Question

For an application where the Reynolds number is to be kept constant, a liquid with a density of 1 g cm\(^-3\) and viscosity of 0.01 Poise results in a characteristic speed of 1 cm s\(^-1\). If this liquid is replaced by another with a density of 1.25 g cm\(^-3\) and viscosity of 0.015 Poise, the characteristic velocity will be ......... cm s\(^-1\) (rounded off to one decimal place).

Hint:

When maintaining constant Reynolds number, the characteristic velocity of a fluid is inversely proportional to the viscosity and directly proportional to the density. Use the Reynolds number equation to solve for velocity when changing fluid properties.

Answer 1

The Reynolds number \( Re \) is a dimensionless quantity that helps predict flow patterns in different fluid flow situations. It is given by the equation:

\[ Re = \frac{\rho u L}{\mu} \] Where:
- \( \rho \) is the density of the fluid,
- \( u \) is the characteristic velocity,
- \( L \) is the characteristic length,
- \( \mu \) is the dynamic viscosity.

In this problem, we are asked to maintain the Reynolds number constant. Since the Reynolds number is constant, we can use a proportionality relationship between the two fluids. Let’s set up the following equation based on the given fluids:

\[ \frac{\rho_1 u_1}{\mu_1} = \frac{\rho_2 u_2}{\mu_2} \] Where:
- \( \rho_1 = 1 \, {g/cm}^3 \),
- \( \mu_1 = 0.01 \, {Poise} \),
- \( u_1 = 1 \, {cm/s} \) (characteristic velocity of the first fluid),
- \( \rho_2 = 1.25 \, {g/cm}^3 \),
- \( \mu_2 = 0.015 \, {Poise} \) (viscosity of the second fluid).

Now, solving for the characteristic velocity \( u_2 \) of the second fluid:

\[ u_2 = u_1 \frac{\rho_1}{\rho_2} \frac{\mu_2}{\mu_1} \] Substituting the known values:

\[ u_2 = 1 \times \frac{1}{1.25} \times \frac{0.015}{0.01} = 1.2 \, {cm/s} \] Thus, the characteristic velocity of the second fluid is \( \mathbf{1.2 \, {cm/s}} \), which is slightly higher than that of the first fluid due to the increased viscosity and density of the second fluid.