Steam at atmospheric pressure condenses over a horizontal tube of D = 18 mm maintained at a temperature of Tw = 90 °C over a length of L = 2 m. Determine the rate of condensation based on Nusselt analysis.

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Accepted Answer

The required properties are taken either from steam tables or table of properties of water.

Properties of steam

Saturation pressure: [latex]p_{s}=1.014[/latex] bar absolute

Enthalpy of vaporization: [latex]h_{f g}=2.256 imes 10^{6} J / kg[/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]

Geometric parameters specified are

Diameter of tube: D = 18 mm = 0.018 m

Tube length: L = 2 m

The temperatures are specified as

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

Tube wall temperature: [latex]T_{w}=90^{\circ} C[/latex]

The average heat transfer coefficient is calculated using Eq. 17.69 as

[latex]\bar{h}_{h}=0.728\left[\frac{ ho_{f}^{2} g h_{f g} k_{f}^{3}}{\mu_{f}\left(T_{s}-T_{w} ight) D} ight][/latex] (17.69)

[latex]\bar{h}_{h}=0.728\left[\frac{958.1^{2} imes 9.81 imes 2.256 imes 10^{6} imes 0.680^{3}}{0.000279 imes(100-90) imes 0.018} ight]=13749.9 W / m ^{2}{ }^{\circ} C[/latex]

Heat transfer area is calculated as

[latex]A=\pi D L=\pi imes 0.018 imes 2=0.1131 m ^{2}[/latex]
The total heat transfer at the tube wall is calculated as

[latex]Q_{w}=\bar{h}_{h} A\left(T_{s}-T_{w} ight)=13749.9 imes 0.1131 imes(100-90)=15550.8 W[/latex]

This should equal the heat released by condensing steam. Hence, the condensate mass flow rate is
[latex]\dot{m}=\frac{Q_{w}}{h_{f g}}=\frac{15550.8}{2.256 imes 10^{6}}=0.00689 kg / s =24.8 kg / h[/latex]

Question 17.9: Steam at atmospheric pressure condenses over a horizontal tu...

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