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)bestthing.info 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.
Selected type: Added to Your Shopping Cart. Evaluation Copy Request an Evaluation Copy. Koretsky ISBN: Student View Student Companion Site. About the Author Milo D. Permissions Request permission to reuse content from this site. Vapor Pressure 23 The Critical Point 24 1. Transfer of Energy Between the System and the Surroundings 42 2.
Problem Formulation Learning Objectives 6. The Clausius—Clapeyron Equation 6. Fugacity Learning Objectives 7. Applications Learning Objectives 8. LLE 8. VLLE 8. 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. 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.
Engineering and Chemical Thermodynamics by Milo Koretsky, and is
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.
Engineering and Chemical Thermodynamics Solutions Manual
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 If we substitute the first expression into the second. Can you calculate C? An illustrative plot of z-1 v vs. If we plot z-1 v vs. The slope of this region would yield the third virial coefficient.. At very low pressures.
A more careful examination of Equation 2 suggests another possibility. We may choose to report this value as B.
What value would you be more apt to use? T B Level B Linear B AVG B 1st value -9 -3 Temperatures of Note that the average value is indicative of the 1st method above while the value at the lowest pressure used is indicative of the second value used.
The first value uses much more data while the second method uses limited data in a better range. Also reported were the average value and the value at the lowest pressure used.
The results are reported in the table and figure below. Oxford University Press. Dymond and E. Alternative 1: Rewrite the virial equation: Values of B reported in the CRC are also shown on the summary plot. They agree most closely with the first level method. A Critical Compilation. Alternative 2: From the virial equation: Calculate molar volume solutions for: If the values of the expressions are not equal. Guess P sat 2. Calculate values of molar volumes that result at the chosen P sat: The general method is as follows 1.
Guess P sat: Guess Psat: Repeat this process until the areas are equal. The liquid density is 7. Now use the EOS to find the molar volume: Critical data for nitrogen obtained in Table A. Critical data for oxygen obtained in Table A. If we use the ideal gas law. It is To find the thermal expansion coefficient. To calculate the isothermal compressibility.
Specific Volume of Liquid Water vs.
Pressure 0. The alkenes have low percentile errors due to their nonpolar nature. Using data from Table A. The molar volume is required for the calculations. Now that these values are known. Using a numerical technique. Critical data for propane obtained in Table A. To estimate the molar volume. The solution using one possibility is illustrated below. CO2 is a suitable substance.
Calculate the necessary parameters for the EOS: Multiple possibilities exist for which substance to use in the vial. From the Appendix A. P critical point liquid part d not enough part c CO2 vapor liquid. As the substance is heated. The following quantities are required to calculate the molar volume with the Peng-Robinson equation.
Following the procedure outlined above. The pressure is greater than the critical pressure. From the mixing rules. The b parameter is directly related to the size of the molecule. From Equations 4. But first. 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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Request permission to reuse content from this site. Undetected country. NO YES. Engineering and Chemical Thermodynamics, 2nd Edition. Read an Excerpt Excerpt 1: PDF Excerpt 2:Since the glass is adiabatic.
To test the conservation of mass. Q can be solved. Mole Frac. The liquid is at a temperature higher than the freezing point and the solid at lower temperature. Carmen Herrmann et al- Ghost transmission: Both m2 and m1 can be calculated by dividing the tank volume by the specific volume The following equation is used: