1. Problem 1: Calculate viscosity of honey flowing through a capillary tube. Given: - Flow rate, $Q = 1$ liter/minute $= \frac{1}{1000\times 60} = 1.6667 \times 10^{-5}$ m³/s - Diameter, $D = 2$ cm $= 0.02$ m - Length, $L = 50$ cm $= 0.5$ m - Pressure drop, $\Delta P = 40$ kPa $= 40000$ Pa Step 1: Find the mean velocity $v_m$ using the relation $$v_m = \frac{Q}{A} = \frac{Q}{\pi (\frac{D}{2})^2}$$ Calculate cross-sectional area: $$A = \pi (0.01)^2 = 3.1416 \times 10^{-4} \text{ m}^2$$ Calculate mean velocity: $$v_m = \frac{1.6667 \times 10^{-5}}{3.1416 \times 10^{-4}} = 0.05305 \text{ m/s}$$ Step 2: Calculate viscosity $\mu$ using Hagen-Poiseuille equation: $$\mu = \frac{D^2 \Delta P}{32 L v_m}$$ Substitute values: $$\mu = \frac{(0.02)^2 \times 40000}{32 \times 0.5 \times 0.05305} = \frac{0.0004 \times 40000}{0.8488} = \frac{16}{0.8488} = 18.85 \text{ Pa}\cdot\text{s}$$ \newpage 2. Problem 2: Tomato concentrate laminar flow in a pipe Given: - Diameter, $D = 0.0475$ m - Length, $L = 10$ m - Volumetric flow rate, $Q = 3$ m³/hr $= \frac{3}{3600} = 0.0008333$ m³/s - Consistency index, $K = 18.7$ Pa s$^{0.4}$ - Flow behavior index, $n = 0.4$ (a) Calculate mean velocity $v_m$: $$v_m = \frac{Q}{A} = \frac{Q}{\pi (D/2)^2}$$ Calculate cross-sectional area: $$A = \pi (0.02375)^2 = \pi \times 0.00056406 = 0.001771 \text{ m}^2$$ Calculate mean velocity: $$v_m = \frac{0.0008333}{0.001771} = 0.4705 \text{ m/s}$$ (b) Calculate maximum velocity $v_{max}$ for laminar flow of power-law fluid: For laminar flow of power-law fluid: $$v_{max} = \frac{n+1}{n} v_m = \frac{0.4+1}{0.4} \times 0.4705 = \frac{1.4}{0.4} \times 0.4705 = 3.5 \times 0.4705 = 1.6467 \text{ m/s}$$ (c) Calculate pressure drop $\Delta P$ using power-law fluid pressure drop formula: Pressure drop for power-law fluid: $$\Delta P = \frac{2 K L (8 v_m / D)^n}{D}$$ Calculate shear rate factor: $$\frac{8 v_m}{D} = \frac{8 \times 0.4705}{0.0475} = 79.34 \text{ s}^{-1}$$ Calculate $\Delta P$: $$\Delta P = \frac{2 \times 18.7 \times 10 \times (79.34)^{0.4}}{0.0475}$$ Calculate $(79.34)^{0.4}$: $$79.34^{0.4} = e^{0.4 \ln(79.34)} = e^{0.4 \times 4.374} = e^{1.7496} = 5.75$$ Calculate $\Delta P$: $$\Delta P = \frac{2 \times 18.7 \times 10 \times 5.75}{0.0475} = \frac{2150.5}{0.0475} = 45274 \text{ Pa} = 45.3 \text{ kPa}$$ Comparison: The pressure drop of tomato concentrate is approximately $45.3$ kPa. Refer to Example 5.2 (not provided here) for sucrose solution pressure drop comparison.