A fluid is flowing steadily under laminar conditions over a thin rectangular plate at temperature \(T_s\) as shown. The free-stream velocity and temperature are \(u_\infty\) and \(T_\infty\), respectively. When the fluid flow is only in the \(x\)-direction, \(h_x\) is the local 64ac0493b52af67589bd410c coefficient. Similarly, when the fluid flow is only in the \(y\)-direction, \(h_y\) is the corresponding local 64ac0493b52af67589bd410c coefficient. Use the correlation \(\mathrm{Nu}=0.332\,\mathrm{Re}^{1/2}\,\mathrm{Pr}^{1/3}\) for the local 64ac0493b52af67589bd410c coefficient, where Nu, Re and Pr are the appropriate Nusselt, Reynolds and Prandtl numbers. The average 64ac0493b52af67589bd410c coefficients are defined as \(\displaystyle \overline{h}_l=\frac{1}{l}\int_0^l h_x\,dx\) and \(\displaystyle \overline{h}_w=\frac{1}{w}\int_0^w h_y\,dy\). If \(w=1\ \text{m}\) and \(l=4\ \text{m}\), the value of the ratio \(\overline{h}_w/\overline{h}_l\) is ____________ (in integer).
Show Hint
With \(\mathrm{Nu}_x\propto \mathrm{Re}_x^{1/2}\), the local \(h\) varies as \(x^{-1/2}\) for laminar boundary layers over flat plates.
Averages over a length scale \(L\) then scale as \(1/\sqrt{L}\); ratios reduce to square roots of the respective lengths.
Step 1: For laminar flow over an isothermal flat plate, the local correlation is
\[
\mathrm{Nu}_x=0.332\,\mathrm{Re}_x^{1/2}\,\mathrm{Pr}^{1/3}
=0.332\left(\frac{u_\infty x}{\nu}\right)^{1/2}\mathrm{Pr}^{1/3}.
\]
Thus
\[
h_x=\frac{k\,\mathrm{Nu}_x}{x}
=0.332\,k\,\mathrm{Pr}^{1/3}\left(\frac{u_\infty}{\nu}\right)^{1/2} x^{-1/2}
\equiv C\,x^{-1/2},
\]
where \(C=0.332\,k\,\mathrm{Pr}^{1/3}\left(\dfrac{u_\infty}{\nu}\right)^{1/2}\) (independent of \(x\)).
Similarly for \(y\)-direction flow: \(h_y=C\,y^{-1/2}\).
Step 2: Compute the area-averaged coefficients.
\[
\overline{h}_l=\frac{1}{l}\int_0^l Cx^{-1/2}dx=\frac{1}{l}\left[2C x^{1/2}\right]_0^l=\frac{2C}{\sqrt{l}},
\quad
\overline{h}_w=\frac{1}{w}\int_0^w C y^{-1/2}dy=\frac{2C}{\sqrt{w}}.
\]
Step 3: Take the ratio:
\[
\frac{\overline{h}_w}{\overline{h}_l}
=\frac{2C/\sqrt{w}}{2C/\sqrt{l}}=\sqrt{\frac{l}{w}}
=\sqrt{\frac{4}{1}}=2.
\]
\[
\boxed{2}
\]
