among the sets of experimental data upon which the various correlations are based [2]. Fortunately, tests with commercial tube bundles indicate that the effect of surface condition is much less pronounced in the operation of industrial equipment [11]. Nevertheless, significant conservatism in equipment design is warranted by the level of uncertainty in the fundamental boiling  correlations.