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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 150 -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 room air in 24 min, \((b)\) the entropy change of the steam, \((c)\) the entropy change of the air in the room, and (d) the exergy destruction for this process, in \(\mathrm{kJ}\). Assume the air pressure in the room remains constant at \(100 \mathrm{kPa}\) at all times, and take \(T_{0}=10^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: To find the average temperature of the room after 24 minutes, follow these steps: 1. Determine the initial state of the steam in the radiator using given pressure and temperature and find the specific volume, internal energy, and entropy. 2. Calculate the total energy transferred to the room by considering the energy balance equation and the energy transferred by the fan. 3. Calculate the mass of the air in the room using the Ideal Gas Law and the given initial temperature and dimensions. 4. Calculate the average temperature change of the room using the energy transferred, mass of the air, and specific heat capacity of the air. Then, find the new average temperature. 5. Determine the entropy changes of the steam and the air using the final states and specific heat capacities. 6. Calculate the exergy destruction using the entropy changes and a reference temperature. By following these steps, you can find the average temperature of the room, the entropy changes of the steam and the air, and the exergy destruction for this process.

Step by step solution

01

Find the initial state of the steam in the radiator

First, we need to find the initial state of the steam in the radiator. We are given the pressure and temperature of the steam as \(P_{1} = 200 \mathrm{kPa}\) and \(T_{1} = 200^{\circ} \mathrm{C}\). Using the steam tables, we can find the specific volume, internal energy, and entropy of the steam at this initial state.
02

Calculate the total energy transferred to the room

Since the room is well-insulated, the only heat transfer is through the steam radiator. The energy balance equation between the radiator and the room can be represented as: \(Q_{1-2} = U_{1} - U_{2}\) The internal energy of the steam changes as the steam cools down and transfers its heat to the room. The energy transferred due to the fan can be found by multiplying the fan power by the time it has been operating: \(Q_{fan} = W_{fan} \times t = 150\, \mathrm{W} \times 24 \times 60 \, \mathrm{sec}\) So, the total energy transferred to the room will be: \(Q_{total} = Q_{1-2} + Q_{fan}\)
03

Calculate the average temperature of the room air after 24 minutes

To find the average temperature of the room air, we first need to know the mass of air in the room. Assuming the air is an ideal gas, we can use the Ideal Gas Law: \(m = \frac{P_{room}V_{room}}{R_{air}T_{room}}\) Since we are given the initial temperature of the room and the dimensions, we can calculate the mass of the air. With the energy transferred to the room and the specific heat capacity of the air, we can estimate the average temperature change and subsequently the new average temperature of the air in 24 minutes: \(\Delta T = \frac{Q_{total}}{m \times c_p}\) \(T_{avg} = T_{room} + \Delta T\)
04

Calculate the entropy change of the steam and the air

To calculate the entropy change of the steam and the air, we need to find the final state of the steam and the air. For the steam, the final pressure is given as \(P_{2} = 100\, \mathrm{kPa}\), and we can use the steam tables to find the new specific volume, internal energy, and entropy. For the air, we can use the specific heat capacity and the temperature change to find the entropy change: \(\Delta S_{steam} = S_{2} - S_{1}\) \(\Delta S_{air} = m \times c_p \ln \left(\frac{T_{avg}}{T_{room}}\right)\)
05

Calculate the exergy destruction

Finally, we can calculate the exergy destruction for this process. Exergy destruction can be found using the entropy change and the reference temperature: \(X_{destruction}= T_0 \cdot (\Delta S_{steam} + \Delta S_{air})\) By following these steps, we can determine the average temperature of the room after 24 minutes, the entropy changes of the steam and the air, and the exergy destruction for this process.

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Most popular questions from this chapter

Oxygen gas is compressed in a piston-cylinder device from an initial state of \(12 \mathrm{ft}^{3} / \mathrm{lbm}\) and \(75^{\circ} \mathrm{F}\) to a final state of \(1.5 \mathrm{ft}^{3} / \mathrm{lbm}\) and \(525^{\circ} \mathrm{F}\). Determine the reversible work input and the increase in the exergy of the oxygen during this process. Assume the surroundings to be at 14.7 psia and \(75^{\circ} \mathrm{F}\).

Consider a process during which no entropy is generated \(\left(S_{\mathrm{gen}}=0\right) .\) Does the exergy destruction for this process have to be zero?

Two rigid tanks are connected by a valve. Tank \(A\) is insulated and contains \(0.2 \mathrm{m}^{3}\) of steam at \(400 \mathrm{kPa}\) and 80 percent quality. Tank \(B\) is uninsulated and contains \(3 \mathrm{kg}\) of steam at \(200 \mathrm{kPa}\) and \(250^{\circ} \mathrm{C}\). The valve is now opened, and steam flows from tank \(A\) to \(\tan k B\) until the pressure in \(\tan k A\) drops to 300 kPa. During this process \(900 \mathrm{kJ}\) of heat is transferred from tank \(B\) to the surroundings at \(0^{\circ} \mathrm{C}\). Assuming the steam remaining inside tank \(A\) to have undergone a reversible adiabatic process, determine \((a)\) the final temperature in each \(\tan \mathrm{k}\) and \((b)\) the work potential wasted during this process.

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In order to cool 1 ton of water at \(20^{\circ} \mathrm{C}\) in an insulated tank, a person pours \(80 \mathrm{kg}\) of ice at \(-5^{\circ} \mathrm{C}\) into the water. Determine ( \(a\) ) the final equilibrium temperature in the \(\operatorname{tank}\) and \((b)\) the exergy destroyed during this process. The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg}\), respectively. Take \(T_{0}=20^{\circ} \mathrm{C}\).

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