Question 9.4: The so-called skin friction coefficient cf is defined as the...

The so-called skin friction coefficient c_{f} is defined as the wall shear stress divided by the mean dynamic pressure. Find a formula for c_{f} for a steady flow in a rigid tube.

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The dynamic pressure is defined as \rho \overline{\nu }^{2}/2, where \overline{\nu } is a scalar measure of the mean velocity [i.e., the speed; see Bernoulli’s equation (8.80)]. For the case of a tube flow, therefore, the mean dynamic pressure is Some Exact Solutions

                               P_{dyn}=\frac{1}{2}\rho \overline{\nu }^{2}=\frac{1}{2}\rho \left(-\frac{a^{2}}{8\mu }\frac{dp}{dz} \right)^{2},

where the mean velocity is given by Eq. (9.50). Hence, in this case (Fung 1993),

                  \overline{\nu }=\frac{Q}{A}=\left(-\frac{\pi a^{4}}{8\mu }\frac{dp}{dz} \right)\left(\frac{1}{\pi a^{2}} \right)=-\frac{a^{2}}{8\mu }\frac{dp}{dz}.        (9.50)

 

                             c_{f}=-\frac{a}{2}\left(\frac{dp}{dz} \right)\left\{\frac{\rho }{2}\left[\frac{a^{4}}{64\mu ^{2}}\left(\frac{dp}{dz} \right)^{2} \right] \right\}^{-1}=-64\mu ^{2}\left[\rho a^{3}\left(\frac{dp}{dz} \right) \right]^{-1},

which, via Eq. (9.54), can be written as

\frac{dp}{dz}=-\frac{8\mu Q}{\pi a^{4}}     or    \frac{dp}{dz}=-\frac{8\mu \overline{\nu } }{a^{2}};            (9.54)

 

                           c_{f}=\frac{-64\mu ^{2}}{\rho a^{3}(-8\mu \overline{\nu }/a^{2} )}=\frac{16}{\rho (2a)\overline{\nu }/\mu }=\frac{16}{Re},

where the Reynolds’ number is

                                                                               Re=\frac{\rho \overline{\nu }d }{\mu },

with d=2a the diameter of the tube. Re is a very important nondimensional parameter in fluid mechanics; it will be discussed in greater depth in Chap. 10. In summary, the wall shear stress in this tube flow can also be written as

                                                                \tau _{w}=\frac{1}{2}\rho \overline{\nu }^{2}c_{f}=\frac{1}{2}\rho \overline{\nu }^{2}\left(\frac{16}{Re} \right).

For example, in the human aorta,

\overline{\nu }=0.15  m/s,            d=0.03  m
\rho =1,060   kg/m^{3},           \mu =3.3\times 10^{-3}  Ns/m^{2};

hence,

                                     Re=\frac{(1,060  kg/m^{3})(0.15  m/s)(0.03  m)}{3.3\times 10^{-3}  N  s/m^{2}}=1,445,

where 1   kg   m/s^{2}=1 N. This value is less than that expected for turbulent flow (Re\gt 2,100); hence, our laminar assumption applies. Note, too, that the asso-ciated value of \tau _{w}=0.13  kg/m  s^{2}=0.13   Pa, which is within the reported range albeit lower than that which is usually reported (1.5 Pa).

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