Consider Example 14.1 again but assuming that the tube has become rough with build up of scale and is characterized by a roughness of ε =0.05mm. Calculate the pressure drop per meter length in this case if all other data remains the same.

Question

Consider Example 14.1 again but assuming that the tube has become rough with build up of scale and is characterized by a roughness of [latex]\epsilon=0.05 mm[/latex]. Calculate the pressure drop per meter length in this case if all other data remains the same.

Accepted Answer

The friction factor alone changes in this case as compared to the previous example. With [latex]\epsilon=0.05 mm[/latex] , we have [latex]\frac{\epsilon}{D}=\frac{0.05}{19}=0.0026[/latex]. The friction factor calculation is based on the Colebrook–White equation. The initial value is calculated using Eq. 14.26 as

[latex]f_{0}=\frac{1}{4}\left[\log \left(\frac{\epsilon}{D}+\frac{21.238}{R e_{D}^{0.9}} ight)-0.5682 ight]^{-2}[/latex] (14.26)

[latex]f_{0}=\frac{1}{4}\left[\log \left(\frac{0.05}{19}+\frac{21.238}{111124^{0.9}} ight)-0.5682 ight]^{-2}=0.0267[/latex]

Using this in Eq. 14.25, we get a better value of friction factor given by

[latex]\frac{1}{\sqrt{f}}=1.14-2 \log \left(\frac{\epsilon}{D}+\frac{9.35}{R e_{D} \sqrt{f}} ight)[/latex] (14.25)

[latex]\frac{1}{\sqrt{f}}=1.14-2 \log \left(\frac{0.05}{19}+\frac{9.35}{111124 imes \sqrt{0.0267}} ight)=6.1396[/latex]

or

[latex]f=\frac{1}{6.1396^{2}}=0.0265[/latex]

The reader may note that this also agrees with Moody chart. The pressure drop per meter of tube may hence be calculated as

[latex]-\frac{\Delta p}{L}=\frac{f ho U^{2}}{2 D}=\frac{0.0265 imes 2.791 imes 37.91^{2}}{2 imes 0.019}=2797 Pa / m[/latex]

Q. 14.2

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