Question

A \(4-m \times 5-m \times 6-m\) room is to be heated by an electric resistance heater placed in a short duct in the room. Initially, the room is at \(15^{\circ} \mathrm{C}\), and the local atmospheric pressure is \(98 \mathrm{kPa} .\) The room is losing heat steadily to the outside at a rate of \(150 \mathrm{kJ} / \mathrm{min} .\) A \(200-\mathrm{W}\) fan circulates the air steadily through the duct and the electric heater at an average mass flow rate of \(40 \mathrm{kg} / \mathrm{min} .\) The duct can be assumed to be adiabatic, and there is no air leaking in or out of the room. If it takes 20 min for the room air to reach an average temperature of \(25^{\circ} \mathrm{C}\), find \((a)\) the power rating of the electric heater and ( \(b\) ) the temperature rise that the air experiences each time it passes through the heater.

Short answer

Answer: The power rating of the electric heater is 4010 W, and the temperature rise of the air when it passes through the heater is approximately 6.02 K.

Step-by-step solution

Calculate total heat loss from the room

To find the power rating of the electric heater, we first need to determine the total heat loss from the room. Given that the room is losing heat to the outside at a rate of 150 kJ/min, we can convert this to Watts: Total heat loss = \(150 \mathrm{kJ/min} \times \frac{1000 \mathrm{W}}{1 \mathrm{kJ}} \times \frac{1 \mathrm{min}}{60 \mathrm{s}} = 2500 \mathrm{W}\)

Calculate the total heat added by the electric heater

We know that the fan is circulating air through the room at a rate of 40 kg/min and has a power of 200 W. The mass flow rate of air should be in kg/s: Mass flow rate of air = \(\frac{40 \mathrm{kg}}{60 \mathrm{s}} = \frac{2}{3} \mathrm{kg/s}\) Since it takes 20 minutes for the room's air to reach an average temperature of 25°C from an initial temperature of 15°C, we can calculate the rate of heat addition using the specific heat capacity of air (\(c_p = 1.005 \mathrm{kJ/(kg·K)}\)): \(Q_{in} = mc_p\Delta T\), where: \(Q_{in}\) - total heat added \(m\) - mass flow rate of air \(c_p\) - specific heat capacity of air \(\Delta T\) - temperature difference Plug in the given values and convert \(Q_{in}\) to Watts: \(Q_{in} = \frac{2}{3} \mathrm{kg/s} \times 1.005 \frac{\mathrm{kJ}}{\mathrm{kg·K}} \times (25 - 15) K \times \frac{1000 W}{1 \mathrm{kJ}} = 6710 \mathrm{W}\)

Calculate the power rating of the electric heater

Now that we have both the total heat loss from the room and the total heat added, we can find the power rating of the electric heater by balancing the heat energy: Power rating of electric heater = Total heat added - Total heat loss - Fan power Power rating of electric heater = \(6710 \mathrm{W} - 2500 \mathrm{W} - 200 \mathrm{W} = 4010 \mathrm{W}\) So, the power rating of the electric heater is 4010 W.

Calculate the temperature rise of the air in the heater

Now, we will calculate the temperature rise that the air experiences each time it passes through the heater. We will use the power rating of the electric heater and the mass flow rate of air to find this value: \(Q_{heater} = mc_p\Delta T_{heater}\), where: \(Q_{heater}\) - power rating of the electric heater \(\Delta T_{heater}\) - temperature rise of the air in the heater Rearranging the equation to solve for \(\Delta T_{heater}\) and using the given values, we have: \(\Delta T_{heater} = \frac{Q_{heater}}{mc_p} = \frac{4010 \mathrm{W} \times \frac{1 \mathrm{kJ}}{1000 \mathrm{W}}}{\frac{2}{3} \mathrm{kg/s} \times 1.005 \frac{\mathrm{kJ}}{\mathrm{kg·K}}} = 6.018 \mathrm{K}\) Thus, the air experiences a temperature rise of approximately 6.02 K each time it passes through the heater.