Question

In a dairy processing plant, milk enters a 30 m long and 2 cm diameter tube at 60°C and leaves at 57°C. The total heat loss over the tube length is 381.15 W. The specific heat capacity, density, and viscosity of milk are 3.85 kJ kg$^{-1}$ K$^{-1}$, 1020 kg m$^{-3}$, and 1.20 cP, respectively. The Reynolds number for the flow is _________ (rounded off to the nearest integer). Given: Acceleration of gravity \( g = 9.81 \ \text{m/s}^2 \) and \( \pi = 3.14 \)

Hint:

The Reynolds number is crucial for determining whether the flow is laminar or turbulent. A higher Reynolds number indicates a more turbulent flow.

Answer 1

Correct Answer: 1700

First, calculate the Reynolds number using the formula: \[ Re = \frac{\rho \cdot v \cdot D}{\mu} \] Where:
- \( \rho = 1020 \ \text{kg/m}^3 \) (density),
- \( D = 0.02 \ \text{m} \) (diameter),
- \( \mu = 1.20 \times 10^{-3} \ \text{kg/(m·s)} \)
(viscosity), - \( v \) is the velocity of the milk.
From the heat loss equation: \[ Q = \dot{m} \cdot c_p \cdot \Delta T \] Where:
- \( \dot{m} \) is the mass flow rate,
- \( c_p = 3.85 \ \text{kJ/kg·K} \),
- \( \Delta T = 60^\circ C - 57^\circ C = 3^\circ C \).
The mass flow rate is: \[ \dot{m} = \frac{Q}{c_p \cdot \Delta T} = \frac{381.15}{3.85 \times 3} = 33.06 \ \text{kg/s} \] Now, calculate the velocity \( v \) using the mass flow rate formula: \[ \dot{m} = \rho \cdot v \cdot A \] Where \( A \) is the cross-sectional area of the tube: \[ A = \pi \times \left(\frac{D}{2}\right)^2 = \pi \times \left(\frac{0.02}{2}\right)^2 = 3.14 \times 0.0001 = 3.14 \times 10^{-5} \ \text{m}^2 \] Now, solve for \( v \): \[ 33.06 = 1020 \times v \times 3.14 \times 10^{-5} \] \[ v = \frac{33.06}{1020 \times 3.14 \times 10^{-5}} = \frac{33.06}{0.03198} = 1035.1 \ \text{m/s} \] Finally, calculate the Reynolds number: \[ Re = \frac{1020 \cdot 1035.1 \cdot 0.02}{1.20 \times 10^{-3}} = \frac{21142.08}{0.0012} = 17618.4 \] Thus, the Reynolds number for the flow is: \[ \boxed{17619} \]