Question:

A viscous liquid is pumped through a pipe network in a chemical plant. The annual pumping cost per unit length of pipe is given by \[ C_{\text{pump}} = \frac{48.13 q^2 \mu}{D^4} \] The annual cost of the installed piping system per unit length of pipe is given by \[ C_{\text{piping}} = 45.92 D \] Here, D is the inner diameter of the pipe in meter, q is the volumetric flowrate of the liquid in m\(^3\)s\(^{-1}\) and \(\mu\) is the viscosity of the liquid in Pa.s. If the viscosity of the liquid is \(20 \times 10^{-3}\) Pa.s and the volumetric flow rate of the liquid is \(10^{-4}\) m\(^3\)s\(^{-1}\), the economic inner diameter of the pipe is \(\underline{\hspace{1cm}}\) meter (round off to 3 decimal places).

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For economic pipe diameter, equate pumping and installation costs, then solve for diameter.
Updated On: Jan 2, 2026
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Correct Answer: 0.014

Solution and Explanation

First, equate the pumping cost and piping cost: \[ C_{\text{pump}} = C_{\text{piping}} \] \[ \frac{48.13 q^2 \mu}{D^4} = 45.92 D \] Substitute given values \(q = 10^{-4}\) m\(^3\)s\(^{-1}\), \(\mu = 20 \times 10^{-3}\) Pa.s: \[ \frac{48.13 (10^{-4})^2 (20 \times 10^{-3})}{D^4} = 45.92 D \] Solve for \(D\): \[ D^5 = \frac{48.13 \times (10^{-8}) \times (20 \times 10^{-3})}{45.92} \] \[ D^5 \approx 1.98 \times 10^{-11} \] Taking the fifth root: \[ D \approx 0.014 \text{ to } 0.016 \text{ m} \] Final answer: 0.014 to 0.016.
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