Question

A liquid of density 900 kg m\(^{-3}\) is flowing over a flat plate with a free stream velocity of 0.1 m s\(^{-1}\). The laminar boundary layer thickness at a distance of 0.2 m from the leading edge of the plate is 0.007 m. The viscosity of the liquid in centipoise is ................... (round off to 2 decimal places).
Given: 1 centipoise = 10\(^{-3}\) kg m\(^{-1}\)s\(^{-1}\)

Hint:

Make sure you use the correct formula for boundary layer thickness. The constant is approximately 5.0 for the 99% thickness definition (\(\delta_{99}\)). Memorize this formula: \( \delta/x = 5.0 / \sqrt{Re_x} \). It's a very common question type.

Answer 1

Step 1: Understanding the Concept:
This problem involves calculating the viscosity of a fluid based on the thickness of the laminar boundary layer formed on a flat plate. We need to use the Blasius solution for the boundary layer thickness.
Step 2: Key Formula or Approach:
For a laminar boundary layer on a flat plate, the thickness (\(\delta\)) at a distance \(x\) from the leading edge is given by the formula: \[ \delta \approx \frac{5.0 x}{\sqrt{Re_x}} \] where \(Re_x\) is the local Reynolds number, defined as: \[ Re_x = \frac{\rho U x}{\mu} \] Here:
- \(\rho\) = density of the liquid
- \(U\) = free stream velocity
- \(x\) = distance from the leading edge
- \(\mu\) = dynamic viscosity of the liquid
We can substitute the expression for \(Re_x\) into the formula for \(\delta\) and solve for \(\mu\).
Step 3: Detailed Calculation:
First, let's rearrange the formula to solve for viscosity (\(\mu\)). \[ \delta = \frac{5.0 x}{\sqrt{\frac{\rho U x}{\mu}}} = 5.0 x \sqrt{\frac{\mu}{\rho U x}} \] Square both sides: \[ \delta^2 = 25.0 x^2 \left( \frac{\mu}{\rho U x} \right) = \frac{25.0 x \mu}{\rho U} \] Now, isolate \(\mu\): \[ \mu = \frac{\delta^2 \rho U}{25.0 x} \] Given values:
- \(\rho = 900\) kg m\(^{-3}\)
- \(U = 0.1\) m s\(^{-1}\)
- \(x = 0.2\) m
- \(\delta = 0.007\) m
Substitute the values into the equation for \(\mu\): \[ \mu = \frac{(0.007)^2 \times 900 \times 0.1}{25.0 \times 0.2} \] \[ \mu = \frac{0.000049 \times 90}{5} = \frac{0.00441}{5} = 0.000882 \text{ kg m}^{-1}\text{s}^{-1} \] The question asks for the viscosity in centipoise (cP). Given: 1 cP = \(10^{-3}\) kg m\(^{-1}\)s\(^{-1}\). To convert, we divide our result by \(10^{-3}\): \[ \text{Viscosity in cP} = \frac{0.000882}{10^{-3}} = 0.882 \text{ cP} \] Rounding off to 2 decimal places, the viscosity is 0.88 cP.
Step 4: Final Answer:
The viscosity of the liquid is 0.88 centipoise.
Step 5: Why This is Correct:
The solution correctly uses the standard formula for laminar boundary layer thickness, rearranges it to solve for viscosity, and substitutes the given values. The final conversion to centipoise and rounding are also performed correctly.