initial commit / add all books

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diff --git a/542/CH16/EX16.1/Example_16_1.sci b/542/CH16/EX16.1/Example_16_1.sci
new file mode 100755
index 000000000..edef797b7
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+++ b/542/CH16/EX16.1/Example_16_1.sci
@@ -0,0 +1,30 @@
+clear;

+clc;

+printf("\n Example 16.1");

+//For the first drying operation

+w1 = 0.25;                 //in kg/kg

+w = 0.10;                  //in kg/kg

+wc = 0.15;                 //in kg/kg

+we = 0.05;                 //in kg/kg

+f1 = w1-we;                //in kg/kg

+fc = wc - we;              //in kg/kg

+f = w - we;                //in kg/kg

+

+//x = mA

+function[x]=TotalDryingTime(t)

+    x = (1/t)*[(f1-fc)/fc + log(fc/f)];

+    funcprot(0);

+endfunction

+printf("\n     mA = %.3f kg/sec",TotalDryingTime(15));

+

+//For the second drying operation

+w12 = 0.30;                  //in kg/kg

+w2 = 0.08;                   //in kg/kg

+wc2 = 0.15;                  //in kg/kg

+we2 = 0.05;                  //in kg/kg

+f12 = w12 - we2;             //in kg/kg

+fc2 = wc2 - we2;             //in kg/kg

+f2 = w2 - we2;               //in kg/kg

+

+t1=(1/TotalDryingTime(15))*[(0.25-0.10)/0.10 + log(0.10/0.03)];

+printf("\n The total drying time is then %.1f ksec or %.2f hr ",t1,t1*1000/3600);

diff --git a/542/CH16/EX16.2/Example_16_2.sci b/542/CH16/EX16.2/Example_16_2.sci
new file mode 100755
index 000000000..4f0f1ba41
--- /dev/null
+++ b/542/CH16/EX16.2/Example_16_2.sci
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+clear;

+clc;

+printf("\n Example 16.2");

+E = [1 0.64 0.49 0.38 0.295 0.22 0.14];

+J = [0 0.1 0.2 0.3 0.5 0.6 0.7];

+plot(J,E,rect=[0,1,0,1]);

+xtitle("Plot for drying data","J = kt/L^2","E");

+

+//For the 10 mm strips

+mi = (0.28 - 0.07);        //Initial free moisture content in kg/kg

+mf = (0.13-0.07);          //Final free moisture content in kg/kg

+//at

+t = 25;                    //time is in ksecs

+E = (0.06/0.21);

+//at E = 0.286 ,J = 0.52 from the plot of given data and J = kt/L^2

+k = poly([0],'k');

+k1 = roots(0.52 - (k*t)/(10/2)^2);

+printf("\n k = %f",k1);

+

+//for the 60 mm planks

+m1i = (0.22 - 0.07);       //Initial free moisture content in kg/kg

+m1f = (0.13 - 0.07);       //Final free moisture content in kg/kg

+E = (m1f/m1i);

+//at E = 0.20 from yhe plot of the given data J = 0.63 and J = kt/L^2

+t1 = 0.63*(60/2)^(2)/k1;

+printf("\n t1 = %d ksecs or %.1f days",t1,(t1*1000/(3600*24)));

+

diff --git a/542/CH16/EX16.3/Example_16_3.sci b/542/CH16/EX16.3/Example_16_3.sci
new file mode 100755
index 000000000..cc88b6783
--- /dev/null
+++ b/542/CH16/EX16.3/Example_16_3.sci
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+clear;

+clc;

+printf("\n example 16.3");

+//(a) Air

+//G kg/s dry air enter with 0.006G kg/s water vapour and hence the heat content of this stream= 

+//[(1.00G) + (0.006G × 2.01)](385 − 273) = 113.35G kW

+

+//(b) Wet solid

+//0.125 kg/s enter containing 0.40 kg water/kg wet solid, assuming the moisture is expressed on a wet basis.

+flowWater = (0.125*0.40);               //in kg/sec

+flowDrySolid = (0.125-0.050);           //in kg/sec

+//Hence heat content of this stream

+q = [(0.050*4.18)+(0.075*0.88)]*(295-273);

+printf("\n The heat content of this stream = %.2f kW",q);

+

+//Heat out 

+//(a) Air

+//Heat in exit air = [(1.00 G) + (0.006 G × 2.01)](310 − 273) = 37.45G kW.

+fd = 0.075;               //mass flow rate of dry solids in kg/sec

+w = 0.05*0.075/(1+0.05);  //water in the dried solids leaving in kg/secs

+we = (0.050 - w);         //The water evaporated into gas stream in kg/secs

+

+//Assuming evaporation takes place at 295 K,then:

+qout = 0.0464*[2.01*(310-295)+2449+4.18*(295-273)];

+printf("\n Heat in the water vapour = %.1f kW",qout);

+

+//the total heat in this stream = (119.30 + 37.45G) kW.

+//(b) Dried solids

+

+

+   //The dried solids contain 0.0036 kg/s water and hence heat conten     t of this stream is:

+   q1 = [(0.075*0.88)+(0.0036*4.18)/(305-273)];

+   printf("\n The dried solids contain 0.0036 kg/s water and hence heat content of this stream is: %.2f kW",q1);

+   

+   

+//(c) Losses

+//These amount to 20 kJ/kg dry air or 20m kW.

+//Heat Balance

+G = poly([0],'G');

+G1 = roots(113.35*G + 6.05 - 119.30 - 37.45*G - 2.59 - 20*G);

+printf("\n G = %.2f kg/secs",G1);

+printf("\n Water in the outlet stream %.4f kg/secs",0.006*2.07+0.0464);

+printf("\n The humidity H = %.4f kg/kg dry air",0.0588/2.07);

+

+

+

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diff --git a/542/CH16/EX16.4/Example_16_4.sci b/542/CH16/EX16.4/Example_16_4.sci
new file mode 100755
index 000000000..0df32648e
--- /dev/null
+++ b/542/CH16/EX16.4/Example_16_4.sci
@@ -0,0 +1,40 @@
+clear;

+clc;

+printf("\n Example 16.4");

+//In 100 kg of feed

+

+   //mass of water = 

+   mw = 100*30/100;                  //mass of water in kg

+   //mass of dry solids = 

+   md = 100-30;                      //mass of dry solids

+   

+//and:

+//For b kg water in the dried solids: 100b/(b + 70) = 15.5

+b = poly([0],'b');

+b1 = roots(100*b - 15.5*(b+70));

+printf("\n water in the product ,b = %.1f kg",b1);

+

+   //Initial water content 

+   w1 = 30/70;            //Intial moisture content in kg/kg dry solids

+   //Final moisture content

+   w2 = (12.8/70);        //Final moisture content in kg/kg dry solids

+   //water to be removed

+   w3 = (30-12.8);        //water to be removed in kg

+

+//Surface of drying

+S = (0.03*70);            //Surface for drying in m^2

+rate = (0.0007*2.1);      //Rate of drying in kg/sec

+

+//As the final moisture content is above the critical value, all the drying is at this constant rate and the time of drying is:

+t = 17.2/0.00147;

+printf("\n As the final moisture content is above the critical value, all the drying is at this constant rate and the time of drying is: %d ksecs or %.2f hr",t,t/3600);

+

+

+

+

+

+

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+

+

diff --git a/542/CH16/EX16.5/Example_16_5.sci b/542/CH16/EX16.5/Example_16_5.sci
new file mode 100755
index 000000000..a7e03099a
--- /dev/null
+++ b/542/CH16/EX16.5/Example_16_5.sci
@@ -0,0 +1,77 @@
+clear;

+clc;

+printf("\n Example 16.5");

+//Inlet air temperature

+Tair = 400;            //Inlet air temperature in kelvins

+H = 0.01;              //Humidity is in kg/kg dry air

+//*therefore wet bulb temperature = 

+Twetbulb = 312;        //Inlet wet-bulb temperature

+NTU = 1.5;              //Number of transfer units

+

+//Then for adiabatic drying the outlet air temperature,To is given by

+To = poly([0],'To');

+To1 = roots(exp(1.5)*(To-312)-(400-312));

+printf("\n For adiabatic drying the outlet air temperature = %.1f K",To1);

+

+//Solids outlet temperature will be taken to be maximum allowable,325K

+//From the steam tables in the Appendix, the latent heat of vaporisation of water at 312 K is2410 kJ/kg. Again from steam tables, the specific heat capacity of water vapour = 1.88 kJ/kg K and that of the solids will be taken as 2.18 kJ/kg K.

+

+//For a mass flow of solids of 0.35 kg/s and inlet and outlet moisture contents of 0.15 and 0.005 kg/kg dry solids respectively, the mass of water evaporated = 0.35(0.15 − 0.005) = 0.0508 kg/s.

+

+//For unit mass of solids ,the heat duty includes:

+       //Heat to the solids

+       qsolids = 2.18*(325-300);       //heat to solids in kJ/kg

+       //Heat to raise the moisture to the dew point

+       qdew = (0.15*4.187*(312-300));  //in kJ/kg

+       //Heat of vaporisation

+       qvap = 2410*(0.15-0.005);       //in kJ/kg

+       //Heat to raise remaining moisture to solids outlet temperature

+       qremaining = (0.05*4.187)*(325-312);

+       //Heat to raise evaporated moisture to the air outlet temp.

+       qevapo = (0.15-0.005)*1.88*(331.5-312);

+       qtotal = qsolids + qdew + qvap + qremaining + qevapo;

+       printf("\n Total heat = %.1f kJ/kg or %d kW",qtotal,qtotal*0.35);

+

+//The humid heat of entering air is 1.03 kJ/kg K

+       //G1 (1 +H1) = Q/Cp1(T1 − T2)

+       //where: G1 (kg/s) is the mass flowrate of inlet air,

+       //H1 (kg/kg) is the humidity of inlet air,

+       //Q (kW) is the heat duty,

+       //Cp1 (kJ/kg K) is the humid heat of inlet air

+       //and: T1 and T2 (K) are the inlet and outlet air temperatures         respectively.

+       G1 = poly([0],'G1');

+       G = roots(G1*(1+0.01)-146/(1.03*(400-331.5)));

+       printf("\n Mass flow rate of inlet air = %.2f kg/secs",G);

+       printf("\n Mass flow rate of dry air ,Ga = %.2f kg/secs",G/1.01);

+       printf("\n the humidity of the outlet air H2 = %.4f kg/kg",0.01+0.0508/2.05);

+

+//At a dry bulb temperature of 331.5 K, with a humidity of 0.0347 kg/kg, the wet-bulb temperatureof the outlet air, from Figure 13.4 in Volume 1, is 312 K, the same as the inlet, which is the case for adiabatic drying.

+

+//The dryer diameter is then found from the allowable mass velocity of the air and the entering air flow and for a mass velocity of 0.95 kg/m^2.secs, the cross sectional area of the dryer is

+printf("\n The cross sectional area of the dryer is %.2f m^2",2.07/0.95);

+printf("\n The equivalent diameter of the dryer = %.2f m",[(4*2.18)/%pi]^0.5);

+

+//With a constant drying temperature of 312 K:

+        //at the inlet

+        deltaT1 = 400-312;           //inlet temperature is in deg K

+        //at the outlet

+        deltaT2 = 331.5-312;         //outlet temperature is in deg K

+        Tlogmean = (deltaT1 - deltaT2)/log(deltaT1/deltaT2);

+        printf("\n Logarithmic mean temperature difference = %.1f deg K",Tlogmean);

+//The length of the dryer, L is then: L = Q/(0.0625πDG^(0.67)*Tm)

+    //where D (m) is the diameter

+    //and G(kg/m^2.secs) is the air mass velocity.

+    L = 146/[0.0625*(%pi)*1.67*(0.95)^(0.67)*45.5];

+    printf("\n The length of the dryer = %.1f m",L);

+    printf("\n Length/diameter ratio = %d ",10.1/1.67);

+

+

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diff --git a/542/CH16/EX16.6/Example_16_6.sci b/542/CH16/EX16.6/Example_16_6.sci
new file mode 100755
index 000000000..f55ccd97b
--- /dev/null
+++ b/542/CH16/EX16.6/Example_16_6.sci
@@ -0,0 +1,41 @@
+clear;

+clc;

+printf("\n Example 16.6");

+H = 0.036;               //Humidity is in kg/kg at 811 K

+//Taking R as 90 per cent and P as 101.3 kN/m2, then, for assumed values of Tb of 321, 333 and 344 K

+     //Pw = 13,20 and 32 kN/m2, respectively

+     //G = 27.8, 12.9 and 6.02 kg/s, respectively.

+     

+//for Tb = 321, 333 and 344 K,

+//G = 7.16, 7.8 and 7.54 kg/s respectively.

+Tb = [321 333 344];

+G1 = [27.8 12.9 6.02];           //Temperature is in kelvins

+G = [7.16 7.8 7.54];             //flow rate in kg/secs

+plot2d(Tb,G,style=3);

+plot2d(Tb,G1,style=2);

+xtitle("Temperature vs Flow rate","Temperature Tb(K)","Flow rate G(kg/secs)");

+

+

+//Plotting G against Tb for each equation on the same axis, then

+Go = 8.3;                         //Gas flow rate is in kg/secs

+Tb = 340;                         //temperature is in Kelvins

+uf = 0.61;                        //velocity is in m/secs

+

+D = sqrt(340*(8.3+(1.58*1.26))/(278*0.61));

+printf("\n D = %.2f m",D);

+

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