Question 9.6: Calculate the void fraction and two-phase density for the re...
Calculate the void fraction and two-phase density for the reboiler return line of Example 9.5 using:
(a) The homogeneous flow model.
(b) The Lockhart–Martinelli correlation.
(c) The Chisholm correlation.
(d) The CISE correlation
(a) For the homogeneous flow model, the void fraction is given by Equation (9.60):
\epsilon_{V}=\frac{x}{x + (1 – x) \rho_{V} / \rho_{L}}=\frac{0.2}{0.2 + 0.8(0.4787 / 38.94)}
\epsilon_{ V }=0.9531
The two-phase density can be calculated using either Equation (9.54) or (9.51). The latter was used in Example 9.5 to obtain \rho_{ tp } = 2.2813 lbm/ft³. Using Equation (9.54) we obtain
\rho_{tp} = [x/ \rho_{V} + (1 – x)/\rho_{L}]^{-1} (9.51)
\rho_{ tp }=\epsilon_{V} \rho_{V}+\left(1-\epsilon_{V}\right) \rho_{L}=0.9531 \times 0.4787+(1-0.9531) \times 38.94\rho_{t p}=2.2825 \cong 2.28 lbm / ft ^{3}
The difference between the two values of \rho_{t p} is due to round-off error in the value of \epsilon_{V}, and is negligible in the present context.
(b) A value of \phi_{L}^{2} = 43.07 was calculated in Example 9.5, giving \phi_{L}= 6.5628. The void fraction is computed from Equation (9.61):
The two-phase density is calculated using Equation (9.62):
\rho_{t p}=\frac{\rho_{L}+\left(\phi_{L}-1\right) \rho_{V}}{\phi_{L}}=\frac{38.94+(6.5628-1) \times 0.4787}{6.5628}\rho_{ tp }=6.3392 \cong 6.34 lbm / ft ^{3}
(c) From Example 9.5, the Lockhart–Martinelli parameter is X_{ tt} = 0.521 < 1. Therefore, Equation (9.63) gives:
S R=\left(\rho_{L} / \rho_{V}\right)^{0.25}=(38.94 / 0.4787)^{0.25}=3.0032Substitution into Equation (9.59) gives the void fraction:
\epsilon_{V}=\frac{x}{x+S R(1-x) \rho_{V} / \rho_{L}}=\frac{0.2}{0.2+3.0032 \times 0.8 \times 0.4787 / 38.94}\epsilon_{V}=0.8713
The two-phase density is obtained from Equation (9.54).
\rho_{t p}=\epsilon_{V} \rho_{V}+\left(1-\epsilon_{V}\right) \rho_{L}=0.8713 \times 0.4787+(1-0.8713) \times 38.94\rho_{ tp }=5.4287 \cong 5.43 lbm / ft ^{3}
(d) The following values are obtained from Example 9.5:
D_{i}=0.835 ft\sigma=7.812 \times 10^{-4} lbf / ft
G=547,846 lbm / h \cdot ft ^{2}
\operatorname{Re}_{ LO }=1,068,405
The Weber number is calculated as follows:
W e_{L O}=\frac{G^{2} D_{i}}{g_{c} \rho_{L} \sigma}=\frac{(547,846)^{2} \times 0.835}{4.17 \times 10^{8} \times 38.94 \times 7.812 \times 10^{-4}}=19,756The parameters in the CISE correlation are calculated using Equations (9.65) to (9.67). From part (a), the homogeneous void
fraction is \left(\epsilon_{V}\right)_{\text {hom }}= 0.9531.
E_{1}=1.578 R e_{L O}^{-0.19}\left(\rho_{L} / \rho_{V}\right)^{0.22}=1.578(1,068,405)^{-0.19}(38.94 / 0.4787)^{0.22}
E_{1}=0.2973
E_{2}=0.0273 W e_{L O} R e_{L O}^{-0.51}\left(\rho_{L} / \rho_{V}\right)^{-0.08}
=0.0273 \times 19,756(1,068,405)^{-0.51}(38.94 / 0.4787)^{-0.08}
E_{2}=0.3194
The slip ratio is calculated from Equation (9.64):
S R=1+E_{1}\left[\left(\frac{\gamma}{1+\gamma E_{2}}\right)+\gamma E_{2}\right]^{0.5}=1+0.2973\left[\frac{20.322}{1+20.322 \times 0.3194}+20.322 \times 0.3194\right]^{0.5}
SR = 1.9013
The void fraction is calculated using Equation (9.59):
\epsilon_{V}=\frac{x}{x+S R(1-x) \rho_{V} / \rho_{L}}=\frac{0.2}{0.2+1.9013 \times 0.8 \times 0.4787 / 38.94}\epsilon_{V}=0.9145
Finally, the two-phase density is calculated using Equation (9.54).
\rho_{t p}=\epsilon_{V} \rho_{V}+\left(1-\epsilon_{V}\right) \rho_{L}=0.9145 \times 0.4787+(1-0.9145) \times 38.94\rho_{ t p}=3.7671 \cong 3.77 lbm / ft ^{3}
Notice that while the void fractions predicted by the four methods are within about 12% of one another, the two-phase densities differ by large percentages