Question 9.5: The return line from a thermosyphon reboiler consists of 10-...

(a) The friction loss for the liquid fraction flowing alone in the pipe is calculated first.

G_{L}=\frac{\dot{m}_{L}}{(\pi / 4) D_{i}^{2}}=\frac{240,000}{(\pi / 4)(0.835)^{2}}=438,277  lbm / h \cdot ft ^{2}

R e_{L}=\frac{D_{i} G_{L}}{\mu_{L}}=\frac{0.835  \times  438,277}{0.177  \times  2.419}=854,724

s_{L}=\rho_{L} / \rho_{\text {water }}=38.94 / 62.43=0.6237

Equation (4.8) is used to calculate the friction factor:

f=0.3673  Re^{-0.2314}        (4.8)

f_{L} = 0.3673  Re_{L}^{-0.2314}=0.3673(854,724)^{-0.2314}=0.01557

The friction loss per unit length is given by:

\left(\frac{\Delta P_{f}}{L}\right)_{L}=\frac{f_{L} G_{L}^{2}}{7.50  \times  10^{12} D_{i} s_{L} \phi_{L}}=\frac{0.01557(438,277)^{2}}{7.5  \times  10^{12}  \times  0.835  \times  0.6237  \times  1.0}

\left(\frac{\Delta P_{f}}{L}\right)_{L}=0.000766  psi / ft

Since the flow is turbulent, the Lockhart–Martinelli parameter is calculated using Equation (9.37).

X_{ tt } \cong\left(\frac{1-x}{x}\right)^{0.9}\left(\frac{\rho_{V}}{\rho_{L}}\right)^{0.5}\left(\frac{\mu_{L}}{\mu_{V}}\right)^{0.1}

= (0.8 / 0.2)^{0.9}(0.4787 / 38.94)^{0.5}(0.177 / 0.00885)^{0.1}

X_{ tt } \cong 0.521

For comparison, the reader can verify that using Equation (9.36) with n = 0.2314 to calculate X_{tt} gives a value of 0.534, which differs from the value calculated above by less than 3%. For turbulent flow, the two-phase multiplier is given by Equation (9.28):

\phi_{L}^{2}=1+\frac{20}{X_{tt}}+\frac{1}{X_{tt}^{2}}  =  1+\frac{20}{0.521}+\frac{1}{(0.521)^{2}}=43.07

X = \left(\frac{1  –  x}{x}\right)^{(2  –  n)/2} (\mu_{L}/\mu_{V})^{n/2} (\rho_{V}/\rho_{L})^{0.5}                (9.36)

The two-phase friction loss is then:

\left(\frac{\Delta P_{f}}{L}\right)_{ tp }=\phi_{ L }^{2}\left(\frac{\Delta P_{f}}{L}\right)_{L}=43.07 \times 0.000766=0.033  psi / ft

(b) For the Chisholm method, the friction loss is calculated assuming the total flow has the properties of the liquid. Thus,

G=\frac{\dot{m}}{(\pi / 4) D_{i}^{2}}=\frac{300,000}{(\pi / 4)(0.835)^{2}}=547,846  lbm / h \cdot ft ^{2}

\operatorname{Re}_{L O}=\frac{D_{i} G}{\mu_{ L }}=\frac{0.835(547,846)}{0.177  \times  2.419}=1,068,405

f_{L O}=0.3673 R e_{L O}^{-0.2314} =0.3673(1,068,405)^{-0.2314}=0.01479

\left(\frac{\Delta P_{f}}{L}\right)_{L O}=\frac{f_{L O} G^{2}}{7.50  \times  10^{12} D_{i} s_{L} \phi_{L}}=\frac{0.01479(547,846)^{2}}{7.5  \times  10^{12}  \times  0.835  \times  0.6237  \times  1.0}

\left(\frac{\Delta P _{ f }}{ L }\right)_{L O}=0.001136  psi / ft

Since the flow is turbulent, the Chisholm parameter is calculated using Equation (9.42) with n =0.2314.

Y=\left(\rho_{L} / \rho_{V}\right)^{0.5}\left(\mu_{V} / \mu_{L}\right)^{n / 2} = (38.94 / 0.4787)^{0.5}(0.00885 / 0.177)^{0.1157}

Y \cong 6.38

Since Y < 9.5, Equation (9.44) gives:

The two-phase multiplier is calculated using Equation (9.43):

\phi_{L O}^{2}=1+\left(Y^{2}-1\right)\left\{B[x(1-x)]^{(2-n) / 2}+x^{2-n}\right\}

\phi_{L O}^{2} \cong 19.22

Finally, the two-phase friction loss is given by Equation (9.38):

\left(\frac{\Delta P_{f}}{L}\right)_{t p}=\phi_{L O}^{2}\left(\frac{\Delta P_{f}}{L}\right)_{L O}

 = 19.22 \times 0.001136 \cong 0.022  psi / ft

(c) For the Friedel method, we need only recalculate \phi_{L O}^{2} according to Equation (9.45). The parameters E, F, H, Fr , and We are first calculated using Equations (9.47)–(9.52).

\phi_{L O}^{2} = E + \frac{3.24 FH}{Fr^{0.045} We^{0.035}}                       (9.45)

F=x^{0.78} (1-x)^{0.24}           (9.47)

E=(1-x)^{2}+x^{2}\left(\mu_{V} / \mu_{L}\right)^{n}\left(\rho_{L} / \rho_{V}\right)

=  (0.8)^{2}+(0.2)^{2}(0.00885 / 0.177)^{0.2314}(38.94 / 0.4787)

E = 2.2668

F=x^{0.78}(1-x)^{0.24}=(0.2)^{0.78}(0.8)^{0.24}=0.2701

H=\left(\rho_{L} / \rho_{V}\right)^{0.91}\left(\mu_{V} / \mu_{L}\right)^{0.19}\left(1-\mu_{V} / \mu_{L}\right)^{0.7}

=(38.94 / 0.4787)^{0.91}(0.00885 / 0.177)^{0.19}(1-0.00885 / 0.177)^{0.7}

H = 29.896

\rho_{t p}=\left[x / \rho_{V}+(1-x) / \rho_{L}\right]^{-1}=[0.2 / 0.4787+0.8 / 38.94]^{-1}

\rho_{t p}=2.2813  lbm / ft ^{3}

F r=\frac{G^{2}}{g D_{i} \rho_{t p}^{2}}=\frac{(547,846)^{2}}{4.17 \times 10^{8} \times 0.835(2.2813)^{2}}=165.63

The surface tension must be converted to English units for consistency.

\sigma=11.4  dyne / cm  \times  6.8523  \times  10^{-5}\left(\frac{ lbf / ft }{ dyne / cm }\right)=7.812  \times  10^{-4} lbf / ft

W e=\frac{G^{2} D_{i}}{g_{c} \rho_{ tp } \sigma}=\frac{(547,846)^{2}  \times  0.835}{4.17  \times  10^{8}  \times  2.2813  \times  7.812  \times  10^{-4}}

W e=337,227

\phi_{L O}^{2}=E+\frac{3.24 FH }{ Fr ^{0.045} We^{0.035}}

=2.2668+\frac{3.24  \times  0.2701  \times  29.896}{(165.63)^{0.045}(337,227)^{0.035}}

\phi_{L O}^{2}=15.58

The two-phase friction loss is calculated using the value of \left(\Delta P_{f} / L\right)_{L O} found in part (b):

\left(\frac{\Delta P_{f}}{L}\right)_{t p}=\phi_{L O}^{2}\left(\frac{\Delta P_{f}}{L}\right)_{L O}=15.58  \times  0.001136 \cong 0.018  psi / ft

(d) For the MSH method, the two-phase multiplier is calculated using Equation (9.53).

\phi_{L O}^{2}=Y^{2} x^{3}+\left[1+2 x\left(Y^{2}-1\right)\right](1-x)^{1 / 3}

=  (6.38)^{2}(0.2)^{3}+\left[1+2 \times 0.2\left((6.38)^{2}-1\right)\right](1-0.2)^{1 / 3}

\phi_{L O}^{2} \cong 16.00

The two-phase friction loss is again given by Equation (9.38):

\left(\frac{\Delta P_{f}}{L}\right)_{t p}=\phi_{L O}^{2}\left(\frac{\Delta P_{f}}{L}\right)_{L O}=16.00 \times 0.001136 \cong 0.018  psi / ft

The predictions of the Friedel and MSH correlations are in very close agreement, while the Chisholm method predicts a value about 20% higher and the Lockhart–Martinelli prediction is about 80% higher.