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Chapter 14

Q. 14.2

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 \epsilon=0.05 mm. Calculate the pressure drop per meter length in this case if all other data remains the same.

Step-by-Step

Verified Solution

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

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

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

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

\frac{1}{\sqrt{f}}=1.14-2 \log \left(\frac{\epsilon}{D}+\frac{9.35}{R e_{D} \sqrt{f}}\right)  (14.25)

\frac{1}{\sqrt{f}}=1.14-2 \log \left(\frac{0.05}{19}+\frac{9.35}{111124 \times \sqrt{0.0267}}\right)=6.1396

or

f=\frac{1}{6.1396^{2}}=0.0265

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

-\frac{\Delta p}{L}=\frac{f \rho U^{2}}{2 D}=\frac{0.0265 \times 2.791 \times 37.91^{2}}{2 \times 0.019}=2797 Pa / m