# Air at 20 degrees C flows with a velocity of 8 m/s over a 1.5 m x 6 m flat plate whose...

## Question:

Air at 20{eq}^\circ {/eq}C flows with a velocity of 8 m/s over a 1.5 m x 6 m flat plate whose temperature is 140{eq}^\circ {/eq}C. Determine the rate of heat transfer from the plate if

a) the air flows parallel to the 6-m-long side and {eq}[ {/eq}6{eq}] {/eq}.

b) the air flows parallel to the 1.5-m side. {eq}[ {/eq}4{eq}] {/eq} Neglect radiation.

The properties of air at the appropriate film temperature are k = 0.02953 W/mK and v = 2.548 x 10{eq}^{-5} {/eq} m{eq}^2 {/eq}/s.

## Convective heat transfer:

It is the process of transfer of thermal energy due to the motion of molecules from one point to another point in a fluid. Heat transfer depends on the material's property and the change in temperature.

## Answer and Explanation:

**Given**

- Velocity of air is {eq}V = 8\;{{\rm{m}} {\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} {/eq}.

- Area of flat plate is {eq}A = 1.5\;{\rm{m}} \times 6\;{\rm{m}} {/eq}.

- Coefficient of thermal conductivity is {eq}k = 0.02953\;{{\rm{W}} {\left/ {\vphantom {{\rm{W}} {{\rm{m}} \cdot {\rm{K}}}}} \right. } {{\rm{m}} \cdot {\rm{K}}}} {/eq}.

- Kinematic viscosity is {eq}\nu = 2.548 \times {10^{ - 5}}\;{{{{\rm{m}}^{\rm{2}}}} {\left/ {\vphantom {{{{\rm{m}}^{\rm{2}}}} {\rm{s}}}} \right. } {\rm{s}}} {/eq}.

- Air temperature is {eq}{T_1} = 20^\circ {\rm{C}} {/eq}.

- Plate temperature is {eq}{T_2} = 140^\circ {\rm{C}} {/eq}.

By using table Prandtl number is {eq}\Pr = 0.7154 {/eq}.

(a) When air flow in parallel to {eq}6\;{\rm{m}} {/eq} side.

Reynold number is given by,

{eq}{\mathop{\rm Re}\nolimits} = \dfrac{{VL}}{\nu } {/eq}

Here, {eq}L = 6\;{\rm{m}} {/eq}.

Substituting values in above equation.

{eq}\begin{align*} {\mathop{\rm Re}\nolimits} &= \dfrac{{6 \times 8}}{{2.548 \times {{10}^{ - 5}}}}\\ {\mathop{\rm Re}\nolimits} &= 1883830.45 \end{align*} {/eq}

Nusselt number for laminar flow is calculated as,

{eq}\begin{align*} Nu &= 0.664{\left( {{\mathop{\rm Re}\nolimits} } \right)^{0.5}}{\left( {\Pr } \right)^{\dfrac{1}{3}}}\\ \dfrac{{hL}}{k} &= 0.664{\left( {1883830.45} \right)^{0.5}} \times {\left( {0.7154} \right)^{\dfrac{1}{3}}}\\ h &= \dfrac{{0.02953 \times 815.999}}{6}\\ h &= 4.0160\;{{\rm{W}} {\left/ {\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}} \cdot {\rm{K}}}}} \right. } {{{\rm{m}}^{\rm{2}}} \cdot {\rm{K}}}} \end{align*} {/eq}

Heat transfer is given by,

{eq}Q = hA\Delta T {/eq}

Here, {eq}\Delta T {/eq} is temperature difference.

Temperature difference is calculated as,

{eq}\begin{align*} \Delta T &= {T_2} - {T_1}\\ \Delta T &= 140 - 20\\ \Delta T &= 120^\circ {\rm{C}} \end{align*} {/eq}

Substituting the values in above equation.

{eq}\begin{align*} Q &= 4.016 \times 1.5 \times 6 \times \left( {120} \right)\\ Q &= 4337.3661\;{\rm{W}} \end{align*} {/eq}

Thus, heat transfer is {eq}4337.3661\;{\rm{W}} {/eq}.

(b) When heat flows in parallel to {eq}1.5\;{\rm{m}} {/eq} side.

Reynold number is calculated as,

{eq}{\mathop{\rm Re}\nolimits} = \dfrac{{V{L_1}}}{\nu } {/eq}

Here, {eq}{L_1} = 1.5\;{\rm{m}} {/eq}.

Substituting values in above equation.

{eq}\begin{align*} {\mathop{\rm Re}\nolimits} &= \dfrac{{1.5 \times 8}}{{2.548 \times {{10}^{ - 5}}}}\\ {\mathop{\rm Re}\nolimits} &= 470957.613 \end{align*} {/eq}

Nusselt number for laminar flow is given as,

{eq}Nu = 0.664{\left( {{\mathop{\rm Re}\nolimits} } \right)^{0.5}}{\left( {\Pr } \right)^{\dfrac{1}{3}}} {/eq}

Substituting the values in above equation.

{eq}\begin{align*} \dfrac{{h{L_1}}}{k} &= 0.664{\left( {470957.613} \right)^{0.5}} \times {\left( {0.7154} \right)^{\dfrac{1}{3}}}\\ h &= \dfrac{{0.02953 \times 407.99}}{{1.5}}\\ h &= 8.032\;{{\rm{W}} {\left/ {\vphantom {{\rm{W}} {{{\rm{m}}^{\rm{2}}} \cdot {\rm{K}}}}} \right. } {{{\rm{m}}^{\rm{2}}} \cdot {\rm{K}}}} \end{align*} {/eq}

Heat transfer is given by,

{eq}Q = hA\Delta T {/eq}

Substituting the values in above equation.

{eq}\begin{align*} Q &= 8.032 \times 1.5 \times 6 \times \left( {120} \right)\\ Q &= 8674.561\;{\rm{W}} \end{align*} {/eq}

Thus, heat transfer is {eq}8674.561\;{\rm{W}} {/eq}.