Saturated steam at 1 atmosphere pressure passes through a smooth copper tube of inside diameter D = 18 mm with a mean velocity of Ug = 20 m/s. The copper tube is maintained at a temperature of Tw = 90 °C. The tube length is L = 0.5 m. What is the condensate mass flow rate at x = L if film

Question

Saturated steam at 1 atmosphere pressure passes through a smooth copper tube of inside diameter D = 18 mm with a mean velocity of [latex]U_{g}=20 m / s[/latex]. The copper tube is maintained at a temperature of [latex]T_{w}=90^{\circ} C[/latex]. The tube length is L = 0.5 m. What is the condensate mass flow rate at x = L if film condensation takes place at the tube surface. Take in to account the shear exerted by the flow on the condensate film. Express the results in suitable non-dimensional form. What will be your answers in case the vapor is stationary.

Accepted Answer

Step 1 The required properties are taken either from steam tables or table of properties of water. From steam tables we have

Properties of steam

Saturation tempe: [latex]T_{s}=100{ }^{\circ} C[/latex]

Enthalpy of vaporization: [latex]h_{f g}=2.257 imes 10^{6} J / kg[/latex]

Density of vapor: [latex] ho_{g}=0.5982 kg / m ^{3}[/latex]

Viscosity of steam: [latex]\mu_{g}=1.2 imes 10^{-5} kg / m s[/latex]

Properties of water

Density of water: [latex] ho_{f}=958.1 kg / m ^{3}[/latex]

Thermal conductivity of water: [latex]k_{f}=0.680 W / m ^{\circ} C[/latex]

Dynamic viscosity of water: [latex]\mu_{f}=0.000279 kg / m s[/latex]

Prandtl number of water: [latex]\operatorname{Pr}_{f}=1.73[/latex]

Step 2 With steam velocity of [latex]U_{g}=20 m / s[/latex] the Reynolds number of steam may be calculated as

[latex]R e_{g}=\frac{ ho_{g} D U_{g}}{\mu_{g}}=\frac{0.5982 imes 0.018 imes 20}{1.2 imes 10^{-5}}=17946[/latex]

The steam flow is turbulent.

Step 3 The friction factor is calculated based on Eq. 14.24(a) as

[latex]f=\frac{0.316}{R e_{D}^{0.25}} ext { for } R e_{D}<2 imes 10^{4}[/latex] (14.24)(a)

[latex]f=0.316 imes 17946^{-0.25}=0.0273[/latex]

The shear stress imposed by steam flow on the condensate may then be obtained as

[latex] au_{\delta}=\frac{f ho_{g} U_{g}^{2}}{8}=\frac{0.0273 imes 0.5982 imes 20^{2}}{8}=0.817 Pa[/latex]

Step 4 With x = L = 0.5 m the equation for [latex]\delta[/latex] given by Eq. 17.76 takes the form

[latex]\delta^{4}+\frac{4}{3} \frac{ au_{\delta} \delta^{3}}{ ho_{f} g}=\frac{4 k_{f} \mu_{f}\left(T_{s}-T_{w} ight) x}{ ho_{f}^{2} g h_{f g}}[/latex] (17.76)

[latex]\delta_{L}^{4}+0.0001118 \delta_{L}^{3}=3.7632 imes 10^{-16}[/latex]

This equation is easily solved to get [latex]\delta_{L}=9.59 imes 10^{-5} m[/latex].

Step 5 The mean flow velocity of the condensate is calculated using Eq. 17.75 as

[latex]U=\int_{0}^{1}\left[\frac{ ho_{f} g \delta^{2}}{\mu_{f}}\left(\eta-\frac{\eta^{2}}{2} ight)+\frac{ au_{\delta} \delta}{\mu_{f}} \eta ight] d \eta=\frac{ ho_{f} g \delta^{2}}{3 \mu_{f}}+\frac{ au_{\delta} \delta}{2 \mu_{f}}[/latex] (17.75)

[latex]U_{l}=\frac{958.1 imes 9.81 imes\left(9.59 imes 10^{-5} ight)^{2}}{3 imes 2.79 imes 10^{-4}}+\frac{0.817 imes 9.59 imes 10^{-5}}{2 imes 2.79 imes 10^{-4}}=0.2437 m / s[/latex]

The condensate mass flow rate at x = L may then be calculated as

[latex]\dot{m}_{L}= ho_{f} \pi D \delta_{L} U_{l}=958.1 imes \pi imes 0.018 imes 9.59 imes 10^{-5} imes 0.2427=0.00127 kg / s[/latex]

Condensate Reynolds number is obtained as

[latex]R e_{l}=\frac{\dot{m}_{L}}{\pi D \mu_{f}}=\frac{0.00127}{\pi imes 0.018 imes 2.79 imes 10^{-4}}=80.5[/latex]

Since the condensate Reynolds number is less than [latex]256 imes 1.73^{-0.47}=197.9[/latex], the condensate flow is laminar. Hence, the use of the quantities derived earlier are appropriate.

Step 6 Heat removed at the wall between x = 0 and x = L is then given by

[latex]Q_{L}=\dot{m}_{L} h_{f g}=0.00127 imes 2.257 imes 10^{6}=2866.4 W[/latex]

The mean heat transfer coefficient between x = 0 and x = L is then obtained as

[latex]\bar{h}_{L}=\frac{Q_{L}}{\pi D L\left(T_{s}-T_{w} ight)}=\frac{2866.4}{\pi imes 0.018 imes 0.5(100-90)}=10137.8 W / m ^{2}{ }^{\circ} C[/latex]

Step 7 Now we shall introduce the appropriate non-dimensional variables to represent the results. Characteristic length is given by

[latex]L_{c h}=\left(\frac{\mu_{f}^{2}}{g ho_{f}^{2}} ight)^{\frac{1}{3}}=\left(\frac{\left(2.79 imes 10^{-4} ight)^{2}}{9.81 imes 958.1^{2}} ight)^{\frac{1}{3}}=2.0523 imes 10^{-5} m[/latex]

The average Nusselt number is then calculated as

[latex]\overline{N u}_{L}=\frac{\bar{h}_{L} L_{c h}}{k_{f}}=\frac{10137.8 imes 2.0523 imes 10^{-5}}{0.68}=0.306[/latex]

Non-dimensional shear stress is given by

[latex] au_{\delta}^{*}=\frac{ au_{\delta}}{ ho_{f} L_{c h} g}=\frac{0.817}{958.1 imes 2.0523 imes 10^{-5} imes 9.81}=4.24[/latex]

Thus, the results are depicted as

[latex]R e_{l}=80.5, au_{\delta}^{*}=4.24, \overline{N u}_{L}=0.306[/latex]

Step 8 The reader may calculate the results when the vapor is stationary to get

[latex]R e_{l}=61.6[/latex] , [latex] au_{\delta}^{*}=0[/latex] , [latex]\overline{N u}_{L}=0.234[/latex]

Step 9 Thus, heat transfer is improved by a factor of [latex]\frac{0.306}{0.234}=1.31[/latex] or some 31% when the vapor is flowing as compared to the stationary vapor case.

Question 17.10: Saturated steam at 1 atmosphere pressure passes through a sm...

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