Question

A sealed electronic box is to be cooled by tap water flowing through the channels on two of its sides. It is specified that the temperature rise of the water not exceed \(4^{\circ} \mathrm{C}\) The power dissipation of the box is \(2 \mathrm{kW}\), which is removed entirely by water. If the box operates 24 hours a day, 365 days a year, determine the mass flow rate of water flowing through the box and the amount of cooling water used per year.

Short answer

How much cooling water is used per year if the box operates continuously? Answer: The mass flow rate of water flowing through the box is approximately 0.1194 kg/s, and the amount of cooling water used per year is approximately 3,763,624.64 kg.

Step-by-step solution

Convert the given power dissipation into Watts

Since the power dissipation is given in kW, we need to convert it to Watts for our calculations: $$ P = 2\,\mathrm{kW} \times 1000\,\mathrm{\frac{W}{kW}} = 2000\,\mathrm{W} $$

Calculate the heat transfer rate

The heat transfer rate \(Q\) is given by the power dissipation as follows: $$ Q = m \times c \times \Delta T $$ Where \(m\) is the mass flow rate, \(c\) is the specific heat capacity of water, and \(\Delta T\) is the temperature rise. Rearranging this equation to find \(m\): $$ m = \frac{Q}{c \times \Delta T} $$

Use the known values of \(Q\), \(c\), and \(\Delta T\) to find \(m\)

We know that the temperature rise \(\Delta T\) is \(4^{\circ}\mathrm{C}\), and the specific heat capacity of water \(c\) is approximately \(4186\,\mathrm{\frac{J}{kg\cdot K}}\). Plugging in these values along with the found power dissipation: $$ m = \frac{2000\,\mathrm{W}}{4186\,\mathrm{\frac{J}{kg\cdot K}} \times 4\mathrm{K}} = 0.1194\,\mathrm{\frac{kg}{s}} $$

Calculate the annual water usage

To find the amount of water used per year, we need to consider that the box operates 24 hours a day for 365 days a year. We can calculate the total water usage by converting the mass flow rate to an annual amount: $$ m_{annual} = 0.1194\,\mathrm{\frac{kg}{s}} \times 60\,\mathrm{\frac{s}{min}} \times 60\,\mathrm{\frac{min}{hr}} \times 24\,\mathrm{\frac{hr}{day}} \times 365\,\mathrm{days} $$ $$ m_{annual} = 3763624.64\,\mathrm{kg} $$ Thus, the mass flow rate of water flowing through the box is approximately \(0.1194\,\mathrm{\frac{kg}{s}}\), and the amount of cooling water used per year is approximately \(3763624.64\,\mathrm{kg}\).