Question 9.10: Non-Newtonian Pipe Flow 2 A gelatinous foodstuff is to be co......
Non-Newtonian Pipe Flow 2
A gelatinous foodstuff is to be conveyed from a holding vessel to a packaging line along a 10-m straight pipe run using pressurised sterile air above the material as a propellant to provide a volumetric flow rate to the packaging line of 5 m³h^{–1} . Care is required in the transportation since the texture of the foodstuff is known to be permanently degraded if the bulk of the material is subjected to high shear rates that exceed 250 s^{–1} . Using data that were obtained with a laboratory test using a capillary viscometer with a diameter of 2.5 mm and length of 500 mm, determine the minimum acceptable inside diameter of the transfer pipe and the pressure drop along the pipe if this diameter is to be used.
\begin{matrix} \dot{Q} \ (m^3 \times 10^{-6}) min^{-1} & 1.70 & 3.70 & 9.78 & 15.45 & 19.80 \\ \Delta p \ (Nm^{-2} \times 10^{-5}) & 1.30 & 2.17 & 4.40 & 5.78 & 6.88 \end{matrix}
Assuming that the gelatinous foodstuff is non-Newtonian, the shear rate is given by
{\dot{\gamma}}={\frac{\dot{Q}}{\pi R^{3}}}{\left(3+{\frac{d\ln{\dot{Q}}}{d\ln\Delta p}}\right)} (9.44)
A plot of ln \dot{Q} against ln Δp gives a straight line with a gradient of 1.47. The minimum pipe radius is found to be
R = \sqrt[3]{\frac{\dot{Q}}{\pi \dot\gamma} \left\lgroup 3 + \frac{d \ln \dot Q}{d \ln \Delta p} \right\rgroup } = \sqrt[3]{\frac{5/3600}{\pi \times 250} (3+1.47) } = 0.02 \ m (9.45)
From a force balance on the foodstuff flowing in the pipe, the wall shear stress is
\tau_{\mathrm{w}}={\frac{\Delta p R}{2L}} (9.46)
This value is the same for both the pipe and the capillary viscometer. Using the data for the viscometer, the pressure drop along the capillary would therefore correspond to 1810\times10^{5}~{\mathrm{Nm}}^{-2}, so the pressure drop along the pipe would be
\Delta p_{p i p e}=\frac{L_{p i p e}R_{c a p}}{L_{c a p}R_{p i p e}}\,\Delta p_{c a p} ={\frac{10\times2.5\times10^{-3}}{0.5\times0.02}}\times1810\times10^{5}=452.5\times10^{5}~{\mathrm{Nm}}^{-2} (9.47)