Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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Chapter 15, Problem 15.7P
Pure liquid species 2 and 3 are for practical purposes immiscible in one another. Liquid species 1 is soluble in both liquid 2 and liquid 3. One mole each of liquids 1,2, and 3 are shaken together to form an equilibrium mixture of two liquid phases: an
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4.59
Using the unilateral z-transform, solve the following difference equations with the given initial
conditions.
(a) y[n]-3y[n-1] = x[n], with x[n] = 4u[n], y[− 1] = 1
(b) y[n]-5y[n-1]+6y[n-2]= x[n], with x[n] = u[n], y[-1] = 3, y[-2]= 2
Ans. (a) y[n] = -2+9(3)", n ≥ -1
(b) y[n]=+8(2)" - (3)", n ≥ -2
(30)
6. In a process design, the following process streams must be cooled or heated:
Stream
No
mCp
Temperature In Temperature Out
°C
°C
kW/°C
1
5
350
270
2
9
270
120
3
3
100
320
4
5
120
288
Use the MUMNE algorithm for heat exchanger networks with a minimum approach
temperature of 20°C.
(5)
a. Determine the temperature interval diagram.
(3)
(2)
(10)
(10)
b. Determine the cascade diagram, the pinch temperatures, and the minimum hot and
cold utilities.
c. Determine the minimum number of heat exchangers above and below the pinch.
d. Determine a valid heat exchange network above the pinch.
e. Determine a valid heat exchange network below the pinch.
Use this equation to solve it.
Chapter 15 Solutions
Introduction to Chemical Engineering Thermodynamics
Ch. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Work Prob. 15.2 for the van Laar equation. 15.2. A...Ch. 15 - Consider a binary vapor-phase mixture described by...Ch. 15 - Prob. 15.5PCh. 15 - It has been suggested that a value for GEof at...Ch. 15 - Pure liquid species 2 and 3 are for practical...Ch. 15 - It is demonstrated in Ex. 15.5 that the Wilson...Ch. 15 - Vapor sulfur hexafluoride SF6 at pressures of...Ch. 15 - Prob. 15.10P
Ch. 15 - Toluene(l) and water (2) are essentially...Ch. 15 - n-Heptane(l) and water (2) are essentially...Ch. 15 - Consider a binary system of species 1 and 2 in...Ch. 15 - Prob. 15.14PCh. 15 - 15.15. Work the preceding problem for mole...Ch. 15 - The Case I behavior for SLE (Sec. 15.4) has an...Ch. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Estimate the solubility of naphthalene! 1) in...Ch. 15 - Prob. 15.21PCh. 15 - 15.22. The UNILAN equation for pure-species...Ch. 15 - In Ex. 15.8, Henry's constant for adsorption k,...Ch. 15 - It was assumed in the development of Eq. (15.39)...Ch. 15 - Suppose that the adsorbate equation of state is...Ch. 15 - Suppose that the adsorbate equation of state is...Ch. 15 - Derive the result given in the third step of the...Ch. 15 - Consider a ternary system comprising solute...Ch. 15 - It is possible in principle for a binary liquid...Ch. 15 - With V2=V2 , Eq.(15.66) for the osmotic pressure...Ch. 15 - A liquid-process feed stream F contains 99 mol-%...Ch. 15 - At 25°C the solubility of n-hexane in water is 2...Ch. 15 - A binary liquid mixture is only partially miscible...Ch. 15 - The spinodal curve for a binary liquid system is...Ch. 15 - Prob. 15.36PCh. 15 - Many fluids could be used as solvent species for...
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- Example 8: 900 Kg dry solid per hour is dried in a counter current continues dryer from 0.4 to 0.04 Kg H20/Kg wet solid moisture content. The wet solid enters the dryer at 25 °C and leaves at 55 °C. Fresh air at 25 °C and 0.01Kg vapor/Kg dry air is mixed with a part of the moist air leaving the dryer and heated to a temperature of 130 °C in a finned air heater and enters the dryer with 0.025 Kg/Kg alry air. Air leaving the dryer at 85 °C and have a humidity 0.055 Kg vaper/Kg dry air. At equilibrium the wet solid weight is 908 Kg solid per hour. *=0.0088 Calculate:- Heat loss from the dryer and the rate of fresh air. Take the specific heat of the solid and moisture are 980 and 4.18J/Kg.K respectively, A. =2500 KJ/Kg. Humid heat at 0.01 Kg vap/Kg dry=1.0238 KJ/Kg. "C. Humid heat at 0.055 Kg/Kg 1.1084 KJ/Kg. "C 5:42 Oarrow_forwardQ1: From the Figure below for (=0.2 find the following 1. Rise Time 2. Time of oscillation 3. Overshoot value 4. Maximum value 5. When 1.2 which case will be? 1.6 1.4 1.2 12 1.0 |=0.8- 0.6 0.4 0.8 0.2- 0.6 0.4 0.2 1.2 = 1.0 0 2 4 6 8 10 10 t/Tarrow_forwardPlease, I need solution in detailsarrow_forward
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