Question

If the frequency of the stirrer in a mixing tank is increased by a factor of 2 while all other parameters are kept constant, by what factor is the power requirement increased at high Reynolds number?

• A. 4
• B. 8
• C. 16
• D. 32

Hint:

Mixing Vessel Scaling Rules: - Laminar Regime (\(Re < 10\)): \(N_P \propto \frac{1}{Re} \implies P \propto N^2\) (Power depends on speed squared). - Turbulent Regime (\(Re > 10^4\)): \(N_P = \text{constant} \implies P \propto N^3\) (Power depends on speed cubed).

Answer 1

The Correct Option is B

Concept: In agitated mixing vessels, the relationship between fluid properties, tank geometry, stirrer speed, and power consumption is analyzed using dimensionless groups. The key dimensionless groups involved are:
• Power Number (\(N_P\)): \( N_P = \frac{P}{\rho \cdot N^3 \cdot D^5} \)
• Reynolds Number for Agitation (\(Re_{\text{agitation}}\)): \( Re = \frac{\rho \cdot N \cdot D^2}{\mu} \) Where \(P\) is the power requirement, \(\rho\) is the fluid density, \(N\) is the stirrer rotational frequency, \(D\) is the impeller diameter, and \(\mu\) is the dynamic viscosity.

Step 1: Analyzing power behavior at a high Reynolds number.
At high Reynolds numbers (\(Re > 10^4\)), the mixing tank operates in a fully turbulent flow regime. In this turbulent regime, inertial forces dominate over viscous forces, and the Power Number (\(N_P\)) becomes constant, independent of changes to the Reynolds number: \[ N_P = C \quad \text{(where } C \text{ is a constant depending only on tank geometry)} \]

Step 2: Setting up the proportional relationship for power.
Using the constant Power Number assumption, we can isolate the power parameter \(P\): \[ \frac{P}{\rho \cdot N^3 \cdot D^5} = C \implies P = C \cdot \rho \cdot N^3 \cdot D^5 \] Since the problem states that all other parameters (fluid density \(\rho\), impeller diameter \(D\), and tank geometry \(C\)) are kept strictly constant, the power requirement \(P\) is directly proportional to the cube of the stirrer frequency \(N\): \[ P \propto N^3 \]

Step 3: Calculating the final change factor.
Let the initial frequency be \(N_1\) and the new frequency be \(N_2 = 2N_1\). We can set up a ratio to find the new power requirement \(P_2\): \[ \frac{P_2}{P_1} = \left( \frac{N_2}{N_1} \right)^3 \] Substitute \(N_2 = 2N_1\) into this ratio: \[ \frac{P_2}{P_1} = \left( \frac{2N_1}{N_1} \right)^3 = (2)^3 = 8 \] \[ P_2 = 8 \cdot P_1 \] This shows that doubling the stirrer frequency under fully turbulent conditions increases the total power requirement by a factor of 8, which matches Option (B).