A 0.03-in-thick copper plate (k = 223 Btu/h · ft · °F) is sandwiched between two 0.1-in.-thick epoxy boards (k = 0.15 Btu/h · ft · °F) that are 7 in. × 9 in. in size. Determine the effective thermal conductivity of the board along its 9-in.-long side. What fraction of the heat conducted along that

Question

A 0.03-in-thick copper plate ([latex]k=223 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex]) is sandwiched between two 0.1-in.-thick epoxy boards ([latex]k = 0.15 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex]) that are [latex]7 ext { in. } imes 9 ext { in. }[/latex] in size. Determine the effective thermal conductivity of the board along its 9-in.-long side. What fraction of the heat conducted along that side is conducted through copper?

Accepted Answer

A thin copper plate is sandwiched between two layers of epoxy boards. The effective thermal conductivity of the board along its 9 in long side and the fraction of the heat conducted through copper along that side are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer is one-dimensional since heat transfer from the side surfaces are disregarded 3 Thermal conductivities are constant.

Properties The thermal conductivities are given to be [latex]k=223 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex] for copper and [latex]0.15 Btu / h \cdot ft \cdot{ }^{\circ} F[/latex] for epoxy layers.

Analysis We take the length in the direction of heat transfer to be L and the width of the board to be w. Then heat conduction along this two-layer plate can be expressed as (we treat the two layers of epoxy as a single layer that is twice as thick)

[latex]\dot{Q}=\dot{Q}_{ ext {copper }}+\dot{Q}_{ ext {epoxy }}=\left(k A \frac{\Delta T}{L} ight)_{ ext {copper }}+\left(k A \frac{\Delta T}{L} ight)_{ ext {epoxy }}=\left[(k t)_{ ext {copper }}+(k t)_{ ext {epoxy }} ight] w \frac{\Delta T}{L}[/latex]

Heat conduction along an “equivalent” plate of thick ness [latex]t=t_{ ext {copper }}+t_{ ext {epoxy }}[/latex] and thermal conductivity [latex]k_{ ext {eff }}[/latex] can be expressed as

[latex]\dot{Q}=\left(k A \frac{\Delta T}{L} ight)_{ ext {board }}=k_{ ext {eff }}\left(t_{ ext {copper }}+t_{ ext {epoxy }} ight) w \frac{\Delta T}{L}[/latex]

Setting the two relations above equal to each other and solving for the effective conductivity gives

[latex]k_{ ext {eff }}\left(t_{ ext {copper }}+t_{ ext {epoxy }} ight)=(k t)_{ ext {copper }}+(k t)_{ ext {epoxy }} \longrightarrow k_{ ext {eff }}=\frac{(k t)_{ ext {copper }}+(k t)_{ ext {epoxy }}}{t_{ ext {copper }}+t_{ ext {epoxy }}}[/latex]

Note that heat conduction is proportional to kt. Substituting, the fraction of heat conducted along the copper layer and the effective thermal conductivity of the plate are determined to be

[latex] (k t)_{ ext {copper }}=\left(223 Btu / h \cdot ft \cdot{ }^{\circ} F ight)(0.03 / 12 ft )=0.5575 Btu / h \cdot{ }^{\circ} F[/latex]

[latex] (k t)_{ ext {epoxy }}=2\left(0.15 Btu / h \cdot ft \cdot { }^{\circ} F ight)(0.1 / 12 ft )=0.0025 Btu / h \cdot{ }^{\circ} F[/latex]

[latex] (k t)_{ ext {total }}=(k t)_{ ext {copper }}+(k t)_{ ext {epoxy }}=(0.5575+0.0025)=0.56 Btu / h \cdot{ }^{\circ} F[/latex]

and

[latex]k_{e f f}=\frac{(k t)_{ ext {copper }}+(k t)_{ ext {epoxy }}}{t_{ ext {copper }}+t_{ ext {epoxy }}}[/latex]

[latex] =\frac{0.56 Btu / h \cdot{ }^{\circ} F }{[(0.03 / 12)+2(0.1 / 12)] ft }=29.2 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F [/latex]

[latex] f_{ ext {copper }}=\frac{(k t)_{ ext {copper }}}{(k t)_{ ext {total }}}=\frac{0.5575}{0.56} = 0.996 = 99.6 \%[/latex]

Question 3.38.E: A 0.03-in-thick copper plate (k = 223 Btu/h · ft · °F) is sa...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions