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A room at \(72 \mathrm{~F}\) is heated electrically with \(1500 \mathrm{~W}\) to keep a steady temperature. The outside ambient air is at \(40 \mathrm{~F}\). Find the flux of \(S(=\dot{Q} / T)\) into the room air, into the ambient, and the rate of entropy generation.

Short Answer

Expert verified
Entropy flux into the room is approximately 5.076 J/Ks, to the ambient is 5.404 J/Ks, and entropy generation rate is 0.328 J/Ks.

Step by step solution

01

Convert Temperatures to Kelvin

First, we need to convert the given temperatures from Fahrenheit to Kelvin. The formula to convert Fahrenheit to Kelvin is \( T(K) = (T(°F) + 459.67) \times \frac{5}{9} \). For the room temperature: \( T_{ ext{room}} = (72 + 459.67) \times \frac{5}{9} = 295.372 \text{ K} \) For the ambient temperature: \( T_{ ext{ambient}} = (40 + 459.67) \times \frac{5}{9} = 277.594 \text{ K} \)
02

Calculate Heat Transfer Rate (\( \dot{Q} \))

The heat transfer rate \( \dot{Q} \) is given directly by the power of the heater, which is \( 1500 \text{ W} \) or \( 1500 \text{ J/s} \). This represents the energy being supplied into the room per unit time.
03

Calculate the Entropy Flux into the Room

The entropy flux into the room \( S_{\text{in}} \) can be calculated using the formula \( S_{\text{in}} = \frac{\dot{Q}}{T_{\text{room}}} \). Using the values obtained, we calculate: \( S_{\text{in}} = \frac{1500 \, \text{J/s}}{295.372 \, \text{K}} \approx 5.076 \, \text{J/Ks} \).
04

Calculate the Entropy Flux to the Ambient air

The entropy flux to the ambient \( S_{\text{out}} \) is calculated similarly using the ambient temperature: \( S_{\text{out}} = \frac{\dot{Q}}{T_{\text{ambient}}} \). Substituting the given values, \( S_{\text{out}} = \frac{1500 \, \text{J/s}}{277.594 \, \text{K}} \approx 5.404 \, \text{J/Ks} \).
05

Calculate the Rate of Entropy Generation

The rate of entropy generation \( S_{gen} \) is the difference between the entropy fluxes out and into the system. Therefore, \( S_{gen} = S_{\text{out}} - S_{\text{in}} \). Substituting the calculated values: \( S_{gen} = 5.404 \, \text{J/Ks} - 5.076 \, \text{J/Ks} = 0.328 \, \text{J/Ks} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics, describing the movement of thermal energy from one place to another. It occurs in three primary modes: conduction, convection, and radiation. In the context of this exercise, we are primarily interested in the transfer of heat through electrical means.
In this scenario, a room is being heated electrically with a power of 1500 W. This indicates that electrical energy is being converted into thermal energy at a rate of 1500 J/s to maintain the room's temperature at a steady 72°F.
  • **Conduction**: Heat energy moves through a material without the substance itself moving. Example: Heat moving through the walls or the heating element warming the air.
  • **Convection**: Occurs when a fluid, such as air or water, carries heat energy from one place to another. In the room, this means warmer air circulates, distributing the heat evenly.
  • **Radiation**: Energy emitted in the form of electromagnetic waves. Not directly involved in this heating process but is a general method of heat transfer.
Thus, understanding how heat is transferred helps to identify how heating systems work to maintain a room at a desired temperature.
Entropy Generation
Entropy generation is a vital concept in the study of thermodynamics and indicates the irreversibility of processes. In simple terms, it refers to the increase in entropy within a system due to energy transformations and interactions.
In this exercise, entropy generation occurs because the room is not perfectly isolated, and there is a heat transfer between different temperatures (room and ambient). The calculated rate of entropy generation (0.328 J/Ks) in the room shows that some energy is dispersed as irretrievable heat due to inefficiencies.
  • **Irreversibility**: Any real-world process will have some form of entropy generation because no system is perfectly efficient.
  • **Energy Dispersion**: As heat moves from the room at 72°F to the cooler ambient at 40°F, energy is not completely retained or utilized.
Understanding entropy generation helps us realize the natural tendency towards disorder in any system and the ever-present quest for more efficient energy use.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships and conversions between heat and different forms of energy. It is based on four fundamental laws that explain how energy transfers in physical systems.
The current exercise highlights several key thermodynamic principles:
  • **First Law of Thermodynamics**: This law talks about the conservation of energy, meaning energy cannot be created or destroyed, only transformed. The electrical heater transforms electrical energy into thermal energy to heat the room.
  • **Second Law of Thermodynamics**: This focuses on entropy, asserting that entropy of an isolated system always increases over time. This is observable through the calculation of entropy generation.
  • **System and Surroundings Interaction**: The room (system) exchanges energy with the ambient air (surroundings), an example of an open system where energy but not matter is exchanged.
A thorough understanding of thermodynamics provides a framework to explore how energy systems operate, their efficiency levels, and potential improvements in energy conservation and management.

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Most popular questions from this chapter

An insulated piston/cylinder setup contains carbon dioxide gas at \(800 \mathrm{kPa}, 300 \mathrm{~K}\) that is then compressed to 6 MPa in a reversible adiabatic process. Calculate the final temperature and the specific work using (a) ideal gas Tables \(A .8\) and \((b)\) using constant specific heats in Table A.5.

Water in a piston/cylinder at \(400^{\circ} \mathrm{C}, 2000 \mathrm{kPa}\) is expanded in a reversible adiabatic process. The specific work is measured to be \(415.72 \mathrm{~kJ} / \mathrm{kg}\) out. Find the final \(P\) and \(T\) and show the \(P-v\) and \(T-s\) diagrams for the process.

A cylinder/piston contains air at ambient conditions, \(14.7 \mathrm{lbf} / \mathrm{in} .^{2}\) and \(70 \mathrm{~F}\) with a volume of \(10 \mathrm{ft}^{3} .\) The air is compressed to \(100 \mathrm{lbf} / \mathrm{in} .{ }^{2}\) in a reversible polytropic process with exponent \(n=1.2,\) after which it is expanded back to \(14.7 \mathrm{lbf} / \mathrm{in} .^{2}\) in a reversible adiabatic process. a. Show the two processes in \(P-v\) and \(T-s\) diagrams. b. Determine the final temperature and the net work.

A piston/cylinder system contains \(50 \mathrm{~L}\) air at \(300^{\circ} \mathrm{C}, 100 \mathrm{kPa}\), with the piston initially on a set of stops. A total external constant force acts on the piston, so a balancing pressure inside should be \(200 \mathrm{kPa}\). The cylinder is made of \(2 \mathrm{~kg}\) steel initially at \(1300^{\circ} \mathrm{C}\). The system is insulated, so heat transfer occurs only between the steel cylinder and the air. The system comes to equilibrium. Find the final temperature and the entropy generation.

A rigid, insulated vessel contains superheated vapor steam at \(3 \mathrm{MPa}, 400^{\circ} \mathrm{C}\). A valve on the vessel is opened, allowing steam to escape, as shown in Fig. \(\mathrm{P} 6.57 .\) The overall process is irreversible, but the steam remaining inside the vessel goes through a reversible adiabatic expansion. Determine the fraction of steam that has escaped when the final state inside is saturated vapor.

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