Discussion
Consider steady flow of an incompressible fluid through two long and straight pipes of diameters \(d_{1}\) and \(d_{2}\) arranged in series. Both pipes are of equal length and the flow is turbulent in both pipes. The friction factor for turbulent flow though pipes is of the form, \(f = K(\text{Re})^{-n}\), where K and n are known positive constants and Re is the Reynolds number. Neglecting minor losses, the ratio of the frictional pressure drop in pipe 1 to that in pipe 2, \(\left(\frac{\Delta P_{1}}{\Delta P_{2}}\right)\), is given by
Answer: A
The ratio is \(\left(\frac{d_{2}}{d_{1}}\right)^{(5-n)}\).
The Darcy-Weisbach equation gives \(\Delta P = f\frac{L}{d}\frac{\rho V^{2}}{2}\). With the same flow rate in both pipes, \(V \propto 1/d^{2}\) and \(Re \propto Vd \propto 1/d\), so \(f \propto Re^{-n} \propto d^{n}\):
\[ \begin{aligned} \Delta P &\propto d^{n} \cdot \frac{1}{d} \cdot \frac{1}{d^{4}} = d^{\,n-5} \\ \frac{\Delta P_{1}}{\Delta P_{2}} &= \left(\frac{d_{1}}{d_{2}}\right)^{n-5} = \left(\frac{d_{2}}{d_{1}}\right)^{5-n} \end{aligned} \]
The exponent 5 alone is the constant-friction-factor result, (5 + n) has the Reynolds-number effect in the wrong direction, and (3 - n) mishandles the velocity dependence.
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