Engineering and Chemical Thermodynamics, 2nd Edition - Milo D. Koretsky. Koray Doğunal. FUNDAMENTAL PHYSICAL CONSTANTS Speed of light in. Engineering and Chemical Thermodynamics by Milo Koretsky, and is. 46 Pages · Kerry_Patterson,_Joseph_Grenny,_Ron_McMillan,_Al_( zlibraryexau2g3p_onion)norinkgibipen.cf Fundamentals of Chemical Engineering Thermodynamics. Engineering-Chemical-Thermodynamics-Koretsky-Solutions-Manual .pdf - Ebook download as PDF File .pdf), Text File .txt) or read book online.
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Engineering and Chemical Thermodynamics, 2nd Edition Milo D. Koretsky Introduces concepts of thermodynamics by showing connections to familiar. Koretsky helps students understand and visualize thermodynamics through a qualitative discussion of the role of molecular interactions and a highly visual. Engineering and Chemical Thermodynamics Solutions Manual. Get access 2nd Edition. Author: Milo D Koretsky Why is Chegg Study better than downloaded Engineering and Chemical Thermodynamics PDF solution manuals ? It's easier.
Looks like you are currently in Ukraine but have requested a page in the Algeria site. Would you like to change to the Algeria site? Milo D. Koretsky helps students understand and visualize thermodynamics through a qualitative discussion of the role of molecular interactions and a highly visual presentation of the material. By showing how principles of thermodynamics relate to molecular concepts learned in prior courses, Engineering and Chemical Thermodynamics, 2e helps students construct new knowledge on a solid conceptual foundation.
Specifically designed to accommodate students with different learning styles, this text helps establish a solid foundation in engineering and chemical thermodynamics. Clear conceptual development, worked-out examples and numerous end-of-chapter problems promote deep learning of thermodynamics and teach students how to apply thermodynamics to real-world engineering problems.
View Instructor Companion Site. Contact your Rep for all inquiries. View Student Companion Site. Koretsky received his Ph. His research interests in thin film materials processing, including plasma chemistry and physics, electrochemical processes and semiconductor yield prediction. His teaching interests include integration of microelectronic unit operations into the ChE curriculum and thermodynamics.
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Read an Excerpt Excerpt 1: PDF Excerpt 2: PDF Excerpt 3: Selected type: Added to Your Shopping Cart. Using the given expression for heat transfer and the definition of dU. The system is constant volume. At 6 PM. During this process. Attractive interactions between the compressed gas molecules can also contribute to this phenomena. When the temperature drops below the freezing point of water. This work decreases the internal energy of the gas — lowering the temperature.
Both m2 and m1 can be calculated by dividing the tank volume by the specific volume To solve this problem. The correct temperature is the one where the above relationship holds. Integrating and using the mass balance above: Now we do some math: Energy balance Neglecting ke and pe, the unsteady energy balance, in molar units, is written as:. Integrating both sides with respect to time from the empty initial state to the final state gives:.
The work is given by:. Energy balance Since the potential and kinetic energy effects can be neglected, the open system, unsteady state energy balance is. The process is adiabatic and no shaft work is done.
Furthermore, there is only one inlet stream and not outlet stream. Therefore, the energy balance simplifies to. We can use the value of hin and the fact that the steam in the pipe is at 9 MPa to find the temperature. Since the problem asks how much energy is stored in the battery after 10 hours of operation, the process is not steady-state.
Let the battery be the system. Furthermore, heating of the battery as it is charged can be ignored. The energy balance is. Both of these values remain constant over time, so integration provides. Assume that potential energy changes can be neglected. Furthermore, assume that the temperature of the water does not change in the process, so the change in internal energy is zero.
Also, view the process as adiabatic. The energy balance reduces to. The actual power being provided by the stream can be calculated using the efficiency information.
Therefore, the energy balance becomes.
There are a number of reasons for the low conversion efficiency. A possible potential energy loss inherent in the design of the energy conversion apparatus decreases the efficiency. Heat is lost to the surroundings during conversion.
Some of the energy is also lost due to friction drag effects. Assuming the rate of heat transfer and potential energy effects are negligible and realizing that there is one inlet and one outlet allows the simplification of the above equation to. To find the heat in we will apply the 1st law.
Assuming steady state, the open system energy balance with one stream in and one stream out can be written:. Thus this problem reduces to finding the change in the thermodynamic property, enthalpy from the inlet to the outlet. We know 2 intensive properties at both the inlet and outlet so the values for the other properties like enthalpy! From Table A. Since this expression is limited to ideal gases any change in temperature must be under ideal conditions. From the definition of heat capacity:.
With negligible eK and eP, the 1st law for a steady state process becomes:. We can calculate the change in enthalpy from ideal gas heat capacity data provided in the Appendix. Nothing is mentioned about shaft work, so the term can be eliminated from the energy balance. The potential and kinetic energy effects can also be neglected. Since there is one inlet and one outlet, the energy balance reduces to. N-hexane is composed of 20 atoms, but carbon monoxide has two.
One would expect the heat capacity to be greater for n-hexane since there are more modes for molecular kinetic energy translational, kinetic, and vibrational. Because the heat capacity is greater and the rate of heat transfer is the same, the final temperature will be less. Assume that heat transfer and potential energy effects are negligible.
The shaft work term is also zero. Therefore, the energy balance reduces to. The molar flow rates can be eliminated from the expression since they are equal. It is also important that the units for the molecular weight and universal gas constant are consistent. The following values were used. The molar flow rates can be eliminated from the expression.
Using the definitions of enthalpy and the kinetic energy, the equation can be rewritten as. The temperature and velocity of the outlet stream are unknown, so another equation is needed to solve this problem. From the conservation of mass,. None of the power is lost to the surroundings and the potential and kinetic energy effects must be neglected. To find the work, we still need T2.
Engineering and Chemical Thermodynamics Solutions Manual
We need to pick a reasonable process to estimate T2. Since the heat flow is zero for this open system problem, we choose an adiabatic, reversible piston situation. For this situation,. We will assume the process is also adiabatic. Again, we must calculate the molar flow rates from the ideal gas law.
Upon substitution and evaluation, we obtain. N2 is an ideal gas All power supplied by the power supply is transferred to the N2 Uniform temperature radially throughout sensor tube Kinetic and potential energy effects negligible in energy balance.
We can use temperature and heat load information from the sensor tube to find the molar flow rate through the sensor tube. First, perform an energy balance for the sensor tube:. To find the flow rate in standard cubic centimeters per minute, apply the conversion factor found in Part a. When the water content of the environment is greater.
This is not required with a constant volume. In the constant pressure cylinder work. This process requires latent heat which cools you. In both cases the internal energy of the gas must be increased.
Series1 A B Steam Tables 3. T T2 is constrained. Since the enthalpy and internal energy of liquid water is essentially independent of pressure in this pressure and temperature range.
The change in enthalpy in Part b is 5.
Nuclear Physics: Exploring the Heart of Matter
Table A. Since oxygen molecules are linear. The vibration contributions can also be analyzed for oxygen and ammonia. Ammonia molecules are non- linear. One thermodynamic property is required to constrain the system. Since nothing is mentioned about work in the problem statement.
W can be set to zero. Enough information is known about the initial state to find the volume of the container. The following expression mathematically represents the change in enthalpy. Assumed the heat capacities are independent of temperature to obtain this expression.
From Appendix A. Since the glass is adiabatic. Substitution of values into the above energy balance allows calculation of Tf. At state 2. Q can be solved.
As the liquid evaporates. They are related through the Antoine Equation. P2 and T2 are not independent of each other. Since the heat of reaction at K is desired. Since stoichiometric amount of air is used. The heat of reaction remains the same because only 1 mole of acetylene reacts. The heat of reaction is calculated as shown in Part a. W11 and Q The value of P2 is not known. The step from state 1 to state 2 is a reversible.
States 1 and 2 are at the higher temperature. From this graph and the description of Carnot cycles in Section 2. Since both the temperature and pressure are known for states 1 and 3. States 3 and 4 have the lower temperature. Since it is isothermal. The polytropic relationship can be employed to find P2. A slight modification of Equation 2. W12 can be calculated. For an adiabatic. Wnet 7. The efficiency of the process can be calculated using Equation 2. Since the both the temperature and pressure are known for states 2 and 4.
Such a process is illustrated below: The process from state 1 to state 2 is a reversible. The polytropic relationship can be employed to find P1. The following equations were used to find the necessary properties: The value of P1 is not known. The coefficient of performance is defined in Equation 2. This equation will be used to calculate the work for the remaining processes. From the energy balance developed for the process from state 3 to state 4 Its sign is negative.
The power obtained from the turbine is the area under the curve from state 1 to 2. Its sign is positive.
The area under the latter curve is much larger remember the log scale. After neglecting potential and kinetic energy effects. Since no work is done. Problem 2. P2 is known. Since the process is reversible and adiabatic. T1 and P1 are known. The first law can be applied to constrain state 2. With potential and kinetic energy effects neglected. The molar volumes of most liquids do not change much with pressure at constant temperature. We do not need to consider the molecular configurations over energy since the temperature difference is so slight.
There are many ways to reconcile this difference. As the temperature increases. We wish to use the steam tables to calculate the entropy change of liquid water as it goes from its freezing point to its boiling point.
In process i. The steam tables in Appendices B. If we believe that the entropy of water is weakly affected by pressure. The molecular configuration over space contribution to entropy is drastically increased in this process. Refer to Section 3. In the liquid state. Now consider process ii.
In the vapor state. Let the mixture of ice and water immediately after the ice has been added represent the system. From Equation 3. We also know the entropy change for the universe is zero for reversible processes.
Because the system is insulated. PI where we end up at the temperature in state 2. To find it. State 1 adiabatic State 2 rev. The process in part a can be drawn as follows: The change in entropy can be represented as follows: Since the change in internal energy is greater. If the process were irreversible. Neglecting potential and kinetic energy effects. Since we know the pressure and internal energy at state 2.
Information given in the problem statement also constrains state 1. If you take the system to be the entire tank both sides of the partition. The ideal gas law can be used to calculate the moles of gas present. This can be shown by employing mass and energy balances. After the partition ruptures. O2 since the tank is well-insulated. During the throttling process. Since there is no heat transfer or work.
Now test the first law of thermodynamics by writing an energy balance. For the second law to be valid. To test the conservation of mass. For constant heat capacity. Nozzle b Since the process is reversible and adiabatic. We must be careful. For a reversible process.
Energy Balance: B Using Equation 3. Before this problem is solved. Since the temperature of the surroundings is constant. The following equation shows how the change in internal energy can be calculated.
The energy balance can be used to obtain Q. It has two components: Mass balance: We can eliminate PB. This expression can be simplified by recognizing that PB. Applying the second law. Converting the units of mass flow rate gives: The reversible work is calculated as: The minimum pressure is required for a reversible process.
Substitute these expressions into the entropy balance and solve for P1: The process is adiabatic. We can represent this process in terms of two latches. This depiction is the expansion analog of the compression process depicted for Example 2.
The corresponding reversible process of Problem 3. Problem 3. Another way to view this argument is to look at this process as a closed system.
The initial state is the same as for Problem 3. Clearly the process on the left does less work. Since the water only expands against 1 bar.
The process is initiated by removal of the first latch and ends when the piston comes to rest against the second latch. An entropy balance gives: The change in entropy between state 2 and state 3 is given by the following equation Since it is also adiabatic. Since the rate of heat transfer is negligible and the expansion occurs reversibly in an ideal Rankine cycle. Liquid 0. H2O at 0. H2O l at 0. Equation 3. Now we find the rate of heat transfer for the evaporator.
For the entire Rankine cycle. This process is sketched in the upper left hand Ts diagram below. This effect is desirable since it will prolong the life of the turbine. Lower the condenser pressure. We can achieve this since the fluid operates in a closed loop. This change reduces the moisture content of steam leaving the turbine.
Increase the degree of superheating of steam in the boiler. Lowering the pressure in the condenser will lower the corresponding saturation temperature.
This change will enlarge the area on the Ts diagram. This can be done in a variety of ways: This process is schematically shown on the bottom right Ts diagram. One way is to divide the turbine into two stages.
This process leads to less moisture content at the turbine exit desirable and limits the temperature of the superheat desirable. This process is illustrated below. Increase the boiler pressure. We can be more creative about how we use the energy available in the boiler.
This will increase the boiler temperature which will increase the area as shown on the bottom left Ts diagram. Turbine 1 and Turbine 2. First start with the turbine state 2. Liquid 1 4 Subcooled Liquid The saturation condition constrains state 3.
States 1. Use the definition of isentropic efficiencies: The enthalpy of state 4 can be calculated from Equation 3. Vapor 1. And Vapor 0. For states 2 and 4. Mixture 0. Vapor 0. Refer to Figure 3. Ra l at 0. Ra v at 0. Using Equation 3.
The cost of the turbine is not justified by the increase in COP. An isentropic turbine adds significant level of complexity to the cycle. Is this modification practical? Turbines are expensive and wear over time.
Engineering and Chemical Thermodynamics by Milo Koretsky, and is
R liquid at 0. R vapor at 0. To find the other enthalpies we must use the following relationships: An energy balance shows: Ra vapor at 0. Now that the pressures are known. From the NIST website. The temperatures of each state are not constant. One possible refrigeration cycle is presented below. In order to find h1. For states 3 and 4. The listed saturation temperatures are the temperatures at which the fluid evaporates and condenses.
To define each state. F 4 oC 4 5 5 condenser Wc s compressor A number of refrigerants will work for this system. F reservoir QC. Since the heat duties are equal for the refrigerator and the freezer. The compositions of states 4. Note the axis are shifted from the usual manner. Work is supplied to magnetize the material and to spin the wheel. The four states of the magnetic material are shown of the sT diagram below.
When stretched the polymer chains tend to align. If the process is adiabatic. The alignment decreases the spatial configurations the polymer can have.
The only way this can be accomplished is by increased temperature. The following data was taken from Table A. This problem shows that the entropy change of the system is negative. We must look at the change in entropy of the surroundings to determine if the second law is violated. By looking at the enthalpies. In pure crystals of Cd and Te. The randomness does not increase when CdTe forms. Morris is arguing that since evolution results in more order. In CdTe. A system can decrease in entropy if the entropy of the surroundings increases by at least that much.
The second law states that the entropy of the universe will remain constant or increase for any process. We can qualitatively relate this concept to the possible hands in a game of poker. We consider a hand of poker containing 5 cards randomly draw from a deck of 52 cards. There are 13 different possible ranks of four of a kind. We do this math in a similar way. For a given hand there are five cards we can pick first.
For the first card in the hand. The greater the number of configurations. In contrast. The fifth card in the hand could be any of the other 48 cards. There are a finite number of permutations in which we can arrange a 52 card deck in 5 cards. In fact. Thus the number of permutations of 5 cards is: So the number of ways we can make the same hand is: This can be rewritten: Coulombic potential energy is proportional to r This modification is similar to the van der Waals equation.
Since we are limited to 1 parameter. Since net electric point charges exert very strong forces. The coulombic forces between the gas molecules affect the system pressure. For both O2 and propane. Disregarded the positive value.
The values are equal because the ionization energies are similar. At higher temperatures. For the mixture. The molecules interact less. The intermolecular distance of molecules is greater at lower pressures. Since styrene monomers are essentially non-polar. For the 5-monomer long polymer chain.
Dispersion is the controlling intermolecular force in this system. The b parameter is also related to the size of the molecule since it accounts for the volume occupied by the molecules. CH 3Cl. CH 2Cl2. Polarizability of each atom The polarizability of a molecule scales with the number of atoms. To illustrate the principles in Chapter 4. For chloroform.
Using the first two molecules. Method 1. The polarizability of C-Cl bonds is calculable with the polarizability of chloroform. Bond Polarizability For this method. This value predicts the polarizabilities of the other species in the table reasonably well. More accurate values for the polarizabilities can be calculated using more of the data given in the problem. Diethylether and n-butanol have the same atomic formula and similar spatial conformations.
There is greater charge separation in the double bond of ketone. Since induction and dispersion forces are similar in these molecules. Methyl ethyl ketone has fewer atoms. Dispersion forces depend on the first ionization potential and polarizability. Now we must determine if there is greater attraction in n-butanol or methyl ethyl ketone.
The polarizability scales with molecular size. Ionization energy is approximately equal for each molecule. The size of the molecular electron orbital of methyl ethyl ketone is approximately equal to the sizes of diethyl ether and n-butanol. Because these are non-polar. Intermolecular attractions are greater.
The hydrogen bonding and dipole-dipole interactions are present in isopropanol. In the gas phase. At K and 30 bar. The molecular kinetic energy is identical since the temperature is the same. The intermolecular forces are greater in the isopropanol. The potential energy has a negative value for attractive interactions. The dispersion forces in n-pentane are stronger than the dipole-dipole forces of isopropanol. Internal energy value is the sum of potential and kinetic energies of the molecules.
For real NH3. The intermolecular forces in the real gas cause the molecules to align so that the positive charge in one molecule is adjacent to a negative charge in a neighboring molecule to reduce potential energy.
They outweigh the volume displaced by the physical size of NH3. For ideal NH3. In the real gas. Ammonia has an electric dipole in which positive and negative charge are separated. The absolute values of the kinetic energies are identical at identical temperature: The compressibility factor is greater for Ne. The weak forces present in Ne have a much smaller effect.
The intermolecular attractions present in NH3 reduce the number of possible configurations. Since both species are gases. In NH3. With Ne. The asymmetry of NH3 results in more possible configurations that NH3 can have. Both species are gases at these conditions. The compressibility factor will be slightly greater than one. NH3 is asymmetrical. The molar volume can be found from the compressibility factor. At K and 25 bar. To estimate the distance between each atom. London interactions are much more important.
Using the above criteria. We need to choose reasonable criteria to specify. Other choices may be just as valid. As provided in the text. Equation 4. Potential functions 0. The Lennard-Jones potential increases more steeply at small radii. The two models are in reasonable qualitative agreement The most stable configuration the bottom of the well occurs at a greater separation for the exp model. The interaction between the chlorine and sodium ions is Coulombic attraction.
Polarizabilities are greater in larger molecules. The magnitude of the dipole-dipole interactions is similar so the pertinent intermolecular force in these molecules is dispersion. Dispersion and dipole-dipole interactions are present in all five species listed. The stronger the intermolecular attraction. The molecular size increases from left to right.
From Table 4. The potential energy can be quantified with the Lennard-Jones potential function. The bond length is the r value where the potential is a minimum.
Size of Molecules: All three molecules have comparable dispersion forces.
Table 4. The following table was made: The values for the above equation were taken from Table A. It does not take into account the structure given to the fluid through intermolecular forces. The basic potential result presented in the text assumes that the species are evenly distributed throughout the volume.
The radial distribution function depends on pressure and temperature of the fluid. If we say that the potential energy between two molecules depends on the amount of time that they spend close to each other. Both of the later equations include a temperature dependence in this term. We can use our knowledge of intermolecular forces.
It makes sense that this should be included in the force correction since this is taking into account repulsive forces.
The inclusion of a "b" term in the second term may help relax van der Waal's "hard sphere" model with a more realistic potential function. This form represents a hard sphere model. The second term. One example of a more detailed explanation follows: If we look at the Redlich-Kwong equation. If we compare these equations to the van der Waals equation. Another explanation goes as follows: They simply represent experimental data better. The following sketch illustrates how 2 species could have the same van der Waals attractive forces: We have seen that if attractive forces depend on orientation dipole-dipole.
Thus it has more opportunity for attractive interactions than the larger species. The Peng-Robinson equation exhibits the most complicated form in an attempt to better fit experimental data. The value from Part b is 1. Part a is not as accurate as Part b and Part c because water is not an ideal vapor.
RT Substituting B and C found above..
From Equation P We can find V by multiplying the given expression for molar volume by the total number of moles. If you take the system to be the entire tank both sides of the partition. From the definition of electrical work.
It's easier to figure out tough problems faster using Chegg Study. Dispersion forces depend on the first ionization potential and polarizability.
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