Question

Consider a large classroom with 75 students on a hot summer day. All the lights with \(2.0 \mathrm{~kW}\) of rated power are kept on. The room has no external walls, and thus heat gain through the walls and the roof is negligible. Chilled air is available at \(15^{\circ} \mathrm{C}\), and the temperature of the return air is not to exceed \(25^{\circ} \mathrm{C}\). The average rate of metabolic heat generation by a person sitting or doing light work is \(115 \mathrm{~W}\) ( \(70 \mathrm{~W}\) sensible and \(45 \mathrm{~W}\) latent). Determine the required flow rate of air that needs to be supplied to the room.

Short answer

Based on the given information and calculations, the required flow rate of air to be supplied to the classroom with 75 students is approximately 0.072 kg/s. This takes into account the total sensible heat load produced by the students and the lights, as well as the temperature difference between the supply and return air.

Step-by-step solution

Calculate Total Sensible Heat Load

First, we need to find the total sensible heat load in the room. We will add the heat generated by the students and by the lights: Total Sensible Heat Load \(= \text{Number of students } \times (\text {Sensible heat per student}) + (\text{Light power})\) \(Q_{total} = 75 \times 70\ \mathrm{W} + 2.0 \times 1000\ \mathrm{W}\) \(Q_{total} = 5250\ \mathrm{W} + 2000\ \mathrm{W} = 7250\ \mathrm{W}\)

Convert Watts to Kilowatts

Next, let's convert the total sensible heat load from watts to kilowatts for convenience: \(Q_{total} = 7250\ \mathrm{W} \times \frac{1\ \mathrm{kW}}{1000\ \mathrm{W}} = 7.25\ \mathrm{kW}\)

Calculate the Sensible Heat Load per Degree Temperature Change

Now, we'll find the sensible heat load per degree temperature change: \(\Delta T = T_{return} - T_{supply} = 25^{\circ}\mathrm{C} - 15^{\circ}\mathrm{C} = 10^{\circ}\mathrm{C}\) \(Q_{sensible} = \frac{Q_{total}}{\Delta T} = \frac{7.25\ \mathrm{kW}}{10^{\circ}\mathrm{C}} = 0.725\ \mathrm{kW/}^{\circ}\mathrm{C}\)

Calculate the Required Flow Rate

Finally, we can calculate the required flow rate of air by using the formula for sensible heat load: \(Q_{sensible} = \dot{m} \cdot c_p \cdot \Delta T\) Where \(\dot{m}\) is the mass flow rate of air and \(c_p\) is the specific heat of air at constant pressure (approximately \(1.006\ \mathrm{kJ/kg \cdot K}\)). We need to solve for \(\dot{m}\): \(\dot{m} = \frac{Q_{sensible}}{c_p \cdot \Delta T} = \frac{0.725\ \mathrm{kW}}{1.006\ \mathrm{kJ/kg \cdot K} \cdot 10^{\circ}\mathrm{C}}\) \(\dot{m} = \frac{0.725\ \mathrm{kW} \times 1000\ \mathrm{W/kW} }{1.006\ \mathrm{kJ/kg \cdot K} \times 1000\ \mathrm{J/kJ} \cdot 10^{\circ}\mathrm{C}}\) \(\dot{m} \approx 0.072\ \mathrm{kg/s}\) The required flow rate of air to be supplied to the room is approximately \(0.072\ \mathrm{kg/s}\).