Archives: Solution Manual
Mathematics Chapter 14 only the last of these for is I in the first
Chemical Engineering Chapter 11 For the ethylbenzene process shown in Appendix
Chapter 11 11.1 For the ethylbenzene process shown in Appendix B, check the design specifications for the following three pieces of equipment against the appropriate heuristics, P-301, V-302, T- 302. Comment on any significant differences that you find. P-301 […]
Mathematics Chapter 13 that the change of variables is restricted so
Chemical Engineering Chapter 10 Incremental Yearly Savings Look at incremental cases
10-19 10.25 Look at incremental cases. ROROII = Incremental Yearly Savings / Incremental Investment ROROIIBase +1 – Base = (0.77-0.75)/(5.1-5) = 0.20 See table for complete ROROII calculations. Case FCI ($ million) Cash Flow ($ million) ROROII 10.26 EAOC = […]
Mathematics Chapter 13 it is conceivable that there might exist paths
Chemical Engineering Chapter 10 If improvements are proposed for a process
Chapter 10 Chapter 10 (short answers) 10.1 NPV = 0 at DCFROR. So, if use hurdle rate to calculate NPV, if NPV > 0, rate of return exceeds hurdle rate. 10.2 Choose the project with the greatest NPV. 10.3 If […]
Mathematics Chapter 12 Suppose we try to use a single complex plane
Chemical Engineering Appendix C Problems at the Cumene Production Facility
C-17 C-18 Project 2 New 20,000 tonnes/y Allyl Chloride Plant – Unit 600 C-19 C-20 C-21 C-22 C-23 C-24 C-25 Project 3 Scale-Down of Phthalic Anhydride Production – Unit 700 24.13 22.2 24.13. 24.2 C-26 22.12 22.3. Project 4 Design […]
Mathematics Chapter 11 The point made in this exercise is an important
Chemical Engineering Appendix C Review of Allyl Chloride Production
C-1 Appendix C Project 1 – Review of Allyl Chloride Production C-2 C-3 C-4 C-5 C-6 C-7 C-8 C-9 C-10 C-11 C-12 C-13 C-14 C-15 C-16
Mathematics Chapter 11 one that is worth emphasizing in class in connection
Chemical Engineering Chapter 9 Interest is the return on an investment or
9-1 Chapter 9 Chapter 9 (Short Answers) 9.1. Simple interest is calculated such that the interest is based on the original principal. 9.2 The nominal interest rate is a number based on interest payments once per year; however, if interest […]
Mathematics Chapter 10 Seek all possible dimensionless products of
Chemical Engineering Chapter 8 The simple answer is that the cooling water
Chapter 8 8.1 The term 0.18FCI has nothing to do with the interest on capital investment. From Table 8.2 we see that the following costs are based on the FCI: Maintenance and repairs Operating supplies Local insurance and taxes Plant […]
Mathematics Chapter 10 one cannot render nonconformable matrices
Chemical Engineering Chapter 7 Most The Cost Heat Exchanger Involves Machining
7-1 Chapter 7 7.1 (i) Capacity or size (for heat exchanger this would be heat exchanger area) 7.2 CEPCI is used to adjust purchased costs of equipment for different times. It is a measure for the inflation of costs associated […]
Mathematics Chapter 9 Note that the maxima in and need not occur
Chemical Engineering Chapter 6 Set Based Available Highest Hot Utility For
6-1 Chapter 6 6.1 Methods for setting pressure of a distillation column a. Set based on the pressure required to condense the overhead stream using cooling water 6.2 Run a distillation column above ambient pressure because a. The components to […]
Mathematics Chapter 9 If a and & have the same then each
Chemical Engineering Chapter 5 Process description for ethylbenzene process
5-1 Chapter 5 5.1 For ethylbenzene process in Figure B.2.1 5.2 For styrene process in Figure B.3.1 Feeds: ethylbenzene, steam Products: styrene, benzene/toluene (by-products), hydrogen (by-product), wastewater (waste stream) 5.3 For drying oil process in Figure B.4.1 Feeds: acetylated castor […]
Mathematics Chapter 8 solution, namely, all points in that plane
Chemical Engineering Chapter 3 What is the difference between a zero-wait
3-1 Chapter 3 3.1. What is a flowshop plant? A flowshop plant is a plant in which several batch products are produced using all or a sub- 3.2. What is a jobshop plant? A flowshop plant is a plant in […]
Mathematics Chapter 7 stable proper node or a stable improper mode
Chemical Engineering Chapter 2 One example is the addition of steam to a
2-1 Chapter 2 2.1 The five elements of the Hierarchy of Process Design are: a. Batch or continuous process 2.2 a. Separate/purify unreacted feed and recycle use when separation is feasible. b. Recycle without separation but with purge […]
Mathematics Chapter 7 For a motion to be periodic it is necessary
Chemical Engineering Chapter 1 Clearance for tube bundle removal on a heat
Chapter 1 1.1 Block Flow Diagram (BFD) Process Flow Diagram (PFD) Piping and Instrument Diagrams (P&ID) (a) PFD (b) BFD (c) PFD or P&ID (d) P&ID (e) P&ID 1.2 P&ID 1.3 It is important for a process engineer to be […]
Mathematics Chapter 6 but once it begins to level off longer need such
Mathematics Chapter 5 but the former will suffice, and is even more
Mathematics Chapter 4 That same number of differentiations of the denominator
Mathematics Chapter 4 We need to show that its dropping out
Mathematics Chapter 3 if the are not distinct then the set is LD
Mathematics Chapter 2 in which case a single valued differentiable
Mathematics Chapter 1 but this is a sub case of is a constant and that
Mechanical Engineering Chapter 22 Homework The Block Having Weight Lb Immersed Liquid
1261 *22–72. SOLUTION Since the system is underdamped. F rom Eq. 22–32 Appling the initial condition at and . (1) (2) 11.888 cos f–0.9392 sin u=0 0=De–0 C 11.888 cos (0 +f)–0.9392 sin (0 +f) D since DZ0 Dsin f=0.62 […]
Mechanical Engineering Chapter 22 Homework The Block Given Upward Velocity 06
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Mechanical Engineering Chapter 22 Homework The general solution of the above differential equation
1230 22–41. If the block-and-spring model is subjected to the periodic force , show that the differential equation of motion is , where xis measured from the equilibrium position of the block. What is the general solution of this equation? […]
Mechanical Engineering Chapter 22 Homework The Surface Smooth And The Springs Are
1210 22–21. If the wire AB is subjected to a tension of 20 lb, determine the equation which describes the motion when the 5-lb weight is displaced 2 in. horizontally and released from rest. Ans: x=0.167 cos 6.55t SOLUTION L′KL […]
Mechanical Engineering Chapter 22 Homework An 8-kg block is suspended from a spring
1190 22–1. A spring is stretched 175 mm by an 8-kg block. If the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 1.50 m>s, determine the differential equation which describes the motion. […]
Mechanical Engineering Chapter 21 Homework It is initially spinning about the axis at when a meteoroid
1180 21–69. The top has a mass of 90 g, a center of mass at G,and a radius of gyration about its axis of symmetry.About any transverse axis acting through point Othe radius of gyration is .If the top is […]
Mechanical Engineering Chapter 21 Homework when it is given three rotations defined by the Euler angles
1169 21–58. SOLUTION Hence Ans. ©Fz=m(aG)z;Az+Bz–15 =0 ©Fy=m(aG)y;Ay+By=-a15 32.2 b(1)(12) ©Fx=m(aG)x;Bx=0 ©Mz=Izv # z–(Ix–Iy)vxvy,Ay(1) –By(1) =c1 12 a15 32.2 b A 3(0.5)2+(2)2 B d(12) –0 ©My=Iyv # y–(Iz–Ix)vzvx,Bz(1) –Az(1) =0 ©Mx=Ixv # x–(Iy–Iz)vyvz,0=0 v # x=v # y=0, v # […]
Mechanical Engineering Chapter 21 Homework Has Weight Lbft And Supported Pin
1149 *21–40. SOLUTION In general Substitute and expanding the cross product yields Subsitute Hx,Hyand Hzusing Eq. 21–10. For the icomponent, Ans. One can obtain yand zcomponents in a similar manner. +Æ y(Izvz–Izx vx–Izy vy) ©Mx=d dt (Ixvx–Ixy vy–Ixz vz)-Æ z(Iyvy–Iyz […]
Mechanical Engineering Chapter 21 Homework Determine the moment of inertia of the disk about
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without […]
Mechanical Engineering Chapter 21 Homework where r is the distance from the origin O to dm
1109 21–1. SOLUTION However,,where ris the distance from the origin Oto dm. Since is constant, it does not depend on the orientation of the x,y,zaxis. Consequently, is also indepenent of the orientation of the x,y,zaxis. Q.E.D.Ixx +I yy +Izz ƒrƒ […]
Mechanical Engineering Chapter 20 Homework Rotating About The Axis With Angular
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Mechanical Engineering Chapter 20 Homework In this case the collar appears only to move radially
1085 20–35. Solve Prob. 20–28 if the connection at B consists of a pin as shown in the figure below, rather than a ball-and- socket joint. Hint: The constraint allows rotation of the rod both along the bar (j direction) […]
Mechanical Engineering Chapter 20 Homework Directed Perpendicular The Axis Rod Ab
1065 20–19. SOLUTION The resultant angular velocity is always directed along the instantaneous axis of zero velocity IA. Ans. ={4.35i+12.7j} rad>s v=13.44 sin 18.87° i+13.44 cos 18.87° j v sin 147.09° =8 sin 18.87° v=13.44 rad>s , roF For ,. […]
Mechanical Engineering Chapter 20 Homework Acceleration The Direction Does Not Change With
1045 20–1. The propeller of an airplane is rotating at a constant speed while the plane is undergoing a turn at a constant rate Determine the angular acceleration of the propeller if (a) the turn is horizontal, i.e., and (b) […]
Mechanical Engineering Chapter 19 Homework No portion of this material may be reproduced
1025 *19–40. l ω 0 P SOLUTION (a) Ans. (b) From part (a) Ans.v=1 4v0=1 4(4) =1 rad s v=1 4v0 c1 12 ml2dv0=c1 3ml2dv ©(HP)0=©(HP)1 vl=1.5 m. 0=4 rad>s,m=2 kg, V0 A thin rod of mass m has an […]
Mechanical Engineering Chapter 19 Homework Determine The Angular Velocity The Disk Just
1005 19–21. SOLUTION c 0 +F A(4)(0.9) –5(4)(0.3) =30 32.2(0.45)2 v +) (HG)1+© LMG dt =(HG)2 ( 0 +5(4) –F A(4) =30 32.2 vG Since no slipping occurs Set Ans. v=12.7 rad>s F A=2.33 lb Also, c Ans. v=12.7 rad>s […]
Mechanical Engineering Chapter 19 Homework The plane has amass of 200 Mg, its center of mass
985 19–1. SOLUTION Q.E.D.rP>G=k2 G rG>O However, yG=vrG>Oor rG>O=yG v rP>G=k2 G yG>v rG>O(myG)+rP>G(myG)=rG>O(myG)+(mk2 G)v HO=(rG>O+rP>G)myG=rG>O(myG)+IGv, where IG=mk2 G The rigid body (slab) has a mass mand rotates with an angular velocity about an axis passing through the fixed point […]