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Google Scholar. Square, L. Academic Press, Elsevier Google Scholar. Rabinovich, M. Krouchev, F. Rattay, M. Sawan, and A. Bean, B. Guckenheimer, J. SIAM J. Schwiening, C. Colwell and M. PLoS Comput Biol 5 1 Strassberg, A. Neural Comput. Feudel, U. An automatic valve allows the air to leave at a constant rate of 0. A relief valve attached to the top of the tank allows vapor to discharge when the gage pressure reaches kPa. The pressure is maintained at that value as more heat is transferred.
A Heat transfer equals work for a process. B Net heat transfer equals net work for a cycle. C Net heat transfer minus net work equals internal energy change for a cycle. D Heat transfer minus work equals internal energy for a process. A Heat transfer equals internal energy change for a process.
B Heat transfer and work have the same magnitude for a constant-volume quasiequilibrium process in which the internal energy remains constant. C The total energy input must equal the total work output for an engine operating on a cycle. D The internal energy change plus the work must equal zero for an adiabatic quasiequilibrium process.
How much heat is rejected? Find the paddle wheel work if the rigid volume is insulated. What is T2 when the frictionless piston hits the stops? Estimate the rejected heat. Estimate the heat removal.
How much heat is needed? Neglect all other forms of energy input. Select the correct statement for a piston-cylinder arrangement. Represents the rate of change of energy between the inlet and outlet. Is often neglected in control-volume applications. Includes the work rate due to the pressure forces.
What temperature change is expected? Estimate the temperature change of the cooling water. Estimate the heat loss. B The enthalpy across a valve remains constant. C The internal energy in the tank remains constant.
D The temperature in the tank remains constant. It is the second law of thermodynamics which helps us establish the direction of a particular process. Consider, for example, the situation illustrated in Fig. This, however, would be a violation of the second law of thermodynamics and would thus be an impossibility. In this chapter we will state the second law as it applies to a cycle. Several devices will be analyzed. This will be followed in Chapter 6 by a statement of the second law as it applies to a process.
If the objective of the device is to perform work it is a heat engine; if its objective is to supply energy to a body it is a heat pump; if its objective is to extract energy from a body it is a refrigerator. A schematic diagram of a simple heat engine is shown in Fig. If the cycle of Fig. A heat pump would provide energy as heat QH to the warmer body e. The work would also be given by 5. Here we use magnitudes only. Note that an engine or refrigerator operates between two thermal energy reservoirs, entities which are capable of providing or accepting heat without changing temperatures.
The atmosphere or a lake serves as heat sinks; furnaces, solar collectors, or burners serve as heat sources. Temperatures TH and TL identify the respective temperatures of a source and a sink.
The second law, however, establishes limits that are surprisingly low, limits that cannot be exceeded regardless of the cleverness of proposed designs. One additional note concerning heat engines is appropriate. The internal combustion engine is an example. The second law of thermodynamics can be stated in a variety of ways. Here we present two: the Clausius statement and the Kelvin—Planck statement. Neither is presented in mathematical terms. We will, however, provide a property of the system, entropy, which can be used to determine whether the second law is being violated for any particular situation.
This statement relates to a refrigerator or a heat pump. It states that it is impossible to construct a refrigerator that transfers energy from a cooler body to a hotter body without the input of work; this violation is shown in Fig. In other words, it is impossible to construct a heat engine that extracts energy from a reservoir, does work, and does not transfer heat to a low-temperature reservoir. Note that the two statements of the second law are negative statements.
Neither has ever been proved; they are expressions of experimental observations. No experimental evidence has ever been obtained that violates either statement of the second law. It should also be noted that the two statements are equivalent. This will be demonstrated with an example. Solution: We will show that a violation of the Clausius statement implies a violation of the Kelvin— Planck statement, and vice versa, demonstrating that the two statements are equivalent.
Consider the system shown in Fig. The device on the left transfers heat and violates the Clausius statement, since it has no work input. Let the heat engine transfer the same amount of heat QL. If we simply transfer the heat QL directly from the engine to the device, as shown in Fig. Conversely Problem 5. Such an engine is called a reversible engine. The process obviously has to be a quasiequilibrium process; additional requirements are: 1.
No friction is involved in the process. Unrestrained expansion does not occur. To illustrate that friction makes a process irreversible consider the system of block plus an inclined plane, shown in Fig. Weights are added until the block is raised to the position shown in part b.
Now, to return the system to its original state some weight must be removed so that the block will slide back down the plane, as shown in part c. Also, the block and plane are at a higher temperature due to the friction, and heat must be transferred to the surroundings to return the system to its original state. This will also change the surroundings.
Because there has been a Fig. A reversible process requires that no friction be present. Bringing the blocks together results in a heat transfer process; the surroundings are not involved in this process.
To return the system to its original state, we must refrigerate the block that had its temperature raised. This will require a work input, demanded by the second law, resulting in a change in the surroundings.
For an example of unrestrained expansion, consider the high-pressure gas contained in the cylinder of Fig. Pull the pin and let the piston suddenly move to the stops shown. Note that the only work done by the gas on the surroundings is to move the piston against atmospheric pressure. Now, to reverse this process it is necessary to exert a force on the piston.
This will demand a considerable amount of work, to be supplied by the surroundings. In addition, the temperature will increase substantially, and this heat must be transferred to the surroundings to return the temperature to its original value. Unrestrained expansion cannot occur in a reversible process. It is an ideal engine that uses reversible processes to form its cycle of operation; thus it is also called a reversible engine.
The cycle associated with the Carnot engine is shown in Fig. It is composed of the following four reversible processes: 1! Heat is transferred reversibly from the high-temperature reservoir at the constant temperature TH. The piston in the cylinder is withdrawn and the volume increases.
The cylinder is completely insulated so that no heat transfer occurs during this reversible process. The piston continues to be withdrawn, with the volume increasing. An isothermal compression. Heat is transferred reversibly to the low-temperature reservoir at the constant temperature TL. The piston compresses the working substance, with the volume decreasing. The completely insulated cylinder allows no heat transfer during this reversible process.
The piston continues to compress the working substance until the original volume, temperature, and pressure are reached, thereby completing the cycle. Let the heat transferred from the high-temperature reservoir to the engine be CHAP. Solution: Suppose that a Carnot engine drives a Carnot refrigerator as shown in Fig. Let the heat rejected by the engine be equal to the heat required by the refrigerator. We will assume the working substance to be an ideal gas see Example 5.
We can do this for all reversible engines or refrigerators. Consequently, the relationship 5. The Carnot engine, when operated in reverse, becomes a heat pump or a refrigerator, depending on the desired heat transfer. Since there are 3.
If the desired output of the engine is 15 kW, as shown in Fig. Calculate the minimum percentage increase in work required, by assuming a Carnot refrigerator, for the same amount of energy removed. And this is a minimum percentage increase, since we have assumed an ideal refrigerator. What size electrical motor rated in horsepower is required for the refrigerator? Calculate the rate of heat transfer from the high-temperature reservoir.
Is this possible? For the adiabatic processes we know that see Fig. What COP would a refrigerator operating on the same cycle have?
If the COPs of the two refrigerators are the same, what should the intermediate temperature be? If the compressors require an electrical energy input of 4 kW, calculate the COP. At this speed essentially all the power produced by the engine is used to overcome air drag. Is this a violation of the second law? Determine the temperature of the high-temperature reservoir and the power produced if the rate of energy addition is 40 kW.
It is supposed to produce 43 hp from the kJ of energy extracted each minute. Is the proposal feasible? The corresponding volumes are and 25 in3. If there is 0. The pressure ratio for the adiabatic compression is 15 to 1 and the volume during the heat-addition process is tripled. It is determined that kJ is required each minute to accomplish this. Calculate the minimum horsepower required. What is the COP a if the purpose is to cool the groundwater and b if the purpose is to heat a building?
Suppose that we wish to remove 0. Review Questions for the FE Examination 5. A No process can produce more work than the heat it accepts. B No engine can produce more work than the heat it intakes. C An engine cannot produce work without accepting heat. D An engine has to reject heat. A A paddle wheel. B A burst membrane. C A resistance heater. D A piston compressing gas in a race engine.
This engine is: A impossible B reversible C possible D probable 5. The rejected heat is nearest: A B It operates between two constant-temperature thermal reservoirs.
A There are two adiabatic processes. B There are two constant-pressure processes. C Work occurs for all four processes. D Each process is a reversible process. This is not a cycle. Using 5. For a reversible adiabatic process the entropy change is zero. We often sketch a temperature-entropy diagram for cycles or processes of interest.
The Carnot cycle provides a simple display when plotting temperature vs. It is shown in Fig. The rectangular area in Fig. Since the heat transfer is equal to the work done for a cycle, the area also represents the net work accomplished by the system during the cycle. We arrived at it assuming a reversible process.
However, since it involves only properties of the system, it holds for an irreversible process also. If the entropy change is zero, i. A paddle wheel, inserted in the volume, does kJ of work on the air.
The gases are expanded to kPa with a reversible adiabatic process. The temperature T2 is found from 6. Hence, we integrate 6. For an isentropic process we cannot use 6. However, we can use 6. The gases are expanded to kPa in a reversible, adiabatic process.
For pure substances, such as steam, entropy is included as an entry in the tables. It is only the change in entropy that is of interest; hence, this arbitrary datum for entropy is of no consequence.
In the superheated region it is tabulated as a function of temperature and pressure along with the other properties. For a compressed liquid it is included as an entry in Table C-4, the compressed liquid table, or it can be approximated by the saturated liquid values sf at the given temperature.
The temperature-entropy diagram is of particular interest and is often sketched during the problem solution. A T-s diagram is sketched in Fig. Note that the high-pressure lines in the compressed liquid region are indistinguishable from the saturated liquid line. It is often helpful to visualize a process on a T-s diagram, since such a diagram illustrates assumptions regarding irreversibilites. The general shape of an h-s diagram is sketched in Fig. Returning to 6. Tables usually list values for Cp ; these are assumed to be equal to C.
The pressure is reduced to 10 psia by removing energy via heat transfer. Calculate the entropy change and the heat transfer and sketch a T-s diagram. State 2 is in the quality region. Consider an irreversible cycle operating between the same two reservoirs, shown in Fig. We desire a quantity that can easily be used as a measure of the irreversibilities that exist in a particular device. The ideal performance is often that associated with an isentropic process.
For a compressor the actual work required is greater than the ideal work requirement of an isentropic process. All pressures are kPa. Calculate the rate of entropy production. Thus, the exiting water is also subcooled. Finally, modifying 6. The mixing process between the superheated steam and the subcooled water is indeed an irreversible process. Sketch the process on a T-s diagram.
Sketch the real process on a T-s diagram. In an actual turbine, moisture formation cannot be tolerated because of damage to the turbine blades. Solved Problems 6. Calculate the entropy change of each reservoir and the net entropy change of the two reservoirs after 20 min of operation. Calculate the entropy change if the initial volume is 0.
Using 6. If the temperature is held constant, calculate the heat transfer and the entropy change. Determine the entropy change if the volume is insulated. Since the process is quite rapid, with little chance for heat transfer, we will assume an adiabatic reversible process. For such a process we may use 6. Calculate the heat transfer and the entropy change.
The quality from v2 is used since it is less sensitive to temperature change. The ice warms up and some of it then melts. The original water cools. If the container is insulated, calculate the entropy change of the universe.
Calculate the entropy change of the universe. Since the volume is rigid, we can locate state 2 by trial and error as follows. The rate of entropy production is then found from 6.
If a small hole is drilled in the tank, estimate the velocity of the escaping air. As the tank heats up, the volume remains constant. The exit state is at the same entropy as the inlet. The pressure is constant in the boiler and condenser. The isentropic process from 2 to 3 allows us to locate state 3. Its COP is Calculate the entropy change of the space and the low-temperature reservoir after 10 min of operation. Atmospheric pressure is 13 psia. The initial pressure is 0 psi gage Fig.
Frictionless piston Fig. To accomplish this, heat transfer must occur. One chamber is pressurized with air to kPa and the other is completely evacuated.
The membrane is ruptured and after a period of time equilibrium is restored. What is the entropy change? It is attached to a paddle wheel located in a rigid 2-m3 volume. Determine the entropy change and the heat transfer. Calculate the work necessary and the entropy change. If the steam undergoes an isentropic expansion to 20 kPa, determine the work output. What is the work needed? Compute the heat transfer necessary and the entropy change. The heat addition results in saturated vapor. Heat is transferred to the air from a high-temperature reservoir until the temperature is tripled in value while the pressure is held constant.
Heat is transferred from the air until the pressure reaches kPa. The initial pressure is kPa. What is the entropy change of the universe? Saturated water leaves the preheater. Calculate the entropy production if all pressures are 60 psia. Assuming an isentropic process, calculate the exit velocity if the exit pressure is 85 kPa. The device used for the compression process is well-insulated. Assuming an isentropic process, calculate the maximum power rating of this turbine. The entropy change is nearest: 0.
The total entropy change is nearest: A 0. A The entropy of an isolated system must remain constant or increase. B The entropy of a hot copper block decreases as it cools. C If ice is melted in water in an insulated container, the net entropy decreases.
D Work must be input if energy is transferred from a cold body to a hot body. As stated previously, a reversible process is a process that, having taken place, can be reversed and, having been reversed, leaves no change in either the system or the surroundings. There is no mixing. There is no turbulence. There is no combustion. It can be easily shown that the reversible work or the work output from a reversible process going from state A to state B is the maximum work that can be achieved for the state change from A to B.
It is of interest to compare the actual work for a process to the reversible work for a process. This comparison is done in two ways. It is generally higher and provides a better comparison to the ideal. Once the irreversibilities for devices in an actual engineering system, such as a steam power cycle, have been calculated, attempts to improve the performance of the system can be guided by attacking the largest irreversibilities. Similarly, since the maximum possible work will be reversible work, irreversibility can be used to evaluate the feasibility of a device.
If the irreversibility of a proposed device is less than zero, the device is not feasible. Reactors Bioreactions and Bioprocessing Solids Processing and Particle Technology Waste Management Process Safety Energy Resources, Conversion, and Utilization Materials of Construction Index follows Section. Home Perry's chemical engineers' handbook [Ninth edition, 85th anniversary edition] , Fluid mechanics for chemical engineers [Third edition.
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