Archives: Solution Manual
Chapter 9 because the sample mean is approximately normally
Chapter 9 9.1a. 1/6 b. 1/6 9.2 a )1X(P = =P(1,1)= 1/36 b )6X(P = 9.3a P( X = 1) = (1/6) 5 = .0001286 b P( X = 6) = (1/6) 5 = .0001286 9.4 The variance of X […]
Chapter 8 we calculate probabilities by determining the area
Chapter 8 8.1a. P(X > 45) 1550 2)6075( 1550 2)4560( − + − = .0800 b. P(10 < X < 40) 1550 1550 1550 c. P(X < 25) 1550 7)1525( 1550 17)015( 1550 10])15[0( […]
Chapter 7 Expected number of runs without bunting
7.80a Mean = .0289, standard deviation = .1282 b Mean = .0261, standard deviation = .1406 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted […]
Chemical Engineering Chapter 14 The radial conversion profiles for various order
14-54 I. Variation of Da number Conversion Damköhler number/ Da Closed-vessel Open-vessel Parameter 0.69 0.432 0.378 8*U0 1.38 0.713 0.641 4*U0 II. Variation of Pe number Conversion Peclet number/Pe Closed-vessel Open-vessel Parameter 1.04e3 1 1 100* DAB 1.04e4 1 1 […]
Chapter 7 Portfolio 3 has the smallest standard deviation
7.71a Mean = .0060, standard deviation = .0417 b Mean = .0053, standard deviation = .0471 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted […]
Chemical Engineering Chapter 14 The enthalpy balance allows the identification
14-41 P14-17 The presence of a minimum (run2) imply the presence of a radial temperature profile that effects the reaction rate and determines the deviation from the concentration profile given by Eq. (14.51). The enthalpy balance allows the identification of […]
Chapter 7 it is finite because students cannot earn an
Chapter 7 7.1 a 0, 1, 2, … b Yes, we can identify the first value (0), the second (1), and so on. c It is finite, because the number of cars is finite. d The variable is discrete because […]
Chemical Engineering Chapter 14 Da and hence Per cannot be obtained using tubular
14-21 P14-7 Tubular Reactor 1st order, irreversible, pulse tracer test → 2 ! = 65 2 s and m t = 10 s Calculate, k k = ! X“1 1 ln = 3.91/25 s = 0.156 s-1 Therefore assume closed […]
Chemical Engineering Chapter 14 Approximated formula for Segregation Model
14-1 Solutions for Chapter 14 – Models for Non-ideal Reactors P14-1 Individualized solution P14-2 (a) Approximated formula for Segregation Model (1st reaction) ! k Error 0.1 -0.0001667 1 -0.1667 10 -166.7 P14-2 (b) Parameters Dispersion Model Closed-Closed dispersion model ( […]
Chapter 6 wireless Web user uses it primarily for e-mail
The probability that the plane will crash is P(M and B1 and B2) = [P(M)][ P(B1)][ P(B2)] = (.0001) (.01) (.01) = .00000001 We have assumed that the 3 systems will fail independently of one another. 6.69 P( wireless Web […]
Chemical Engineering Chapter 13 Irreversible, first order, long tubular reactor
13-41 Maximum Mixedness ( ) ( )X F E C r d dX Ao A ! ! ! “ += 1 Rate Law : 2 BAA CkCr =! ( ) XCC AoA !=1 ( ) XCC BoB !=1 ( )32 […]
Chapter 6 If all the teams in major league baseball
Chapter 6 6.1 a Relative frequency approach b If the conditions today repeat themselves an infinite number of days rain will fall on 10% of the next days. 6.2 a Subjective approach b If all the teams in major league […]
Chemical Engineering Chapter 13 This The Same Expression For The Exit
13-21 ( ) ( ) ( ) ( ) ( ) ( ) ! ! ! ! ! ! ! ! ! ! d F eE e C C d F E k d F E k Ao A“# $ […]
Chapter 5 No Because The Sampled Population Consists The
lung cancer. b The study is observational. Experimental data would require the statistics practitioner to randomly assign some people to smoke and others not to smoke. because of the need to hire well-trained interviewers and possibly pay travel-related costs if […]
Chemical Engineering Chapter 13 Solving iteratively Hilder approximate formula
13-1 Solutions for Chapter 13 – Distributions of Residence Times for Chemical Reactors P13-1 No solution will be given. P13-2 (a) The area of a triangle (h=0.044, b=5) can approximate the area of the tail :0.11 P13-2 (b) ~0.11 For […]
Chapter 4 From the least squares line we can more precisely
4.116 a y ˆ = 17.933 + .6041x 0 10 20 30 60 65 70 75 80 Height b The coefficient of determination is .0505, which indicates that only 5.05% of the variation in incomes is explained by the variation […]
Chapter 4 away attendance increases on average by 4746 for each win
4.73 y ˆ = 315.5 + 3.3x; Fixed costs = $315.50, variable costs = $3.30 R² = 0.5925 300 320 340 360 010 20 30 40 50 Labor cost Batch size 4.74 y ˆ = 263.4 + 71.65x; Estimated fixed […]
Chemical Engineering Chapter 12 Sketch of concentration profile for different
12-17 P12-10 (c) Let’s try again with “0 = 10 P12-10 (d) We now need to resolve the problem with the fact that there is a critical value of λ, λc, for which both ψ = 0 and d“ d#=0 […]
Chemical Engineering Chapter 12 For the build up of material that hinders
12-1 Solutions for Chapter 12 – Diffusion and Reaction in Porous Catalysts P12-1 Individualized solution P12-2 (a) (a) “=5 P12-2 (b) (1) First Order Reaction Kinetics d2# d$2“Da#=0 Da =kb2 De #=A cosh Da$+B sinh Da$ d# d$=ADa sinh Da$+BDa […]
Chapter 4 The mean number of days to submit grades
Chapter 4 4.1 a 12 54081333060 441040 152552 n x xi+++++++++++ == = 12 489 = 40.75 Ordered data: 0, 5, 15, 25, 30, 33, 40, 44, 52, 60, 81, 104; Median = (33 + 40)/2 = 36.5 Mode […]
Chemical Engineering Chapter 11 no accumulation in the capillary tube or in either
11-21 M=XAMA+XBMB CB=XBC (*105) (*103) z XB XA CB CA ρΒ ρA ρ ωΒ ωA ω 0 0.329 0.671 1.3 2.65 0.376 2.02 2.4 0.158 0.842 60.7 5 0.434 0.566 1.71 2.24 0.495 1.71 2.21 0.226 0.774 56 10 0.573 […]
Chapter 3 This Plan Should Permit Many
3.87 The value of the British pound has fluctuated quite a bit but the current exchange rate is close to the value in 1987. 1 1.2 1.4 1.6 1.8 1 25 49 73 97 121 145 169 193 217 241 […]
Chemical Engineering Chapter 11 Assume density doesn’t change that much
Solutions for Chapter 11 – External Diffusion Effects on Heterogeneous Reactions P11-1 Individualized solution P11-2 (a) WB=“2WA WA=cDAB 1+yA ( ) dyA dz =“cDABdln 1+yA ( ) dz Integrating with yA=0 at z=“ WA=cDAB ” # z ( ) ln […]
Chapter 3 We need other variables to try to explain
b. It appears that the number of crimes have stayed the same of decreased slightly. c. We need other variables to try to explain these data. 0 2,000 4,000 6,000 1 3 5 7 9 11 13 15 17 19 […]
Chemical Engineering Chapter 10 We know that as the temperature increases
10-93 CDP10-Q CDP10-R 10-94 10-95 10-96 10-97 CDP10-S 10-98 CDP10-T 10-99 CDP10-U 10-100 CDP10-V 10-101 CDP10-W CDP10-X 10-102 10-103 10-104
Chapter 3 There is a strong positive linear relationship
3.42 Exports to Canada 0.0 5000.0 10000.0 15000.0 1 14 27 40 53 66 79 92 105 118 131 144 157 170 183 196 209 222 235 248 261 274 287 300 313 326 Imports from Canada 20000.0 25000.0 30000.0 […]
Chemical Engineering Chapter 10 To determine the mechanism and rate limiting
10-81 CDP10-I 10-82 CDP10-J 10-83 CDP10-K CDP10-L 10-84 10-85 CDP10-M 10-86 10-87 CDP10-N CDP10-O 10-88 10-89 CDP10-P 10-90 10-91 10-92
Chapter 3 The histogram is unimodal and somewhat positively
Chapter 3 3.1 9 or 10 3.2 10 or 11 3.3 a 7 to 9 b Interval width 9.18 8 37188 = − (rounded to 20); upper limits: 40, 60, 80,100,120, 140, 160 180, 200 3.4 a 7 to […]
Chemical Engineering Chapter 10 The thickness on these wafers can be obtained
P10-25 10-61 P10-26 (a) 10-62 10-63 P10-26 (b) Using the same program we can see that the maximum conversion is 0.887 P10-26 (c) CDP10-A 10-64 10-65 CDP10-B 10-66 10-67 CDP10-C 10-68 10-69 The thickness on these wafers can be obtained […]
Chapter 2 Excel is the choice of about half the sample
b c Excel is the choice of about half the sample, one-quarter have opted for Minitab, and a small fraction chose SAS and SPSS. Minitab 24% 2.33 Busch Light 7% Coors Light 22% Michelob Light 4% Miller Lite 21% Excel […]
Chemical Engineering Chapter 10 Elementary reaction with 1st order decay
P10-18 (c) 10-41 10-42 10-43 P10-18 (d) 10-44 P10-18 (e) P10-19(a) D da k dt =! S W U t= S dW U dt= D k dt da dW dt U ! = 1D k W a U =! S […]
Chapter 2 The bar chart provides the frequencies and the pie
Chapter 2 2.1 Nominal: Occupation, undergraduate major. Ordinal: Rating of university professor, Taste test ratings. Interval: age, income 2.2 a Interval b Interval c Nominal d Ordinal 2.3 a Interval b Nominal c Ordinal d Interval e Interval 2.4 a […]
Chemical Engineering Chapter 10 one should make a number of guesses of the rate
10-21 Explicit equations as entered by the user [1] e = 1 [2] Pao = 10 P10-10 (a) [3] Pa = y*Pao*(1-X)/(1+e*X) [4] k1 = 560 [5] k2 = 2.04 [6] Fao = 600 [7] ra = -k1*Pa/((1+(k2*Pa))^2) [8] rate […]
Chemical Engineering Chapter 10 at first appears to work well but not as well as
10-1 Solutions for Chapter 10 – Catalysis and Catalytic Reactors P10-1 Individualized solution P10-2 (a) Example 10-1 (1) Pentane isomerization Pt nP iP!!“ Assume that Pt is the catalyst used. Minimum: ( ) 3 7 1 1 3*10 0.5 7.69 […]
Chemical Engineering Chapter 9 When the upper steady state is used as the initial
[4] vo = 400 [5] UA = 250*150 [6] Ta = 530 [7] rho = 50 [8] To = 530 [9] E = 30000 [10] k = 2*7.08e11*exp(-E/1.987/T) [11] tau = V/vo [12] Ca = Cao/(1+tau*k) [13] Cp = .75 […]
Chemical Engineering Chapter 9 Activation energy and Arrhenius constant from
9-21 P9-10 (c) The unsteady energy balance for a batch reactor is showed in equation 9-11. ( ) ( ) ! “#“$#++ = ii ARxs CpN VrHWQ dt dT & & Adiabatic operation ( Q & =0), neglecting s W […]
Chemical Engineering Chapter 9 The concentration profiles also change
9-1 Solutions for Chapter 9 – Unsteady State Nonisothermal Reactor Design P9-1 Individualized solution P9-2 (a) Example 9-1 P9-2 (b) Example 9-2 To show that no explosion occurred without cooling failure. The new T0 of 20 ˚F (497 ˚R) gives […]
Chemical Engineering Chapter 8 Explicit equations as entered by the user
8-96 T 550 225.22683 550 225.22683 Cao 0.064 0.064 0.064 0.064 To 550 550 550 550 k 1.949E+05 5.047E-05 1.949E+05 5.047E-05 Kc 1.01E+06 1841.4832 1.01E+06 1841.4832 f 1 1 2.4419826 2.4419826 Ca 0.064 0.0547713 0.064 0.0547713 Cb 0.064 0.0547713 0.064 […]
Chemical Engineering Chapter 8 The non-adiabatic profiles show an increase
P8-25 Mole balance: C A B A B C dF dF dF r r r dW dW dW = = = Rate Laws: 1 2 3 1 1 2 3 3 B A B C A A A B B […]
Chemical Engineering Chapter 8 Case Broken Preheater Ineffective Heat Exchanger Case
8-61 P8-16 (a) G(T)=“HRX X= “ k 1+ “ k,k=6.6 #10$3exp E R 1 350 $1 T % & ‘ ( ) * + , – . / 0 R(T)=Cp01+ “ ( ) T#Tc ( ) Cp0=“iCpi =50 # “ […]
Chemical Engineering Chapter 8 If we use a nonlinear equation solver to solve
8-41 9 P0 1.013E+06 1.013E+06 1.013E+06 1.013E+06 10 CA0 270.8283 270.8283 270.8283 270.8283 Differential equations 1 d(X)/d(W) = –rA / v0 / CA0 2 d(T)/d(W) = (Uarho * (Ta – T) + rA * 20000) / v0 / CA0 / […]
Chemical Engineering Chapter 8 Next Increase The Coolant Flow Rate And
8-21 8-22 P8-5 2A B C+! A B C Fio lb mole hr ! ” # $ % & ‘ 10 10 0.0 Tio(F) 80 80 – 20, 000 R Btu H lb mol A != , Energy balance with […]
Chemical Engineering Chapter 8 The slope of energy balance will decrease by a factor
Solutions for Chapter 8 – Steady-State Nonisothermal Reactor Design P8-1 Individualized solution P8-2 (a) Example 8-1 For CSTR X= $ k 1+ $ k= $ Ae“E RT 1+Ae“E RT One equation, two unknowns Adiabatic energy balance T=T0“#HRx X CPA In […]
Chemical Engineering Chapter 7 With the reaction self catalyzed the mole balance
7-59 CDP7-C (e) From part (c), we have: 7-60 7-61 CDP7-D No solution will be given. 7-62 CDP7-E CDP7-F No solution will be given. CDP7-G 7-63 7-64 CDP7-H 7-65 CDP7-I No solution will be given. CDP7-J (a) No solution will […]
Chemical Engineering Chapter 7 Dilution Rate The Maximum Cell Concentration Occurs
7-41 C g dC r dt = V = 5000dm3 , gC rDC = ( ) SSSO rCCD !=! gCSS rYr / =! rg = C SM SC CK C + max µ V v DO = See Polymath program […]
Chemical Engineering Chapter 7 can be made and plugged back into the equation for
7-21 P7-10 (a) E+Sk1 “ # “ E•S E•Sk2 “ # “ E+S E•Sk3 “ # “ P+E P+Ek4 “ # “ E•S E # & [ ] [ ] [ ] [ ] k2+k3 # $ & ‘ E […]
Chemical Engineering Chapter 7 Solutions For Reaction Mechanisms Pathways
C4 0 0 2.665E-07 1.276E-08 C7 0 0 0.0979179 0.0979179 C3 0 0 0.0012475 0.0012475 C5 0 0 0.0979179 0.0979179 C8 0 0 6.237E-04 6.237E-04 CP5 0 0 0.0979123 0.0979123 CP1 0.1 2.166E-04 0.1 2.166E-04 k5 3.98E+09 3.98E+09 3.98E+09 3.98E+09 […]
Chemical Engineering Chapter 6 Cost steadily rises with temperature and reaches
6-76 The concentration of B is highest at t = 100 sec, where its value is 0.1363 gmol/dm3 CDP6-C 6-77 6-78 CDP6-D 6-79 6-80 CDP6-E 6-81 6-82 6-83 CDP6-F (a) Mole balances: dFA/dV = -rD-rU fa 0.06705 0.0018652 0.06705 0.0018652 […]
Chemical Engineering Chapter 6 What factors influence the amplitude and frequency
6-61 P6-21 (a) Isothermal gas phase reaction in a membrane reactor packed with catalyst. A B + C ‘ 1 1 B C C C A C C r k C K ! “ =# $ % y 1 […]
Chemical Engineering Chapter 6 Differential equations as entered by the user
6-41 ODE Report (RKF45) Differential equations as entered by the user [1] d(T)/d(t) = 0.1 Explicit equations as entered by the user [1] E2 = 10000 0 0.005 0.01 0.015 0.02 400 600 800 1000 1200 1400 1600 T [K] […]