One way of cooling off integrated circuits efficiently is by having a film of liquid evaporating from the surface of the chip. Figure P5.23 shows the general situation. Evaporation takes place at the surface of the liquid into an atmosphere with a relative humidity of \(50 \%\). The temperature of the environment is held at \(35^{\circ} \mathrm{C}\) and the dynamics of the environment provide for a mass transfer coefficient \(k_{c}=2.5 \times 10^{-4} \mathrm{~mol} / \mathrm{Pa} \cdot \mathrm{s}\). The computer chip dissipates \(10 \mathrm{~W} / \mathrm{cm}^{2}\).

a. If the heat of vaporization of the liquid is \(35 \mathrm{~kJ} / \mathrm{mol}\), what is the flux of vapor from the surface of the liquid at steady state?

b. The vapor pressure of the liquid can be represented by the following Antoine Equation:

\[\log _{10} P^{\text {sat }}(\mathrm{mmHg})=7.96681-\frac{1668.21}{T+228.0}\]

What is the temperature at the surface of the liquid?

c. What is the temperature at the surface of the chip assuming film thickness of \(100 \mu \mathrm{m}\) and a thermal conductivity for the liquid of \(0.5 \mathrm{~W} / \mathrm{mK}\) ?

d. If we modify our machine by upping the clock speed by \(10 \%\), upping the heat dissipation as well, can we still operate the system without exceeding the chip's maximum temperature of \(125^{\circ} \mathrm{C}\) ?

Transcribed Image Text:

FIGURE P5.23 k = 2.5 x 104 mol/Pas 11% Liquid hvap = 35 kJ/mol IC q" = 10 W/cm