Question

For an application where the Reynolds number is to be kept constant, a liquid with density of 1 g per cubic cm and viscosity of 0.01 Poise results in a characteristic speed of 1 cm per second. If this liquid is replaced by another with density of 1.25 g per cubic cm and viscosity of 0.015 Poise, the characteristic velocity will be ........... cm per second (rounded off to one decimal place). Assume the characteristic length of the flow to be the same in both cases.

Hint:

For constant Reynolds number and same length scale, velocity is proportional to \(\mu / \rho\). So, if viscosity increases or density decreases, velocity must increase to keep Reynolds number unchanged.

Answer 1

Step 1: Recall Reynolds number formula.
The Reynolds number for flow is given by: \[ Re = \frac{\rho \, V \, L}{\mu} \] where - \(\rho\) = density of fluid, - \(V\) = velocity of fluid, - \(L\) = characteristic length, - \(\mu\) = dynamic viscosity of fluid. Step 2: Condition of the problem.
It is stated that the Reynolds number is kept constant. This means: \[ \frac{\rho_1 V_1 L}{\mu_1} = \frac{\rho_2 V_2 L}{\mu_2} \] Since the characteristic length \(L\) is the same in both cases, it cancels out: \[ \frac{\rho_1 V_1}{\mu_1} = \frac{\rho_2 V_2}{\mu_2} \] Step 3: Substitute values for case 1 (original liquid).
- \(\rho_1 = 1 \, \text{g/cm}^3\) - \(\mu_1 = 0.01 \, \text{Poise}\) - \(V_1 = 1 \, \text{cm/s}\) So, \[ \frac{1 \times 1}{0.01} = 100 \] Step 4: Write the equation for case 2 (new liquid).
- \(\rho_2 = 1.25 \, \text{g/cm}^3\) - \(\mu_2 = 0.015 \, \text{Poise}\) - \(V_2 = ?\) From the relation: \[ \frac{\rho_1 V_1}{\mu_1} = \frac{\rho_2 V_2}{\mu_2} \] \[ 100 = \frac{1.25 \, V_2}{0.015} \] Step 5: Solve for \(V_2\).
\[ 100 \times 0.015 = 1.25 \, V_2 \] \[ 1.5 = 1.25 \, V_2 \] \[ V_2 = \frac{1.5}{1.25} = 1.2 \] Step 6: Final answer.
Thus, the characteristic velocity of the second liquid is: \[ \boxed{1.2 \, \text{cm per second}} \]