Question
Download Solution PDFFor flow over a flat plate the hydrodynamic boundary layer thickness is 0.5 mm. The dynamic viscosity is 25 × 10-6 Pa-s, specific heat is 2.0 kJ/kgK and thermal conductivity is 0.05 W/mK. The thermal boundary layer thickness would be
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Prandtl Number: It is defined as the ratio of momentum diffusivity to thermal diffusivity.
\(Pr = \frac{\nu }{\alpha } = \frac{{momentum\;diffusivity}}{{Thermal\;diffusivty}} = \frac{{\frac{\mu }{\rho }}}{{\frac{k}{{{c_p}\rho }}}} = \frac{{\mu {c_p}}}{k}\)
The relationship between the thermal boundary layer and the hydrodynamic boundary layer is given by the Prandtl number.
\({δ _t} = \frac{δ }{{{{\left( {Pr} \right)}^{\frac{1}{3}}}}}\)
where δ = hydrodynamic boundary layer thickness and δt = thermal boundary layer thickness.
Calculation:
Given:
δ = 0.5 mm.
\(Pr = \frac{{\mu {C_p}}}{k} = \frac{{25 \times {{10}^{ - 6}} \times 2 \times 10^3}}{{0.05}} = 1\)
\({δ _t} = \frac{δ }{{{{\left( {Pr} \right)}^{\frac{1}{3}}}}}\)
δt = δ = 0.5 mm.
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