CHAPTER 9 SOLUTIONS 9.1
1000C 1 2 700C
To furnace
cal 6.9 mol K
Heat losses
Air
B12
= H12 ToS12 = 6.9 x (700 1000) 6.9 x 298 ln = -2,070 (-552) = -1,517.4 cal/mol
973 1273
9-1
9.2
Ideal gas 25C 1 atm 150C 50 atm
Use a two-step process iso
CHAPTER 16 SOLUTIONS 16.1 The following is based on information from the 1998 Annual Report of Merck & Co., Inc., as obtained from their web site at: www.Merck.com. (a) From their mission statement:
"Merck & Co., Inc. is a leading research-driven ph
CHAPTER 3 SOLUTIONS 3.1 a. Dew point pressure at 120F = 48.9C = 322 K f { PBP } = xi - 1 = 0
i
K
i
yi
i
-1 = 0
yi P Ps - 1 = 0 i i y P is - 1 = 0 i P i 1 1 P= = y y y Pis P1s + P2s i i 1 2
Vapor pressure data interpolation Pentane T, C 36.1
CHAPTER 2 SOLUTIONS 2.1 Using the GAMS program in Example 2.2, a method could not be found to locate the repeat units having the second and third lowest values of the objective function, Z. Initially, an attempt was made to place a lower bound on Z,
CHAPTER 21 SOLUTIONS
21.1
The RGA is completed using the property that rows and columns must add to unity:
0.8 - 4.7 4.9 = - 4.0 4.3 0.7 4.2 1.4 - 4.6 Based on the RGA coefficients, the most promising pairings are: y1 u1, y2 u3 and y3 u2;
CHAPTER 20 SOLUTIONS
20.1
Reactor with Independent Feed Preheating. The degrees-of-freedom analysis is carried out based on the following assumptions: (a) Single reaction A to B; (b) Constant hold-up in both preheater and reactor; (c) Constant flui
CHAPTER 19 SOLUTIONS 19.1 a) Espresso coffee is prepared in a machine that pumps cold water at high pressure (10 20 bar is the range of commonly used values) into a hot water boiler, which displaces near boiling hot water. This water is then forced
CHAPTER 18 SOLUTIONS 18.1 The unknowns are A and B, the quantities of A and B produced in gal/week. The objective function is: (1) max J = 5A + 3.5B
A,B
The constraints are: A B (2) 2 [Type 1 columns are available 6,000 hrs/week] + 6, 000 100 100 i
CHAPTER 14 SOLUTIONS 14.1 Effect of absorbent rate and number of stages on absorption (a) Double L to 474 kmol/hr. Then, the absorption factor for n-butane is doubled to 2.70. The number of stages remains at 4. From the Kremser equation, Eq. (14.5),
CHAPTER 13 SOLUTIONS 13.1 First, use the results of Example 13.7 to estimate the heat exchanger area for a minimum T approach of 10oF. For Example 13.7: Log-mean T = 46.23oF Heat duty = 9,472,000 Btu/hr Toluene exit temperature = 268.7oF Closest appr
CHAPTER 15 SOLUTIONS Exercise 15.1 Liquid Oxygen Pump Page 1
The following results were obtained with the ChemCad simulator. They do not include the calculation of the NPSH, which follows in a hand calculation. CHEMCAD 5.2.3 FLOWSHEET SUMMARY Equip
CHAPTER 12 SOLUTIONS 12.1 Nomenclature F Kdeg Km mS P S Sf V X YP/S YX/S Mass Balances Cell Mass
d( XV ) = { X , S } XV dt dV dX X +V = { X , S } XV dt dt
Substrate feed rate gsubst/hr Penicillin degradation rate constant hr-1 g/L hr-1 Penicillin
CHAPTER 8 SOLUTIONS Exercise 8.1 Cumene process with drag (purge) streams Page 1
The flowsheet for the process is a modification to that shown in Figure 8.1 as follows: 1. Add a distillation column, C4, whose feed is the bottoms from C3. The distil
CHAPTER 7 SOLUTIONS 7.1 First consider all the possible splits. They are as follows with identification numbers 1 to 20, where: A = C3, B = B1, C = NB, D = B2, and E = C5. The five components are ordered from A to E in decreasing volatility. 5 - Comp
CHAPTER 6 SOLUTIONS 6.1 This exercise is solved using HYSYS.Plant. The HYSYS PFD below indicates how the simulation is put together. Note that each of the three PFRs is defined as a diabatic bed with uniform heat transfer rate. The optimizer is set u