The return line from a thermosyphon reboiler consists of 10-in. schedule 40 pipe with an ID of 10.02 in. (0.835 ft). The total mass flow rate in the line is 300,000 lbm/h, of which 60,000 lbm/h (20%) is vapor and 240,000 lbm/h is liquid. Physical properties of the vapor and liquid fractions are

Question

The return line from a thermosyphon reboiler consists of 10-in. schedule 40 pipe with an ID of 10.02 in. (0.835 ft). The total mass flow rate in the line is 300,000 lbm/h, of which 60,000 lbm/h (20%) is vapor and 240,000 lbm/h is liquid. Physical properties of the vapor and liquid fractions are given in the following table:

Property	Liquid	Vapor
μ (cp)	0.177	0.00885
ρ (lbm/ft³)	38.94	0.4787
σ (dyne/cm)	11.4	–

Calculate the friction loss per unit length in the line using:
(a) The Lockhart–Martinelli correlation.
(b) The Chisholm correlation.
(c) The Friedel correlation.
(d) The MSH correlation.

Accepted Answer

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

[latex]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}[/latex]

[latex]R e_{L}=\frac{D_{i} G_{L}}{\mu_{L}}=\frac{0.835 imes 438,277}{0.177 imes 2.419}=854,724[/latex]

[latex]s_{L}= ho_{L} / ho_{ ext {water }}=38.94 / 62.43=0.6237[/latex]

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

[latex]f=0.3673 Re^{-0.2314} [/latex] (4.8)

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

The friction loss per unit length is given by:

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

[latex]\left(\frac{\Delta P_{f}}{L} ight)_{L}=0.000766 psi / ft[/latex]

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

[latex]X_{ tt } \cong\left(\frac{1-x}{x} ight)^{0.9}\left(\frac{ ho_{V}}{ ho_{L}} ight)^{0.5}\left(\frac{\mu_{L}}{\mu_{V}} ight)^{0.1}[/latex]

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

[latex]X_{ tt } \cong 0.521[/latex]

For comparison, the reader can verify that using Equation (9.36) with n = 0.2314 to calculate [latex]X_{tt}[/latex] 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):

[latex]\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[/latex]

[latex]X = \left(\frac{1 - x}{x} ight)^{(2 - n)/2} (\mu_{L}/\mu_{V})^{n/2} ( ho_{V}/ ho_{L})^{0.5}[/latex] (9.36)

The two-phase friction loss is then:

[latex]\left(\frac{\Delta P_{f}}{L} ight)_{ tp }=\phi_{ L }^{2}\left(\frac{\Delta P_{f}}{L} ight)_{L}=43.07 imes 0.000766=0.033 psi / ft[/latex]

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

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

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

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

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

[latex]\left(\frac{\Delta P _{ f }}{ L } ight)_{L O}=0.001136 psi / ft[/latex]

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

[latex]Y=\left( ho_{L} / ho_{V} ight)^{0.5}\left(\mu_{V} / \mu_{L} ight)^{n / 2} = (38.94 / 0.4787)^{0.5}(0.00885 / 0.177)^{0.1157}[/latex]

[latex]Y \cong 6.38[/latex]

Since Y < 9.5, Equation (9.44) gives:

[latex]\begin{matrix}B &=& 1500 / G^{0.5} \quad \quad (028)\end{matrix} [/latex]

[latex]B = 1500/G^{0.5} = 1500 /(547,846)^{0.5} \cong 2.027[/latex]

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

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

= [latex]1+\left[(6.38)^{2}-1 ight]\left\{2.027[0.2(1-0.2)]^{0.8843}+(0.2)^{1.7686} ight\}[/latex]

[latex]\phi_{L O}^{2} \cong 19.22[/latex]

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

[latex]\left(\frac{\Delta P_{f}}{L} ight)_{t p}=\phi_{L O}^{2}\left(\frac{\Delta P_{f}}{L} ight)_{L O}[/latex]

= [latex]19.22 imes 0.001136 \cong 0.022 psi / ft[/latex]

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

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

[latex]F=x^{0.78} (1-x)^{0.24}[/latex] (9.47)

[latex]E=(1-x)^{2}+x^{2}\left(\mu_{V} / \mu_{L} ight)^{n}\left( ho_{L} / ho_{V} ight)[/latex]

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

E = 2.2668

[latex]F=x^{0.78}(1-x)^{0.24}=(0.2)^{0.78}(0.8)^{0.24}=0.2701[/latex]

[latex]H=\left( ho_{L} / ho_{V} ight)^{0.91}\left(\mu_{V} / \mu_{L} ight)^{0.19}\left(1-\mu_{V} / \mu_{L} ight)^{0.7}[/latex]

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

H = 29.896

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

[latex] ho_{t p}=2.2813 lbm / ft ^{3}[/latex]

[latex]F r=\frac{G^{2}}{g D_{i} ho_{t p}^{2}}=\frac{(547,846)^{2}}{4.17 imes 10^{8} imes 0.835(2.2813)^{2}}=165.63[/latex]

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

[latex]\sigma=11.4 dyne / cm imes 6.8523 imes 10^{-5}\left(\frac{ lbf / ft }{ dyne / cm } ight)=7.812 imes 10^{-4} lbf / ft[/latex]

[latex]W e=\frac{G^{2} D_{i}}{g_{c} ho_{ tp } \sigma}=\frac{(547,846)^{2} imes 0.835}{4.17 imes 10^{8} imes 2.2813 imes 7.812 imes 10^{-4}}[/latex]

[latex]W e=337,227[/latex]

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

[latex]=2.2668+\frac{3.24 imes 0.2701 imes 29.896}{(165.63)^{0.045}(337,227)^{0.035}}[/latex]

[latex]\phi_{L O}^{2}=15.58[/latex]

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

[latex]\left(\frac{\Delta P_{f}}{L} ight)_{t p}=\phi_{L O}^{2}\left(\frac{\Delta P_{f}}{L} ight)_{L O}=15.58 imes 0.001136 \cong 0.018 psi / ft[/latex]

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

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

= [latex](6.38)^{2}(0.2)^{3}+\left[1+2 imes 0.2\left((6.38)^{2}-1 ight) ight](1-0.2)^{1 / 3}[/latex]

[latex]\phi_{L O}^{2} \cong 16.00[/latex]

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

[latex]\left(\frac{\Delta P_{f}}{L} ight)_{t p}=\phi_{L O}^{2}\left(\frac{\Delta P_{f}}{L} ight)_{L O}=16.00 imes 0.001136 \cong 0.018 psi / ft[/latex]

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.

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

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