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diff --git a/479/CH12/EX12.1/Example_12_1.sce b/479/CH12/EX12.1/Example_12_1.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 12

+//Refrigeration

+

+//Example 12.1

+clear;

+clc;

+

+//Given

+m = 500;//mass of ice produced per hour  in Kg

+T1 = 15;//Initial temperature of water

+T2 = -5;//Final temperature of ice

+Ci = 0.5;//Specific  heat of ice in Kcal/Kg deg celsius

+Cw = 1;//Specific heat of water in Kcal/Kg deg celsius

+L_f = 79.71;//Latent heat of fusion in Kcal/Kg

+Tf = 0;//Frezzing point of ice in deg celsius

+

+//To Calculate the theoretical horse power required

+Q2 = m*(Cw*(T1-Tf)+L_f+Ci*(Tf-T2));//Heat to be extracted per hour in Kcal

+//From equation 12.1 (page no 220)

+C.O.P = (T2+273)/((T1+273)-(T2+273));

+W = Q2/C.O.P;//Work in Kcal/hr

+W1 = W*(427/(60*4500));

+mprintf('The therotical horse power required is %f hp',W1);

+//end
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diff --git a/479/CH12/EX12.2/Example_12_2.sce b/479/CH12/EX12.2/Example_12_2.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 12

+//Refrigeration

+

+//Example 12.2

+clear;

+clc;

+

+//Given

+Ta = 298;//Initial temperature in K

+Tb = 203;//Final temperature in k

+T1 = 298;//Water temperature in K

+n = 1;//Kgmole of CO2

+//Cp = 5.89+0.0112T ; Specific heat of CO2 in Kcal/Kgmole K

+

+//To Calculate the compressor load

+//From equation 12.2a and b (page no 221)

+function y = f(T)

+    y = ((T1-T)/T)*n*(5.89+0.0112*T);

+endfunction

+W = intg(Ta,Tb,f);

+mprintf('The compressor load is %f Kcal/Kgmole',W);

+//end
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diff --git a/479/CH12/EX12.2/Example_12_2.txt b/479/CH12/EX12.2/Example_12_2.txt
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+For example 12.2, the answer given in the book is 71.5 Kcal/Kgmole.

+but i am getting -164.797 Kcal/Kgmole.There may be some calculation mistake in the book.
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diff --git a/479/CH12/EX12.3/Example_12_3.sce b/479/CH12/EX12.3/Example_12_3.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 12

+//Refrigeration

+

+//Example 12.3

+clear;

+clc;

+

+//Given

+//Consider the figure 12.4 (page no 226) 

+m = 5;//tonnes of refrigeration

+T1 = 273-10;//temperature of the saturated vapour in K

+T2 = 273+35;//temperature of the super heated vapour in K

+T3 = 273+25;//temperature of the saturated liquid in K

+T4 = 273+25;//temperature of the wet vapour in K

+H1 = 341.8;//enthalpy of the saturated vapour in Kcal/Kg

+H2 = 409.0;//enthalpy of the super heated vapour in Kcal/Kg

+H3 = 350.0;//enthalpy of the saturated liquid in Kcal/Kg

+H4 = 71.3;//enthalpy of the wet vapour in Kcal/Kg

+

+//To Calculate the C.O.P, mass of refrigerant required, compressor horse power required and the C.O.P & compressor horse power for a reversed Carnot cycle

+//(i)Calculation of the C.O.P of the compression cycle

+//From equation 12.6 (page no 226)

+C.O.P = (H1-H4)/(H2-H1);

+mprintf('(i)C.O.P of the compression cycle is %f',C.O.P);

+

+//(ii)Calculation of mass of refrigerant required

+//From equation 12.7 (page no 226)

+M = (m*50.4)/(H1-H4);

+mprintf('\n (ii)The mass of refrigerant required is %f Kg/mt',M);

+

+//(iii)Calculation of the compressor horse power

+//From equation 12.5 (page no 226)

+C_hp = (H2-H1)*M*(427/4500);

+mprintf('\n (iii)The compressor horse power is %f hp',C_hp);

+

+//(iv)Calculation for reversed Carnot cycle

+//From equation 12.1 (page no 220)

+C.O.P = T1/(T3-T1);

+mprintf('\n\n (iv)C.O.P for the reversed Carnot cycle is %f',C.O.P);

+Q2 = m*50.4*(427/4500);//in hp

+C_hp = Q2/C.O.P;

+mprintf('\n    Compressor horse power for the reversed Carnot cycle is %f hp',C_hp);

+//end
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diff --git a/479/CH12/EX12.4/Example_12_4.sce b/479/CH12/EX12.4/Example_12_4.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 12

+//Refrigeration

+

+//Example 12.4

+clear;

+clc;

+

+//Given

+//Water at 20 deg cel is chilled to 10 deg cel by flash evaporation

+Pv = 0.012;//Vapour pressure of water at 10 deg celsius in Kgf/sq.cm

+H1 = 20.03;//Enthalpy of liquid water at 20 deg cel in Kcal/Kg

+H2 = 10.4;//Enthalpy of liquid water at 10 deg cel in Kcal/Kg

+Hv = 601.6;//Enthalpy of saturated vapour at 10 deg cel in Kcal/kg

+

+//To calculate the pressure in the tank and the amount of make up water required

+P = Pv;//pressure in the tank = vapour pressure of water

+mprintf('The pressure in the tank is %f Kgf/sq.cm',P);

+//From equation 12.8 (page no 234)

+x = (H1-H2)/(Hv-H2);

+mprintf('\n The amount of make up water required is %f Kg',x);

+//end
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