University:
Kentucky State University
Course:
COS 232| Programming I
Academic year:

2019
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143

Pages:

23
Author:

diamundgalx45p

8.11 Substituting the parameter values yields 10 ε3 1−ε = 150 + 1.75 1−ε 1000 This can be rearranged and expressed as a roots problem f (ε ) = 0.15(1 − ε ) + 1.75 − 10 ε3 =0 1−ε A plot of the function suggests a root at about 0.45. 3 2 1 0 -1 0 0.2 0.4 0.6 -2 -3 But suppose that we do not have a plot. How do we come up with a good initial guess. The void fraction (the fraction of the volume that is not solid; i.e. consists of voids) varies between 0 and 1. As can be seen, a value of 1 (which is physically unrealistic) causes a division by zero. Therefore, two physically-based initial guesses can be chosen as 0 and 0.99. Note that the zero is not physically realistic either, but since it does not cause any mathematical difficulties, it is OK. Applying bisection yields a result of ε = 0.461857 in 15 iterations with an absolute approximate relative error of 6.5×10−3 % 8.12 The total pressure is equal to the partial pressures of the components: P = Pb + Pt According to Antoine’s equation Pb = e Ab − Bb T +Cb Pt = e At − Bt T +Ct Combining the equations yields f (T ) = e Ab − Bb T +Cb +e At − Bt T +Ct −P=0 The root of this equation can be evaluated to yield T = 350.5. 8.13 There are a variety of ways to solve this system of 5 equations K1 = [H + ][HCO 3− ] [CO 2 ] (1) K2 = [H + ][CO 32− ] [HCO 3− ] (2) K w = [H + ][OH − ] (3) cT = [CO 2 ] + [HCO3− ] + [CO32− ] (4) Alk = [HCO3− ] + 2[CO 32− ] + [OH − ] − [H + ] (5) One way is to combine the equations to produce a single polynomial. Equations 1 and 2 can be solved for [H 2 CO*3 ] = [H + ][HCO 3− ] K1 [CO 23− ] = [H + ]K 2 [HCO 3− ] These results can be substituted into Eq. 4, which can be solved for [H 2 CO*3 ] = F0 cT [HCO 3− ] = F1cT [CO 32− ] = F2 cT where F0, F1, and F2 are the fractions of the total inorganic carbon in carbon dioxide, bicarbonate and carbonate, respectively, where F0 = K 1K 2 [H + ] 2 K1[H + ] F2 = + 2 F = 1 + 2 + + 2 + [H ] + K1[H + ] + K 1K 2 [H ] + K 1[H ] + K1 K 2 [H ] + K 1[H ] + K1 K 2 Now these equations, along with the Eq. 3 can be substituted into Eq. 5 to give 0 = F1cT + 2 F2 cT + K w [H + ] − [H + ] − Alk Although it might not be apparent, this result is a fourth-order polynomial in [H+]. [H + ]4 + ( K1 + Alk )[ H + ]3 + ( K1K 2 + AlkK1 − K w − K1cT )[ H + ]2 + ( AlkK1K 2 − K1K w − 2 K1 K 2 cT )[ H + ] − K1 K 2 K w = 0 Substituting parameter values gives [H + ]4 + 2.001 × 10 −3 [ H + ]3 − 5.012 × 10 −10 [H + ]2 − 1.055 × 10 −19 [ H + ] − 2.512 × 10 −31 = 0 This equation can be solved for [H+] = 2.51x10-7 (pH = 6.6). This value can then be used to compute [OH − ] = 10 −14 = 3.98 × 10 −8 2.51 ×10 −7 (2.51 × 10 ) [H CO ] = ( 2.51 ×10 ) + 10 ( 2.51 × 10 ) + 10 10 (2.51 × 10 ) [HCO ] = (2.51× 10 ) + 10 ( 2.51 × 10 ) + 10 -7 2 * 2 3 -7 2 -7 −6.3 − 3 [CO 23− ] = −6.3 -7 2 ( 2.51× 10 ) −6.3 10 -7 −6.3 −6.3 -7 10 10 −6.310 −10.3 -7 2 + 10 −6.3 −10.3 (2.51× 10 ) + 10 -7 − 6 .3 10 −10.3 −10.3 ∫ Cout Cin K k max C + 1 k max dc = −   C out  1   + C out − Cin   K ln  k max   Cin   Therefore, the problem amounts to finding the root of f (Cout ) = V 1 + F k max   Cout  K ln    Cin    + Cout − Cin    Excel solver can be used to find the root: ) ( ) 3 × 10 −3 = 0.666562 3 × 10 −3 = 0.002 ( ) 3 × 10 −3 = 0. 000133 3 × 10 −3 = 1.33 × 10 −4 M 8.14 The integral can be evaluated as − ( 3 × 10 −3 = 0.33304 3 × 10 −3 = 0. 001 8.24 %Region from x=8 to x=10 x1=[8:.1:10]; y1=20*(x1-(x1-5))-15-57; figure (1) plot(x1,y1) grid %Region from x=7 to x=8 x2=[7:.1:8]; y2=20*(x2-(x2-5))-57; figure (2) plot(x2,y2) grid %Region from x=5 to x=7 x3=[5:.1:7]; y3=20*(x3-(x3-5))-57; figure (3) plot(x3,y3) grid %Region from x=0 to x=5 x4=[0:.1:5]; y4=20*x4-57; figure (4) plot(x4,y4) grid %Region from x=0 to x=10 figure (5) plot(x1,y1,x2,y2,x3,y3,x4,y4) grid title('shear diagram') a=[20 -57] roots(a) a = 20 -57 ans = 2.8500 s hear diagram 60 40 20 0 -20 -40 -60 0 1 2 3 4 5 6 7 8 9 10 8.25 %Region from x=7 to x=8 x2=[7:.1:8]; y2=-10*(x2.^2-(x2-5).^2)+150+57*x2; figure (2) plot(x2,y2) grid %Region from x=5 to x=7 x3=[5:.1:7]; y3=-10*(x3.^2-(x3-5).^2)+57*x3; figure (3) plot(x3,y3) grid %Region from x=0 to x=5 x4=[0:.1:5]; y4=-10*(x4.^2)+57*x4; figure (4) plot(x4,y4) grid %Region from x=0 to x=10 figure (5) plot(x1,y1,x2,y2,x3,y3,x4,y4) grid title('moment diagram') a=[-43 250] roots(a) a = -43 250 ans = 5.8140 m om ent diagram 100 80 60 40 20 0 -20 -40 -60 0 1 2 3 4 5 6 7 8 9 10 8.26 A Matlab script can be used to determine that the slope equals zero at x = 3.94 m. %Region from x=8 to x=10 x1=[8:.1:10]; y1=((-10/3)*(x1.^3-(x1-5).^3))+7.5*(x1-8).^2+150*(x1-7)+(57/2)*x1.^2238.25; figure (1) plot(x1,y1) grid %Region from x=7 to x=8 x2=[7:.1:8]; y2=((-10/3)*(x2.^3-(x2-5).^3))+150*(x2-7)+(57/2)*x2.^2-238.25; figure (2) plot(x2,y2) grid %Region from x=5 to x=7 x3=[5:.1:7]; y3=((-10/3)*(x3.^3-(x3-5).^3))+(57/2)*x3.^2-238.25; figure (3) plot(x3,y3) grid %Region from x=0 to x=5 x4=[0:.1:5]; y4=((-10/3)*(x4.^3))+(57/2)*x4.^2-238.25; figure (4) plot(x4,y4) grid %Region from x=0 to x=10 figure (5) plot(x1,y1,x2,y2,x3,y3,x4,y4) grid title('slope diagram') a=[-10/3 57/2 0 -238.25] roots(a) a = -3.3333 ans = 7.1531 3.9357 -2.5388 28.5000 0 -238.2500 s lope diagram 200 150 100 50 0 -50 -100 -150 -200 -250 0 8.27 1 2 3 4 5 6 7 8 9 10 %Region from x=8 to x=10 x1=[8:.1:10]; y1=(-5/6)*(x1.^4-(x1-5).^4)+(15/6)*(x1-8).^3+75*(x1-7).^2+(57/6)*x1.^3238.25*x1; figure (1) plot(x1,y1) grid %Region from x=7 to x=8 x2=[7:.1:8]; y2=(-5/6)*(x2.^4-(x2-5).^4)+75*(x2-7).^2+(57/6)*x2.^3-238.25*x2; figure (2) plot(x2,y2) grid %Region from x=5 to x=7 x3=[5:.1:7]; y3=(-5/6)*(x3.^4-(x3-5).^4)+(57/6)*x3.^3-238.25*x3; figure (3) plot(x3,y3) grid %Region from x=0 to x=5 x4=[0:.1:5]; y4=(-5/6)*(x4.^4)+(57/6)*x4.^3-238.25*x4; figure (4) plot(x4,y4) grid %Region from x=0 to x=10 figure (5) plot(x1,y1,x2,y2,x3,y3,x4,y4) grid title('displacement curve') a = -3.3333 ans = 7.1531 3.9357 -2.5388 28.5000 0 -238.2500 Therefore, other than the end supports, there are no points of zero displacement along the beam. dis plac em ent c urve 0 -100 -200 -300 -400 -500 -600 0 1 2 3 4 5 6 7 8 9 10 8.39 Excel Solver solution: 8.40 The problem reduces to finding the value of n that drives the second part of the equation to 1. In other words, finding the root of f ( n) = ( ) n ( n −1) / n R −1 − 1 = 0 n −1 c Inspection of the equation indicates that singularities occur at x = 0 and 1. A plot indicates that otherwise, the function is smooth. 0.5 0 -0.5 0 0.5 1 1.5 -1 A tool such as the Excel Solver can be used to locate the root at n = 0.8518. 8.41 The sequence of calculation need to compute the pressure drop in each pipe is A = π (D / 2) 2 v= Q A Re = Dρv µ  1  f = root 4 .0 log Re f − 0.4 −  f   ( ∆P = f ) ρv 2 2D The six balance equations can then be solved for the 6 unknowns. The root location can be solved with a technique like the modified false position method. A bracketing method is advisable since initial guesses that bound the normal range of friction factors can be readily determined. The following VBA function procedure is designed to do this Option Explicit Function FalsePos(Re) Dim iter As Integer, imax As Integer Dim il As Integer, iu As Integer Dim xrold As Single, fl As Single, fu As Single, fr As Single Dim xl As Single, xu As Single, es As Single Dim xr As Single, ea As Single xl = 0.00001 xu = 1 es = 0.01 imax = 40 iter = 0 fl = f(xl, Re) fu = f(xu, Re) Do xrold = xr xr = xu - fu * (xl - xu) / (fl - fu) fr = f(xr, Re) iter = iter + 1 If xr <> 0 Then ea = Abs((xr - xrold) / xr) * 100 End If If fl * fr < 0 Then xu = xr fu = f(xu, Re) iu = 0 il = il + 1 If il >= 2 Then fl = fl / 2 ElseIf fl * fr > 0 Then xl = xr fl = f(xl, Re) il = 0 iu = iu + 1 If iu >= 2 Then fu = fu / 2 Else ea = 0# End If If ea < es Or iter >= imax Then Exit Do Loop FalsePos = xr End Function Function f(x, Re) f = 4 * Log(Re * Sqr(x)) / Log(10) - 0.4 - 1 / Sqr(x) End Function The following Excel spreadsheet can be set up to solve the problem. Note that the function call, =falsepos(F8), is entered into cell G8 and then copied down to G9:G14. This invokes the function procedure so that the friction factor is determined at each iteration. The resulting final solution is 8.42 The following application of Excel Solver can be set up: The solution is: 8.43 The results are 120 100 80 60 40 20 0 1 8.44 2 3 % Shuttle Liftoff Engine Angle % Newton-Raphson Method of iteratively finding a single root format long % Constants LGB = 4.0; LGS = 24.0; LTS = 38.0; WS = 0.230E6; WB = 1.663E6; TB = 5.3E6; TS = 1.125E6; es = 0.5E-7; nmax = 200; % Initial estimate in radians x = 0.25 %Calculation loop for i=1:nmax fx = LGB*WB-LGB*TB-LGS*WS+LGS*TS*cos(x)-LTS*TS*sin(x); dfx = -LGS*TS*sin(x)-LTS*TS*cos(x); xn=x-fx/dfx; %convergence check ea=abs((xn-x)/xn); if (ea<=es) fprintf('convergence: Root = %f radians \n',xn) theta = (180/pi)*x; fprintf('Engine Angle = %f degrees \n',theta) break end x=xn; x end % Shuttle Liftoff Engine Angle % Newton-Raphson Method of iteratively finding a single root % Plot of Resultant Moment vs Engine Anale format long % Constants LGB = 4.0; LGS = 24.0; LTS = 38.0; WS = 0.195E6; WB = 1.663E6; TB = 5.3E6; TS = 1.125E6; x=-5:0.1:5; fx = LGB*WB-LGB*TB-LGS*WS+LGS*TS*cos(x)-LTS*TS*sin(x); plot(x,fx) grid axis([-6 6 -8e7 4e7]) title('Space Shuttle Resultant Moment vs Engine Angle') xlabel('Engine angle ~ radians') ylabel('Resultant Moment ~ lb-ft') x = x = x = x = 0.25000000000000 0.15678173034564 0.15518504730788 0.15518449747125 convergence: Root = 0.155184 radians Engine Angle = 8.891417 degrees x 10 7 S pac e S huttle Res ultant M om ent vs E ngine A ngle 4 Res ultant M om ent ~ lb-ft 2 0 -2 -4 -6 -8 -6 -4 -2 0 2 4 E ngine angle ~ radians 8.45 This problem was solved using the roots command in Matlab. c = 1 -33 -704 -1859 roots(c) ans = 48.3543 -12.2041 -3.1502 Therefore, σ 1 = 48.4 Mpa 1 2 3 T 100 88.31493 80.9082 σ 2 = -3.15 MPa t 20 30.1157 36.53126 σ 3 = -12.20 MPa 6

Chapter 8 Space Shuttle Resultant Moment vs Engine Angle

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