Chapter 7, Problem 7.30P

To determine

The percent shift in d/x compared to textbook’s formula.

Answer to Problem 7.30P

The percent shift in d/x compared to the textbook’s formula is an average 9.87 increase.

Explanation of Solution

Given Information:

The value of constants k & B are 0.421 & 5.62 respectively.

Concept Used:

$\begin{array}{l}Theequation(7.42)is:\\ {(}^{\frac{2}{c_f}}=2.44\ln \left[{\mathrm{Re}}_{\delta }{(\frac{c_f}{2})}^{1/2}\right]+5.0\\ with\mathrm{mod}ification:\\ {(}^{\frac{2}{c_f}}=\frac{1}{.421}\ln \left[{\mathrm{Re}}_{\delta }{(\frac{c_f}{2})}^{1/2}\right]+5.62\\ \end{array}$

Calculation:

$findthevalueof{c}_{f}forthedifferentvalueof{\mathrm{Re}}_{\delta }$

$(a){\mathrm{Re}}_{\delta }=10000$

${(\frac{2}{c_f})}^{1/2}=\frac{1}{0.421}\ln \left[10000{(\frac{c_f}{2})}^{1/2}\right]+5.62$

$onsolving:$

$c_f=0.004828$

$(b){\mathrm{Re}}_{\delta }=100000$

${(\frac{2}{c_f})}^{1/2}=\frac{1}{0.421}\ln \left[100000{(\frac{c_f}{2})}^{1/2}\right]+5.62$

$onsolving:$

$c_f=0.003129$

$(c){\mathrm{Re}}_{\delta }=1000000$

${(\frac{2}{c_f})}^{1/2}=\frac{1}{0.421}\ln \left[100000{(\frac{c_f}{2})}^{1/2}\right]+5.62$

$onsolving:$

$c_f=0.002167$

$(d){\mathrm{Re}}_{\delta }=10000000$

${(\frac{2}{c_f})}^{1/2}=\frac{1}{0.421}\ln \left[1000000{(\frac{c_f}{2})}^{1/2}\right]+5.62$

$onsolving:$

$c_f=0.001586$

$asfromequation(7.42)weget:$

$c_f=0.0203{\mathrm{Re}}_{\delta }{}^{-0.160}=2\frac{d}{dx}(\frac{7}{72}\delta )$

$\Rightarrow {\mathrm{Re}}_{\delta }{}^{-.0160}=9.58\frac{d({\mathrm{Re}}_{\delta })}{d({\mathrm{Re}}_x)}$

$on\mathrm{int}ergrationweget:$

$\frac{\delta }{x}=\frac{0.1628}{{\mathrm{Re}}_x{}^{0.1379}}$

$thisisquitesimillartotheequation:$

$\frac{\delta }{x}=\frac{0.16}{{\mathrm{Re}}_x{}^{(1/7)}}$

$onsolvingfor{\mathrm{Re}}_{x}for{10}^{6}to{10}^{9}theincreasepercentageare7.62,9.375,10.43and12.048respectively.$

$Thentheaveragepercentageincreasesis9.87nearlytenpercent(10\%)$

Conclusion:

The percent shift in d/x compared to the textbook’s formula is an average 9.87 increase.