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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /839 | |
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initial commit / add all books
Diffstat (limited to '839')
104 files changed, 4208 insertions, 0 deletions
diff --git a/839/CH1/EX1.1/Example_1_1.sce b/839/CH1/EX1.1/Example_1_1.sce new file mode 100755 index 000000000..83165a78a --- /dev/null +++ b/839/CH1/EX1.1/Example_1_1.sce @@ -0,0 +1,26 @@ +//clear//
+clear;
+clc;
+
+//Example 1.1
+
+//Solution
+//(a)
+//Using Eq.(1.6,), (1.26), and (1.27)
+//Let N = 1N
+N = 0.3048/(9.80665*0.45359237*0.3048); //[lbf]
+
+//(b)
+//Using (1.38), (1.16), (1.26), and (1.31)
+//Let B = 1 Btu
+B = 0.45359237*1000/1.8; //[cal]
+
+//(c)
+//Using Eq.(1.6), (1.14), (1.15), (1.26), (1.27), and (1.36)
+//Let P = 1 atm
+P = 1.01325*10^5*0.3048/(32.174*0.45359237*12^2); //[lbf/in.^2]
+
+//(d)
+// Using Eq.(1.8), (1.33), (1.37), (1.26), and (1.27)
+//Let hp = 1hp
+hp = 550*32.174*0.45359237*0.3048^2/1000; //[kW]
diff --git a/839/CH10/EX10.1/Example_10_1.sce b/839/CH10/EX10.1/Example_10_1.sce new file mode 100755 index 000000000..c7f39a641 --- /dev/null +++ b/839/CH10/EX10.1/Example_10_1.sce @@ -0,0 +1,18 @@ +//clear//
+clear;
+clc;
+
+//Exmple 10.1
+//Given
+T1 = 32; //[F]
+T2 = 200; //[F]
+k1 = 0.021; //[Btu/ft-h-F]
+k2 = 0.032; //[Btu/ft-h-F]
+A = 25; //[ft^2]
+B = 6/12; //[ft]
+//average temperature and thermal condutivity of the wall
+Tavg = (40+180)/2; //[F]
+kbar = k1+(Tavg-T1)*(k2-k1)/(T2-T1); //[Btu/ft-h-F]
+delta_T = 180-40; //[F]
+//Using Eq.(10.5)
+q = kbar*A*delta_T/B //[Btu/h]
diff --git a/839/CH10/EX10.2/Example_10_2.sce b/839/CH10/EX10.2/Example_10_2.sce new file mode 100755 index 000000000..f1a3690c2 --- /dev/null +++ b/839/CH10/EX10.2/Example_10_2.sce @@ -0,0 +1,35 @@ +//clear//
+clear;
+clc;
+
+//Example 10.2
+//Given
+delta1 = 4.5/12 ;//[ft]
+k1 = 0.08; //[Btu/ft-h-F]
+delta2 = 9/12; //[ft]
+k2 = 0.8; //[ft]
+Tin = 1400 //[F]
+Tout = 170 //[F]
+Rc = 0.5; //[ft^2-h-F/Btu]
+//(a)
+//Considering unit cross sectional area
+A = 1; //[ft^2]
+RA = delta1/k1; //[ft^2-h-F/Btu]
+RB = delta2/k2; //[ft^2-h-F/Btu]
+R = RA+RB; //[ft^2-h-F/Btu]
+delta_T = Tin-Tout; //[F] overall temperature drop
+//Using Eq.(10.9)
+q = A*delta_T/R //[Btu/h]
+
+//(b)
+//The temperature drop in one series of resistances is to the
+//individual resistance as the overall temperature drop is to the
+//overall resistance, or
+delta_TA = RA*delta_T/R; //[F]
+//Temperature at the inteface
+Tf = Tin-delta_TA //[F]
+
+//(c) The total resistance will now include contact resistance
+R = R+Rc; //[ft^2-h-F/Btu]
+//the heat loss from unit square area
+q = delta_T/R //[Btu/h]
diff --git a/839/CH10/EX10.3/Example_10_3.sce b/839/CH10/EX10.3/Example_10_3.sce new file mode 100755 index 000000000..2391d8783 --- /dev/null +++ b/839/CH10/EX10.3/Example_10_3.sce @@ -0,0 +1,20 @@ +//clear//
+clear;
+clc;
+
+//Example 10.3
+//Given
+r1 = 60/2; //[mm]
+r2 = (50+r1); //[mm]
+k2 = 0.055; //[W/m-C]
+r3 = 40+r2; //[mm]
+k3 = 0.05; //[W/m-C]
+To = 30; //[C]
+Ti = 150; //[C]
+//Logrithimic mean for silica layer and cork layer
+rl_s = (r2-r1)/log(r2/r1) //[mm]
+rl_c = (r3-r2)/log(r3/r2) //[mm]
+
+//Using Eq.(10.15) and Eq.(10.14) simulataneously
+//And Adding these two Equations
+qbyL = (Ti-To)/4.13 //[W/m]
diff --git a/839/CH10/EX10.4/Example_10_4.sce b/839/CH10/EX10.4/Example_10_4.sce new file mode 100755 index 000000000..aa28fe88d --- /dev/null +++ b/839/CH10/EX10.4/Example_10_4.sce @@ -0,0 +1,24 @@ +//clear//
+clear;
+clc;
+
+//Example 10.4
+//Given
+k = 0.075; //[Btu/ft-h-F]
+rho = 56.2; //[lb/ft^3]
+Cp = 0.40; //[Btu/lb-F]
+s = 0.5/12; //[ft.]
+Ts = 250; //[F]
+Ta = 70; //[F]
+Tb_bar = 210; //[F]
+
+//(a)
+Temp_diff_ratio = (Ts-Tb_bar)/(Ts-Ta);
+alpha = k/(rho*Cp);
+// From Fig.10.6
+N_Fo =0.52;
+tT = N_Fo*s^2/alpha //[h]
+
+//(b)
+//Substituting in Eq.(10.23)
+QTbyA = s*rho*Cp*(Tb_bar-Ta) //[Btu/ft^2]
diff --git a/839/CH10/EX10.5/Example_10_5.sce b/839/CH10/EX10.5/Example_10_5.sce new file mode 100755 index 000000000..df4ce5e9f --- /dev/null +++ b/839/CH10/EX10.5/Example_10_5.sce @@ -0,0 +1,22 @@ +//clear//
+clear;
+clc;
+
+//Example 10.5
+//Given
+Ts = -20; //[C]
+Ta = 5; //[C]
+T = 0; //[C]
+t = 12; //[h]
+alpha = 0.0011; //[m^2/h]
+
+//(a)
+Temp_diff_ratio = (Ts-T)/(Ts-Ta);
+//From Fig.(10.8),
+Z = 0.91;
+//therefore depth
+x = Z*2*sqrt(alpha*t) //[m]
+
+//(b)
+//From Eq.(10.27), the penetration distance is
+x_rho = 3.64*sqrt(alpha*t) //[m]
diff --git a/839/CH11/EX11.1/Example_11_1.sce b/839/CH11/EX11.1/Example_11_1.sce new file mode 100755 index 000000000..e311a5909 --- /dev/null +++ b/839/CH11/EX11.1/Example_11_1.sce @@ -0,0 +1,20 @@ +//clear//
+clear;
+clc;
+
+//Example 11.1
+//From Appendix 5
+Di = 1.049/12; //[ft]
+Do = 1.315/12; //[ft]
+xw = 0.133/12; //[ft]
+km = 26; //[Btu/ft-h-F]
+//Using Eq.(10.15) for Logrithmic mean diameter DL_bar
+DL_bar = (Do-Di)/log(Do/Di); //[ft]
+//From Table 11.1
+hi = 180; //[Btu/ft^2-h-F]
+ho = 300; //[Btu/ft^2-h-F]
+hdi = 1000; //[Btu/ft^2-h-F]
+hdo = 500; //[Btu/ft^2-h-F]
+
+//Overall heat transfer coefficient
+Uo = 1/(Do/(Di*hdi)+Do/(Di*hi)+(xw*Do)/(km*DL_bar)+1/ho+1/hdo) //[Btu/ft^2-h-F]
diff --git a/839/CH12/EX12.1/Example_12_1.sce b/839/CH12/EX12.1/Example_12_1.sce new file mode 100755 index 000000000..e8cfa158e --- /dev/null +++ b/839/CH12/EX12.1/Example_12_1.sce @@ -0,0 +1,15 @@ +//clear//
+clear;
+clc;
+
+//Example 12.1
+To = 230; //[F]
+Ti = 80; //[F]
+//Using Table 12.1
+hi = 400; //[Btu/ft^2-h-F]
+ho = 500; //[Btu/ft^2-h-F]
+//From Appendix 6
+Di = 0.620; //[in.]
+Do = 0.750; //[in.]
+//Using Eq.(12.39)
+detla_Tt = (1/hi)/(1/hi+(Di/(Do*ho)))*(To-Ti)
diff --git a/839/CH12/EX12.2/Example_12_2.sce b/839/CH12/EX12.2/Example_12_2.sce new file mode 100755 index 000000000..d8ee5f923 --- /dev/null +++ b/839/CH12/EX12.2/Example_12_2.sce @@ -0,0 +1,77 @@ +//clear//
+clear;
+clc;
+
+//Example 12.2
+//Given
+Tb1 = 141; //[F]
+Tb2 = 79; //[F]/
+Tw1 = 65; //[F]
+Tw2 = 75; //[F]
+Vb_bar = 5; //[ft/s]
+rho_b = 53.1; //[lb/ft^3]
+mu_b = 1.16; //[lb/ft-h], Form Appendix 9
+k_b = 0.089; //[Btu/ft-h-F], From Appendix 13
+Cp_b = 0.435; //[Btu/lb-F], From Appendix 16
+//Using Appndix 14
+rho_w = 62.3; //[lb /ft^3]
+mu_w = 2.34; //[lb/ft-h]
+k_w = 0.346; //[Btu/ft-h-F]
+Cp_w = 1; //[Btu/lb-F]
+
+
+//Soultion
+Tavg_b = (Tb1+Tb2)/2; //[F]
+Tavg_w = (Tw1+Tw2)/2; //[F]
+Dit = 0.745/12; //[ft]
+Dot = 0.875/12; //[ft]
+//Using Appendix 5
+//The inside diameter of the jacket
+Dij = 1.610/12; //[ft]
+//From Appendix 6, the inside sectional area of the copper tube (for a 7/8 in. BWG 16 tube)
+S = 0.00303; //[ft^2]
+//Equivalent diameter of the annular jacket space
+De = 4*(%pi/4*(Dij^2-Dot^2)/(%pi*(Dij+Dot))); //[ft]
+mb_dot = Vb_bar*rho_b*S; //[lb/s]
+//The rate of heat flow
+q = mb_dot*Cp_b*(Tb1-Tb2); //[Btu/s]
+//mass flow rate of water
+mw_dot = q/(Cp_w*(Tw2-Tw1)); //[lb/s]
+//Water velocity
+Vw_bar = mw_dot/(%pi/4*(Dij^2-Dot^2)*rho_w); //[ft/s]
+//Reynolds number for benzene and water
+Nre_b = Dit*Vb_bar*rho_b*3600/mu_b;
+Nre_w = De*Vw_bar*rho_w*3600/mu_w;
+//Prandtl Number for benzene and water
+Npr_b = Cp_b*mu_b/k_b;
+Npr_w = Cp_w*mu_w/k_w;
+
+//Preliminary estimates of the coefficients are obtained using Eq.(12.32), omitting the
+//correction for viscosity ratio:
+//Benzene
+hi = 0.023*Vb_bar*3600*rho_b*Cp_b/(Nre_b^0.2*Npr_b^(2/3)); //[Btu/ft^2-h-F]
+//Water
+ho = 0.023*Vw_bar*3600*rho_w*Cp_w/(Nre_w^0.2*Npr_w^(2/3)); //[Btu/ft^2-h-F]
+//Using Eq.(12.39)
+//Temperature drop over the benzene resistance
+delta_Ti = (1/hi)/(1/hi+Dit/(Dot*ho))*(Tavg_b-Tavg_w); //[F]
+Tw = Tavg_b - delta_Ti; //[F]
+
+//The viscosities of the liquids at Tw
+muw_b = 1.45; //[lb/ft-h]
+muw_w = 2.42*0.852; //[lb/ft-h]
+//Using Eq.(12.24), viscosity-correction factors phi are
+phi_b = (mu_b/muw_b)^0.14;
+phi_w = (mu_w/muw_w)^0.14;
+//The corrected coefficients are
+hi = hi*phi_b; //[Btu/ft^2-h-F]
+ho = ho*phi_w; //[Btu/ft^2-h-F]
+//The temperature drop over the benzene resistance and the wall temperature
+delta_Ti = (1/hi)/(1/hi+Dit/(Dot*ho))*(Tavg_b-Tavg_w); //[F]
+Tw = Tavg_b - delta_Ti //[F]
+//This is so close to previously calculated wall temperature that a second approximation
+//is unnecessary
+//Using Eq.(11.29), neglecting the resistance of the tube wall
+Uo = 1/(Dot/(Dit*hi)+1/ho); //[Btu/ft^2-h-F]
+disp('The overall coefficient is');
+disp('Btu/ft^2-h-F',Uo);
diff --git a/839/CH12/EX12.3/Example_12_3.sce b/839/CH12/EX12.3/Example_12_3.sce new file mode 100755 index 000000000..9f1442e18 --- /dev/null +++ b/839/CH12/EX12.3/Example_12_3.sce @@ -0,0 +1,40 @@ +//clear//
+clear;
+clc;
+
+//Example 12.3
+//Given
+L = 15; //[ft]
+k = 0.082; //[Btu/ft-h-F]
+Cp = 0.48; //[Btu/lb-F]
+T1 = 150; //[F]
+T2 = 250; //[F]
+Tw = 350; //[F]
+//From Table 12.3
+mu1 = 6; //[cP]
+mu2 = 3.3; //[cP]
+mu_w = 1.37; //[cP]
+mu = (mu1+mu2)/2; //[cP]
+//From Appendix 5
+D = 0.364/12; //[ft]
+//viscosity-correction factor phi is
+phi = (mu/mu_w)^0.14;
+//Assuming Laminar flow and Graetz number large enough to apply Eq.(12.25)
+//Using Eq.(12.25)
+//h = k/D*2*phi*(Cp*mdot/(k*L))^(1/3);
+//To use Eq.(12.18)
+Log_T = ((Tw-T1)-(Tw-T2))/log((Tw-T1)/(Tw-T2)); //[F]
+//From Eq.(12.18)
+//h = Cp*100*mdot/(%pi*D*L*Log_T)
+//From Eq.(12.25) and Eq.(12.18)
+mdot = (4.69/0.233)^(3/2); //[lb/h]
+//and
+h = 0.233*mdot; //[Btu/ft^2-h-F]
+disp('lb/h',mdot,'oil flow rate')
+
+disp('Btu/ft^2-h-F',h,'Expected Coefficient')
+Ngz = mdot*Cp/(k*L);
+//This is large enough so that Eq.(12.25) applies,
+//Reynolds Number
+Nre = D*mdot/((%pi/4*D^2)*mu*2.42);
+//Nre is in Laminar Range
diff --git a/839/CH12/EX12.4/Example_12_4.sce b/839/CH12/EX12.4/Example_12_4.sce new file mode 100755 index 000000000..63dd872b7 --- /dev/null +++ b/839/CH12/EX12.4/Example_12_4.sce @@ -0,0 +1,53 @@ +//clear//
+clear;
+clc;
+
+//Example 12.4
+//Given
+P = 1; //[atm]
+Vbar = 1.5; //[ft/s]
+Ti = 68; //[F]
+To = 188; //[F]
+Tw = 220; //[F]
+Tbar = (Ti+To)/2; //[F]
+D = 2.067/12; //[ft], from Appendix 5
+mu = 0.019; //[cP], at 128[F], from Appendix 8
+rho = 29/359*(492/(68+460)); //[lb/ft^3], at 68[F]
+G = Vbar*rho*3600; //[lb/ft^2-h]
+Nre = D*G/(mu*2.42);
+g = 32.14;
+//Hence the flow is laminar
+//Applying Eq.(12.25)
+Cp = 0.25; //[Bu/lb-F], at 128[F], Appendix 15
+k = 0.0163; //[Btu/ft-h-F], at 128[F], Appendix 12
+//By linear interpolation
+mu_w = 0.021; //[cP], Appendix 5
+//internal cross sectional area of pipe is
+S = 0.02330; //[ft^2], Appendix 5
+//mass flow rate
+mdot = G*S; //[lb/h]
+//the heat load
+q = mdot*Cp*(To-Ti); //[Btu/h]
+//The logrithmic mean temperature difference is
+delta_T1 = Tw-To; //[F]
+delta_T2 = Tw-Ti; //[F]
+Log_T = (delta_T1-delta_T2)/log(delta_T1/delta_T2); //[F]
+
+//heat transfer coefficient h = q/A*Log_T
+//A = 0.541*L
+//Also from Eq.(12.25), the heat transfer coefficient is
+//h = 2*k/D*(mdot*Cp/k*L)^(1/3)*(mu/mu_w)^(1/4)
+//Equating the two realtionships for h
+L = (6.820/0.9813)^(3/2); // [ft]
+//This result is corrected for the effect of natural convection
+//To use Eq.(12.80)
+beeta = 1/(460+Tbar) ;//[R^-1], at 128[F]
+delta_T = Tw-Tbar; //[F]
+rho = 0.0676; //[lb/ft^3]
+//Grashof number
+Ngr = D^3*rho^2*g*beeta*delta_T/(mu*6.72*10^-4)^2;
+//From Eq.(12.80)
+phi_n = 2.25*(1+0.01*Ngr^(1/3))/log10(Nre);
+//this is factor is used to correct the value of L
+L = L/phi_n; //[ft]
+disp('ft',L,'lenght of heated section is')
diff --git a/839/CH13/EX13.1/Example_13_1.sce b/839/CH13/EX13.1/Example_13_1.sce new file mode 100755 index 000000000..d2acdc477 --- /dev/null +++ b/839/CH13/EX13.1/Example_13_1.sce @@ -0,0 +1,60 @@ +//clear//
+clear;
+clc;
+
+//Example 13.1
+//Given
+Pa = 1; //[atm]
+lambda = 139.7; //[Btu/lb]
+L = 5; //[ft]
+Tw = 175; //[F]
+hi = 400; //[Btu/ft^2-h-F]
+g = 4.17*10^8; //[ft/h^2]
+Th = 270; //[F]
+rho_f = 65.4; //[lb/ft^3]
+kf = 0.083; //[Btu/ft-h-F], from Appendix 13
+muf = 0.726; //[lb/ft-h], from Appendix 9
+Do = 0.75/12; //[ft]
+Di = Do-(2*0.065)/12; //[f]
+//(a)
+Twall = 205; //[F]
+err = 50;
+h = 1.13;
+while(err>10)
+delta_To = Th-Twall;
+//from Eq.(13.11)
+Tf = Th-3*(Th-Twall)/4; //[F]
+h = h*(kf^3*rho_f^2*g*lambda/(delta_To*L*muf))^(1/4); //[Btu/ft^2-h-F]
+//Using Eq.(12.29)
+delta_Ti = 1/hi/(1/hi+Di/(Do*h))*(Th-Tw); //[F]
+Twall_new = Tw + delta_Ti; //[F]
+err = Twall_new-Twall; //[F]
+Twall = Twall_new; //[F]
+end
+//To ckeck whether the flow is actually laminar
+Ao = 0.1963*L; //[ft^2], from Appendix 6
+//the rate of heat transfer
+q = h*Ao*(Th-Twall); //[Btu/h]
+mdot = q/lambda; //[lb/ft-h]
+disp('[Btu/ft^2-h-F]',h,'coefficient of chlorobenzene is')
+
+
+//(b)
+//For a horizontal condenser, Using Eq.(13.16)
+N =6;
+Twall = 215; //[F]
+err = 50;
+h = 0.725;
+muf = 0.68; //[lb/ft-h], from Appendix 6
+while(err>10)
+delta_To = Th-Twall;
+//from Eq.(13.11)
+Tf = Th-3*(Th-Twall)/4; //[F]
+h = h*(kf^3*rho_f^2*g*lambda/(6*delta_To*Do*muf))^(1/4); //[Btu/ft^2-h-F]
+//Using Eq.(12.29)
+delta_Ti = 1/hi/(1/hi+Di/(Do*h))*(Th-Tw); //[F]
+Twall_new = Tw + delta_Ti; //[F]
+err = Twall_new-Twall; //[F]
+Twall = Twall_new; //[F]
+end
+disp('[Btu/ft^2-h-F]',h,'coefficient of chlorobenzene is')
diff --git a/839/CH13/EX13.2/Example_13_2.sce b/839/CH13/EX13.2/Example_13_2.sce new file mode 100755 index 000000000..9c7c857b2 --- /dev/null +++ b/839/CH13/EX13.2/Example_13_2.sce @@ -0,0 +1,34 @@ +//clear//
+clear;
+clc;
+
+//Example 13.2
+//Given
+P = 2; //[atm]
+
+//(a)
+//From Fig. 13.7
+//Critical pressure of benzene
+Pc = 47.7; //[atm]
+PbyPc = P/Pc;
+//From Fig. 13.7 the ordinate (q/A)max/Pc is about 190, and
+qbyA_max = 190*Pc*14.696; //[Btu/h-ft^2]
+disp('Btu/h-ft^2',qbyA_max,'The maximum heat flux is')
+//Also from Fig. 13,7
+delta_Tc = 62; //[F]
+disp('F',delta_Tc,'The critical temperature difference is')
+// film coefficient
+h = qbyA_max/delta_Tc; //[Btu/h-ft^2-F]
+disp('Btu/h-ft^2-F',h,'The film coefficient is')
+
+//(b)
+//Given
+P = 0.2; //[atm]
+PbyPc = P/Pc;
+//Using Eq.(13.20)
+//noting that lambda, sigma and rho_L are nearly constant and rho_L>rho_V
+// qbyA_max~rho_V^(1/2)~P^(1/2)
+qbyA_max = qbyA_max*(0.2/2)^(1/2); //[Btu/h-ft^2]
+disp('Btu/h-ft^2',qbyA_max,'The maximum heat flux is')
+//The critical temperature difference would be greater than 100 [F] and
+//the film coefficient would be less than 410 [Btu/h-ft^2-F]
diff --git a/839/CH14/EX14.1/Example_14_1.sce b/839/CH14/EX14.1/Example_14_1.sce new file mode 100755 index 000000000..a3bc589ae --- /dev/null +++ b/839/CH14/EX14.1/Example_14_1.sce @@ -0,0 +1,30 @@ +//clear//
+clear;
+clc;
+
+//Example 14.1
+//Given
+d = 150; //[mm]
+T1 = 300+272; //[K]
+T3 = 25+273; //[K]
+eps1 = 0.56;
+eps2 = 1.0;
+eps3 = eps1;
+sigma = 5.672
+
+//(a)
+//Using Eq.(14.38)
+//q12 = sigma*A1*F12*(T1^4-T2^4)
+//q23 = sigma*A2*F23*(T2^4-T3^4)
+//At equilibrium, q12=q23
+//From Eq.(14.39)
+F12 = 1/(1/eps1+1/eps2-1)
+F23 = F12;
+//A1 = A2
+T2 = (100*((T1/100)^4+(T3/100)^4)^(1/4))/2^(1/4); //[K]
+disp('F',T2,'the temperature of lacquered sheet is')
+
+//(b)
+//From Eq.(14.38), heat flux
+q12byA = sigma*F12*((T1/100)^4-(T2/100)^4); //[W/m^2]
+disp('W/m^2',q12byA,'the heat flux is')
diff --git a/839/CH15/EX15.1/Example_15_1.sce b/839/CH15/EX15.1/Example_15_1.sce new file mode 100755 index 000000000..19ec00188 --- /dev/null +++ b/839/CH15/EX15.1/Example_15_1.sce @@ -0,0 +1,34 @@ +//clear//
+clear;
+clc;
+
+//Example 15.1
+//Given
+Ds = 35/12; //[ft]
+Do = 0.75/12; //[ft]
+p = 1/12; //[ft]
+P = 1; //[ft]
+mdot = 10^5; //[lb/h]
+mu_60 = 0.70; //[cP], at 60 [F], from Appendix 9
+mu_140 = 0.38; //[cP], at 140 [F], from Appendix 9
+Cp = 0.41; //[Btu/lb-F], from Appendix 16
+k = 0.092; //[Btu/ft-h-F], from Appendix 13
+
+//Shell side coefficient is found using Donohue Eq.(15.4)
+//From Eq.(15.2), the area for crossflow is
+Sc = 2.9167*P*(P-Do/p); //[ft^2]
+//The number of tubes in the baffle window is approximately equal to the fractional
+//area of the window f times the total nmber of tubes. For a 25 percent baffle
+f = 0.1955
+Nb = f*828;
+//Nb~161
+Nb = 161;
+//Using Eq.(15.1), area of the baffle window
+Sb = (f*%pi*Ds^2/4)-(Nb*%pi*Do^2/4); //[ft^2]
+//Using Eq.(15.3), the mass velocities are
+Gc = mdot/Sc; //[lb/ft^2-h]
+Gb = mdot/Sb; //[lb/ft^2-h]
+Ge = sqrt(Gc*Gb); //[lb/ft^2-h]
+//Using Eq.(15.4)
+ho = k/Do*(0.2*(Do*Ge/(mu_60*2.42))^0.6*(Cp*mu_60*2.42/k)^0.33*(mu_60/mu_140)^0.14);//[Btu/ft^2-h-F]
+disp('Btu/ft^2-h-F',ho,'The individual heat transfer coefficent of benzene is')
diff --git a/839/CH15/EX15.2/Example_15_2.sce b/839/CH15/EX15.2/Example_15_2.sce new file mode 100755 index 000000000..3dbe02fbb --- /dev/null +++ b/839/CH15/EX15.2/Example_15_2.sce @@ -0,0 +1,26 @@ +//clear//
+clear;
+clc;
+
+//Example 15.2
+//Given
+Tca = 70; //[C]
+Tcb = 130; //[C]
+Tha = 240; //[C]
+Thb = 120; //[C]
+//Solution
+//Using Eq.(15.7) and (15.8)
+neta_h = (Tcb-Tca)/(Tha-Tca);
+Z = (Tha-Thb)/(Tcb-Tca);
+//From Fig 15.7a, the correction factor is found
+Fg = 0.735;
+//the temperature drops are
+//At shell inlet:
+deltaT_i = Tha-Tcb; //[C]
+//At shell outlet:
+deltaT_o = Thb-Tca; //[C]
+Log_T = (deltaT_i-deltaT_o)/log(deltaT_i/deltaT_o);
+// the correct value of Log_T is
+Log_T = Fg*Log_T; //[C]
+disp('C',Log_T,'The correct mean emperature drop is')
+//Because of low value of Fg, a 1-2 heat exchanger is not suitable for this duty
diff --git a/839/CH15/EX15.3/Example_15_3.sce b/839/CH15/EX15.3/Example_15_3.sce new file mode 100755 index 000000000..492e6c56a --- /dev/null +++ b/839/CH15/EX15.3/Example_15_3.sce @@ -0,0 +1,25 @@ +//clear//
+clear;
+clc;
+
+//Exapmle 15.3
+//Given
+Tca = 70; //[C]
+Tcb = 130; //[C]
+Tha = 240; //[C]
+Thb = 120; //[C]
+//Solution
+//Using Eq.(15.7) and (15.8)
+neta_h = (Tcb-Tca)/(Tha-Tca);
+Z = (Tha-Thb)/(Tcb-Tca);
+//Using Fig 15.7b, the correction factor is
+Fg = 0.945;
+//the temperature drops are
+//At shell inlet:
+deltaT_i = Tha-Tcb; //[C]
+//At shell outlet:
+deltaT_o = Thb-Tca; //[C]
+Log_T = (deltaT_i-deltaT_o)/log(deltaT_i/deltaT_o);
+// the correct value of Log_T is
+Log_T = Fg*Log_T; //[C]
+disp('C',Log_T,'The correct mean emperature drop is')
diff --git a/839/CH15/EX15.4/Example_15_4.sce b/839/CH15/EX15.4/Example_15_4.sce new file mode 100755 index 000000000..2a1724600 --- /dev/null +++ b/839/CH15/EX15.4/Example_15_4.sce @@ -0,0 +1,54 @@ +//clear//
+clear;
+clc;
+
+//Example 15.4
+//Given
+N = 28;
+xF = 0.5/12; // [ft]
+yF = 0.035/12; //[ft]
+km = 26; // [Btu/ft-h-F]
+AT = 2.830; //[ft^2/ft]
+Ab = 0.416; //[ft^2/ft]
+hi = 1500; //[Btu/ft^2-h-F]
+G = 5000; //[lb/h-ft^2]
+Tavg = 130; //[F]
+Tw = 250; //[F]
+mu = 0.046; //[lb/ft-h], from Appendix 8
+Cp = 0.25; //[Btu/lb-F], from Appendix 15
+k = 0.0162; //[Btu/ft-h-F], from Appendix 12
+ID_shell = 3.068/12; //[ft], from Appendix 5
+OD_pipe = 1.9/12; //[ft], from Appendix 5
+//cross sectional area of shell space
+Ac = %pi/4*(ID_shell^2-OD_pipe^2)-N*xF*yF //[ft^2]
+//The perimeter of air space
+Ap = %pi*ID_shell+AT; //[ft]
+//hydraulic radius
+rh = Ac/Ap; //[ft]
+//equivalent diameter
+De = 4*rh; //[ft]
+//Reynolds Number
+Nre = De*h/mu
+//In computing mu_w the resistance of the wall and the steam film
+//are considered negligible, so
+mu_w = 0.0528; //[lb/ft-h]
+Npr = mu*Cp/k
+//Using Fig. 15.17, the heat transfer factor is
+jh = 0.0031;
+ho = jh*Cp*G*(mu/mu_w)^0.14/Npr^(2/3); //[Btu/ft^2-h-F]
+
+//For rectangular fins, disreagrding the contribution of the ends of the fins to
+//the perimeter, Lp = 2L and S = Lyf, where yf is the fin thickness and L is the
+//length of the fin. Then, from Eq.(15.11)
+aFxF = xF*sqrt(2*ho/(km*yF));
+//From Fig. 15.16
+netaF = 0.93;
+Dt = 1.610/12; //[ft], from Appendix 5
+DLbar = (OD_pipe-Dt)/log(OD_pipe/Dt); //[ft]
+Ai = %pi*Dt*1.0; //[ft^2]
+AF = AT-Ab; //[ft^2/ft]
+xw = (OD_pipe-Dt)/2; //[ft]
+
+//Using Eq.(15.10), the overall coefficient
+Ut = 1/(Ai/(ho*(netaF*AF+Ab))+(xw*Dt/(km*DLbar))+1/hi);//[Btu/ft^2-h-F]
+disp('Btu/ft^2-h-F',Ut,'The overall heat transfer coefficent is')
diff --git a/839/CH16/EX16.1/Example_16_1.sce b/839/CH16/EX16.1/Example_16_1.sce new file mode 100755 index 000000000..c25334b10 --- /dev/null +++ b/839/CH16/EX16.1/Example_16_1.sce @@ -0,0 +1,52 @@ +//clear//
+clear;
+clc;
+
+//Example 16.1
+//Given
+mdot = 20000; //[lb/h]
+xin = 0.20;
+xout = 0.50;
+Pg = 20; //[lbf/in.^2]
+Pabs = 1.93; //[lbf/in.^2]
+U = 250; //[Btu/ft^2-h-F]
+Tf = 100; //[F]
+
+//Solution
+//the amount of water in feed and thick liquor, from material balance
+w_feed = 80/20; //[lb/per pound of solid]
+w_liquor = 50/50; //[lb/per pound of solid]
+//water evaporated
+w_eva = w_feed-w_liquor; //[lb/per pound of solid]
+//or
+w_eva = w_eva*mdot*xin; //[lb/h]
+//Flow raye of thick liquor is
+ml_dot = mdot-w_eva //[lb/h]
+
+//Steam consumed
+//Since with strong solutions of NaOH the heat of dilution is not negligible,
+//the rate of heat transfer is found from Eq.(16.4) and Fig. 16.8.
+//The vaporiztion temperature of the 50 percent solution at a pressure of 100 mmHg
+//is found as follows
+Tb_w = 124; //[F], at 100 mmHg, from Appendix 7
+Tb_s = 197; //[F], from Fig. 16.8
+BPE = Tb_s-Tb_w; //[F]
+//From Fig. 16.8, the enthalpies of the feed and thick liquor are found
+Hf = 55; //[Btu/lb], 20% solids, 100 [F]
+H = 221; //[Btu/lb], 50% solids, 197 [F]
+//Enthalpy of the leaving water vapor is found from the steam table
+Hv = 1149; //[Btu/lb], At 197 [F] and 1.93 [lbf/in.^2]
+//Enthalpy of the vapor leaving the evaporator
+lambda_s = 939;//[Btu/lb], At 20 [lbf/in.^2], from Appendix 7
+//Using Eq.(16.4), the rate of heat transfer and steam consumption
+q = (mdot-ml_dot)*Hv + ml_dot*H - mdot*Hf; //[Btu/h]
+ms_dot = q/lambda_s; //[lb/h]
+disp('lb/h',ms_dot,'steam consumed is')
+//Economy
+Economy = ml_dot/ms_dot
+disp(Economy,'Economy')
+//Heating Surface
+//The condensation temperature of the steam is 259 [F], the heating area required is
+
+A = q/(U*(259-197)) //[ft^2]
+disp('ft^2',A,'heating area required is')
diff --git a/839/CH16/EX16.2/Example_16_2.sce b/839/CH16/EX16.2/Example_16_2.sce new file mode 100755 index 000000000..478b47c66 --- /dev/null +++ b/839/CH16/EX16.2/Example_16_2.sce @@ -0,0 +1,25 @@ +//clear//
+clear;
+clc;
+
+//Example 16.2
+//Given
+Ti = 108; //[C]
+Tl = 52; //[C]
+U1 = 2500; //[W/m^2]
+U2 = 2000; //[W/m^2]
+U3 = 1000; //[W/m^2]
+
+//Solution
+//Total temperature drop
+delta_T = Ti-Tl; //[C]
+//From Eq.(16.13), the temperature drops in several effects will be
+//approximaely inversely proportional to the coeficients. Thus
+delta_T1 = 1/U1/(1/U1+1/U2+1/U3)*delta_T; //[C]
+delta_T2 = 1/U2/(1/U1+1/U2+1/U3)*delta_T; //[C]
+delta_T3 = 1/U3/(1/U1+1/U2+1/U3)*delta_T; //[C]
+//Consequently the boiling points will be
+Tb1 = Ti-delta_T1; //[C]
+Tb2 = Tb1-delta_T2; //[C]
+disp('C',Tb1,'The boiling point in the first effect is')
+disp('C',Tb2,'The boiling point in the second effect is')
diff --git a/839/CH16/EX16.3/Example_16_3.sce b/839/CH16/EX16.3/Example_16_3.sce new file mode 100755 index 000000000..53e78a4aa --- /dev/null +++ b/839/CH16/EX16.3/Example_16_3.sce @@ -0,0 +1,28 @@ +//clear//
+clear;
+clc;
+
+//Example 16.3
+//Given
+mdot_ft = 60000; //[lb/h]
+xin = 0.10;
+Tin = 180; //[F]
+xout = 0.50
+Ps = 50; //[lbf/in.^2]
+Tc = 100; //[F]
+
+//Solution
+//From Table 16.2
+U1 = 700; //[Btu/ft^2-h-F]
+U2 = 1000; //[Btu/ft^2-h-F]
+U3 = 800; //[Btu/ft^2-h-F]
+//The total rate of evaporation is calculated from an overall material balance
+//assuming the solds go through the evaporator without loss
+//Table 16.3
+mdot_fs = 6000; //[lb/h]
+mdot_fw = 54000; //[lb/h]
+mdot_lt = 12000; //[lb/h]
+mdot_ls = 6000; //[lb/h]
+mdot_lw = 6000; //[lb/h]
+w_evap = mdot_ft-mdot_fs; //[lb/h]
+
diff --git a/839/CH17/EX17.1/Example_17_1.sce b/839/CH17/EX17.1/Example_17_1.sce new file mode 100755 index 000000000..95f209799 --- /dev/null +++ b/839/CH17/EX17.1/Example_17_1.sce @@ -0,0 +1,40 @@ +//clear//
+clear;
+clc;
+
+//Example 17.1
+//Given
+yb = 0.30;
+
+//Let
+Vb = 100; //[mol]
+Ace_in = yb*Vb; //[mol]
+Air_in = Vb-Ace_in; //[mol]
+//97 percent acetone aborbed, Acetone leaving is
+Ace_out = 0.03*Ace_in; //[mol]
+ya = Ace_out/(Air_in+Ace_out);
+//Acetone absorbed
+Ace_abs = Ace_in-Ace_out; //[mol]
+//10 percent acetone in the leaving solution and no acetone in the entering oil
+Lb = Ace_abs/0.1; //[mol]
+La = Lb-Ace_abs; //[mol]
+//To find out as intermediate point on the operating line, making an acetone balance
+//around the top part of the tower, assuming a particular value of yV the moles of
+//acetone left in the gas.
+for i=1:30
+ y(i) = i/(i+Air_in);
+//The moles of acetone lost by the gas in the secion, must equal to the moles gained by //the liquid
+Ace_lost = i-Ace_out; //[mol]
+//Hence
+x(i) = Ace_lost/(La+Ace_lost);
+end
+xe = linspace(0.001,0.15,100);
+ye = 1.9*xe;
+
+plot(x,y)
+plot(xe,ye,'r')
+xlabel('x')
+ylabel('y')
+legend('Operating line','Equilibrium line')
+title('Diagram Example 17.1')
+//The number of ideal stages determined from Fig is 4
diff --git a/839/CH17/EX17.2/Example_17_2.sce b/839/CH17/EX17.2/Example_17_2.sce new file mode 100755 index 000000000..71a69f644 --- /dev/null +++ b/839/CH17/EX17.2/Example_17_2.sce @@ -0,0 +1,44 @@ +//clear//
+clear;
+clc;
+
+//Example 17.2
+//Given
+Nreal = 7;
+VbyL = 1.5;
+m = 0.8;
+yb = 0;
+xb_star = 0;
+//xb=0.1*xa;
+
+//(a)
+//Stripping Factor
+S = m*VbyL;
+//From an ammonia balance,
+//ya =0.9*xa/VbyL;
+//Also
+//xa_star = ya/m
+//Using Eq.(17.28)
+//N = ln((xa-0.75*xa)/(0.1*xa-0))/ln(S)
+N = log(0.25/0.1)/log(S);
+disp(N,'Number of ideal trays required are')
+stage_eff = N/Nreal*100;
+disp('%',stage_eff,'Stage Efficiency is')
+
+//(b)
+VbyL = 2;
+S = m*VbyL;
+//Then,
+//Let A = (xa-xa_star)/xb
+A = exp(5.02);
+//Let 'f' be the fraction of NH3 removed. Then xb = (1-f)*xa.
+//By a material balance
+//y = L/V*(xa-xb) = 1/2*(xa-(1-f)*xa)= 1/2*f*xa
+//xa_star = ya/m = 0.5*f*xa/0.8 = 0.625*f*xa
+//Thus,
+//xa-xa_star = (1-0.625*f)*xa
+//Also,
+//xa-xa_star = 10.59*xb = 10.59*(1-f)*xa
+//from these
+f = 0.962
+disp('%',f,'percentage removal obtained in this case is')
diff --git a/839/CH18/EX18.1/Example_18_1.sce b/839/CH18/EX18.1/Example_18_1.sce new file mode 100755 index 000000000..9a438281a --- /dev/null +++ b/839/CH18/EX18.1/Example_18_1.sce @@ -0,0 +1,23 @@ +//clear//
+clear;
+clc;
+
+//Example 18.1
+//Given
+xF = 0.50;
+P = 1; //[atm]
+f =0.0001:0.2:1.2;
+A = -(1./f-1);
+x = [0.01:0.01:1];
+for i =1:length(f)
+ y(i,:) =-A(i)*x+xF/f(i)
+end
+//From Fig. 18.2
+xB = [0.50,0.455,0.41,0.365,0.325,0.29];
+yD = [0.71,0.67,0.63,0.585,0.54,0.5];
+//From Fig 18.3
+T = [92.2,93.7,95.0,96.5,97.7,99];
+plot(f,T./100,f,xB,f,yD)
+xlabel('f-moles vaporized per mole of feed')
+ylabel('Concentration, mole fraction Benzene')
+legend('Temperature(C)*100','Con. of Bnzene in liquid','Con. of Bnzene in vapor')
diff --git a/839/CH18/EX18.2/Example_18_2.sce b/839/CH18/EX18.2/Example_18_2.sce new file mode 100755 index 000000000..b47fe7128 --- /dev/null +++ b/839/CH18/EX18.2/Example_18_2.sce @@ -0,0 +1,83 @@ +//clear//
+clear;
+clc;
+
+//Example 18.2
+//Given
+mdot = 30000; //[kg/h]
+wF_b = 40;
+wD = 97;
+wB = 2;
+R = 3.5;
+lambda_b = 7360; //[cal/g mol]
+lambda_t = 7960; //[cal/g mol]
+alpha = 2.5;
+TB = 95; //[C]
+TF = 20; //[C]
+P = 1; //[atm]
+Mb = 78;
+Mt = 92;
+Cp = 0.44; //[cal/g-C]
+//(a)
+//The concentrations of feed, overhead and bottoms in mole fraction of benzene are
+xF = (wF_b/Mb)/(wF_b/Mb+((100-wF_b)/Mt));
+xD = (wD/Mb)/(wD/Mb+((100-wD)/Mt));
+xB = (wB/Mb)/(wB/Mb+((100-wB)/Mt));
+//The average molecular weight of the feed is
+Mavg = 100/(wF_b/Mb+(100-wF_b)/Mt);
+//the average heat of vaporization
+lambda_avg = xF*lambda_b+(1-xF)*lambda_t; //[cal/g mol]
+//Feed rate
+F = mdot/Mavg; //[kg mol/h]
+//Using Eq.(18.5), by overall benzene balance
+D = F*(xF-xB)/(xD-xB); //[kg mol/h]
+B = F-D; //[kg mol/h]
+disp('respectively','kg mol/h',B,'kg mol/h',F,'the mole of overhead and bottom products are')
+
+
+//(b)Detemination of number of ideal plates and position of feed plate
+//(i)
+//Using Fig.18.16
+//Drawing the feed line with f = 0 on equilibrium diagram,
+//Plotting the operating lines with intercept from Eq.(18.19)is 0.216
+//By counting the rectangular steps it is found that, besides the reboiler,
+//11 ideal plates are neded and feed should be introduced on the 7th plate from
+//the top.
+
+//(ii)
+//The latent heat of vaporization of the feed
+lambda = lambda_avg/Mavg; //[cal/g]
+//Using Eq.(18.24)
+q = 1+Cp*(TB-TF)/lambda;
+//From Eq.(18.31)
+slope = -q/(1-q);
+//From Fig. 18.17
+//It is found that a reboiler and 10 ideal plates are needed and feed is to be introduced
+//on the fifth plate
+
+//(iii)
+q = 1/3;
+slope = -q/(1-q);
+//From Fig. 18.18
+//It calls for a reboiler and 12 plates, with the feed entering on the 7th plate
+
+//(c)
+//vapor flow in the rectifying section is
+V = 4.5*D; //[kg mol/h]
+lambda_s = 522; //[cal/g], From Appendix 7
+q = [1, 1.37, 0.333]
+//Using Eq.(18.27)
+Vbar = V-F*(1-q)
+//Using Eq.(18.32), steam required
+ms_dot = lambda_t/lambda_s*Vbar; //[kg/h]
+disp('respectively','kg/h',ms_dot(3),'kg/h',ms_dot(2),'kg/h',ms_dot(1),'the steam consumption in the above three cases is')
+
+
+//(d)
+Tw1 = 25; //[C]
+Tw2 = 40; //[C]
+//The cooling water needed is same in all cases, Using Eq.(18.33)
+mw_dot = V*lambda_t/(Tw2-Tw1); //[kg/h]
+rho_25 = 62.24*16.018; //[kg/m^3]
+vw_dot = mw_dot/rho_25; //[m^3/h]
+disp('m^3/h',vw_dot,'cooling water needed is')
diff --git a/839/CH18/EX18.3/Example_18_3.sce b/839/CH18/EX18.3/Example_18_3.sce new file mode 100755 index 000000000..cd43921a9 --- /dev/null +++ b/839/CH18/EX18.3/Example_18_3.sce @@ -0,0 +1,47 @@ +//clear//
+clear;
+clc;
+
+//Example 18.3
+//Given
+mdot = 30000; //[kg/h]
+wF_b = 40;
+wD = 97;
+wB = 2;
+R = 3.5;
+lambda_b = 7360; //[cal/g mol]
+lambda_t = 7960; //[cal/g mol]
+alpha = 2.5;
+TB = 95; //[C]
+TF = 20; //[C]
+P = 1; //[atm]
+Mb = 78;
+Mt = 92;
+Cp = 0.44; //[cal/g-C]
+//Solution
+xF = (wF_b/Mb)/(wF_b/Mb+((100-wF_b)/Mt));
+xD = (wD/Mb)/(wD/Mb+((100-wD)/Mt));
+xB = (wB/Mb)/(wB/Mb+((100-wB)/Mt));
+//The average molecular weight of the feed is
+Mavg = 100/(wF_b/Mb+(100-wF_b)/Mt);
+//the average heat of vaporization
+lambda_avg = xF*lambda_b+(1-xF)*lambda_t; //[cal/g mol]
+//Feed rate
+F = mdot/Mavg; //[kg mol/h]
+//Using Eq.(18.5), by overall benzene balance
+D = F*(xF-xB)/(xD-xB); //[kg mol/h]
+B = F-D; //[kg mol/h]
+//Using Table 18.3, in all three cases respectively
+xprime = [0.44,0.521,0.3];
+yprime = [0.658,0.730,0.513];
+
+//(a)
+//Using Eq.(18.43)
+RDm = (xD-yprime)./(yprime-xprime)
+disp('respectively',RDm(3),RDm(2),RDm(1),'Minimum Reflux Ratio for three cases is')
+
+//(b)
+//For minimum umber of plates the, the reflux ratio is infinite, the operating lines
+//coincides with the diagonal, and there are no differences between the three cases.
+//The plot is given by Fig 18.22. A reboiler and eight plates are needed.
+
diff --git a/839/CH18/EX18.4/Example_18_4.sce b/839/CH18/EX18.4/Example_18_4.sce new file mode 100755 index 000000000..61c99d153 --- /dev/null +++ b/839/CH18/EX18.4/Example_18_4.sce @@ -0,0 +1,26 @@ +//clear//
+clear;
+clc;
+
+//Example 18.4
+//Given
+xa = 0.02;
+Vbar = 0.2; //[mol/mol of Feed]
+xb = 0.0001;
+yb = 0;
+xe = 0:0.01:1;
+m = 9
+ye = m*xe;
+//Let
+F = 1; //[mol]
+Lbar = F; //[mol]
+
+//Solution
+ya_star = m*xa;
+yb_star = m*xb;
+//By overall ethonal balance
+ya = Lbar/Vbar*(xa-xb)+ yb
+//Using Eq.(17.27), As both operting lines and equilibrium lines are straight
+N = log((ya-ya_star)/(yb-yb_star))/log((yb_star-ya_star)/(yb-ya));
+
+disp(N,'Ideal plates needed are' )
diff --git a/839/CH18/EX18.6/Example_18_6.sce b/839/CH18/EX18.6/Example_18_6.sce new file mode 100755 index 000000000..df9872730 --- /dev/null +++ b/839/CH18/EX18.6/Example_18_6.sce @@ -0,0 +1,79 @@ +//clear//
+clear;
+clc;
+
+//Example 18.6
+//Given
+xF = 0.40;
+P = 1; //[atm]
+D = 5800; //[kg/h]
+R = 3.5;
+LbyV = R/(1+R);
+//Solution
+//Physical properties of methanol
+M = 32;
+Tnb = 65; //[C]
+rho_v = M*273/(22.4*338); //[kg/^3]
+rho_l_0 = 810; //[kg/m^3], At 0C, from Perry, Chemical Engineers' Handbook
+rho_l_20 = 792; //[kg/m^3], At 20C, from Perry, Chemical Engineers' Handbook
+rho_l = 750; //[kg/m^3], At 65C
+sigma = 19; //[dyn/cm], from Lange's Handbook of Chemistry
+//(a)
+//Vapor velocity and column diameter
+//Using Fig. 18.28, the abscissa is
+abscissa = LbyV*(rho_v/rho_l)^(1/2);
+//for 18-in. plate spacing
+Kv = 0.29;
+//Allowable vapor velocity
+uc = Kv*((rho_l-rho_v)/rho_v)^(1/2)*(sigma/20)^(0.2); //[ft/s]
+//Vapor flow rate
+V = D*(R+1)/(3600*rho_v); //[m^3/s]
+//Cross setional area of the column
+Bubbling_area = V/2.23; //[m^2]
+//If the bubble area is 0.7 of the total column area
+Column_area = Bubbling_area/0.7; //[m^2]
+//Column diameter
+Dc = sqrt(4*Column_area/%pi); //[m]
+disp('respectively','m',Dc,'and','ft/s',uc,'the allowable velocity and colmn diameter are')
+
+//(b)
+//Pressure drop:
+//Area of one unit of three holes on a trangular 3/4-in. pitch is
+//1/2*3/4*(3/4*sqrt(3/2)) in.^2. The hole area in this section (half a hole)is
+//1/2*%pi/4*(1/4)^2 in.^2. Thus the hole area is %pi/128*64/9*sqrt(3), or 10.08 percent
+//of the bubbling area.
+//Vapor velocity through holes:
+uo = 2.23/0.1008; //[m/s]
+//Using Eq.(18.58),
+//From Fig. 18.27
+Co = 0.73;
+hd = 51.0*uo^2*rho_v/(Co^2*rho_l); //[mm methanol]
+//Head of liquid on plate:
+//Weir height
+hw = 2*25.4; //[mm]
+//Height of the liquid above weir:
+//Assuming the downcomer area is 15 percent of the column
+//area on each side of th column. From Perry, the chord
+//length for sucha segmental downcomer is 1.62 times the radius
+//of the colmn, so
+Lw = 1.62*2.23/2; //[m]
+//Liqiud Flow rate:
+qL = D*(R+1)/(rho_l*60); //[m^3/min]
+//From Eq.(18.60)
+how = 43.4*(qL/Lw)^(2/3) //[mm]
+//From Eq.(18.59), with
+beeta = 0.6;
+hI = beeta*(hw+how); //[mm]
+//Total height of liquid, from Eq.(18.62)
+hT = hd+hI; //[mm]
+disp('mm methanol',hT,'pressure drop per plate is')
+
+//(c)
+//Froth height in th downcomer :
+//Using Eq.(18.62).,Estimating
+hf_L = 10; //[mm methanol]
+//Then,
+Zc = (2*hI)+hd+hf_L; //[mm]
+//from Eq.(18.63)
+Z = Zc/0.5; //[mm]
+disp('mm methanol',Z,'Froth height in the downcomer is')
diff --git a/839/CH18/EX18.7/Example_18_7.sce b/839/CH18/EX18.7/Example_18_7.sce new file mode 100755 index 000000000..60f38e239 --- /dev/null +++ b/839/CH18/EX18.7/Example_18_7.sce @@ -0,0 +1,32 @@ +//clear//
+clear;
+clc;
+
+//Example 18.7
+//Given
+xF = 0.40;
+P = 1; //[atm]
+D = 5800; //[kg/h]
+R = 3.5;
+LbyV = R/(1+R);
+//Solution
+//Physical properties of methanol
+M = 32;
+Tnb = 65; //[C]
+rho_v = M*273/(22.4*338); //[kg/^3]
+rho_l_0 = 810; //[kg/m^3], At 0C, from Perry, Chemical Engineers' Handbook
+rho_l_20 = 792; //[kg/m^3], At 20C, from Perry, Chemical Engineers' Handbook
+rho_l = 750; //[kg/m^3], At 65C
+sigma = 19; //[dyn/cm], from Lange's Handbook of Chemistry
+//(a)
+//Vapor velocity and column diameter
+//Using Fig. 18.28, the abscissa is
+abscissa = LbyV*(rho_v/rho_l)^(1/2);
+//for 18-in. plate spacing
+Kv = 0.29;
+//Allowable vapor velocity
+uc = Kv*((rho_l-rho_v)/rho_v)^(1/2)*(sigma/20)^(0.2)/(3.2825112); //[ft/s]
+//From Eq.(18.71), the F factor is
+F = uc*sqrt(rho_v);
+disp(F,'the value of F factor is')
+
diff --git a/839/CH18/EX18.8/Example_18_8.sce b/839/CH18/EX18.8/Example_18_8.sce new file mode 100755 index 000000000..df622aee4 --- /dev/null +++ b/839/CH18/EX18.8/Example_18_8.sce @@ -0,0 +1,28 @@ +//clear//
+clear;
+clc;
+
+//Example 18.8
+//Given
+xOA = 0.15;
+xAi = 0.015;
+
+P = 1; //[atm]
+
+//Solution
+
+Pv = 3.4; //[atm]
+alpha_o = 3.4; //at 36 C
+Tbi = 27; //[C]
+alpha_i = 3.6
+alpha = (alpha_o+alpha_i)/2;
+//Basis 1 mol Feed
+nOA = 0.15; //[mol]
+nA = 0.015; //[mol]
+nOB = 0.85; //[mol]
+//Using Eq.(18.79)
+nB = nOB*(nA/nOA)^(1/alpha); //[mol]
+n = nA+nB; //[mol]
+xA = nA/n;
+disp('mol',nB,'pentane removed is')
+disp((1-xA),'xB',xA,'xA','composition of the remaining liquid is')
diff --git a/839/CH19/EX19.2/Example_19_2.sce b/839/CH19/EX19.2/Example_19_2.sce new file mode 100755 index 000000000..4ecbba749 --- /dev/null +++ b/839/CH19/EX19.2/Example_19_2.sce @@ -0,0 +1,44 @@ +//clear//
+clear;
+clc;
+
+//Example 19.2
+//Given
+P = 1.2; //[atm]
+Tb = 97; //[C]
+Td = 105; //[C]
+f = 0.6;
+
+xF(1) = 0.33;
+xF(2) = 0.37;
+xF(3) = 0.30;
+
+//Solution
+//(a)
+
+//From Fig. 19.1
+K(1) = 2.68/P;
+K(2) = 1.21/P;
+K(3) = 0.554/P;
+//In Eq.(19.12), the right hand side of the equation becomes
+RHS = (xF./(f*(K-1)+1));
+RHS2 = sum(RHS)
+disp('C',Td,'flash temperature is');
+disp('percent',RHS(3),'n-octaneexane','percent',RHS(2),'n-heptane','percent',RHS(1),'n-hexane','Composition of the liquid product is');
+y = RHS.*K;
+disp('percent',y(3),'n-octane','percent',y(2),'n-heptane','percent',y(1),'n-hexane','Composition of the vapor product is');
+
+//(b)
+//To determine the temperature of the feed before flashing,
+//an enthalpy balance is made using 105 C as the reference temperature.
+//The heats of vaporization at 105 C and the average heat capacities of the
+//liquid from 105 to 200 C are obtained from the literature.
+Cp = [62,70,78]'; //[cal/mol-C], Cp(1) = n-hexane, Cp(2) = n-heptane, and Cp(3) = n-octane
+delta_Hv = [6370,7510,8560]'; //[cal/mol], delta_hv(1) = n-hexane, delta_hv(2) = n-heptane, and delta_hv(3) = n-octane
+//Based on liquid at 105 C, the enthalpies of the product are
+H_vapor = f*sum((y.*delta_Hv)) //[cal]
+H_liquid = 0;
+//For the feed
+Cp_bar = sum(xF.*Cp) //[cal/mol-C]
+T0 = H_vapor/Cp_bar+Td;
+disp('C',T0,'preheat temperature is')
diff --git a/839/CH19/EX19.3/Example_19_3.sce b/839/CH19/EX19.3/Example_19_3.sce new file mode 100755 index 000000000..75d08ddb1 --- /dev/null +++ b/839/CH19/EX19.3/Example_19_3.sce @@ -0,0 +1,45 @@ +//clear//
+clear;
+clc;
+
+//Example 19.3
+//Given
+xF = [0.33,0.37,0.30]'; //[mole fraction] xF(1) = n-hexane, xF(2) = n-heptane, and xF(3) = n-octane
+P = 1.2; //[atm]
+f = 0.60;
+xD_hex = 0.99; //[mole fraction]
+xB_hex = [0.01]; //[mole fraction]
+K(1) = 2.68/P;
+K(2) = 1.21/P;
+K(3) = 0.554/P;
+//Solution
+//The n-hexane is the light key(LK), the n-hepane is the heavy key(HK), and the
+//n-octane is a heavy nonkey(HNK)
+//Aplying mass balance and assuming no n-octane and 0.99 mole fraction n-hexane in the
+//distillate.
+//Basis:
+F = 100; //[mol/h]
+//B+D = 100;
+//For hexane,
+//F*xF = D*xD+B*xB
+//from the above two equaiton
+A_BD = [1,1;xD_hex xB_hex];
+B_BD = [F;F*xF(1)];
+//A_BD*x_BD = B_BD
+x_BD = inv(A_BD)*B_BD;
+D = x_BD(1);
+B = x_BD(2);
+xD = [0.99,0.01,0.0]';
+xB = [0.01,0.544,0.446]';
+comp_D = xD.*D;
+comp_B = xB.*B;
+
+disp('mol/h',comp_D(3),'n-octane','mol/h',comp_D(2),'n-heptane','mol/h',comp_D(1),'n-hexane','The composition of the overhead product is');
+disp('mol/h',comp_B(3),'n-octane','mol/h',comp_B(2),'n-heptane','mol/h',comp_B(1),'n-hexane','The composition of the bottom product is');
+
+//To find out minimum number of plates, using Eq.(19.13)[Fenske Equation]
+//using relative volativity of the light key to the heavy key, which is the
+//ratio of the K factors. The K values at the flash temperatue are taken from Example 19.2
+alpha_LK_HK = K(1)/K(2);
+Nmin = log((xD(1)/xD(2))/(xB(1)/xB(2)))/log(alpha_LK_HK)-1;
+disp('plus a reboiler',Nmin,'The minimum number of ideal stages is');
diff --git a/839/CH19/EX19.4/Example_19_4.sce b/839/CH19/EX19.4/Example_19_4.sce new file mode 100755 index 000000000..2742b7af2 --- /dev/null +++ b/839/CH19/EX19.4/Example_19_4.sce @@ -0,0 +1,76 @@ +//clear//
+clear;
+clc;
+
+//Example 19.4
+//Given
+//x(1) = n-pentane, x(2) = n-hexane, x(3) = n-heptane and x(4) = n-octane
+//xF = feed, xD = distillate and xB = bottom
+xF = [4 40 50 6]'./100 //[mole fraction]
+P = 1; //[atm]
+xD1(2) = 0.98;
+xD1(3) = 0.01;
+
+//Solution
+//The keys are n-hexane and n-heptane, and the other components are
+//sufficiently different in volatility to be distributed.
+//Basis:
+F = 100; //[mol]
+xD1(1) = 1;
+xD1(4) = 0;
+D = sum(F*xF.*xD1); //[mol]
+xD = (F*xF.*xD1)./(D)
+B = F-D; //[mol]
+xB = (F*xF-D*xD)/B;
+K_80 = [3.62,1.39,0.56,0.23]';
+K_81 = [3.72,1.43,0.58]';
+K_81_2 = [3.74,1.44,0.584]';
+KxF = [0.145,0.556,0.280,0.014]';
+
+//(a)
+//The bubble point is 80 C, and at this temperature
+alphaLK_HK = K_80./K_80(3);
+//For an approximate solution,using Eq.(19.15)
+RDm = (F/D)*(((D*xD(2)/(F*xF(2)))-alphaLK_HK(2)*(D*xD(3)/(F*xF(3))))/(alphaLK_HK(2)-1))
+
+//To use Underwood method, the K values at 80 C are converted to relative
+//volatilities and the root of Eq.(19.29) between 1 and 2.48 is found by trial.
+//Since q = 1.0, the terms must sum to zero.
+phi = 1.5
+f = 0;
+err = 1;
+while(err>0.1)
+ fnew = sum(((alphaLK_HK.*xD)./(alphaLK_HK-phi)));
+ err = abs(f-fnew);
+ if (f>fnew)
+ phi=phi+0.01;
+ else
+ phi=phi-0.01;
+ end
+ f = fnew;
+end
+RDm = f-1;
+
+//(b)
+//To get the conditions in the upper invariant zone, using Eq.(19.24) with
+VbyD = RDm+1;
+DbyV = inv(VbyD);
+VbyF = VbyD*D/F;
+LbyV = RDm/(RDm+1);
+y_80 = DbyV*xD(1:3)./(1-LbyV./K_80(1:3))
+y_81_1 = [0.046,0.637,0.317]';
+x_81_1 = y_81_1./K_81 ;
+//The vapor composition for lower inavariant zone is
+//using Eq.(19.28), for q = 1.0
+BbyVb = 0.552;
+LbbyVb = 1.55;
+K_83 = [1.52,0.618,0.258]';
+y_83 = BbyVb*xB(2:4)./(LbbyVb./K_83-1);
+y_83_3 = [0.662,0.326,0.012]';
+x_83_3 = y_83_3./K_83 ;
+
+disp('respectively','C',81.1,'C',83.3,'The tempeature in Lower zone and Upper zone is')
+disp('respectively',y_83_3(1),'y =',x_83_3(1),'x = ','The LK composition in Lower zone is')
+disp('respectively',y_83_3(2),'y =',x_83_3(2),'x =','The HK composition in Lower zone is')
+disp('respectively',y_81_1(2),'y =',x_81_1(2),'x =','The LK composition in Upper zone is')
+disp('respectively',y_81_1(3),'y =',x_81_1(3),'x =','The HK composition in Upper zone is')
diff --git a/839/CH19/EX19.5/Example_19_5.sce b/839/CH19/EX19.5/Example_19_5.sce new file mode 100755 index 000000000..5275fca97 --- /dev/null +++ b/839/CH19/EX19.5/Example_19_5.sce @@ -0,0 +1,36 @@ +//clear//
+clear;
+clc;
+
+//Example 19.5
+//Given
+Nmin = 9.4+1;
+//From Example 19.3
+xF = [0.33,0.37,0.30]';
+xD = [0.99,0.01,0]';
+K = [2.23,1.01,0.462]';
+alpha = [2.21,1.0,0.457]';
+
+//For a liquid feed
+q = 1;
+phi = 1.45;
+f = 0;
+err = 1;
+while(err>0.1)
+ fnew = sum(((alpha.*xD)./(alpha-phi)));
+ err = abs(f-fnew);
+ if (f>fnew)
+ phi=phi+0.01;
+ else
+ phi=phi-0.01;
+ end
+ f = fnew;
+ end
+ RDm = f-1;
+ RD = RDm*1.5;
+
+//A = (RD-RDm)/RD+1
+//from Fig. 19.5
+N = (Nmin+0.41)/(1-0.41);
+
+disp(N,'The number of ideal plate required are')
diff --git a/839/CH2/EX2.1/Example_2_1.sce b/839/CH2/EX2.1/Example_2_1.sce new file mode 100755 index 000000000..8df33f66c --- /dev/null +++ b/839/CH2/EX2.1/Example_2_1.sce @@ -0,0 +1,16 @@ +//clear//
+clear;
+clc;
+
+//Example 2.1
+rho_A = 13590;
+rho_B = 1260;
+Pa = 14000;
+gc = 1;//[ft-lb/lbf-s^2]
+
+//Using Eq.(2.5); Zb = 250 mmHg
+Pb = -(250/1000)*(9.80665/1)*13590;
+
+//Using Eq.(2.10)
+Rm = (14000+33318)/(9.80665*(13590-1260))
+disp('mm',Rm,'The reading in the mamometer is (Rm) =')
diff --git a/839/CH2/EX2.2/Example_2_2.sce b/839/CH2/EX2.2/Example_2_2.sce new file mode 100755 index 000000000..9100963af --- /dev/null +++ b/839/CH2/EX2.2/Example_2_2.sce @@ -0,0 +1,19 @@ +//clear//
+clear;
+clc;
+
+//Example 2.2
+
+//(a)
+//Using Eq.(2.15)
+t = (100*1.1)/(1153-865)
+rate_each_stream = (1500*42)/(24*60)
+total_liquid_holdup = 2*43.8*23
+vol = total_liquid_holdup/0.95
+disp('gal',vol,'vessel size =')
+
+//(b) tank diameter
+Zt = 0.90*4
+ZA1 = 1.8 //[ft];
+ZA2 = 1.8 + (3.6-1.8)*(54/72)
+disp('ft',ZA2,'tank diameter =')
diff --git a/839/CH20/EX20.1/Example_20_1.sce b/839/CH20/EX20.1/Example_20_1.sce new file mode 100755 index 000000000..79727ea96 --- /dev/null +++ b/839/CH20/EX20.1/Example_20_1.sce @@ -0,0 +1,46 @@ +//clear//
+clear;
+clc;
+
+//Example 20_1
+//Given
+Fin = 2*10^3; //[kg/day]
+//w(1) = paraffin wax, w(2) = paper pulp
+wi = [0.25,0.75]'; //[wieght percent]
+
+//Solution
+//Using convenient units in Eq.(17.24)
+//As the ratio of kerosene to pulp is constant, flow rates should be
+//expressed in pounds of kerosene. Then, all the concentrations must
+//be in pound of wax-free kerosene. The unextracted paper had no kerosene
+//so the first cell must be treated separately.
+//Refering to the Fig.20.3
+//Basis:
+F = 100; //[lb wax + kerosene-free pulp ]
+//By making a mass balance over wax
+//wax_in = F*(wi(1)/wi(2))+ 0.0005*s (s is the wax input with solvent)
+//wax_out = F*(0.002)+(s-200)*0.05
+//by wax_in = wax_out
+s_in = (33.33+9.8)/(0.05-0.0005); //[lb]
+//The concentration of this stream is, therefore
+s_out = 200; //[lb]
+s_stsol = s_in-s_out; //[lb]
+wax_sol = s_stsol*0.05; //[lb]
+//The concentration in the underflow to the second unit equals that
+//of the overflow from the first stage, or 0.05 lb of wax per pound
+//of kerosene. The wax in the underflow to unit 2 is
+wax_uflow_2 = s_out*0.05; //[lb]
+wax_oflow_21 = wax_uflow_2+wax_sol-F*(wi(1)/wi(2)) //[lb]
+
+//The concentrations of this stream is, therefore,
+ya = wax_oflow_21/871;
+yastar = 0.05;
+xa = yastar;
+ybstar = 0.2/s_out;
+xb = ybstar;
+yb = 0.0005;
+
+//Since 1 stage has already ben taken into account,
+//Eq.(17.24), will give N-1 stages, Hence
+N = log((yb-ybstar)/(ya-yastar))/log((yb-ya)/(ybstar-yastar));
+disp(N+1,'The total number of ideal stages is');
diff --git a/839/CH20/EX20.2/Example_20_2.sce b/839/CH20/EX20.2/Example_20_2.sce new file mode 100755 index 000000000..6a2aba118 --- /dev/null +++ b/839/CH20/EX20.2/Example_20_2.sce @@ -0,0 +1,75 @@ +//clear//
+clear;
+clc;
+
+//Example 20.2
+//Given
+F = 1000; //[kg]
+solv_O = 10; //[kg]
+solv_B = 655; //[kg]
+w_out = 60; //[kg]
+//Solution
+//Let solution retained is SR, from Table 20.1
+SR = [0.5,0.505,0.515,0.530,0.550,0.571,0.595,0.620]';
+xb = 0:0.1:0.7;
+//Let x and y be the mass fraction of oil in the underflow and
+//overflow solutions.
+
+//At the solvent inlet,
+Vb = solv_O + solv_B; //[kg solution/h]
+yb = solv_O/Vb;
+err = 1;
+i = 1;
+sr = SR(2);
+xb1 = 0.0;
+while(err>0.001)
+ Lb = sr*F;
+ xbnew = w_out/Lb;
+ err = abs(xb1-xbnew);
+ xb1 = xbnew;
+ sr = SR(i)+(xb1-xb(i))/(xb(i+1)-xb(i))*(SR(i+1)-SR(i));
+ i =i+1;
+end
+Lb = sr*F;
+//Benzene in the underflow at Lb is
+Underlow_B = Lb-w_out; //[kg solutions/h]
+
+// At the solid inle
+La = 400+25; //[kg solutions/h]
+xa = 400/La;
+w_in = 10+400; //[kg/h]
+Extract_O = w_in - w_out; //[kg/h]
+Extract_B = 655+25-447; //[kg/h]
+Va = Extract_O+Extract_B; //[kg/h]
+ya = Extract_O/Va;
+
+//The answers to parts (a) to (d) are
+//(a)
+disp(ya,'The concentration of strong solution is');
+//(b)
+disp(xb1,'The concentration of the soultion adhering to the extracted solids is');
+//(c)
+disp('kg/h',Lb,'The mass of solution leaving with the extracted meal is');
+//(d)
+disp('kg/h',Va,'The mass of extract is');
+
+//(e)
+//To determine an intermediate point on the operating line, choosing,
+xn = 0.5;
+//Soulion retained
+Ln = 0.571*F; //[kg/h]
+//By overall balance, Eq.(20.1)
+V_n_1 = Va+Ln-La; //[kg/h]
+//By oil balance
+y_n_1 = (Ln*xn+Va*ya-La*xa)/V_n_1;
+y =0:0.1:1;
+x = y;
+plot(x,y,[xb1,xn,xa],[yb,y_n_1,ya])
+xgrid()
+xlabel('x')
+ylabel('y')
+title('Figure 20.4')
+legend('y=x','operating line')
+//Using Figure 20.4, number of ideal stages required are
+N = 4;
+disp(N,'Number of stages required are')
diff --git a/839/CH20/EX20.3/Example_20_3.sce b/839/CH20/EX20.3/Example_20_3.sce new file mode 100755 index 000000000..7247e743a --- /dev/null +++ b/839/CH20/EX20.3/Example_20_3.sce @@ -0,0 +1,84 @@ +//clear//
+clear;
+clc;
+
+//Example 20.3
+//Given
+T = 25; //[C]
+//x(1) = Acetone, x(2)= water and x(3)= MIK
+//F = feed
+xF = [0.40, 0.60,0.0]';
+xMIK_i = [0.0,0.0,1.0]';
+
+//Solution
+//Using data from Fig. 20.10, to plot equilibrium curve
+//Fig. 20.13.
+//Basis:
+F = 100; //[mass units/h]
+//Let n = mass flow rate of H2O in extarct
+//m = mass flow rate of MIK in raffinate
+//For 99 percent recovery of A, the extarct has
+E_A = 0.99*xF(1)*F;
+//And the Raffinate has
+R_A = xF(1)*F-E_A;
+//The total flows are
+//At the top,
+//La = F = 40*A+60*H2O
+//Va = 39.6*A+n*H20+(100-m)*MIK = 139.6 + n-m
+//At the bottom,
+Vb = 100; // MIK
+//Lb = 0.4*A+(60-n)*H2O+m*MIK = 60.4 +m-n
+//Since n and m are small and tend to cancel in the summatios for Va and La,
+//the total extract flow Va is about 140, which would make
+yA_a = 39.6/140;
+xA = 0.4/60;
+//From Fig 20.10, for
+yA = 0.283, yH2O = 0.049
+xA = 0.007, xMIK = 0.02
+nm = [6;2];
+err = 1;
+while(err>0.1)
+ nmold = nm;
+ nm(1) = yH2O/(1-yH2O)*(39.6+100-nm(2));
+ nm(2) = xMIK/(1-xMIK)*(0.4+60-nm(1));
+ err = norm(nm-nmold);
+end
+n = nm(1);
+m = nm(2);
+Va = 139.6+n-m;
+yA_a = 39.6/Va;
+Lb = 60.4+m-n;
+xA_b = 0.4/Lb;
+
+//For an intermediate point on the operating line, picking
+yA = 0.12;
+//From Fig. 20.10,
+yH2O = 0.03;
+yMIK = 0.85;
+//Since the raffinate phase has only 2 to 3 pecent MIK, assuming
+//that the amount of MIK in the extract is 100, the same as the solvent
+//fed:
+V = 100/yMIK;
+//By an overall balance from the solvent inlet (bottom) to the intermediate
+//point,
+xb = xA_b;
+L = Lb+V-Vb;
+yb = 0;
+//A balance on A over the same section gives xA;
+xA = (0.4+117.6*0.12-0)/L;
+//For xA and xMIK = 0.03, A balance on MIK from the solvent
+//inlet to the intermediate point gives
+V_revised = 101.1/0.85;
+L_revised = 54.4+118.9-100;
+xA_revised = (0.4+118.9*0.12)/73.3;
+y =0:0.1:1;
+x = y;
+plot(x,y,[0.00074,0.2,0.4,],[0,0.12,0.272,])
+xgrid()
+xlabel('x')
+ylabel('y')
+title('Figure 20.13')
+legend('y=x','operating line')
+
+//From Fig. 20.13
+disp(3.4,'Number of stages')
diff --git a/839/CH21/EX21.1/Example_21_1.sce b/839/CH21/EX21.1/Example_21_1.sce new file mode 100755 index 000000000..a3c3e5fca --- /dev/null +++ b/839/CH21/EX21.1/Example_21_1.sce @@ -0,0 +1,26 @@ +//clear//
+clear;
+clc;
+
+//Exapmle 21.1
+//Given
+yA = 0.20;
+yAi = 0.10;
+
+//Solution
+//(a)
+//Let A = Dv*rho_M/BT
+A = 1; //assumed
+
+//Using Eq.(21.19), for euilmolal diffusion,
+JA = A*(yA-yAi);
+//Form Eq.(21.24), for one way diffusion,
+NA = A*log((1-yAi)/(1-yA));
+NAbyJA = NA/JA;
+disp('In this case the transfer rate with one-way diffusion is',NAbyJA-1,'percent greater than that with equimolal diffusion');
+
+//(b)
+//Whwn, b = BT/2
+A = A*2;
+yA = 1-exp(NA/2)*(1-yA)
+disp(yA,'The value of yA halfway through the layer for one-way diffusion is');
diff --git a/839/CH21/EX21.2/Example_21_2.sce b/839/CH21/EX21.2/Example_21_2.sce new file mode 100755 index 000000000..09c79152c --- /dev/null +++ b/839/CH21/EX21.2/Example_21_2.sce @@ -0,0 +1,23 @@ +//clear//
+clear;
+clc;
+
+//Example 21.2
+//Given
+K = 273.16
+T = 100+K ; //[K]
+P = 10; //[atm]
+//From Table 21.1
+TcA = 198+K; //[K]
+TcB = -147+K; //[K]
+rho_cA = 0.552; //[g/cm^3]
+rho_cB = 0.311; //[g/cm^3]
+MA = 137.5;
+MB = 28;
+
+//Solution
+VcA = MA/rho_cA //[cm^3/g mol]
+VcB = MB/rho_cB //[cm^3/g mol]
+//Substituing in Eq.(21.25)
+Dv = (0.01498*T^1.81*(1/MA+1/MB)^0.5)/(P*(TcA*TcB)^0.1405*(VcA^0.4+VcB^0.4)^2); //[cm^2/s]
+disp('cm^2/s',Dv,'Volumetric Diffusivity (Dv) = ')
diff --git a/839/CH21/EX21.3/Example_21_3.sce b/839/CH21/EX21.3/Example_21_3.sce new file mode 100755 index 000000000..52b624bef --- /dev/null +++ b/839/CH21/EX21.3/Example_21_3.sce @@ -0,0 +1,26 @@ +//clear//
+clear;
+clc;
+
+//Example 21.3
+//Given
+//1 = benzene and 2 = toluene
+M1 = 78.11;
+M2 = 92.13;
+T1_bp = 80.1+273; //[K]
+T2_bp = 110.6+273; //[K]
+VA1 = 96.5; //[cm^3/mol]
+VA2 = 118.3; //[cm^3/mol]
+mu1 = 0.24; //[cP]
+mu2 = 0.26; //[cP]
+T = 110+273; //[K]
+//Solution
+//From Eq.(21.26)
+//For benzene in toulene,
+Dv1 = 7.4*10^-8*(M2)^0.5*T/(mu2*VA1^0.6); //[cm^2/s]
+
+//For toluene in benzene,
+Dv2 = 7.4*10^-8*(M1)^0.5*T/(mu1*VA2^0.6); //[cm^2/s]
+
+disp('cm^2/s',Dv1,'Diffusivity of benzene in toluene is');
+disp('cm^2/s',Dv2,'Diffusivity of toluene in benzene is');
diff --git a/839/CH21/EX21.4/Example_21_4.sce b/839/CH21/EX21.4/Example_21_4.sce new file mode 100755 index 000000000..bbc99599b --- /dev/null +++ b/839/CH21/EX21.4/Example_21_4.sce @@ -0,0 +1,32 @@ +//clear//
+clear;
+clc;
+
+//Example 21.4
+//Given
+Nre = 20000;
+T = 40; //[C]
+D = 2; //[in.]
+Dv1 = 0.288; //[cm^2/s], for water-air
+Dv2 = 0.145; //[cm^2/s], for ethanol-air
+//Solution
+//For air at 40 C
+rho = 29/22410*273.16/313.16; //[g/cm^3]
+mu = 0.0186; //[cP], from Appendix 8
+mubyrho = mu*10^-2/rho; //[cm^2/s]
+
+//(a)
+// For the air-water system,
+Nsc = mubyrho/Dv1;
+//Form Eq.(21.54)
+Nsh = 0.023*(Nre/2)^0.81*Nsc^0.44;
+//In the film theory kc = D/BT and since Nsh = kc*D/Dv
+BT1 = D/Nsh; //[in.]
+disp('in.',BT1,'Effective thickness of the gas film is')
+
+//(b)
+//For the system air-ethanol,
+Nsc = mubyrho/Dv2;
+Nsh = 0.023*(Nre/2)^0.81*Nsc^0.44;
+BT2 = D/Nsh; //[in.]
+disp('in.',BT2,'Effective thickness of the gas film is')
diff --git a/839/CH21/EX21.5/Example_21_5.sce b/839/CH21/EX21.5/Example_21_5.sce new file mode 100755 index 000000000..1ab862d43 --- /dev/null +++ b/839/CH21/EX21.5/Example_21_5.sce @@ -0,0 +1,42 @@ +//clear//
+clear;
+clc;
+
+//Example 21.5
+//Given
+
+T = 110; //[C]
+P = 1; //[atm]
+mu = 0.26; //[cP]
+Dvx = 6.74*10^-5; //[cm^2/s]
+rho_mx = 8.47; //[mol/L]
+Dvy = 0.0494; //[cm^2/s]
+rho_my = 0.0318; //[mol/L]
+
+//(a)
+//Using Eq.(21.78)
+kybykx = (Dvy/Dvx)^0.5*(rho_my/rho_mx);
+//The gas-film coefficient predicted is only 10 percent
+//and if m=1, 90 percent of the overall resistance to mass
+//transfer would be in the gas film.
+disp(kybykx*100,'fraction of the overall resistance in the gas phase is');
+
+//(b)
+//Assuming the column is operated at the same factor F
+//Gas film:
+rho_myprime = 0.00894; //[mol/L]
+Dvyprime = (341/383)^1.81*(Dvy/0.25);
+deltakyprime = sqrt(Dvyprime/Dvy)*rho_myprime/rho_my;
+//Liquid film:
+rho_mxprime = 8.93; //[mol/L]
+muprime = 0.35; //[cP]
+Dvxprime =(341/383)*0.26*Dvx/muprime;
+deltakxprime = sqrt(Dvxprime/Dvx)*(rho_mxprime/rho_mx);
+//kyprime = deltakyprime*ky;
+//kxprime = deltakxprime/0.102*ky;
+//At 1 atm and ky = 0.102kx and Ky = 0.907/ky
+//Kyprime = 0.476*ky
+//For overall transfer units
+NOy = 2*0.476/0.53;
+neta = 1-exp(-NOy);
+disp(neta,'The efficieny will be')
diff --git a/839/CH21/EX21.6/Example_21_6.sce b/839/CH21/EX21.6/Example_21_6.sce new file mode 100755 index 000000000..4fabaae25 --- /dev/null +++ b/839/CH21/EX21.6/Example_21_6.sce @@ -0,0 +1,17 @@ +//clear//
+clear;
+clc;
+
+//Example 21.6
+//Given
+Dvprime = 10^-7; //[cm^2/s]
+rp = 0.04/2; //[cm]
+t = 30*60; //[s]
+//Then,
+beeta = Dvprime*t/rp^2;
+//form Fig. 10.6
+phi = 0.26;
+// Murphree efficiency
+neta_M = 1-phi;
+//Here the average efficieny is nearly equal to the Murphree efficiency.
+disp(4/neta_M,'The actual number of stages is')
diff --git a/839/CH22/EX22.1/Example_22_1.sce b/839/CH22/EX22.1/Example_22_1.sce new file mode 100755 index 000000000..458d24709 --- /dev/null +++ b/839/CH22/EX22.1/Example_22_1.sce @@ -0,0 +1,39 @@ +//clear//
+clear;
+clc;
+
+//Example 22.1
+//Given
+Dp = 1; //[in.]
+vdot = 25000; //[ft^3/h]
+T = 68; //[F]
+P = 1; //[atm]
+ya = 0.02;
+Mair = 29;
+Mg = 17;
+//Solution
+//The average molecular weiht of the entering gas
+M = (1-ya)*Mair+ya*Mg;
+rho_y = M*492/(359*(460+68)); //[lb/ft^3]
+
+//(a)
+//Using Fig. 22.5, when Gy =Gx;
+Gy = 0.472; //[lb/ft^2-s]
+Gx = Gy; //[lb/ft^2-h]
+des_value = Gy/2; //[lb/ft^2-h]
+mdot = vdot*rho_y/3600; //[lb/s]
+//Cross-sectional area of the tower
+S = mdot/des_value //[ft^2]
+//the diameter of the tower is
+Dtower = sqrt(4*S/%pi); //[ft]
+disp('ft',Dtower,'The tower diameter is');
+
+//(b)
+h = 20; //[ft]
+//Using Fig 22.4, the pressure drop for
+Gy = 850; //[lb/f^2-h]
+Gx = Gy;
+delta_P = 0.35; //[in.] (H2O/ft)
+//The total pressure drop
+Pt = delta_P*h; //[in.] H2O
+disp('in. H2O' ,Pt,'The pressure drop would be');
diff --git a/839/CH22/EX22.2/Example_22_2.sce b/839/CH22/EX22.2/Example_22_2.sce new file mode 100755 index 000000000..b827fa813 --- /dev/null +++ b/839/CH22/EX22.2/Example_22_2.sce @@ -0,0 +1,41 @@ +//clear//
+clear;
+clc;
+
+//Example 22.2
+//Given
+Dp = 1; //[in.]
+vdot = 25000; //[ft^3/h]
+T = 68; //[F]
+P = 1; //[atm]
+ya = 0.02;
+Mair = 29;
+Mg = 17;
+//Solution
+//The average molecular weiht of the entering gas
+M = (1-ya)*Mair+ya*Mg;
+rho_y = M*492/(359*(460+68)); //[lb/ft^3]
+rho_x = 62.3; //[lb/ft^3]
+//(a)
+//Using Fig.(22.8), from Example 22.1 A = Gx/Gy = 1 and
+//Let
+A = 1;
+B = A*sqrt(rho_y/rho_x);
+//Form Fig 22.8, the superficial vapor velocity at flooding
+//is uof*sqrt(rho_y/(rho_x-rho_y))=0.11, therefore
+uof = 0.11/sqrt(rho_y/(rho_x-rho_y)); //[m/s]
+//The allowable vapor velocity
+uo = uof*0.5; //[m/s]
+uo = uo*3.28; //[ft/s]
+//the corresponding mass velocity
+Gy = uo*rho_y; //[lb/ft^2-s]
+//The allowable mass velocity in the example was 0.236 lb/ft^2-s.
+//The increase by using structured packing is
+increase = (Gy/0.236)-1;
+disp(increase*100,'The percent increase in mass velocity is');
+
+//(b)
+//The pressure drop
+delta_P = 20*1.22*(0.5/0.9)^1.8; //[in. H2O]
+//This is 1.2 times the pressure drop of 7 in.H2O in the Intolax saddles.
+disp('The pressure drop will be greater than Intolax Saddles')
diff --git a/839/CH22/EX22.3/Example_22_3.sce b/839/CH22/EX22.3/Example_22_3.sce new file mode 100755 index 000000000..d71021ae7 --- /dev/null +++ b/839/CH22/EX22.3/Example_22_3.sce @@ -0,0 +1,90 @@ +//clear//
+clear;
+clc;
+
+//Example 22.3
+//Given
+vdot = 4500; //[SCFM]
+yin = 0.06;
+yout = 0.0002;
+P = 1; //[atm]
+Tiy = 20; //[C]
+Tix = 25; //[C]
+
+//Solution
+//From Perry
+x = [0.0308,0.0406,0.0503,0.0735]';
+y20 = [0.0239,0.0328,0.0417,0.0658]';
+y30 = [0.0389,0.0528,0.0671,0.1049]';
+y40 = [0.0592,0.080,0.1007,0.1579]';
+deltaH = -8.31*10^3; //[cal/g mol], fro NH3=NH3(aq)
+//Basis:
+gas_in = 100; //[g mol dry]
+air_in = (1-yin)*gas_in; //[mol]
+NH3_in = yin*gas_in; //[mol]
+H2O_in = 2.4; //[mol]
+air_out = air_in; //[mol]
+//The moles of NH3 in the outlet gas,
+NH3_out = air_out*(yout/(1-yout)); //[mol NH3]
+//The amount of NH3 absorbed
+NH3_abs = NH3_in-NH3_out; //[mol]
+//Heat Effects:
+//The heat of absorption
+Qa = -NH3_abs*deltaH; //[cal]
+//Sensible heat changes in the gas are
+Qair = air_in*7*5; //[cal]
+QH2O =H2O_in*8*5; //[cal]
+Qsy = 3290+96; //[cal]
+//The amount of vaporization of water from the liquid
+pH2O_20 = 17.5; //[mm Hg], at 20C
+pH2O_25 = 23.7; //[mm Hg], at 25C
+H2O_inlet = gas_in*(pH2O_20/742.5); //[mol]
+H2O_outlet = 94.02*(pH2O_25/736.3); //[mol]
+//The amount of water vaporized
+H2O_vaporized = H2O_outlet-H2O_inlet; //[mol]
+deltaHv = 583; //[cal/g]
+Qv = deltaHv*H2O_vaporized*18.02; //[cal]
+//From Eq.(22.31)
+Qsx = Qa-(Qv+Qsy); //[cal]
+
+Cp = 18; //[cal/g-mol-C]
+xmax = 0.031;
+Tb = 40; //[C]
+Ta = 25; //[C]
+err =1;
+while(err>0.01)
+ Lb = NH3_abs/xmax;
+ Tbnew = Qsx/(Lb*Cp)+Ta;
+ err = Tb-Tbnew;
+ Tb=Tbnew;
+ xmax = xmax+0.002;
+end
+Lmin = Lb-NH3_in; //[mol H2O]
+La = 1.25*Lmin; //[mol]
+Lb = La+NH3_in; //[mol]
+//The temperature rise of the liquuid is
+Tb = Qsx/(Lb*Cp)+Ta; //[C]
+xb = NH3_in/La; //[C]
+ystar = 0.044;
+//Assuming temperature to be linear function of x, so
+T = 30;
+//x = 0.0137;
+//Using the data given for 30C and interpolating to get the
+//initial slope for 25 and the final value ystar for 35, the
+//euilibrium line is drawn
+y = [0.06, 0.03,0.01,0.0002]';
+ystar = [0.048,0.017,0.0055,0]';
+delta_y = y-ystar;
+delta_yL = [0.0125, 0.0080,0.00138]';
+delta_NOy = [2.4,2.5,7.1]';
+NOy = sum(delta_NOy);
+disp(NOy,'The value of NOy is');
+
+
+
+plot(x,y20,x,y30,x,y40);
+xgrid();
+xlabel('x');
+ylabel('y');
+legend('20C','30C','40C');
+title('x vs y of NH3 at different temperatures');
diff --git a/839/CH22/EX22.4/Example_22_4.sce b/839/CH22/EX22.4/Example_22_4.sce new file mode 100755 index 000000000..46c73f521 --- /dev/null +++ b/839/CH22/EX22.4/Example_22_4.sce @@ -0,0 +1,60 @@ +//clear//
+clear;
+clc;
+
+//Example 22.4
+//Given
+ieee();
+H = 0.0075; //[TCE]
+T = 20; //[C]
+P = 1; //[atm]
+wa = 6*10^-6; //[g]
+Ca = 6; //[ppm]
+wb = 4.5*10^-9 //[g]
+M = 18;
+
+//Solution
+m = H/P*10^6/M;
+//With this large value of m, the desorption is liquid-phase controlled.
+//At the minimum air rate, the exit gas will be in equilibrium with the
+//incoming solution.
+MTCE = 131.4;
+j = 1.5;
+for i = 1:7
+xa = wa/MTCE*M;
+ya = m*xa;
+//Per cubic meter of solution fed, the TCE removed is
+VTCE = 10^6*(wa-wb)/MTCE; //[mol]
+//The total amount of gas leaving is
+V = VTCE/ya; //[mol]
+Fmin = V*0.0224; //[std m^3], as 1 gmol = 0.0224 std m^3
+Vmin = Fmin*j;
+//Density at the standard conditions,
+rho = 1.259; //[kg/m^3],
+//so the minimum rate on a mass basis is,
+//Let A = (Gy/Gx)min
+A = Vmin*rho/1000; //[kg air/kg water]
+//If the air rate is 1.5 times the minimum value, then
+ya = ya/j;
+xastar = ya/m;
+Castar = xastar*MTCE/M *10^6; //[ppm]
+delta_Ca = Ca-Castar;
+
+//At bottom
+Cb = 0.0045; //[ppm]
+Cbstar = 0; //[ppm]
+delta_Cb = Cb-Cbstar; //[ppm]
+delta_CL = (delta_Ca-delta_Cb)/log(delta_Ca/delta_Cb); //[ppm]
+Nox(i) = (Ca-Cb)/delta_CL;
+j = j+0.5;
+end
+
+Hox = 3; //[ft]
+Z = Hox*Nox; //[ft]
+//Going from 1.5 to 2Vmin or from 2 to 3Vmin decreases the tower height
+//considerably, and the reduction in pumping work for water is more than
+//the additional energy needed to force air through the column. Further
+//increase in V does not change Z very much, and the optimum air rate is
+//probably in the range 3 to 5Vmin./
+
+disp(Nox,'Number of Transfer units with minimum air rates')
diff --git a/839/CH22/EX22.5/Example_22_5.sce b/839/CH22/EX22.5/Example_22_5.sce new file mode 100755 index 000000000..76d8d4dfa --- /dev/null +++ b/839/CH22/EX22.5/Example_22_5.sce @@ -0,0 +1,99 @@ +//clear//
+clear;
+clc;
+
+//Example 22.5
+//Solution
+//Equlibrium data are shown in Fig.22.22
+//By a heat balance similar to that of Eample 22.3
+//The temperature rise of the liqui was estimated
+//to be
+delta_T = 12.5; //[C]
+//Basis:
+dry_gas_in = 100; //[mol]
+sol_in = 140; //[mol]
+N2_in = 87; //[mol]
+CO2_in = 10; //[mol]
+EO_in = 3; //[mol]
+N2_out = 87; //[mol]
+CO2_out = 10; //[mol]
+EO_out = 3*0.02; //[mol]
+IN = N2_in+CO2_in+EO_in; //[mol]
+OUT = N2_out+CO2_out+EO_out; //[mol]
+//Assuming negligible CO2 absorption and neglect effect of H2O on
+//gas composition.
+//At top:
+xt = 0.004;
+yt = EO_out/OUT;
+//Moles of EO absorbed
+EO_abs = 3*0.98; //[mol]
+//Moles of EO absorbed in water
+EO_H2O = 140*0.0004; //[mol]
+//At bottom:
+xb = (EO_abs+EO_H2O)/(140+EO_abs);
+yb = 0.03;
+//From Fig 22.22
+y = [0.03,0.015,0.005,0.0006]';
+delta_y1 = [0.008,0.0006,0.0024,0.0003]';
+
+for i = 1:length(y)-1
+ delta_y = y(i)-y(i+1);
+ delta_yL = (delta_y1(i)-delta_y1(i+1))/log(delta_y1(i)/delta_y1(i+1));
+ Noy1(i) = delta_y/delta_yL;
+end
+Noy = sum(Noy1);
+
+//Column diameter:
+//Using generalize pressure-drop correlation, Fig.22.6
+//Based on the inlet gas,
+Mbar = 0.87*28+0.1*44+0.03*44;
+//At 40C,
+rho_y = 30.1/359*20*273/313 //[lb/ft^3]
+rho_x = 62.2; //[lb/ft^3]
+//Let A = Gx/Gy*sqrt(rho_y/(rho_x-rho_y))
+A = 1.4*18/(1*30.1)*sqrt(rho_y/(rho_x-rho_y));
+//From Fig. 22.6, for
+delta_P = 0.5; //[in.H2O/ft]
+//Let B = Gy^2*Fp*mux^0.1/(rho_y*(rho_x-rho_y)*gc)
+B = 0.045;
+//From Table 22.1,
+Fp = 40;
+mu = 0.656; //[cP]
+//so
+Gy = sqrt(B*(rho_y)*(rho_x-rho_y)*32.2/(Fp*mu^0.1)); //[lb/ft^2-h]
+//or
+Gy = Gy*3600; //[lb/ft^2-s]
+Gx = 1.4*18/(1*Mbar)*Gy; //[lb/f^2-s]
+//For a feed rate
+F = 10000*Mbar; //[lb/h]
+S = F/Gx; //[ft^2]
+D = sqrt(S*4/%pi); //[ft]
+//Column heigth:
+//From Fig. 22.20 at Gy = 500 and Gx = 1500
+Hy_NH3 = 1.4; //[ft]
+mu_40 =0.0181*10^-2; //[P], Appendix 8
+Dv = 7.01*10^-3; //[cm^2/s], from Eq.(21.25)
+rho = 2.34*10^-2; //[lb/ft^3]
+Nsc = mu_40/(rho*Dv);
+//Form Table 22.1,
+fp = 1.36;
+Hy_EO = 1.4*(1.1/0.66)^0.5*1/1.36*(Gy/500)^0.3*(1500/Gx)^0.4; //[ft]
+//Form Fig. 22.19,
+Hx_O2 = 0.9; //[ft]
+Gx1 = 1500;
+mu1 = 0.00656; //[P]
+rho1 = 1; //[lb/ft^3]
+//Using Eq.(21.28)
+Dv1 = 2.15*10^-5; //[cm^2/s]
+Nsc1 = mu1/(rho1*Dv1);
+//Using Eq.(22.35), with the correction factor fp and Nsc = 381,
+//for O2 in water at 25 C
+Hx_EO = Hx_O2*(Gx/(mu1*100)/(Gx1/0.894))^0.3*(Nsc1/381)^0.5/1.36; //[ft]
+//From Fig 22.22, the average value of m
+m = 1.0;
+//From Eq.(22.30)
+HOy = 1.71+(1*0.96)/1.4; //[ft]
+
+disp(NOy,'number of transfer units required')
+disp('ft',D,'diameter of the column')
+disp('ft',HOy,'packing height')
diff --git a/839/CH22/EX22.6/Example_22_6.sce b/839/CH22/EX22.6/Example_22_6.sce new file mode 100755 index 000000000..326e3f8e9 --- /dev/null +++ b/839/CH22/EX22.6/Example_22_6.sce @@ -0,0 +1,56 @@ +//clear//
+clear;
+clc;
+
+//Example 22.6
+//Solution
+rho_m = 62.2/18; //[mol/ft^3]
+//kya = 0.025*Gy^0.7*Gx^0.25
+H2ObySO2 = 2*0.98964/0.01036;
+//and
+xb = 1/(H2ObySO2+1);
+//The molal mass velocity of the feed gas Gm is
+Gm_in = 200/29*(1/0.8); //[mol/ft^2-h]
+SO2_in = Gm_in*0.2; //[mol/ft^2-h]
+Air_in = Gm_in*0.8; //[mol/ft^2-h]
+Air_out = Air_in; //[mol/ft^2-h]
+SO2_out = Air_out*(0.005/(1-0.005)); //[mol/ft^2-h]
+SO2_abs = SO2_in-SO2_out; //[mol/ft^2-h]
+H2O_in = H2ObySO2*SO2_abs; //[mol/ft^2-h]
+//Operating line
+x = 0:6;
+x = x/10^3;
+A = x./(1-x);
+B = H2O_in/Air_in*A+(0.005/0.995);
+y = B./(B+1);
+plot(x,y)
+xgrid();
+xlabel('x');
+ylabel('y');
+//legend('20C','30C','40C');
+title('x vs y');
+Gxbar = H2O_in*18.02+SO2_abs*64.1/2; //[lb/ft^2-h]
+kxa = 0.131*Gxbar^0.82; //[mol/ft^3-h]
+//The gas film coefficients are calculated for the bottom
+//and the top of the tower:
+//At bottom:
+Gy_B = (Air_in*29)+(SO2_in*64.1); //[lb/ft^2-h]
+kya_B = 0.025*Gy_B^0.7*Gx^0.25; //[mol/ft^3-h]
+//At top:
+Gy_T = (Air_out*29)+(SO2_out*64.1); //[lb/ft^2-h]
+kya_T = 0.025*Gy_T^0.7*Gx^0.25; //[mol/ft^3-h]
+//Assuming
+yLbar = 0.82
+C = kxa*yLbar/kya_B;
+//a line from (yb,xb) with a slope of -C, gives
+yi = 0.164;
+yLbar = 0.818;
+m = 20.1
+Kya_prime = 1/(yLbar/kya_B+m/kxa); //[mol/ft^3-h]
+//The fraction of the total resistance that is in the liquid is
+Rf = m/kxa/(1/Kya_prime);
+//For different values of y1
+y1 =[0.2,0.15,0.1,0.05,0.02,0.005]';
+delta_y1 = [0.103,0.084,0.062,0.034,0.015,0.005]';
+y1i = [0.164,0.118,0.074,0.034,0.012,0.002]';
+delta_yi = y1-y1i;
diff --git a/839/CH23/EX23.1/Example_23_1.sce b/839/CH23/EX23.1/Example_23_1.sce new file mode 100755 index 000000000..943b82960 --- /dev/null +++ b/839/CH23/EX23.1/Example_23_1.sce @@ -0,0 +1,63 @@ +//clear//
+clear;
+clc;
+
+//Example 23.1
+//Given
+T = 320; //[F]
+P = 1 ; //[atm]
+//(1)=CO2, (2)=H2O, (3)=O2, (4)=N2
+y_in = [0.14,0.07,0.03,0.76]';
+Tw = 80; //[F]
+//Solution
+//(a)
+//Basis
+F = 100; //[mol], of gas
+Ts = 120; //[F]
+Cps = [9.72,8.11,7.14,6.98]';
+n_in = F*y_in; //[mol]
+nCp = n_in.*Cps; //
+sum_nCp = sum(nCp);
+sum_n_in = sum(n_in); //[mol]
+Tavg = (Ts+T)/2; //[F]
+lambda_s = 1025.8*18; //[Btu/lb mol], at Ts, from Appendix 7
+//Making a heat balance for z moles of water evaporated
+z = sum_nCp*(T-Ts)/(lambda_s+18*(Ts-Tw));
+//Total moles of water in exit gas
+n_out(2) = z+n_in(2); //[mole]
+//Partial pressure of the water in the exit gas
+PH2O = n_out(2)/107.76*760; //[mm Hg]
+//But at 120 F, PH2Oprime = 87.5 mm Hg (Appendix 7). Saturation
+//temperature Ts must be greater than 120 F. Trying
+Ts = 126; // [F]
+Tavg = (Ts+T)/2; //[F]
+lambda_s = 1022.3*18; //[Btu/lb mol], at Ts, from Appendix 7
+//Making a heat balance for z moles of water evaporated
+z = sum_nCp*(T-Ts)/(lambda_s+18*(Ts-Tw));
+//Total moles of water in exit gas
+n_out(2) = z+n_in(2); //[mole]
+//Partial pressure of the water in the exit gas
+PH2O = n_out(2)/107.76*760; //[mm Hg]
+//This is close enough to the value of PH2Oprime
+disp('F',Ts,'Adiabatic saturation temperature');
+
+//(b)
+//for Tin = Ts, by heat balance
+z = sum_nCp*(T-Ts)/(lambda_s);
+n_out(2) = z + n_in(2); //[mole]
+//Partial pressure of the water in the exit gas
+PH2O = n_out(2)/107.85*760; //[mm Hg]
+//This is higher than the vapor pressure of water at 126 F,
+//103.2 mm Hg, and Ts>126 F. Trying
+Ts = 127; //[F]
+Tavg = (Ts+T)/2; //[F]
+lambda_s = 1021.7*18; //[Btu/lb mol], at Ts, from Appendix 7
+//Making a heat balance for z moles of water evaporated
+z = sum_nCp*(T-Ts)/(lambda_s);
+//Total moles of water in exit gas
+n_out(2) = z+n_in(2); //[mole]
+//Partial pressure of the water in the exit gas
+PH2O = n_out(2)/107.76*760; //[mm Hg]
+//Thus 127 is too high and 126 is too low. Hence,
+Ts = (126+127)/2; //[F]
+disp('F',Ts,'Adiabatic saturation temperature');
diff --git a/839/CH23/EX23.3/Example_23_3.sce b/839/CH23/EX23.3/Example_23_3.sce new file mode 100755 index 000000000..c3e676c04 --- /dev/null +++ b/839/CH23/EX23.3/Example_23_3.sce @@ -0,0 +1,31 @@ +//clear//
+clear;
+clc;
+
+//Example 23.3
+//Given
+Hair_in = 0.022;
+Tair_inpre = 70; //[F]
+mdot = 15000; //[lb/h]
+//Solution
+//Using Fig. 23.10
+Tair_inreh = 85; //[F]
+Tair_outreh = 130; //[F]
+Hin = 0.0030;
+hya = 85;
+Ts = 81; //[F]
+Tair_outpre = 168; //[F]
+humid_heat1 = 0.241; //[Btu/lb-F]
+//Heat required to preheat the air is
+Qpre = humid_heat1*mdot*(Tair_outpre-Tair_inpre); //[Btu/h]
+humid_heat2 = 0.250; //[Btu/lb-F]
+//Heat required in the reheater is
+Qreh = humid_heat2*mdot*(Tair_outreh-Tair_inreh); //[Btu/h]
+//Total heat required
+Qt = Qpre+Qreh; //[Btu/h]
+//To caluculate the volume of the sprqy chamber, Eq.(23.41) may
+//be used. The average humid heat is
+csbar = (humid_heat1+humid_heat2)/2; //[Btu/lb dry air-F]
+//Substituing in Eq.(23.41) gives
+VT = log((Tair_outpre-Ts)/(Tair_inreh-Ts))*mdot*csbar/hya; //[ft^3]
+disp('ft^3',VT,'The volume of the spray chamber is')
diff --git a/839/CH24/EX24.1/Example_24_1.sce b/839/CH24/EX24.1/Example_24_1.sce new file mode 100755 index 000000000..e6d398080 --- /dev/null +++ b/839/CH24/EX24.1/Example_24_1.sce @@ -0,0 +1,43 @@ +//clear//
+clear;
+clc;
+
+//Example 24.1
+//Given
+Twb = 80; //[F]
+Tdb = 120; //[F]
+v = 3.5; //[ft/s]
+rho = 120; //[lb/ft^3]
+Xe = 0;
+Xc = 0.09;
+lambda = 1049; //[Btu/lb]
+M = 29;
+B = 24; //[in.]
+D = 2; //[in.]
+Dc = 2; //[ft]
+//Solution
+//(a)
+//mass velocity
+G = v*M*492*3600/(359*(460+120)); //[lb/ft^2-h]
+//the coefficent, by Eq.(24.13), in fps units, is
+h = 0.01*G^0.2/2^0.2; //[Btu/ft^2-h-F]
+//Substituting in Eq.(21.15) gives
+Rc = 1.94*(Tdb-Twb)/(lambda); //[lb/ft^2-h]
+disp('lb/ft^2-h',Rc,'Drying rate during the constant period is')
+
+//(b)
+//Since drying is from both faces, area
+A = Dc*(B/12)^2; //[ft^2]
+//The rate of drying
+mvdot = Rc*A; //[lb/h]
+//Volume of the cake
+Vc = (B/12)^2*D/12; //[ft^3]
+//mass of the bone-dry solid is
+mdot_bd = rho*Vc; //[lb]
+//The quantity of moisture to be vaporized is
+X2 = 0.20;
+X1 = 0.10;
+Q = mdot_bd*(X2-X1); //[lb]
+//Drying time
+tT = Q/mvdot; //[h]
+disp('h',tT,'drying time')
diff --git a/839/CH24/EX24.2/Example_24_2.sce b/839/CH24/EX24.2/Example_24_2.sce new file mode 100755 index 000000000..14fd4eaef --- /dev/null +++ b/839/CH24/EX24.2/Example_24_2.sce @@ -0,0 +1,15 @@ +//clear//
+clear;
+clc;
+
+//Example 24.2
+//Given
+X1 = 0.25;
+X = 0.05;
+Dvprime = 8.3*10^-6; //[cm^2/s]
+D = 25.4; //[mm]
+
+//Solution
+s = D/(2*10); //[cm]
+tT = 4*s^2/(%pi^2*Dvprime)*log(8*X1/(%pi^2*X))/3600; //[h]
+disp('h',tT,'drying time is')
diff --git a/839/CH24/EX24.3/Example_24_3.sce b/839/CH24/EX24.3/Example_24_3.sce new file mode 100755 index 000000000..2a9dd2297 --- /dev/null +++ b/839/CH24/EX24.3/Example_24_3.sce @@ -0,0 +1,62 @@ +//clear//
+clear;
+clc;
+
+//Example 24.3
+//Given
+Tw = 80; //[F]
+Tdb = 120; //[F]
+v = 3.5; //[ft/s]
+rho = 120; //[lb/ft^3]
+Xe = 0;
+Xc = 0.09;
+lambda = 1049; //[Btu/lb]
+M = 29;
+B = 24; //[in.]
+D = 2; //[in.]
+Dc = 2; //[ft]
+X2 = 0.20;
+X1 = 0.10;
+Dcyl = 1/4; //[in.]
+L = 4; //[in.]
+Vbar = 3.5; //[ft/s]
+Thb = 120;
+
+//Solution
+//Since the Xc is less than 10 percent, all drying takes place
+//in the constant-rate period and the vaporrization temperature,
+//as before, is 80 F.
+//From Exapmle 24.1, mass of water to be evaporated
+mdot = 8*(X2-X1); //[lb]
+//The quantity of heat to be transferred
+QT = mdot*lambda; //[Btu]
+//mass of the dry soild in one cylinder is
+mp = %pi/4*(Dcyl/12)^2*(L/12)*rho; //[lb]
+//surface area of one cylinder is
+Ap = %pi*(Dcyl/12)*(L/12); //[ft^2]
+//Total area exposed by 8 lb solids
+A = 8/mp*Ap; //[ft^2]
+//The heat transfer coefficient is found from the
+//equivalent form of Eq.(21.62)
+//hDbyk = 1.17*Nre^0.585*Npr^(1/3)
+//For air at 1 atm and 120F, the properties are
+rho_a = M/359*492/580; //[lb/ft^3]
+mu_a = 0.019; //[cP], from Appendix 8
+k_a = 0.0162; //[Btu/ft-h-F], from Appendix 12
+Cp_a = 0.25; //[Btu/lb-F], from Appendix 15
+Nre = 1/48*Vbar*rho_a/(mu_a*6.72*10^-4);
+Npr = mu_a*2.42*Cp_a/k_a;
+//Form Eq.(21.62)
+h = (k_a*1.17*Nre^0.585*Npr^(1/3))/(1/48); //[Btu/ft^2-h-F]
+mdot_g = v*3600*rho_a; //[lb]
+//From Fig. 23.2
+cs = 0.25;
+delta_Thb = Thb-Tw; //[F]
+delta_Tha = 8.24; //[F]
+//The heat transferred form the gas to a thin section of the bed
+delta_TL = (delta_Thb-delta_Tha)/log(delta_Thb/delta_Tha); //[F]
+//rate of heat transfer
+qT = h*A*delta_TL; //[Btu/h]
+//drying time
+tT = QT/qT; //[h]
+disp('h',tT,'Required drying time is')
diff --git a/839/CH24/EX24.4/Example_24_4.sce b/839/CH24/EX24.4/Example_24_4.sce new file mode 100755 index 000000000..3f365ead6 --- /dev/null +++ b/839/CH24/EX24.4/Example_24_4.sce @@ -0,0 +1,48 @@ +//clear//
+clear;
+clc;
+
+//Example 24.4
+//Given
+msdot = 2800; //[lb/h]
+Xa = 0.15;
+Xb = 0.005;
+Ti = 80; //[F]
+To = 125; //[F]
+Thb = 260; //[F]
+Hb = 0.01; //[lb water/lb dry air]
+G = 700; //[lb/ft^2-h]
+Cps = 0.52; //[Btu/lb-F]
+
+//Solution
+//Counter current operation will be used.
+//Assuming
+Nt = 1.5; //NTU
+//From Fig. 23.2
+Twb = 102; //[F]
+//From Eq. (2.48)
+Tha = (Thb-Twb)/exp(Nt)+Twb; //[F]
+Tsb = To; //[F]
+lambda = 1036; //[Btu/lb], at 102 F, from Appendix 7
+Cpv = 0.45; //[Btu/lb-F], from Appendix 15
+Cpl = 1.0; //[Btu/lb-F]
+//From Eq.(24.9)
+mvdot = msdot*(Xa-Xb); //[lb/h]
+//The heat duty is found form substitution in Eq.(24.1)
+qTdot = Cps*(To-Ti)+Xa*Cpl*(Twb-Ti)+(Xa-Xb)*lambda+Xb*Cpl*(To-Twb)+(Xa-Xb)*Cpv*(Tha-Twb); //[Btu/lb]
+qT = qTdot*msdot; //[Btu/h]
+//The flow rate of the entering air is found from a heat balance and the humid heat csb.
+//From Fig. 23.2
+csb = 0.245; //[Btu/lb-F],
+mgdot = qT/(csb*(Thb-Tha)*(1+Hb)); //[lb/h of dry air]
+//From Eq.(24.10), The outlet humidity
+Ha = Hb+mvdot/mgdot; //[lb/lb]
+
+//For a given flow rate, the cross-sectional area of the dryer must be
+Ac = qT/(csb*(Thb-Tha))/G; //[ft^2]
+//The dryer diameter is
+D = (4*Ac/%pi)^0.5; //[ft]
+delta_TL = ((Thb-Twb)-(Tha-Twb))/log((Thb-Twb)/(Tha-Twb)); //[F]
+//Using Eq.(24.29), the dryer length
+L = qT/(0.125*%pi*D*G^0.67*delta_TL); //[ft]
+disp('respectively','ft',L,'ft',D,'Required diameter and length of the dryer is')
diff --git a/839/CH25/EX25.1/Example_25_1.sce b/839/CH25/EX25.1/Example_25_1.sce new file mode 100755 index 000000000..9d3d63fb5 --- /dev/null +++ b/839/CH25/EX25.1/Example_25_1.sce @@ -0,0 +1,37 @@ +//clear//
+clear;
+clc;
+
+//Example 25.1
+//Given
+ya = 0.002;
+T = 20+273; //[K]
+
+//Solution
+//(a)
+M = 86.17;
+//from Perry's Chemical Engineers' Handbook, 6th ed.
+Pprime = 120; //[mm Hg]
+fs = Pprime; //[mm Hg]
+rho_L = 0.615; //[g/cm^3], at normal boiling point(68.7 C)
+P = 760; //[mm Hg]
+p = ya*P; //[mm Hg]
+f = p; //[mm Hg]
+V = M/rho_L; //[cm^3/g mol]
+//Let
+A = T/V*log10(fs/f);
+//From Fig. 25.4, volume adsorbed
+V_ads = 31/100; //[cm^3 liquid/g carbon]
+W = V_ads*rho_L; //[g/g carbon]
+disp('g/g carbon',W,'The equilibrium capacity for the bed is')
+
+//(b)
+T = 40+273; //[K]
+Pprime = 276; //[mm Hg]
+fs = Pprime; //[mm Hg]
+A = T/V*log10(fs/f);
+//From Fig. 25.4, volume adsorbed
+V_ads = 27/100; //[cm^3 liquid/g carbon]
+W = V_ads*rho_L; //[g/g carbon]
+disp('g/g carbon',W,'The equilibrium capacity for the bed is')
+
diff --git a/839/CH25/EX25.2/Example_25_2.sce b/839/CH25/EX25.2/Example_25_2.sce new file mode 100755 index 000000000..5d7eae7be --- /dev/null +++ b/839/CH25/EX25.2/Example_25_2.sce @@ -0,0 +1,77 @@ +//clear//
+clear;
+clc;
+
+//Example 25.2
+//Solution
+cbyc0 =0.05;
+u0 = 58; //[cm/s]
+Dv = 0.37; //[m^2/g]
+c0 = 365; //[ppm]
+S = 1194; //[m^2/g]
+T = 25; //[C]
+rho_b = 0.461; //[g/cm^3]
+P = 737; //[mm Hg]
+M = 74.12; //[g/mol]
+eps = 0.457;
+t = 1:0.5:8.5;
+t(4) = 2.4; t(5) = 2.8; t(6) = 3.3;
+cbyc0 =[0.005,0.01,0.027,0.05,0.1,0.2,0.29,0.56,0.0019,0.003,0.0079,0.018,0.039,0.077,0.15,0.24];
+t1 = t(1:8);
+t2 = t(9:16);
+cbyc01 = cbyc0(1:8);
+cbyc02 = cbyc0(9:16);
+plot(t1,cbyc01,t2,cbyc02);
+xgrid();
+xlabel('t, Hours');
+ylabel('c/c0');
+title('Brakthrough curves for Example 25.2');
+legend('L = 8cm','L = 16cm');
+
+//(a)
+FA = u0*c0*10^-6/22400*273/298*737/760*M*3600; //[g/cm^2-h]
+// The total solute adsorbed is the area above the graph multiplied
+//by FA. For the 8-cm bed, the area is
+Area_bed = 4.79; //[h]
+//This area corresponds to the ideal time that would be required to adsorb
+//the same amount if the breakthrough curve were a vertical line. The mass
+//of carbon per unit cross-sectional area of the bed is
+Ac = 8*rho_b; //[g/cm^2]
+//Thus,
+Wsol = FA*Area_bed/Ac; //[g solute/g carbon]
+//At the break point, where
+cbyc0_break = 0.05;
+//and
+t_break =2.4; //[h]
+Area_bed_break = 2.37; //[h]
+//The amount adsorbed up to the break point is then
+Wb = FA*t_break/Ac; //[g solute/ g carbon]
+ratio_W = Wb/Wsol;
+//Thus 50 percent of the bed capacity is unused, which can be representd
+//by a length 4 cm.
+//For the 16-cm bed the breakthrough curve has the same initial slope as the cuve
+//for 8-cm bed, and although data were not taken beyond cbyc0 = 0.25,
+//the curves are assumed to be parallel
+//For the entire bed,
+tT = 9.59; //[h]
+Wsat = FA*tT/(16*rho_b); //[g solute/ g carbon]
+//At
+cbyc0_break = 0.05;
+t_break =7.1; //[h]
+Area_break = 7.07; //[h]
+Wb = FA*Area_break/(16*rho_b); //[g solute/g carbon]
+ratio_W = Wb/Wsat;
+//At the break point, 74 percent of the bed capacity is used,
+//which corresponds to an unused section of length 0.26*16 cm.
+//Within experimental error, the lengths of unused bed agree,
+//and 4.1 cm is expected value for a still longer bed.
+disp('cm',4.2,'length of the bed used','percent',ratio_W,'saturation capacity of the carbon')
+
+//(b)
+L = 32; //[cm]
+L_exp = L-4.1; //[cm]
+//Fraction of the bed used
+fra_bed = L_exp/L;
+//The break-point time is,
+tb = L_exp*rho_b*Wsat/FA; //[h]
+disp('h',tb,'break point-time ')
diff --git a/839/CH25/EX25.3/Example_25_3.sce b/839/CH25/EX25.3/Example_25_3.sce new file mode 100755 index 000000000..238fdaa0f --- /dev/null +++ b/839/CH25/EX25.3/Example_25_3.sce @@ -0,0 +1,49 @@ +//clear//
+clear;
+clc;
+
+//Example 25.3
+//Solution
+cbyc0 =0.05;
+u0 = 58; //[cm/s]
+Dv = 0.37; //[m^2/g]
+c0 = 365; //[ppm]
+S = 1194; //[m^2/g]
+T = 25; //[C]
+rho_b = 0.461; //[g/cm^3]
+P = 737; //[mm Hg]
+M = 74.12; //[g/mol]
+eps = 0.457;
+L = 8; //[cm]
+
+//(a)
+//From Example 25.2
+ratio_W = 0.495;
+tou = 0.495;
+//From Fig. 25.10
+N = -1.6/(tou-1); //at c/c0 = 0.05
+Kca = N*u0/L; //[s^-1]
+disp('s^-1',Kca,'Kca = ',N,'N = ')
+//plot(t1,cbyc01,t2,cbyc02)
+
+//(b)
+Dp = 0.37; //[cm]
+mubyrho = 0.152; //[cm^2/s], at 25C, 1atm
+Dv = 0.0861; //[cm^2/s]
+Nre = Dp*u0/mubyrho;
+Nsc = mubyrho/Dv;
+//From Eq.(21.62),
+Nsh = 1.17*Nre^0.585*Nsc^(1/3);
+kc = Nsh*Dv/Dp; //[cm/s]
+a = 6*(1-eps)/Dp; //[cm^2/cm^3]
+kca = kc*a; //[s^-1]
+//Since Kca is slightly less than half the predicted value of kca,
+//the external resistance is close to half the total resistance, and
+//the calculated value of N need not be revised. The internal
+//coefficient can be obtained from
+Kc = Kca/a; //[cm/s]
+kc_int1 = 1/(1/Kc-1/kc); //[cm/s]
+//If the diffusion into the particle occured only in the gas phase, the
+//maximum possible value of De would be about Dv/4, which leads to
+kc_int2 = 10*Dv/(4*Dp); //[cm/s]
+disp('Kca is slightly less than half the predicted value of kca');
diff --git a/839/CH25/EX25.4/Example_25_4.sce b/839/CH25/EX25.4/Example_25_4.sce new file mode 100755 index 000000000..7369eb582 --- /dev/null +++ b/839/CH25/EX25.4/Example_25_4.sce @@ -0,0 +1,71 @@ +//clear//
+clear;
+clc;
+
+//Example 25.4
+y = 0.0012;
+vdot = 16000; //[ft^3/min]
+P = 760; //[mm Hg]
+rho_b = 30; //[lb/ft^3]
+Lun = 0.5; //[ft]
+
+//Solution
+//(a)
+//Form the hand book
+Pprime = 151; //[mm Hg]
+fs = Pprime; //[mm Hg]
+rho_L = 0.805; //[g/cm^3], at 20C
+Tnb = 79.6; //[C]
+rho_e = 0.75; //[g/cm^3]
+M = 72.1;
+V = M/rho_e;
+p = y*P; //[mm Hg]
+f = p; //[mm Hg]
+//At 35C
+T = 35+273; //[K]
+A = T/V*log10(fs/f);
+//Form Fig. 25.4,
+//the volume adsorbed
+V_ads = 24; //[cm^3/100 g carbon]
+Wsat = V_ads*rho_e; //[g/100 g carbon]
+W0 = 1/3*Wsat; //[g/100 g carbon]
+Working_capacity = Wsat-W0; //[g/100 g carbon]
+//or
+Working_capacity = Working_capacity/100; //[lb/lb carbon]
+disp(Working_capacity,'Working capacity of the bed is')
+
+//(b)
+u0 = 1; //[ft/s]
+A = vdot/u0; //[ft^2]
+D = sqrt(4*A/%pi); //[ft]
+Abed = 10*27; //[ft^2]
+L1 = 4; //[ft]
+c0 = y/359*273/298*72.1; //[lb/ft^3]
+//Form Eq.(25.3)
+tstar = L1*rho_b*(Working_capacity)/(u0*c0*3600); //[h]
+Lu1 = L-Lun; //[ft]
+tb1 = Lu1/L*tstar; //[h]
+
+//if
+L2 = 3; //[ft]
+Lu2 = L2-Lun;
+tb2 = Lu2/L*tstar; //[h]
+//checking for delta_P
+//Using Eq.(7.22)
+phi_s = 0.7; //from Table 28.1
+eps = 0.35; //from Table 7.1
+mu = 1.21*10^-5; //[lb/ft-s]
+rho = 0.074; //[lb/ft^3]
+//For a 4*10-mesh carbon
+Dp = 1.108*10^-2; //[ft]
+deltaPbyL = 150*1*mu*(1-eps)^2/(32.2*phi_s^2*Dp^2*eps^3)+(1.75*rho*1^2*(1-eps)/(32.2*0.7*Dp*eps^3)); //[lbf/ft^2-ft]
+deltaPbyL = deltaPbyL*12/62.4; //[in. H2O/ft]
+//for
+L = 3;
+deltaP = 3*deltaPbyL; //[in. H2O]
+//which satisfactory.
+mc = 2*(10*27*3)*30; //[lb]
+
+disp('ft',L2,'Allowing for uncertainties in the calculations, satisfactory bed length will be')
+disp('ft/s',u0,'gas velocity needed')
+disp('lb',mc,'carbon needed')
diff --git a/839/CH26/EX26.1/Example_26_1.sce b/839/CH26/EX26.1/Example_26_1.sce new file mode 100755 index 000000000..0137dd78b --- /dev/null +++ b/839/CH26/EX26.1/Example_26_1.sce @@ -0,0 +1,82 @@ +//clear//
+clear;
+clc;
+
+//Example 26.1
+//Given
+alpha = 5;
+per = 0.2; //[scf/ft^2-h-atm]
+Pf = 150; //[lbf/in.^2]
+Pp = 15; //[lbf/in.^2]
+
+//Solution
+//(a)
+R = Pp/Pf;
+//At the feed inlet
+xin = 0.209;
+//Using Eq.(26.17)
+A = alpha-1;
+B = 1-alpha-1/R-xin*(alpha-1)/R;
+C = alpha*xin/R;
+yi_in = (-B-sqrt(B^2-4*A*C))/(2*A);
+//At the discharge end
+xd = 0.05;
+//Using Eq.(26.17)
+A = alpha-1;
+B = 1-alpha-1/R-xd*(alpha-1)/R;
+C = alpha*xd/R;
+yi_d = (-B-sqrt(B^2-4*A*C))/(2*A);
+
+//For an approximate solution, these terminal compositions are
+//averaged to give
+ybar = (yi_in+yi_d)/2;
+//From an overall material balance
+//Basis
+Lin = 100; //[scfh]
+V = (Lin*xin-Lin*xd)/(ybar-xd);
+//disp(ybar,'and permeate composition is','percent',V/Lin*100,'The permeate in the feed is');
+
+
+//For more accurate calculation
+j = 2;
+yi_in(1) = 0.5148;
+x(1) = 0.209;
+y(1)= 0.5148;
+L = Lin;
+deltaV = [];
+deltaVybar = [];
+ybar = [];
+for i = 0.2:-0.01:xd
+x(j) = i;
+A = alpha-1;
+B = 1-alpha-1/R-x(j)*(alpha-1)/R;
+C = alpha*x(j)/R;
+yi_in(j) = (-B-sqrt(B^2-4*A*C))/(2*A);
+ybar(j-1) = (yi_in(j-1)+yi_in(j))/2;
+deltaV(j) = L*(x(j-1)-x(j))/(ybar(j-1)-x(j));
+V = sum(deltaV);
+L = Lin - V;
+deltaVybar(j) = deltaV(j-1)*ybar(j-1);
+deltaVybarsum = sum(deltaVybar);
+y(j-1) = deltaVybarsum/V;
+j = j+1;
+end
+disp(y($),'and permeate composition is','percent',V/Lin*100,'The permeate recovered');;
+
+
+//(b)
+//The membrane area obtained from the flux of A using
+//Eq.(26.29) and (26.13)
+//for the first increment x = 0.209 to x = 0.2
+deltaybar1 = 1.4856; //[scfh], for Lin = 100 scfh
+//At x = 0.209
+A1 = 0.209-0.1*0.5148;
+//At x = 0.2
+A2 = 0.2-0.1*(0.50);
+Aavg = (A1+A2)/2
+QAP1 = 0.2*10; //scfh/ft^3
+//for specified flow of 300 scfh
+deltaA = 1/2*1.486/Aavg*180; //[ft^2]
+//The calculation continued with increments of 0.01
+A = 211/2.0*180; //[ft^2]
+disp('ft^2',A,'The membrane area needed is')
diff --git a/839/CH26/EX26.4/Example_26_4.sce b/839/CH26/EX26.4/Example_26_4.sce new file mode 100755 index 000000000..5da677be8 --- /dev/null +++ b/839/CH26/EX26.4/Example_26_4.sce @@ -0,0 +1,28 @@ +//clear//
+clear;
+clc;
+
+//Example 26.4
+//Given
+F = 10; //[gal/day-ft^3]
+Do = 300*10^-6; //[m]
+Di = 200*10^-6; //[m]
+vi = 0.5; //[cm/s]
+rho = 1; //[g/cm^3]
+mu = 0.01; //[g/cm-s], assumed
+f = 0.97;
+
+//Solution
+//For 10 gal/day-ft^2
+Jw = F*231*16.3871/(24*3600*929); //[cm/s]
+Nre = Do*100*vi*rho/mu;
+Ds = 1.6*10^-5; //[cm^2/s]
+Nsc = mu/(rho*Ds);
+
+//Using Eq.(12.69), Analogously to mass transfer
+Nsh = (0.35+0.56*Nre^0.52)/Nsc^-0.3;
+kc = Nsh*Ds/(Do*100); //[cm/s]
+//From Eq.(26.49)
+gama = Jw*f/kc;
+disp('A concentration differnce of 12 percent will not be significant till good flow distribution is maintained');
+
diff --git a/839/CH26/EX26.5/Example_26_5.sce b/839/CH26/EX26.5/Example_26_5.sce new file mode 100755 index 000000000..4656b61de --- /dev/null +++ b/839/CH26/EX26.5/Example_26_5.sce @@ -0,0 +1,33 @@ +//clear//
+clear;
+clc;
+
+//Example 26.5
+//Given (from Example 26.4)
+F = 10; //[gal/day-ft^2], based on external area
+Do = 300*10^-6; //[m]
+Di = 200*10^-6; //[m]
+vi = 0.5; //[cm/s]
+rho = 1; //[g/cm^3]
+mu = 10^-3; //[Pa-s], assumed
+f = 0.97;
+L = 3; //[m]
+
+//Solution
+//(a)
+//Jw based on area
+Jw = 4.72*10^-4*Do/Di*10^-2; //[m/s]
+dt = 200*10^-6; //[m]
+D = dt; //[m]
+//From Eq.(26.53)
+Vbar = 4*(Jw)*L/Di; //[m/s]
+//From Eq.(26.56)
+delta_ps = (Vbar*32*mu*L)/(D)^2*(1/2)/10^5; //[atm]
+disp('atm',delta_ps,'pressure drop = ','m/s',Vbar,'exit velocity = ');
+
+//(b)
+//If the fibres are open at both ends, the effective length is 1.5m and
+//the exit velocity is half as great. The pressure drop is one-fourth as
+//large as it was:
+deltaP = delta_ps/4; //[atm]
+disp('atm',deltaP,'pressure drop (if both ends are open) = ')
diff --git a/839/CH27/EX27.1/Example_27_1.sce b/839/CH27/EX27.1/Example_27_1.sce new file mode 100755 index 000000000..856e8ce70 --- /dev/null +++ b/839/CH27/EX27.1/Example_27_1.sce @@ -0,0 +1,29 @@ +//clear//
+clear;
+clc;
+
+//Example 27.1
+//Given
+T = 60; //[F]
+wA = 0.30; //[MgSO4]
+wB = 0.70; //[H2O]
+
+//Solution
+//From Fig. 27.3 it is noted that the crystals are MgSO4.7H2O
+//and that the concentration of the mother liquid is
+xA = 0.245; //[anhydrous MgSO4]
+xB = 0.755; //[H2O]
+//Bases:
+F_in = 1000; //[kg]
+H2O_in = F_in*wB; //[kg]
+H2O_evp = 0.05*H2O_in; //[kg]
+M1 = 120.4; //[MgSO4 molecular weight]
+M2 = 246.5; //[MgSO4.7H2O molecular weight]
+M2_in = wA*F_in*M2/M1; //[kg]
+H2O_free = F_in-H2O_evp-M2_in; //[kg]
+ML = 100; //[kg]
+M2_in100 = ML*xA*M2/M1; //[kg]
+H2O_free100 = ML - M2_in100; //[kg]
+M2_ML = M2_in100/H2O_free100*H2O_free; //[kg]
+FC = M2_in - M2_ML; //[kg]
+disp(FC,'kilograms of crystals obtained per kilogram of original mixture = ')
diff --git a/839/CH27/EX27.2/Example_27_2.sce b/839/CH27/EX27.2/Example_27_2.sce new file mode 100755 index 000000000..c50ab771e --- /dev/null +++ b/839/CH27/EX27.2/Example_27_2.sce @@ -0,0 +1,27 @@ +//clear//
+clear;
+clc;
+
+//Example 27.2
+//Given
+//A = MgSO4, B = MgSO4.7H2O and C = H2O
+T = 120; //[F]
+wA = 0.325;
+
+//Solution
+//From Fig 27.4
+//Enthalpy coordinate of the point wA
+H1 = -33; //[Btu/lb]
+//Enthalpy coordinate of the final magma at concentration wA
+H2 = -78.4; //[Btu/lb]
+//Per hundred pouds of original solution the change in enthalpy
+F = 100; //[lb]
+delta_H = F*(H1-H2); //[Btu]
+//Applying "center-of gravity principle" to 70 F isotherm in Fig. 27.3
+C_ML = 0.259;
+C_CRY = 0.488;
+//Crystals are
+Cry = F*(wA-C_ML)/(C_CRY-C_ML); //[lb/100lb slurry]
+//The heat evolved per ton of crystals is
+H = delta_H/Cry*2000; //[Btu/ton]
+disp('Btu/ton',H,'The heat evolved per ton of crystals is')
diff --git a/839/CH27/EX27.3/Example_27_3.sce b/839/CH27/EX27.3/Example_27_3.sce new file mode 100755 index 000000000..b8608e2d9 --- /dev/null +++ b/839/CH27/EX27.3/Example_27_3.sce @@ -0,0 +1,27 @@ +//clear//
+clear;
+clc;
+
+//Example 27.3
+//Given
+sigma = 2.5; //[erg/cm^3]
+T = 300; //[K]
+N = 6.0222*10^23;
+R = 8.3134*10^7; //[erg/g mol-K]
+//Solution
+M = 74.56; //[Molecular weight]
+rho = 1.988; //[g/cm^3]
+nu = 2;
+VM = M/rho //[cm^3/g mol]
+//Using Eq.(27.11)
+//Exponential term, excluding 's'
+A = 16*%pi*VM^2*N*sigma^3*10/(3*(T*R)^3*nu^2)
+B0 = 1;
+s(1) = sqrt(-A/log(B0/10^25));
+//For B0;
+s = s(1):0.0001:0.029;
+B0 = exp(57.565)*exp(-A./s.^2);
+plot(s,B0)
+title('B0 vs s')
+xlabel('s')
+ylabel('B0')
diff --git a/839/CH27/EX27.4/Example_27_4.sce b/839/CH27/EX27.4/Example_27_4.sce new file mode 100755 index 000000000..bd01d481b --- /dev/null +++ b/839/CH27/EX27.4/Example_27_4.sce @@ -0,0 +1,20 @@ +//clear//
+clear;
+clc;
+
+//Example 27.4
+//Given
+alpha = 1+0.029;
+//From Example 27.3
+sigma = 2.5; //[erg/cm^3]
+T = 300; //[K]
+N = 6.0222*10^23;
+R = 8.3134*10^7; //[erg/g mol-K]
+M = 74.56; //[Molecular weight]
+rho = 1.988; //[g/cm^3]
+nu = 2;
+VM = M/rho; //[cm^3/g mol]
+
+//Using Eq.(27.9)
+L = 4*VM*sigma/(2*R*T*log(alpha))*10^7; //[nm]
+disp('nm',L,'size of nuclues (L) = ');
diff --git a/839/CH27/EX27.5/Example_27_5.sce b/839/CH27/EX27.5/Example_27_5.sce new file mode 100755 index 000000000..07942177e --- /dev/null +++ b/839/CH27/EX27.5/Example_27_5.sce @@ -0,0 +1,37 @@ +//clear//
+clear;
+clc;
+
+//Example 27.5
+//Let: A = MgSO4; B = MgSO4.7H2O; C = H2O
+//Given
+xA = 0.31;
+T = 86; //[F]
+Tb = 2; //[F]
+vbys = 0.15;
+//PB =
+rho_cr = 105; //[lb/ft^3]
+rho_ml = 82.5; //[lb/ft^3]
+
+//Solution
+//Basis:
+F = 10000; //[lb/h]
+//From Fig 27.13 and Fig 27.4
+crbyml = vbys*rho_cr/((1-vbys)*rho_ml);
+ml_prod = F/crbyml; //[lb/h]
+magma_prod = F+ml_prod //[lb/h]
+xA_avg = (crbyml*0.488+0.285)/1.224;
+//The enthalpy of the magam
+Hmag = (crbyml*(-149)+(-43))/1.224; //[Btu/lb]
+//These are the concenrations of the point e. The point for the feed must
+//lie on the straight line ae.
+//The enthalpy of the feed
+Hf = -21; //[Btu/lb]
+//Temperature of the feed
+Tf = 130; //[F]
+//By COG principle, the evaporation rate
+evap_rate = magma_prod*(Hf-Hmag)/(1098-Hf); //[lb/h]
+Total_feed = magma_prod+evap_rate; //[lb/h]
+disp('F',Tf,'Temperature of the feed is');
+disp('lb/h',Total_feed,'Total feed rate');
+disp('lb/h',evap_rate,'Total evaporation rate');
diff --git a/839/CH27/EX27.6/Example_27_6.sce b/839/CH27/EX27.6/Example_27_6.sce new file mode 100755 index 000000000..9af6b91e6 --- /dev/null +++ b/839/CH27/EX27.6/Example_27_6.sce @@ -0,0 +1,68 @@ +//clear//
+clear;
+clc;
+
+//Example 27.6
+//Given
+G = 0.0018; //[ft/h]
+//Solution
+//Screen opening of 20-mesh standard screen is,
+L = 0.00273; //[ft], Appendix 20
+a = 1; //[Eq.27.16]
+//From Example 27.5
+//The volume flow rate of mother liquor in the product magma
+Q = 44520/82.5; //[ft^3/h]
+//Since, when z=3,
+Lpr = L; //[ft]
+//Using Eq.(27.28)
+//drawdown time
+tou = Lpr/(3*G); //[h]
+//volume of the liquid in the crystallizer
+Vc = tou*Q; //[ft^3]
+//Total magma volume
+Vmagma = Vc/0.85*7.47; //[gal]
+disp('gal',Vmagma,'The magma volume in the crystallizer be');
+//Using Eq.(27.44)
+//The nucleation rate is
+C = 10000; //[lb/h]
+rho_c = 105;
+B0 = 9*C/(2*rho_c*Vc*Lpr^3); //[nuclei/ft^3-h]
+disp('nuclei/ft^3-h',B0,'The nucleation rate necessary is');
+//Using Eq.(27.40), the zero-size particle density is
+n0 = B0/0.0018; //[nuclei/ft^4]
+L1 = (0:8)*10^-3;
+//Using Eq.(27.27)
+//Let A = log10(n), B = log10(n0)
+B = log10(n0);
+A = B - 1.1*10^3*L1/(2.3026);
+figure(1);
+plot(L1*10^3,A);
+xgrid();
+xlabel('L x 10^3 ft');
+ylabel('log n');
+title('Population density vs length');
+
+//From Fig. 27.15c for values of z corresponding to mesh openings.
+L1 = [11,14,16,19,23,27,33,38,46,54,65,78]'*10^-2;
+z = L1/(tou*G*100); //[mm]
+t = 0;
+function f = fun(z,xm)
+ f = z^3*exp(-z)/6;
+endfunction
+[xm]=ode(0,0,z,fun);
+for i=1:length(xm)
+ Diff(i) = z(i)^3*exp(-z(i))/6;
+end
+figure(2);
+subplot(2,1,1);
+plot(z,xm);
+xgrid();
+xlabel('z');
+ylabel('xm');
+title('cumulative mass distribution');
+subplot(2,1,2);
+plot(z,Diff)
+xgrid();
+xlabel('z');
+ylabel('dxm/dz');
+title('differential mass distribution');
diff --git a/839/CH28/EX28.1/Example_28_1.sce b/839/CH28/EX28.1/Example_28_1.sce new file mode 100755 index 000000000..f4a64b1a3 --- /dev/null +++ b/839/CH28/EX28.1/Example_28_1.sce @@ -0,0 +1,46 @@ +//clear//
+clear;
+clc;
+
+//Example 28.1
+//Given
+rho_p = 0.002650; //[g/mm^3]
+a = 2;
+phi_s = 0.571;
+//Solution
+//(a)
+//For the 4/6-mesh increment, from Table 28.2
+x = [0,2.51,12.5,32.07,25.7,15.9,5.38,2.10,1.02,0.77,0.58,0.41,0.31,0.75]'*10^-2; //[mass fraction]
+Dp = [4.699,3.327,2.362,1.651,1.168,0.833,0.589,0.417,0.295,0.208,0.147,0.104,0.074,0.0]'; //[mm]
+Dpbar(1) = 10^-5;
+for i =2:length(Dp)
+ Dpbar(i) = (Dp(i-1)+Dp(i))/2;
+end
+
+//(a)
+//Using Eq.(28.4)
+Aw = 6/(phi_s*rho_p)*sum(x(1:$-1)./Dpbar(1:$-1))/(1-x($)); //[mm^2/g]
+Nw = 1/(a*rho_p)*sum(x(1:$-1)./Dpbar(1:$-1)^3)/(1-x($)); //[particles/g]
+disp('particles/g',Nw,'Nw = ','mm^2/g',Aw,'Aw = ');
+
+//(b)
+//Using Eq.(28.9)
+Dvbar = (1/sum(x(1:$-1)./Dpbar(1:$-1)^3)/(1-x($)))^(1/3); //[mm];
+disp('mm',Dvbar,'Dvbar = ');
+
+//(c)
+//Using Eq.(28.6)
+Dsbar = 1/sum(x(1:$-1)./Dpbar(1:$-1))/(1-x($)); //[mm]
+disp('mm',Dsbar,'Dsbar = ');
+
+//(d)
+//Using Eq.(28.8) and Table 28.3
+Dwbar = sum(x.*Dpbar); //[mm]
+disp('mm',Dwbar,'Dwbar = ');
+
+//(e)
+//Using Eq.(28.11)
+N2 = x($-1)/(a*rho_p*Dpbar($-1)^3); //[particles/g]
+disp('particles/g',N2,'Nt = ');
+fra = N2/Nw;
+disp(fra,'Fraction of the particles in te top 12 increments = ');
diff --git a/839/CH28/EX28.2/Example_28_2.sce b/839/CH28/EX28.2/Example_28_2.sce new file mode 100755 index 000000000..b16ada1fa --- /dev/null +++ b/839/CH28/EX28.2/Example_28_2.sce @@ -0,0 +1,20 @@ +//clear//
+clear;
+clc;
+
+//Example 28.2
+//Given
+x = 0.14;
+xavg = 0.10;
+t = 3; //[min]
+x =[10.24,9.3,7.94,10.24,11.08,10.03,11.91,9.72,9.20,10.76,10.97,10.55]/100;
+
+//Solution
+mu = xavg;
+N =12;
+xbar = mean(x);
+//Substituing in Eq.(28.20)
+Ip = sqrt((N-1)*mu*(1-mu)/(sum(x^2)-xbar*sum(x)));
+//Using Eq.(28.18)
+s = stdev(x);
+disp(s,'s =',Ip,'Ip =')
diff --git a/839/CH29/EX29.1/Example_29_1.sce b/839/CH29/EX29.1/Example_29_1.sce new file mode 100755 index 000000000..8bbceb12c --- /dev/null +++ b/839/CH29/EX29.1/Example_29_1.sce @@ -0,0 +1,16 @@ +//clear//
+clear;
+clc;
+
+//Example 29.1
+//Given
+mdot = 100; //[ton/h]
+w1 = 0.80;
+w2 = 0.80;
+//Solution
+Wi = 12.74; //From Table 29.1
+Dpa = 2*25.4; //[mm]
+Dpb = 0.125*25.4; //[mm]
+//Using Eq.(29.10)
+P = mdot*0.3162*Wi*(1/Dpb^0.5-1/Dpa^0.5); //[kW]
+disp('kW',P,'Power required (P) = ');
diff --git a/839/CH29/EX29.2/Example_29_2.sce b/839/CH29/EX29.2/Example_29_2.sce new file mode 100755 index 000000000..9ed68b5f3 --- /dev/null +++ b/839/CH29/EX29.2/Example_29_2.sce @@ -0,0 +1,74 @@ +//clear//
+clear;
+clc;
+
+//Example 29.2
+//Given
+n = 1:7;
+beeta = 1.3;
+//From Table 29.2
+Dpn = [3.327,2.362,1.651,1.168,0.833,0.589,0.417]'; //[mm]
+Dpu = Dpn; //[mm]
+xn0 = [0.0251,0.125,0.3207,0.2570,0.1590,0.0538,0.0210]';
+Su(1) = 10*10^-4; //[s^-1]
+//B(1) = 1;
+//Solution
+
+//(a)
+//For the 4/6-mesh materials there is no input from coarser
+//material and applying Eq.(29.11). At the end of time tT
+x1 = xn0(1)*0.9;
+tT = 1/Su(1)*log(xn0(1)/x1); //[s]
+disp('s',tT,'Required time is');
+
+//(b)
+
+//Assuming Su varies with Dp^3
+for i = 1:length(Dpn)-1
+ Su(i+1) = Su(i)*(Dpn(i+1)/Dpn(i))^3; //[s^-1]
+end
+for i = 1:length(Dpn)
+ for j = 1:length(Dpu)
+//Using Eq.(29.13)
+ if (j<i)
+ B(i,j)=0;
+ else
+ B(i,j) = (Dpn(j)/Dpn(i))^beeta;
+ end
+end
+end
+
+for i = 1:length(Dpn)-1
+ for j = 1:length(Dpu)-1
+ if (j<i)
+ delta_B(i,j)=0;
+ else
+ delta_B(i,j) = B(i,j)-B(i,j+1);
+ end
+ end
+end
+disp(delta_B,'individual breakage functions');
+
+//(c)
+deltaT = 30; //[s]
+//Using Eq.(29.15)
+x=[];
+x(:,1) = xn0;
+for n = 1:length(xn0)
+ for t = 1:720
+ if (n==1)
+ x(n,t+1) = x(n,t)*(1-Su(n)*deltaT);
+ else
+ x(n,t+1) = x(n,t)*(1-Su(n)*deltaT)+ deltaT*Su(n-1)*delta_B(n-1,n-1)*x(n-1,t);
+ end
+ end
+end
+time = linspace(0,6,721);
+for i =1:length(xn0)
+ plot2d(time,x(i,:),style = i);
+ xgrid();
+ xlabel('time (h)');
+ ylabel('mass fraction (xa)');
+ title('Mass fractions');
+ legend('x1','x2','x3','x4','x5','x6','x7');
+end
diff --git a/839/CH30/EX30.1/Example_30_1.sce b/839/CH30/EX30.1/Example_30_1.sce new file mode 100755 index 000000000..a4cd2c6da --- /dev/null +++ b/839/CH30/EX30.1/Example_30_1.sce @@ -0,0 +1,34 @@ +//clear//
+clear;
+clc;
+
+//Example 30.1
+//Given
+//From Table 30.1
+Dp = [4.699,3.327,2.362,1.651,1.168,0.833,0.589,0.417,0.208,0.0000001]'; //[mm]
+F = [0,0.025,0.15,0.47,0.73,0.885,0.94,0.96,0.98,1.0]';
+O = [0,0.071,0.43,0.85,0.97,0.99,1.00]'; //[1 to 7]
+U = [0.0,0.195,0.58,0.83,0.91,0.94,0.975,1.00]'; //[3 to 10]
+
+//Solution
+plot(Dp,F)
+plot(Dp(1:7),O,'r')
+plot(Dp(3:$),U,'g')
+xgrid();
+xlabel('Dp mm');
+ylabel('Cumulative mass fraction larger than Dp');
+title('Analysis for Example 30.1');
+legend('Feed','Oversize','Undersize');
+
+//Cut-point diameter from the Table 30.1
+Dcp = 1.651; //[mm]
+xF = 0.47;
+xD = 0.85;
+xB = 0.195;
+//From Eq.(30.3)
+DbyF = (xF-xB)/(xD-xB);
+BbyF = 1-DbyF;
+//Using Eq.(30.7), overall effectiveness
+E = (xF-xB)*(xD-xF)*(1-xB)*(xD)/((xD-xB)^2*((1-xF)*xF));
+disp('respectively',BbyF,DbyF,'mass ratio of overflow and underflow is');
+disp(E,'Overall Effectiveness (E) =');
diff --git a/839/CH30/EX30.2/Example_30_2.sce b/839/CH30/EX30.2/Example_30_2.sce new file mode 100755 index 000000000..ec22bbabb --- /dev/null +++ b/839/CH30/EX30.2/Example_30_2.sce @@ -0,0 +1,60 @@ +//clear//
+clear;
+clc;
+
+//Example 30.2
+//Given
+//From Table 30.2
+V = linspace(0.5,6,12)'; //[L]
+t1 = [17.3,41.3,72,108.3,152.1,201.7]'; //[s]
+t2 = [6.8,19,34.6,53.4,76,102,131.2,163]'; //[s]
+t3 = [6.3,14,24.2,37,51.7,69,88.8,110,134,160]'; //[s]
+t4 = [5,11.5,19.8,30.1,42.5,56.8,73,91.2,111,133,156.8,182.5]'; //[s]
+t5 = [4.4,9.5,16.3,24.6,34.7,46.1,59,73.6,89.4,107.3]'; //[s]
+figure(1);
+plot(V(1:length(t1)),t1./V(1:length(t1)));
+plot(V(1:length(t2)),t2./V(1:length(t2)),'r');
+plot(V(1:length(t3)),t3./V(1:length(t3)),'g');
+plot(V(1:length(t4)),t4./V(1:length(t4)),'k');
+plot(V(1:length(t5)),t5./V(1:length(t5)),'y');
+xgrid();
+xlabel('V (L)');
+ylabel('t/V (s/L)');
+legend('deptaP = 6.7','deptaP = 16.2','deptaP = 28.2','deptaP = 36.3','deptaP = 49.1');
+title('t/V vs V');
+
+deltaP = [965,2330,4060,5230,7070]'; //[lbf/ft^2]
+//From Fig. 30.15
+//Slope(Kc/2)
+slope = [10440,5800,3620,3060,2400]'; //[s/ft^6]
+Kc = slope*2; //[s/ft^6]
+//Intercept(1/q0)
+Inter = [800,343,267,212,180]'; //[s/ft^3]
+//Viscosity of water
+muw = 5.95*10^-4; //[lb/ft-s], from Appendix 14
+//Filter area
+A = 440/30.48^2; //[ft^2]
+//concentration
+c = 23.5*28.31/454; //[lb/ft^3]
+gc = 32.14;
+//Using Eq.(30.22)
+Rm = A*gc/muw*deltaP.*(Inter)/10^10; //[ft^-1*10^10]
+//Using Eq.(30.24)
+alpha = A^2*gc/(c*muw)*deltaP.*(Kc)/10^11; //[ft/lb *10^-11]
+figure(2);
+plot2d(deltaP,Rm);
+xgrid();
+xlabel('deltaP (lbf/ft^2)');
+ylabel('Rm (ft^-1*10^-10)');
+title('Rm vs deltaP');
+figure(3);
+plot2d(log(deltaP),log(alpha));
+xgrid();
+xlabel('deltaP (lbf/ft^2)');
+ylabel('alpha (lb/ft*10^-11)');
+title('alpha vs deltaP');
+//Form 30.17
+disp(Rm,'Rm (ft^-1*10^-10) =');
+disp(alpha,'alpha (lb/ft*10^-11) =');
+alpha0 = 1.75*10^11/1000^0.26;
+disp('alpha = 2.9*10^10*deltaP^2.6','Emperical Equation for the cake');
diff --git a/839/CH30/EX30.3/Example_30_3.sce b/839/CH30/EX30.3/Example_30_3.sce new file mode 100755 index 000000000..73b02f672 --- /dev/null +++ b/839/CH30/EX30.3/Example_30_3.sce @@ -0,0 +1,24 @@ +//clear//
+clear;
+clc;
+
+//Example 30.3
+//Given
+f = 0.30;
+tc = 5*60; //[s]
+n = 1/tc; //[s^-1]
+cF = 14.7; //[lb/ft^3]
+deltaP = 1414;
+mFbymC = 2
+//Solution
+alpha0 = 2.9*10^10; //[ft/lb], From Example 30.2
+s = 0.26;
+mu = 6.72*10^-4; //[lb/ft-s]
+rho = 62.3; //[lb/ft^3]
+gc =32.17;
+//Using Eq.(30.19)
+c = cF/(1-(mFbymC-1)*(cF/rho)); //[lb/ft^3]
+mcdot = 10/(60*7.48)*(1/(cF/168.8+1))*cF; //[lb/s]
+//Solving Eq.(30.34)
+AT = mcdot*(alpha0*mu/(2*c*1414^(1-s)*gc*f*n))^(0.5);
+disp('ft^2',AT,'Filter Area(AT) =');
diff --git a/839/CH30/EX30.4/Example_30_4.sce b/839/CH30/EX30.4/Example_30_4.sce new file mode 100755 index 000000000..253ea6122 --- /dev/null +++ b/839/CH30/EX30.4/Example_30_4.sce @@ -0,0 +1,83 @@ +//clear//
+clear;
+clc;
+
+//Example 30.4
+//Given
+D = 2; //[cm]
+Vbar = 150; //[cm/s]
+rho = 1; //[g/cm^3]
+mu = 0.01; //[g/cm-s]
+Dv = 4*10^-7; //[cm^2/s]
+
+//Solution
+//(a)
+Nre = Vbar*D*rho/mu;
+Nsc = mu/(rho*Dv);
+//Using Eq.(21.55)
+Nsh = 0.0096*Nre^0.913*Nsc^0.346;
+kc = Nsh*Dv/D; //[cm/s]
+pi = poly([0,4.4*10^-3,-1.7*10^-6,7.9*10^-8],'c',"coeff");
+//For
+c1 = 10; //[g/L]
+v = 10^-3; //[cm/s]
+//Using Eq.(30.53)
+cs = c1*exp(v/kc); //[g/L]
+deltaPi = horner(pi,cs);
+Qm = 250/36000; //[cm/s-atm]
+//Using Eq.(30.50)
+deltaP = v/Qm+deltaPi; //[atm]
+//Using Eq.(30.53)
+cs = 400;
+vmax = kc*log(cs/c1); //[cm/s]
+deltaP = vmax/Qm+horner(pi,cs); //[tm]
+c = [10,20,40];
+V=[];
+deltaP=[];
+for j = 1:length(c)
+c1 = c(j);
+i = 1;
+vmax = kc*log(cs/c1)*10^4;
+h = (vmax-1)/1000;
+ for v = 1:h:vmax
+ cs = c1*exp(v*10^-4/kc); //[g/L]
+ deltaPi = horner(pi,cs); //[atm]
+ deltaP(j,i) = v*10^-4/Qm+deltaPi; //[atm]
+ V(j,i) = v*10^-4;
+ i = i+1;
+ end
+end
+V = V*36000;
+for l=1:length(c)
+ figure(1)
+ plot2d(deltaP(l,:),V(l,:),style=l);
+ xgrid();
+ xlabel('deltaP (atm)');
+ ylabel('Permeate flux (L/m^2-h)');
+ title('Effective pressure drop and concentration on flux')
+ legend('Cf=10,','Cf=20','Cf=40');
+end
+
+//(b)
+Qmb = Qm/5; //[cm/s-atm]
+vb = 10^-3; //[cm/s]
+c = 40; //[g/L]
+c1 = 40;
+csb = c1*exp(vb/kc);
+deltaPi = horner(pi,csb);
+deltaPb = vb/Qmb+deltaPi;
+disp('The largest effect of the lower membrane permeability is a 30 percent reduction in low pressure drop');
+i = 1;
+vmax = kc*log(400/c1)*10^4;
+h = (vmax-1)/1000;
+for vb = 1:h:vmax
+ csb = c1*exp(vb*10^-4/kc); //[g/L]
+ deltaPi = horner(pi,csb); //[atm]
+ deltaPb(i) = vb*10^-4/Qmb+deltaPi; //[atm]
+ Vb(i) = vb*10^-4;
+ i = i+1;
+end
+Vb = Vb*36000;
+plot2d(deltaPb,Vb, style = l+1)
+legend('Cf=10,','Cf=20','Cf=40','Cf = 40(Qm = 250/5)' );
+
diff --git a/839/CH30/EX30.5/Example_30_5.sce b/839/CH30/EX30.5/Example_30_5.sce new file mode 100755 index 000000000..8a90e9e7c --- /dev/null +++ b/839/CH30/EX30.5/Example_30_5.sce @@ -0,0 +1,60 @@ +//clear//
+clear;
+clc;
+
+//Example 30.5
+//Given
+D = 1.5; //[cm]
+Nre = 25000;
+Qm = 40; //[L/m62-h]
+Mw = 30000;
+Dv = 5*10^-7; //[cm^2/s]
+R = 0.75;
+
+//Solution
+//(a)
+//Base case:
+v = Qm*2.78*10^-5; //[cm/s]
+Nsc = 0.01/Dv;
+//Using Eq.(21.55)
+Nsh = 0.0096*Nre^0.913*Nsc^0.346;
+kc = Nsh*Dv/D; //[cm/s]
+//Let A = K/(1-K)
+A = (1-R)/R*exp(-v/kc);
+K = A/(1+A);
+//If the flux is reduced to 0.556*10^-3 cm/s
+//Let B = (1-R)/R
+B = K/(1-K)*exp(0.556*10^-3/kc);
+R = 1/(1+B);
+//As flux approaches zero R appraoches 1-K:
+Rmax = 1-K;
+disp(R,'fraction rejected (R) =');
+disp(Rmax,'maximum rejection (Rmax) =');
+
+//(b)
+//Using Fig. (30.24)
+kc1 = kc;
+M2 = 10000;
+R2 = 0.35;
+K1 = K;
+lambda1 = 1-K1^0.5;
+lambda2 = lambda1*(10000/Mw)^(1/3);
+K2 = (1-lambda2)^2;
+kc2 = kc1*3^0.22; //[cm/s]
+//Let B2 = (1-R2)/R2
+B2 = K2/(1-K2)*exp(v/kc2);
+R2 = 1/(1+B2);
+disp(R2,'fraction rejected (R2) =');
+
+//(c)
+Dpore = 10^-7; //[cm^2/s]
+eps = 0.5;
+tou = 2;
+De = 2.5*10^-8; //[cm^2/s]
+L = 2*10^-5; //[cm]
+v = 5.56*10^-4; //[cm/s]
+vLbyDe = v*L/De;
+//Using Eq.(30.63)
+K = 0.101;
+c2bycs = K*exp(vLbyDe)/(K-1+exp(vLbyDe));
+disp('Diffusion in the membrane makes the premeate concentrations about twice as high as it would be if c2=Kcs=0.101cs, indicating that the partition coefficient is lower than that estimated in part(a) ');
diff --git a/839/CH4/EX4.1/Example_4_1.sce b/839/CH4/EX4.1/Example_4_1.sce new file mode 100755 index 000000000..fb784fd1e --- /dev/null +++ b/839/CH4/EX4.1/Example_4_1.sce @@ -0,0 +1,48 @@ +//clear//
+clear;
+clc;
+
+//Example 4.1
+
+// (a)
+// density of the fluid
+rho = 0.887*62.37;// [lb/ft^3]
+// total volumetric flow rate
+q = 30*60/7.48; //[ft^3/hr]
+// mass flow rate in pipe A and pipe B is same
+mdot = rho*q //[lb/hr]
+// mass flow rate in each pipe of C is half of the total flow
+mdot_C = mdot/2 //[lb/hr]
+disp('lb/hr',mdot,'mass flow rate pipe A = ')
+disp('lb/hr',mdot,'mass flow rate pipe B = ')
+disp('lb/hr',mdot_C,'mass flow rate pipe C = ')
+
+// (b)
+// Using Eq.(4.4),
+// velocity through pipe A
+V_Abar = 240.7/(3600*0.0233) //[ft/s]
+
+// velocity through pipe B
+V_Bbar = 240.7/(3600*0.0513) //[ft/s]
+
+// velocity through each pipe of C
+V_Cbar = 240.7/(2*3600*0.01414) //[ft/s]
+
+disp('ft/s',V_Abar,'velocity through pipe A = ')
+disp('ft/s',V_Bbar,'velocity through pipe B = ')
+disp('ft/s',V_Cbar,'velocity through pipe C = ')
+
+// (c)
+// Using Eq.(4.8),
+// mass velocity through pipe A
+GA = mdot/0.0233 // [kg/m^2-s]
+
+// mass velocity through pipe B
+GB = mdot/0.0513 //[kg/m^2-s]
+
+// mass velocity through each pipe of C
+GC = mdot/(2*0.01414) //[kg/m^2-s]
+
+disp('kg/m^2-s',GA,'mass velocity through pipe A = ')
+disp('kg/m^2-s',GB,'mass velocity through pipe B = ')
+disp('kg/m^2-s',GC,'mass velocity through pipe C = ')
diff --git a/839/CH4/EX4.2/Example_4_2.sce b/839/CH4/EX4.2/Example_4_2.sce new file mode 100755 index 000000000..e20880827 --- /dev/null +++ b/839/CH4/EX4.2/Example_4_2.sce @@ -0,0 +1,12 @@ +//clear//
+clear;
+clc;
+
+//Example 4.2
+//Applying Eq.(4.25)
+//Pa = Pb, Ua = 0
+// Zb = 0, Za = 5m
+
+//The velocity at streamline discharge
+Ub = sqrt(5*2*9.80665)//[m/s]
+disp('m/s',Ub,'streamline discharge velocity (Ub) =')
diff --git a/839/CH4/EX4.3/Example_4_3.sce b/839/CH4/EX4.3/Example_4_3.sce new file mode 100755 index 000000000..7ed7fcbe6 --- /dev/null +++ b/839/CH4/EX4.3/Example_4_3.sce @@ -0,0 +1,45 @@ +//clear//
+clear;
+clc;
+
+//Example 4.3
+rho = 998; // [kg/m^3]
+Da = 50; //[mm]
+Db = 20; //[mm]
+pa = 100; //[N/m^2]
+
+//(a)
+Va_bar = 1.0; //[m/s]
+Vb_bar = Va_bar*(Da/Db)^2 //[m/s]
+//Using Eq.(4.29)
+//Za = Zb, hf = 0
+pb = pa-rho*(Vb_bar^2-Va_bar^2)/(2*1000) //[kN/m^2]
+disp('kN/m^2',pb,'pb =')
+
+//(b)
+// Combining Eqs.(4.14) & (4.15)
+//For x direction,
+//since Fg = 0, we get Eq.(4.30)
+theta = %pi/4;
+Va_xbar = Va_bar;
+Sa = (%pi/4)*(Da/1000)^2; //[m^2]
+Sax = Sa;
+//From FIg 4.5
+Vb_xbar = Vb_bar*cos(theta);//[m/s]
+Sb = %pi/4*(Db/1000)^2; //[m^2]
+Sbx = Sb*sin(theta);// [m^2]
+//Using Eq.(4.6)
+mdot = Va_bar*rho*Sa; //[kg/s]
+//Substituting in Eq.(4.30)
+//Solving for Fw,x
+beta_a = 1; beta_b = 1;
+Fw_x = mdot*(beta_b*Vb_xbar-beta_a*Va_xbar)-Sax*pa*1000+Sbx*pb*1000 //[N]
+
+//For y direction,
+//Va_ybar = 0, Say = 0
+Vb_ybar = Vb_bar*sin(theta); //[m/s]
+Sby = Sb*cos(theta); //[m^2]
+Va_ybar = 0; //[m/s]
+Say = 0;// [m/s]
+
+Fw_y = mdot*(beta_b*Vb_ybar-beta_a*Va_ybar)-Say*pa*1000+Sby*pb*1000 //[N]
diff --git a/839/CH4/EX4.4/Example_4_4.sce b/839/CH4/EX4.4/Example_4_4.sce new file mode 100755 index 000000000..068f08250 --- /dev/null +++ b/839/CH4/EX4.4/Example_4_4.sce @@ -0,0 +1,34 @@ +//clear//
+clear;
+clc;
+
+//Example 4.4
+gc = 32.17; //[ft-lb/lbf-s^2]
+rho_w = 62.37;//[lb/ft^3], density of water
+sp_gravity = 1.84;
+neta = 0.60;
+hf = 10; //[ft-lbf/lb], friction losses
+Va_bar = 3; //[ft/s]
+Da = 3; //[in.]
+Db = 2; //[in.]
+//From Appendix corss secional area respective to 3in. and 2in. diameter
+Sa = 0.0513; //[ft^2]
+Sb = 0.0233; //[ft^2]
+Za = 0 ;//[ft]
+Zb = 50 ;//[ft]
+Vb_bar = Va_bar*(Sa/Sb); //[ft/s]
+g =gc
+//Using Eq.(4.32)
+Wp = ((Zb*g/gc)+Vb_bar^2/(2*gc)+hf)/neta;//[ft-lbf/lb]
+
+//Using Eq.(4.32) on pump itself
+//station a is the suction connection and station b is the discharge
+//Za = Zb
+//Eq.(4.32) becomes
+//the pressure developed by pume is deltaP = pb-pa
+deltaP = sp_gravity*rho_w*(((Va_bar^2-Vb_bar^2)/(2*gc))+neta*Wp) //[lbf/ft^3]
+
+mdot = Sa*Va_bar*sp_gravity*rho_w;
+
+//the Power
+P = mdot*Wp/550 //[hp]
diff --git a/839/CH5/EX5.1/Example_5_1.sce b/839/CH5/EX5.1/Example_5_1.sce new file mode 100755 index 000000000..141ba9a06 --- /dev/null +++ b/839/CH5/EX5.1/Example_5_1.sce @@ -0,0 +1,25 @@ +//clear//
+clear;
+clc;
+
+//Example 5.1
+// Given
+mu = 0.004; //[kg/m-s]
+D = 0.0779;// [m]
+rho = 0.93*998; //[kg/m^3]
+L = 45; //[m]
+
+//For fittings, form Table 5.1
+sum_Kf = 0.9 + 2*0.2;
+//From Eq.(4.29), assuming alpha_a = 1,
+// since pa = pb, and Va_bar = 0
+//A = Vb_bar^2/2 + hf = g*(Za-Zb)
+A = 9.80665*(6+9); //[m^2/s^2]
+//Using Fig 5.9
+f = 0.0055;
+//Using Eq.(5.68), There is no exapnsion loss and Ke = 0.
+//From Eq.(5.66), since Sa is very lage, Kc = 0.4. Hence
+Vb_bar = sqrt(294.2/(2.7+2311*f));//[m/s]
+//From Appendix 5, cross sectional area of the pipe
+S = 0.00477; //[m^2]
+flow_rate = S*Vb_bar*3600 //[m^3/hr]
diff --git a/839/CH6/EX6.1/Example_6_1.sce b/839/CH6/EX6.1/Example_6_1.sce new file mode 100755 index 000000000..ed2be9638 --- /dev/null +++ b/839/CH6/EX6.1/Example_6_1.sce @@ -0,0 +1,61 @@ +//clear//
+clear;
+clc;
+
+//Example 6.1
+//Given
+gama = 1.4;
+M = 29;
+R = 82.0568*10^-3; //[atm-m^3/Kg mol-K]
+Nma = 0.8;
+gc = 1; //[ft-lb/lbf-s^2]
+//At Entrance
+p0 = 20; //[atm]
+T0 = 555.6; //[K]
+
+//(a)
+// Using Eq.(6.28)
+//Pressure at throat
+pt = (1/(1+((gama-1)/2)*Nma^2)^(1/(1-1/gama)))*p0 //[atm]
+//From Eq.(6.10)
+rho0 = (p0*M)/(R*T0); //[kg/m^3]
+// Using Eq.(6.10) and Eq.(6.26), the velocity in the throat
+ut = sqrt((2*gama*gc*R*T0)/(M*(gama-1))*(1-(pt/p0)^(1-1/gama))); // [m^3-am/kg]^0.5
+//In terms of [m/s], Using Appendix 2, 1 atm = 1.01325*10^ [N/m^2]
+ut = ut*sqrt(1.01325*10^5) //[m/s]
+//Using Eq.(6.23), density at throat
+rho_t = rho0*(pt/p0)^(1/gama) //[kg/m^3]
+//The mass velocity at the throat,
+Gt = ut*rho_t //[kg/m^2-s]
+//Using Eq.(6.24), The temperature at throat
+Tt = T0*(pt/p0)^(1-1/gama) // [K]
+
+//(b)
+// From Eq.(6.29)
+pstar = ((2/(gama+1))^(1/(1-1/gama)))*p0 //[atm]
+//From Eq.(6.24) and (6.29)
+Tstar = T0*(pstar/p0)^(1-1/gama) //[K]
+//From Eq.(6.23)
+rho_star = rho0*(pstar/p0)^(1/gama) //[Kg/m^3]
+//From Eq.(6.30)
+G_star = sqrt(2*gama*gc*rho0*p0*101.325*10^3/(gama-1))*(pstar/p0)^(1/gama)*sqrt(1-(pstar/p0)^(1-1/gama)) //[Kg-m^2/s]
+u_star = G_star/rho_star //[m/s]
+
+//(c)
+// By continuity, G inversely proportional to S, the mass velocity at dischage is
+G_r = G_star/2 // [Kg/m^3-s]
+//Using Eq.(6.30)
+// Let x = pr/p0
+err = 1;
+eps = 10^-3;
+x = rand(1,1);
+
+while(err>eps)
+ xnew = ((0.1294)/sqrt(1-x^(1-1/1.4)))^1.4;
+ err = x-xnew;
+ x=xnew;
+end
+
+//Using Eq.(6.27)
+//The Mach Number at dischage is
+Nmr = sqrt((2/(gama-1))*(1/x^(1-1/gama)-1))
diff --git a/839/CH6/EX6.2/Example_6_2.sce b/839/CH6/EX6.2/Example_6_2.sce new file mode 100755 index 000000000..4abb1d09a --- /dev/null +++ b/839/CH6/EX6.2/Example_6_2.sce @@ -0,0 +1,57 @@ +//clear//
+clear;
+clc;
+
+//Example 6.2
+//Given
+Tr = 1000; //[R]
+pr = 20; //[atm]
+Ma_a = 0.05;
+gama = 1.4;
+gc = 32.174; //[ft-lb/lbf-s^2]
+M = 29;
+R = 1545;
+//(a)
+//Using Eq.(6.45)
+A = 2*(1+((gama-1)/2)*Ma_a^2)/((gama+1)*Ma_a^2);
+fLmax_rh = (1/Ma_a^2-1-(gama+1)*log(A)/2)/gama
+
+//(b)
+//Using Eq.(6.28), the pressure at the end of the isentropic nozzle pa
+A = (1+(gama-1)*(Ma_a^2)/2);
+pa = pr/(A^(gama/(gama-1))) // [atm]
+//From Example 6.1, the density of air at 20atm and 1000R is 0.795 lb/ft^3
+//Using Eq.(6.17), the acoustic velocity
+Aa = sqrt(gc*gama*Tr*R/M) //[m/s]
+//The velocity at the entrance of the pipe
+ua = Ma_a*Aa //[m/s]
+//When L_b = L_max, the gas leaves the pipe at the asterisk conditions, where
+Ma_b = 1;
+// Using Eq.(6.43)
+A = (gama-1)/2;
+Tstar = Tr *(1+A*Ma_a^2)/(1+A*Ma_b^2) // [K]
+// Using Eq.(6.44)
+rho_star = 0.795*Ma_a/sqrt(2*(1+(gama-1)*Ma_a^2/2)/(2.4)) //[lb/ft^3]
+//Using Eq.(6.39)
+pstar = p0*Ma_a/sqrt(1.2) // [atm]
+//Mass velocity through the entire pipe
+G = 0.795*ua //[lb/ft^2-s]
+ustar = G/rho_star //[ft/s]
+
+//(c)
+//Using Eq.(6.45) with f_Lmax_rh = 400
+
+err = 1;
+eps = 10^-3;
+Ma_ac = rand(1,1);
+i =1;
+while((err>eps))
+ A = 2*(1+((gama-1)/2)*Ma_ac^2)/((gama+1)*Ma_ac^2);
+ B = gama*400+1+(gama+1)*log(A)/2;
+ Ma_anew = sqrt(1/B);
+ err = Ma_ac-Ma_anew;
+ Ma_ac = Ma_anew;
+end
+Ma_ac;
+uac = Ma_ac*ua/Ma_a //[ft/s]
+Gc = uac*0.795 //[lb/ft^2-s]
diff --git a/839/CH6/EX6.3/Example_6_3.sce b/839/CH6/EX6.3/Example_6_3.sce new file mode 100755 index 000000000..c9884b1ca --- /dev/null +++ b/839/CH6/EX6.3/Example_6_3.sce @@ -0,0 +1,37 @@ +//clear//
+clear;
+clc;
+
+//Example 6.3
+//Given
+pa = 2.7; //[atm]
+T = 288; //[K]
+D = 0.075; //[m]
+L = 70; //[m]
+Vbar = 60; //[m/s]
+M = 29;
+rh = D/4; //[m]
+mu = 1.74*10^-5 //[kg/m-s] Appendix 8
+rho_a = (29/22.4)*(2.7/1)*(273/288) // [kg/m^3]
+R = 82.056*10^-3;
+G = Vbar*rho_a //[kg/m^2-s]
+Nre = D*G/mu;
+kbyd = 0.00015*(0.3048/0.075);
+f = 0.0044; //[from Fig. 5.9]
+
+//Using Eq.(6.52)
+//pbar = 1.982; //[atm]
+//pb = 1.264; //[atm]
+err = 1;
+eps = 10^-3;
+pb = 1.5;
+
+while(err>eps)
+pbar = (pa+pb)/2;
+A = ((f*L/(2*rh))+log(pa/pb));
+pb_new = pa-(R*T*G^2/(pbar*29*101325))*A;
+err = pb-pb_new;
+pb = pb_new;
+end
+pb; //[atm]
+pbar = (pa+pb)/2 // [atm]
diff --git a/839/CH7/EX7.1/Example_7_1.sce b/839/CH7/EX7.1/Example_7_1.sce new file mode 100755 index 000000000..ec094655d --- /dev/null +++ b/839/CH7/EX7.1/Example_7_1.sce @@ -0,0 +1,39 @@ +//clear//
+clear;
+clc;
+
+//Example 7.1
+// Given
+rho_p = 2800; // [kg/m^3]
+g = 9.80665; //[m/s^2]
+ac = 50*g; // [m/s^2]
+//(a)
+//From appendix 20
+Dp_100 = 0.147; //[mm]
+Dp_80 = 0.175; //[mm]
+Dp = (Dp_100+Dp_80)/2;//[mm]
+
+//From Appendix 14
+mu = 0.801; //[cP]
+rho = 995.7; //[kg/m^2]
+// Using Eq.(7.45)
+K = Dp*10^-3*(g*rho*(rho_p-rho)/(mu*10^-3)^2)^(1/3);
+//This is slightly above the Stoke's-law range
+//Assuming
+N_rep = 4.4;
+//From Fig. 7.6
+Cd = 7.9;
+//From Eq.(7.37)
+mu_ta = sqrt(4*g*(rho_p-rho)*Dp*10^-3/(3*Cd*rho)) // [m/s]
+
+// (b)
+//Using 'ac' in place of 'g' in Eq.(7.45)
+K = K*50^(1/3); // Since only acceleration changes
+//Etimating
+N_rep = 80; //From Fig. (7.6)
+Cd = 1.2;
+mu_tb1 = sqrt(4*ac*(rho_p-rho)*Dp*10^-3/(3*Cd*rho)) // [m/s]
+// For irregular particles Cd is about 20 percent greater
+//than that for spheres
+Cd = 1.2*1.2;
+mu_tb2 = sqrt(4*ac*(rho_p-rho)*Dp*10^-3/(3*Cd*rho)) // [m/s]
diff --git a/839/CH7/EX7.2/Example_7_2.sce b/839/CH7/EX7.2/Example_7_2.sce new file mode 100755 index 000000000..ea53ed8c5 --- /dev/null +++ b/839/CH7/EX7.2/Example_7_2.sce @@ -0,0 +1,29 @@ +//clear//
+clear;
+clc;
+
+//Example 7.2
+//Given
+g = 32.174; //[ft-lb/lbf-s^2]
+eps = 0.8;
+speg_s = 4.0;
+speg_c = 1.594;
+Ds = 0.004; //[in.]
+rho_w = 62.37; //[lbf/ft^3]
+delta_speg = speg_s-speg_c;
+delta_rho = rho_w*delta_speg; //[lbf/ft^3]
+rho_c = rho_w*speg_c; //[lbf/ft^3]
+//From Appendix 9
+mu = 1.03; //[cP]
+//Using Eq.(7.45)
+K = Ds/12*(g*rho_c*(delta_rho)/(mu*6.72*10^-4)^2)^(1/3);
+//Using Eq.(7.40)
+ut = g*(Ds/12)^2*delta_rho/(18*mu*6.72*10^-4) //[ft/s]
+
+//The terminal velocity in hindered settling
+//Calculating Reynolds Number
+Nre = ut*rho_c*Ds/(12*mu*6.72*10^-4);
+//From Fig.(7.7)
+n = 4.1;
+//Using Eq.(7.46)
+us = ut*eps^n //[ft/s]
diff --git a/839/CH7/EX7.3/Example_7_3.sce b/839/CH7/EX7.3/Example_7_3.sce new file mode 100755 index 000000000..83125f933 --- /dev/null +++ b/839/CH7/EX7.3/Example_7_3.sce @@ -0,0 +1,22 @@ +//clear//
+clear;
+clc;
+
+//Example 7.3
+//The quantities needed are
+mu = 0.01; //[P]
+delta_rho = 0.24; //[g/cm^3]
+//Using Eq.(7.51),solving the quadratic equation for Vom_bar
+a = 1.75*1/(0.11*0.4^3);
+b = 150*0.01*0.6/(0.11^2*0.4^3);
+c = - 980*0.24;
+Vom_bar = (-b+sqrt(b^2-4*a*c))/(2*a); //[cm/s]
+//Corresponding Reynolds number
+Nre = 0.11*0.194*0.124/0.01;
+//From Fig 7.13
+m = 3.9;
+//For 25 percent exapnsion
+LbyLm = 1.25;
+eps = 0.52;
+//From Eq.(7.59)
+Vo_bar = 1.94*(0.52/0.40)^3.9 //[mm/s]
diff --git a/839/CH8/EX8.1/Example_8_1.sce b/839/CH8/EX8.1/Example_8_1.sce new file mode 100755 index 000000000..2a30d5e1b --- /dev/null +++ b/839/CH8/EX8.1/Example_8_1.sce @@ -0,0 +1,39 @@ +//clear//
+clear;
+clc;
+
+//Example 8.1
+//Given
+vdot = 40; //[gal/min]
+pb = 50; //[lbf/in.^2]
+Za = 4; //[ft]
+Zb = 10; //[ft]
+hfs = 0.5; //[lbf/in.^2]
+hfd = 5.5; //[lbf/in.^2]
+neta = 0.6;
+rho = 54; //[lb/ft^3]
+pv = 3.8; //[lbf/in.^2]
+g = 9.8; //[m/s^2]
+gc = 32.17 //[ft-lb/lbf-s^2]
+hf = hfs+hfd; // [lbf/in.^2]
+//(a)
+//Using data from Appendix 5
+Vb_bar = vdot/6.34; //[ft/s]
+//Using Eq.(4.32)
+
+Wp_neta = ((14.7+pb)*144/rho)+(g/gc*10)+(Vb_bar^2/(2*gc))+(hf*144/54)-(14.7*144/54); // [ft-lbf/lb]
+delta_H = Wp_neta;
+
+//(b)
+mdot = vdot*rho/(7.48*60); // [lb/s]
+//Using Eq.(8.7), the input power is
+Pb = mdot*delta_H/(550*neta) // [hp]
+
+//(c)
+padash = 14.7*144/rho;
+//The vapor pressure corresponding to a head
+hv = pv*144/rho; // [ft-lbf/lb]
+//friction in the suction line
+hfs = 0.5*144/rho ; // [ft-lbf/lb]
+//Using Eq.(8.7), value of available
+NPSH = padash-hv-hfs-Za // [ft]
diff --git a/839/CH8/EX8.2/Example_8_2.sce b/839/CH8/EX8.2/Example_8_2.sce new file mode 100755 index 000000000..7857e7a9e --- /dev/null +++ b/839/CH8/EX8.2/Example_8_2.sce @@ -0,0 +1,32 @@ +//clear//
+clear;
+clc;
+
+// Example 8_2
+// Given
+pa = 29; //[in.Hg]
+pb = 30.1; //[in.Hg]
+va = 0; //[ft/s]
+vb = 150; //[ft/s]
+Ta = 200; //[F]
+vdot = 10000; //[ft^3/min]
+neta = 0.65;
+M = 31.3;
+R = 29.92;
+gc = 32.17;//[ft-lb/lbf-s^2]
+//actual suction density
+rho_a = M*pa*(460+60)/(378.7*30*(460+Ta)); //[lb/ft^3]
+//acual discharge density
+rho_b = rho_a*pb/pa; //[lb/ft^3]
+// average density of the flowing gas
+rho = (rho_a+rho_b)/2; //[lb/ft^3]
+//mass flow rate
+mdot = vdot*M/(378.7*60) //[lb/s]
+//developed pressure
+dev_p = (pb-pa)*144*14.7/(R*rho); //[ft-lbf/lb]
+//velocity head
+vel_head = vb^2/(2*gc); //[ft-lbf/lb]
+//Using Eq.(8.1), alpha_a = alpha_b = 1,va = 0, Za = Zb,
+Wp = (dev_p+vel_head)/neta // [ft-lbf/lb]
+//Using Eq.(8.4)
+Pb = mdot*Wp/550 //[hp]
diff --git a/839/CH8/EX8.3/Example_8_3.sce b/839/CH8/EX8.3/Example_8_3.sce new file mode 100755 index 000000000..68cf1745a --- /dev/null +++ b/839/CH8/EX8.3/Example_8_3.sce @@ -0,0 +1,38 @@ +//clear//
+clear;
+clc;
+
+//Example 8.3
+//Given
+vdot = 180; //[ft^3/min]
+pa = 14; //[lbf/in.^2]
+pb = 900; //[lbf/in.^2]
+Ta = 80+460; //[K]
+q0 = 0.063; //[m^3/s]
+Cp = 9.3; //[Btu/lbmol-F]
+gama = 1.31;
+delta_Tw = 20; // [F]
+//(a)
+neta = 0.80;
+//For a multistage compressor the total power is a minimum if each stage doed the same amount of work
+//Hence using same copression ration for each stage
+//Using Eq.(8.25)
+//For one stage
+comp_ratio = (900/14)^(1/3);
+//Using Eq.(8.29), the power required by each stage
+Pb = (Ta*q0*gama*vdot)*(comp_ratio^(1-1/gama)-1)/(520*(gama-1)*neta); // [hp]
+//Total Power
+Pt = 3*Pb // [hp]
+
+//(b)
+//Using Eq.(8.22), the temperature at the exit of each stage
+Tb = Ta*comp_ratio^(1-1/gama) // [R]
+
+//(c) Since 1 lb mol = 378.7 std ft^3, the flow rate is
+vdot = vdot*60/378.7; //[lb mol/h]
+// Heat load in each cooler is
+Hl = vdot*Cp*(Tb-Ta) // [Btu/h]
+//Total heat loss
+Htotal = 3*Hl; //[Btu/h]
+//Cooling water requirement
+cwr = Htotal/delta_Tw // [lb/h]
diff --git a/839/CH8/EX8.4/Example_8_4.sce b/839/CH8/EX8.4/Example_8_4.sce new file mode 100755 index 000000000..dfffa98dc --- /dev/null +++ b/839/CH8/EX8.4/Example_8_4.sce @@ -0,0 +1,24 @@ +//clear//
+clear;
+clc;
+
+//Example 8.4
+//Given
+q = 75/3600 ; // [m^3/s]
+rho = 62.37*16.018; //[kg/m^3] From Appendix 4
+Cv = 0.98;
+g = 9.80665; //[m/s^2]
+Sw = 1;
+Sm = 13.6;
+h = 1.25; //[m]
+//(a)
+//Using Eq.(2.10)
+delta_p = g*h*(Sm-Sw)*rho ; //[N/m^2]
+//Using Eq.(8.36), neglecting the effect of beta
+Sb = q/(Cv*sqrt(2*delta_p/rho));
+Db = sqrt(4*Sb/%pi)*100 // [mm]
+
+//(b)
+press_loss = 0.1*delta_p; //[N/m^2]
+// Power required at full flow
+P = q*press_loss/1000 // [kW]
diff --git a/839/CH8/EX8.5/Example_8_5.sce b/839/CH8/EX8.5/Example_8_5.sce new file mode 100755 index 000000000..fea275723 --- /dev/null +++ b/839/CH8/EX8.5/Example_8_5.sce @@ -0,0 +1,34 @@ +//clear//
+clear;
+clc;
+
+//Example 8.5
+//Given
+T = 100; //[F]
+mu_O = 5.45; //[cP]
+spg_O = 0.8927;
+spg_m = 13.6;
+spg_gl = 1.11;
+q = 12000; //[bbl/d]
+rho_ratio = 0.984;
+rho_w = 62.37; //[lb/ft^3]
+h = 30; //[in.]
+gc = 32.174; //[ft-lb/lbf-s^2]
+//(a)
+//Using Eq.(8.42)
+rhoB_60 = spg_O*rho_w; //[lb/ft^3]
+rho_100 = spg_O*rho_w*rho_ratio; //[lb/ft^3]
+mdot = q*42*rhoB_60/(24*3600*7.48); // [lb/s]
+Da = 4.026/12; //[ft]
+delta_p = h/12*(spg_m-spg_gl)*rho_w*(1); //[lbf/ft^2]
+//Using Eq.(8.42)
+beeta = sqrt(4*mdot/(0.61*%pi*Da^2*sqrt(2*gc*delta_p*rho_100)));
+Do = Da*beeta; //[ft]
+// the orifice diameter
+D = 12*Do //[in.]
+
+//(b)
+//Using Fig. 8.20, the fraction of differential pressure loss is
+fra_prss_loss = 0.68;
+//Maximum power consumption
+P = mdot*delta_p*fra_prss_loss/(rho_ratio*rho_w*spg_O*550) //[hp]
diff --git a/839/CH8/EX8.6/Example_8_6.sce b/839/CH8/EX8.6/Example_8_6.sce new file mode 100755 index 000000000..fdd23b034 --- /dev/null +++ b/839/CH8/EX8.6/Example_8_6.sce @@ -0,0 +1,31 @@ +//clear//
+clear;
+clc;
+
+//Example 8.4
+//Given
+Cpt = 0.98;
+Ta = 200; //[F]
+Da = 36; //[in.]
+pa = 15.25; //[in.]
+h = 0.54; //[in.]
+P = 29.92; //[in.]
+spg_m =13.6; //[specific gravity of mercury]
+rho_w = 62.37; //[lb/ft^3]
+gc = 32.174; //[ft-lb/lbf-s^2]
+//Using Eq.(8.52)
+Pabs = P+pa/spg_m; //[in.]
+rho = 29*492*31.04/(359*(200+460)*29.92); //[lb/ft^3]
+//From manometer reading
+delta_p = h/12*rho_w //[lbf/ft^3]
+
+//Using Eq.(8.53, m*aximum velocity, assuming Nma is negligible
+umax = Cpt*sqrt(2*gc*delta_p/rho) // [ft/s]
+//The reynolds number based on maximum velocity
+mu_air = 0.022 ; //[cP] form Appendix 8
+Nre_max = (Da/12)*umax*rho/(mu_air*0.000672);
+//Using Fig 5.7, to obtain average velocity
+Vbar = 0.86*umax // [ft/s]
+Nre = Nre_max*0.86;
+//The volumetric flow rate
+q = Vbar*(Da/12)^2*%pi/4*520/660*Pabs/P*60 //[ft^3/min]
diff --git a/839/CH9/EX9.1/Example_9_1.sce b/839/CH9/EX9.1/Example_9_1.sce new file mode 100755 index 000000000..733b15c17 --- /dev/null +++ b/839/CH9/EX9.1/Example_9_1.sce @@ -0,0 +1,20 @@ +//clear//
+clear;
+clc;
+
+//Exapmle 9.1
+//Given
+Dt = 6; //[ft]
+h = 2; //[ft]
+n = 90/60; //[rps]
+mu = 12*6.72*10^-4; //[lb/ft-s]
+g = 32.17; //[ft/s^2]
+rho = 93.5; //[lb/ft^3]
+Da = 2; // [ft]
+
+Nre = Da^2*n*rho/mu;
+//From curve A of Fig. 9.12
+Np = 5.8
+//Form Eq.(9.20)
+P = Np*rho*n^3*Da^5/g //[ft-lbf/s]
+P = P/550 //[hp]
diff --git a/839/CH9/EX9.2/Example_9_2.sce b/839/CH9/EX9.2/Example_9_2.sce new file mode 100755 index 000000000..f14ebb69d --- /dev/null +++ b/839/CH9/EX9.2/Example_9_2.sce @@ -0,0 +1,30 @@ +//clear//
+clear;
+clc;
+
+//Example 9.2
+//Given
+Dt = 6; //[ft]
+h = 2; //[ft]
+n = 90/60; //[rps]
+mu = 12*6.72*10^-4; //[lb/ft-s]
+g = 32.17; //[ft/s^2]
+rho = 93.5; //[lb/ft^3]
+Da = 2; // [ft]
+
+Nre = Da^2*n*rho/mu;
+//Froude number
+Nfr = n^2*Da/g;
+//From Table 9.1
+a = 1;
+b = 40.0;
+//Using Eq.(9.19)
+m = (a-log(Nre)/2.303)/b;
+//Using Fig. 9.12, curve D,
+Np = 1.07;
+//Corrected valus of Np
+Np = Np*Nfr^m;
+
+//Form Eq.(9.20)
+P = Np*rho*n^3*Da^5/g //[ft-lbf/s]
+P = P/550 //[hp]
diff --git a/839/CH9/EX9.3/Example_9_3.sce b/839/CH9/EX9.3/Example_9_3.sce new file mode 100755 index 000000000..19cca12fc --- /dev/null +++ b/839/CH9/EX9.3/Example_9_3.sce @@ -0,0 +1,21 @@ +//clear//
+clear;
+clc;
+
+//Example 9.3
+//Given
+Dt = 6; //[ft]
+h = 2; //[ft]
+n = 90/60; //[rps]
+mu = 1200*6.72*10^-2; //[lb/ft-s]
+g = 32.17; //[ft/s^2]
+rho = 70 //[lb/ft^3]
+Da = 2; // [ft]
+
+Nre = Da^2*n*rho/mu;
+//From Table 9.3
+KL = 65;
+//From Eq.(9.21)
+Np = KL/Nre;
+P = Np*rho*n^3*Da^5/g //[ft-lbf/s]
+P = P/550 //[hp]
diff --git a/839/CH9/EX9.4/Example_9_4.sce b/839/CH9/EX9.4/Example_9_4.sce new file mode 100755 index 000000000..5da5391ff --- /dev/null +++ b/839/CH9/EX9.4/Example_9_4.sce @@ -0,0 +1,17 @@ +//clear//
+clear;
+clc;
+
+//Example 9.4
+//Given
+Dt = 6; //[ft]
+Da = 2; //[ft]
+n = 80/60; //[rps]
+T = 70; //[F]
+rho = 62.3; //[lb/ft^3], From Appendix 14
+mu = 6.6*10^-4; // [lb/ft-s], From Appendix 14
+
+Nre = Da^2*n*rho/mu;
+//From Fig. 9.15
+ntT = 36;
+tT = ntT/1.333 //[s]
diff --git a/839/CH9/EX9.5/Example_9_5.sce b/839/CH9/EX9.5/Example_9_5.sce new file mode 100755 index 000000000..155063c99 --- /dev/null +++ b/839/CH9/EX9.5/Example_9_5.sce @@ -0,0 +1,33 @@ +//clear//
+clear;
+clc;
+
+//Example 9.5
+//Given
+Dt = 6; //[ft]
+H = 8; //[ft]
+T = 70; //[F]
+sp_gr = 3.18;
+w_fr = 0.25;
+Da = 2; //[ft]
+h = 1.5; //[ft]
+gc = 32.17; //[ft-lb/lbf-s^2]
+// (a)
+//Using data of Buurman et al. in Fig.(9.19)
+//change in nc
+delta_nc = (104/200)^0.2*(2.18/1.59)^0.45*(33.3/11.1)^0.13;
+//change in P
+dalta_P = delta_nc^3;
+
+//Using Fig. 9.19
+V = %pi/4*Dt^2*H*7.48 ; //[gal]
+P = 3.3*V/1000 //[hp]
+
+//(b)
+//From Table 9.3, for a cour blade turbine,
+KT = 1.27;
+Np = KT;
+//slurry density
+rho_m = 1/((w_fr/sp_gr)+(1-w_fr))*62; // [lb/ft^3]
+
+nc = (P*gc*550/(Np*rho_m*Da^5))^(1/3) // [r/s]
diff --git a/839/CH9/EX9.6/Example_9_6.sce b/839/CH9/EX9.6/Example_9_6.sce new file mode 100755 index 000000000..ae7b7837f --- /dev/null +++ b/839/CH9/EX9.6/Example_9_6.sce @@ -0,0 +1,53 @@ +//clear//
+clear;
+clc;
+
+//Example 9.6
+//Given
+Dt = 2; //[m]
+Da = 0.667; //[m]
+n = 180/60; //[rps]
+T = 20; //[C]
+qg = 100; //[m^3/h]
+rho = 1000; //[kg/m^3]
+mu = 10^-3; //[kg/m-s]
+ut = 0.2; //[m/s]
+//(a)
+//The power input is calculated and followed by correction of gas effect
+Nre = n*Da^2*rho/mu;
+//For a flat blade turbine, from Table 9.3
+KT = 5.75;
+//Using Eq.(9.24)
+Po = KT*n^3*Da^5*rho/1000; //[kW]
+At = %pi/4*Dt^2; //[m^2]
+//Superficial gas velocity
+Vs_bar = At*qg/3600/10 //[m/s]
+//From Fig. 9.20 Pg/Po = 0.60
+Pg = Po*0.6; //[kW]
+//From Fig.9.7, depth of liquid is equal to diameter of the tank
+//Hence, liquid volume
+V = %pi/4*Dt^2*Dt; //[m^3]
+//The input power per unit volume
+PgbyV = Pg/V ; //[kW/m^3]
+
+//(b)
+sigma = 72.75; //[g/s^2]
+rho_L = 10^-3; //[g/mm]
+PgbyV = PgbyV*10^3 ; //[g/mm-s^2]
+//Using Eq.(9.46)
+//Let x = shi^(0.5)
+//solving the equation as quadratic equation
+a = 1;
+b = -(Vs_bar/ut)^0.5;
+c = -0.216*((PgbyV)^0.4)*(rho_L^0.2)/(sigma^0.6)*(Vs_bar/ut)^(0.5);
+x = (-b+sqrt(b^2-4*a*c))/(2*a);
+shi = x^2;
+
+//(c)
+//To find out mean bubble diameter
+//Using Eq.(9.44)
+Ds_bar = 4.15*sigma^0.6/(PgbyV^0.4*rho_L^0.2)*shi^0.5+0.9 // [mm]
+
+//(d)
+//From Eq.(9.40)
+aprime = 6*shi/Ds_bar //[mm^-1]
diff --git a/839/CH9/EX9.7/Example_9_7.sce b/839/CH9/EX9.7/Example_9_7.sce new file mode 100755 index 000000000..ab36bb0f1 --- /dev/null +++ b/839/CH9/EX9.7/Example_9_7.sce @@ -0,0 +1,24 @@ +//clear//
+clear;
+clc;
+
+//Exapmle 9.7
+//Given
+Dt = 2; //[m]
+Da = 0.667; //[m]
+n = 180/60; //[rps]
+T = 20; //[C]
+qg = 100; //[m^3/h]
+rho = 1000; //[kg/m^3]
+mu = 10^-3; //[kg/m-s]
+ut = 0.2; //[m/s]
+At = %pi/4*Dt^2; //[m^2]
+//Using values form Example 7.6
+//Assuming Pg/Po decresaes to 0.25
+PgbyV = 0.25*20490/6.28; //[W/m^3]
+//Using Eq.(9.47)
+Vs_barc = 0.114*(PgbyV)*(Dt/1.5)^0.17/1000 //[m/s]
+qg = Vs_barc*At*3600 //[m^3/h]
+//The calculated flooding velocity is beyond the range of the data on which Eq.(9.47)
+//was based, so it may not be relaible. Based on Vs_barc, the highest measured value, qg
+//would be 850 m^3/h.
diff --git a/839/CH9/EX9.8/Example_9_8.sce b/839/CH9/EX9.8/Example_9_8.sce new file mode 100755 index 000000000..44af9cf3b --- /dev/null +++ b/839/CH9/EX9.8/Example_9_8.sce @@ -0,0 +1,27 @@ +//clear//
+clear;
+clc;
+
+//Example 9.8
+//Given
+D1 = 1; //[ft]
+D6 = 6
+Nre_i = 10^4;
+Da = 4; //[in.]
+t1 = 15; //[s]
+P = 2; //[hp/gal]
+
+//(a)
+//Using Fig. 9.15
+//the mixing factor ntT is constant and time tT is asumed constant,
+//speed n will be the same in both vessels.
+//Using Eq.(9.24) with consant density
+PbyD_ratio = (D6/D1)^2;
+//The Power input required in the 6-ft vessel is then
+Pin = 2*PbyD_ratio //[hp/1000gal]
+
+//(b)
+//Using Eq.(9.54) with same input power per unit volume in both vessels
+n6byn1 = (D6/D1)^(2/3)
+//blending in the 6-ft vessel would be
+t6 = t1*n6byn1 // [s]
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