Question

For flow over a flat plate, the hydrodynamic boundary layer thickness is 0.5 mm. If the dynamic viscosity is $ 25 \times 10^{-16} \, \text{Pa-s} $, specific heat is 2000 J/kg·K, and the thermal conductivity is 0.05 W/m·K, then what would be the thermal boundary layer thickness?

• A. 0.7 mm
• B. 0.5 mm
• C. 0.3 mm
• D. 0.1 mm

Hint:

In boundary layer analysis, use the relation \( \delta_t = \delta_h \left( \frac{\nu}{\alpha} \right)^{1/2} \) to estimate the thermal boundary layer thickness from the hydrodynamic thickness.

Answer 1

The Correct Option is B

Step 1: Relation between Hydrodynamic and Thermal Boundary Layer Thickness
For flow over a flat plate, the thermal boundary layer thickness (\( \delta_t \)) is related to the hydrodynamic boundary layer thickness (\( \delta_h \)) by the following relation: \[ \delta_t = \delta_h \left( \frac{\nu}{\alpha} \right)^{1/2} \] Where:
\( \delta_h \) is the hydrodynamic boundary layer thickness.
\( \nu \) is the kinematic viscosity (\( \nu = \frac{\mu}{\rho} \)).
\( \alpha \) is the thermal diffusivity (\( \alpha = \frac{k}{\rho C_p} \)).
Step 2: Given Data
\( \delta_h = 0.5 \, \text{mm} = 0.0005 \, \text{m} \),
\( \mu = 25 \times 10^{-16} \, \text{Pa-s} \),
\( k = 0.05 \, \text{W/m·K} \),
\( C_p = 2000 \, \text{J/kg·K} \).
Step 3: Calculate Thermal Boundary Layer Thickness
Since we don't have the density (\( \rho \)), we can use the approximate relation: \[ \frac{\nu}{\alpha} = \frac{\mu C_p}{k} \] Substituting the values: \[ \frac{\mu C_p}{k} = \frac{25 \times 10^{-16} \times 2000}{0.05} = 1 \times 10^{-12} \] Now, substitute into the formula for \( \delta_t \): \[ \delta_t = 0.0005 \times (1 \times 10^{-12})^{1/2} = 0.0005 \times 10^{-6} = 0.5 \, \text{mm} \] Thus, the thermal boundary layer thickness is \( \boxed{0.5 \, \text{mm}} \).