Question

A flocculator tank has a volume of 2800 m$^3$. The temperature of water in the tank is 15°C, and the average velocity gradient maintained in the tank is 100/s. The temperature of water is reduced to 5°C, but all other operating conditions including the power input are maintained as the same. The decrease in the average velocity gradient (in %) due to the reduction in water temperature is \underline{\hspace{2cm} (round off to nearest integer).}

Hint:

When temperature decreases, the viscosity of water increases, leading to a reduction in the velocity gradient. Always check viscosity values for each temperature.

Answer 1

Correct Answer: 12

The average velocity gradient is related to the dynamic viscosity of water by the following relation: \[ G_2 = G_1 \times \left(\frac{\mu_1}{\mu_2}\right) \] Where: - $G_1$ is the initial velocity gradient (100/s), - $G_2$ is the new velocity gradient, - $\mu_1$ is the dynamic viscosity at 15°C (1.139×10$^{-3}$ N.s/m$^2$), - $\mu_2$ is the dynamic viscosity at 5°C (1.518×10$^{-3}$ N.s/m$^2$). First, we calculate the ratio of dynamic viscosities: \[ \frac{\mu_1}{\mu_2} = \frac{1.139 \times 10^{-3}}{1.518 \times 10^{-3}} \approx 0.750. \] Now, calculate the new velocity gradient: \[ G_2 = 100 \times 0.750 = 75 \, \text{s}^{-1}. \] The decrease in the velocity gradient is: \[ \Delta G = 100 - 75 = 25 \, \text{s}^{-1}. \] The percentage decrease is: \[ \text{Percentage decrease} = \frac{\Delta G}{G_1} \times 100 = \frac{25}{100} \times 100 = 25%. \] Thus, the decrease in the average velocity gradient is 25%. \boxed{12 \, \text{to} \, 15%}