Question

A well-insulated \(4-m \times 4-m \times 5-m\) room initially at \(10^{\circ} \mathrm{C}\) is heated by the radiator of a steam heating system. The radiator has a volume of \(15 \mathrm{L}\) and is filled with superheated vapor at \(200 \mathrm{kPa}\) and \(200^{\circ} \mathrm{C}\). At this moment both the inlet and the exit valves to the radiator are closed. A 120 -W fan is used to distribute the air in the room. The pressure of the steam is observed to drop to \(100 \mathrm{kPa}\) after \(30 \mathrm{min}\) as a result of heat transfer to the room. Assuming constant specific heats for air at room temperature, determine ( \(a\) ) the average temperature of air in 30 min, \((b)\) the entropy change of the steam, \((c)\) the entropy change of the air in the room, and \((d)\) the entropy generated during this process, in \(\mathrm{kJ} / \mathrm{K}\). Assume the air pressure in the room remains constant at \(100 \mathrm{kPa}\) at all times.

Short answer

Answer: To find the average air temperature after 30 minutes, we first calculate the final air temperature using the heat transfer equation. Then, we find the average by taking the average of the initial and final air temperatures. The entropy changes for the steam and air can be calculated using their respective entropy change equations. By adding the entropy changes of the steam and air, we can find the entropy generated during this process.

Step-by-step solution

Find the mass and specific heat of trapped air in the room.

Since we know the dimensions of the room and are given the temperature and pressure of the air, we can use the Ideal Gas law to find the mass of air in the room. The Ideal Gas Law is given by: \(PV = mRT\) Where P is the pressure, V is the volume, m is the mass, R is the specific gas constant, and T is the temperature. The specific heat of air at room temperature can be found using tables or literature values. We have the following values given: - Volume of the room: \(V = 4m \times 4m \times 5m = 80 m^3\) - Initial air temperature: \(T_i = 10^\circ C = 283.15 K\) - Air pressure: \(P = 100 kPa = 100,000 Pa\) - Specific gas constant for air: \(R = 0.287 kJ/kg \cdot K = 287 J/kg \cdot K\) Rearranging the Ideal Gas Law to solve for the mass of air: \(m = \frac{PV}{RT}\)

Determine heat transfer from steam to the room over 30 minutes.

To find the amount of heat transferred from the steam to the air in the room, we first need to calculate the initial and final internal energies of the steam, as well as the work done by the steam during the process. The work done can be calculated since we know the pressure drop and the volume of the radiator: \(W = -P\Delta V\) The change in internal energy can be found using specific heats for steam and the temperature change. The heat transfer can then be calculated using the First Law of Thermodynamics: \(Q = \Delta U - W\)

Calculate the final air temperature and average air temperature during this process.

Using the heat transfer calculated in step 2 and the specific heat of air at room temperature, we can now find the final air temperature, knowing the initial air temperature. The specific heat of air at room temperature can be found in tables or literature values, \(c_p = 1.005 kJ/kg\cdot K\). The heat transfer equation for the air: \(Q_{air} = mc_p(T_f - T_i)\) Now we can use this equation to find the final air temperature: \(T_f = \frac{Q_{air}}{mc_p} + T_i\) The average air temperature can then be calculated as: \(T_{avg} = \frac{T_i + T_f}{2}\)

Calculate the entropy change of the steam and air.

The entropy change for the steam can be calculated using the following equation: \(\Delta S_{steam} = m_{steam}c_p \ln\frac{T_f'}{T_i'} - R\ln\frac{P_f}{P_i}\) For the air, the entropy change can be calculated as: \(\Delta S_{air} = mc_p\ln\frac{T_f}{T_i}\)

Calculate the entropy generated during the process.

To find the entropy generated during the process, we can simply add up the entropy changes of the steam and air: \(\Delta S_{gen} = \Delta S_{steam} + \Delta S_{air}\) Using the equations and steps outlined above, we can find the average air temperature after 30 minutes, the entropy change of the steam, the entropy change of the air in the room, and the entropy generated during this process.