Question

A \(5-\mathrm{m} \times 6-\mathrm{m} \times 8-\mathrm{m}\) room is to be heated by an electrical 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 \(200 \mathrm{~kJ} / \mathrm{min}\). A \(300-\mathrm{W}\) fan circulates the air steadily through the duct and the electric heater at an average mass flow rate of \(50 \mathrm{~kg} / \mathrm{min}\). The duct can be assumed to be adiabatic, and there is no air leaking in or out of the room. If it takes \(18 \mathrm{~min}\) 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

1. Calculate the energy loss from the room over 18 minutes. Energy loss = Heat loss rate × Time Energy loss = 200 kJ/min × 18 min Energy loss = 3600 kJ 2. Calculate the energy needed to increase the temperature of the room to 25 degrees Celsius. Mass of air = V × ρ ≈ 240 × 1.184 ≈ 284.16 kg Energy required = mc_pΔT = 284.16 × 1.005 × (25 - 15) Energy required ≈ 2847.4 kJ 3. Calculate the energy provided by the electric heater. Energy provided by heater = Energy required + Energy loss Energy provided by heater = 2847.4 kJ + 3600 kJ Energy provided by heater ≈ 6447.4 kJ 4. Calculate the power rating of the electric heater. Power rating = Energy provided by heater / Time Power rating = 6447.4 kJ / 18 min Power rating ≈ 358.19 kW 5. Calculate the temperature rise of the air in the heater. Net power = Power rating - Fan power Net power = 358.19 kW - 5 kW Net power = 353.19 kW ΔT = Net power / (m * c_p) ΔT = (353.19 × 10^3 J/s) / (0.9 kg/s × 1.005 × 10^3 J/kg·K) ΔT ≈ 435.53 K

Step-by-step solution

Calculate the energy loss from the room over 18 minutes

We know that the room loses heat at a rate of 200 kJ/min, and we are given that it takes 18 minutes for the room to reach the desired temperature. The energy loss over 18 minutes can be calculated as: \(\text{Energy loss} = \text{Heat loss rate} \times \text{Time}\) \(\text{Energy loss} = 200 \,\text{kJ/min} \times 18\,\text{min}\)

Calculate the energy needed to increase the temperature of the room to 25 degrees Celsius

We'll use the ideal gas law and the specific heat of air at constant pressure \(c_p = 1.005 \,\text{kJ/kg}\cdot\text{K}\) to find the energy needed to increase the temperature of the room. First, we need to calculate the mass of the air in the room: \(V = 5 \times 6 \times 8 = 240 \,\text{m}^3\) \(\rho = \frac{P}{R T_i} = \frac{98 \times 10^3}{287 \times (15+273)} \approx 1.184 \,\text{kg/m}^3\) (where \(R = 287 \,\text{J/kg}\cdot\text{K}\) is the gas constant for air) \(\text{Mass of air} = V \times \rho \approx 240 \times 1.184 \approx 284.16 \,\text{kg}\) Now, we can calculate the energy needed to increase the room's temperature: \(\text{Energy required} = mc_p \Delta T = 284.16 \times 1.005 \times (25 - 15)\)

Calculate the energy provided by the electric heater

Since the room's temperature increases, it means that the electric heater has provided enough energy to compensate for the energy loss and heat the room. We can write an energy balance for the room: \(\text{Energy provided by heater} = \text{Energy required} + \text{Energy loss}\) Substitute the values from steps 1 and 2, and solve for the energy provided by the heater.

Calculate the power rating of the electric heater

Now that we have the total energy provided by the heater over 18 minutes, we can find the power rating: \(\text{Power rating} = \frac{\text{Energy provided by heater}}{\text{Time}}\) Substitute the value from step 3 and divide by 18 minutes to get the power rating.

Calculate the temperature rise of the air in the heater

We can find the temperature rise of the air passing through the heater using the power rating of the heater, fan power, and the mass flow rate: \(\text{Net power} = \text{Power rating} - \text{Fan power}\) \(\Delta T = \frac{\text{Net power}}{m \cdot c_p}\) Substitute the values from step 4 and the given mass flow rate and \(c_p\) to find the temperature rise in the heater.