Question
Download Solution PDFThe ratio of the thickness of the thermal boundary layer to the thickness of the hydrodynamic boundary layer is equal to (Prandtl number)n, where n is______.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFδ = the thickness of the hydrodynamic boundary layer; the region of flow where the velocity is less than 99% of the far-field velocity.
δT = the thickness of the thermal boundary layer; the region of flow where the local temperature nearly reaches the value (99%) of the bulk flow temperature
\(\begin{array}{l} \frac{\delta }{{{\delta _T}}} = {\left( {{\rm{Pr}}} \right)^{\frac{1}{3}}}\;\\ \frac{{{\delta _T}}}{\delta } = {\left( {{\rm{Pr}}} \right)^{ - \frac{1}{3}}} \Rightarrow n = - \frac{1}{3} \end{array}\)
Last updated on May 28, 2025
-> SSC JE ME Notification 2025 will be released on June 30.
-> The SSC JE Mechanical engineering application form will be available from June 30 to July 21.
-> SSC JE 2025 CBT 1 exam for Mechanical Engineering will be conducted from October 2 to 31.
-> SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.