Question

A logic chip used in a computer dissipates \(3 \mathrm{~W}\) of power in an environment at \(120^{\circ} \mathrm{F}\) and has a heat transfer surface area of \(0.08 \mathrm{in}^{2}\). Assuming the heat transfer from the surface to be uniform, determine \((a)\) the amount of heat this chip dissipates during an eight-hour workday in \(\mathrm{kWh}\) and (b) the heat flux on the surface of the chip in W/in \({ }^{2}\).

Short answer

Answer: The logic chip dissipates 0.024 kWh of heat during an eight-hour workday, and the heat flux on its surface is 9000 W/in².

Step-by-step solution

Conversion of power dissipation to energy

We are given that the logic chip dissipates 3 W of power, and we need to find how much heat the chip dissipates during an eight-hour workday. To do this, we use the formula: Energy = Power * Time The power is given in watts, and time should be in hours, so we will convert the 8-hour workday to seconds (1 hour = 3600 seconds). Time = 8 hours * 3600 seconds/hour = 28800 seconds Energy = 3 W * 28800 seconds = 86400 J Now, we need to convert the energy from joules (J) to kilowatt-hours (kWh). The conversion factor is: 1 kWh = 3.6 * 10^6 J

Calculate heat dissipation in kWh

Now, we will use the conversion factor to find the heat dissipation in kWh: Heat dissipation = (86400 J) / (3.6 * 10^6 J/kWh) = 0.024 kWh So, the amount of heat this chip dissipates during an eight-hour workday is 0.024 kWh.

Calculate heat flux on the surface of the chip

We are given that the heat transfer surface area of the chip is 0.08 in², and the power dissipation is 3 W. The heat flux is given as the power dissipation per unit area: Heat flux = Power dissipation / Surface area First, we convert the surface area to m² (1 in² = 6.4516 × 10⁻⁴ m²): Surface area = 0.08 in² * (6.4516 × 10⁻⁴ m²/in²) = 5.16128 × 10⁻⁴ m² Now, we can calculate the heat flux: Heat flux = (3 W) / (5.16128 × 10⁻⁴ m²) = 5814 W/m² Since we are asked for the heat flux in W/in², we need to convert it back to in²: Heat flux = (5814 W/m²) * (1 m² / 6.4516 × 10⁻⁴ in²) = 9000 W/in² So, the heat flux on the surface of the chip is 9000 W/in².