initial commit / add all books

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diff --git a/839/CH16/EX16.1/Example_16_1.sce b/839/CH16/EX16.1/Example_16_1.sce
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+//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
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+++ b/839/CH16/EX16.2/Example_16_2.sce
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+//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
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index 000000000..53e78a4aa
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+++ b/839/CH16/EX16.3/Example_16_3.sce
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+//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]

+