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In the case of turbulent flow through a horizontal isothermal cylinder of diameter ‘D’, the free convection heat transfer coefficient for the cylinder will


1. be independent of diameter
2. vary as D3/4
3. vary as D1/4
4. vary as D1/2

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Correct Answer - Option 1 : be independent of diameter

Explanation:

For Turbulent flow in natural convection, the Nusselt number (Nu) is given by:

Nu = C [Gr, Pr]1/3

The Grashoff number (Gr) is given by

\(\left( {Gr} \right) = \frac{{g \beta \left( {{\rm{\Delta }}T} \right)L_e^3}}{{{\nu ^2}}}\)

i.e. Gr ∝ L3

where L = characteristic length.

For various setups, the characteristic length changes and some are given below:

For vertical plate = L (length of the plate)

For Vertical cylinder = L (length of the cylinder)

For Horizontal cylinder = D (diameter of the cylinder)

For square plate - 0.25 a (a is side of the square plate)

Now;

Nu = C [Gr, Pr]1/3

\(\frac{{hD}}{k} = C{\left[ {\frac{{g \beta \left( {{\rm{\Delta }}T} \right){D^3}}}{{{\nu ^2}}}} \right]^{\frac{1}{3}}}{\left( {Pr} \right)^{1/3}}\)

\(\therefore \frac{{hD}}{k} \propto {D}\)

i.e. h is independent of characteristic length or diameter of the cylinder.

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