initial commit / add all books

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diff --git a/479/CH1/EX1.1/Example_1_1.sce b/479/CH1/EX1.1/Example_1_1.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 1

+//Introduction

+

+//Example 1.1

+clear;

+clc;

+

+//Given 

+m = 100;//m is the mass of the object in kg

+a = 10;//a is the acceeleration due to gravity in m/s^2

+

+//To determine the force exerted

+F = m*a;//F is the force exerted by the object in kg

+mprintf('Force exerted by the object= %f N',F);

+F = (1/9.8065)*m*a;//F is the force exerted by the object in kgf

+mprintf('\n Force exerted by the object= %f N',F);

+//end
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diff --git a/479/CH1/EX1.2/Example_1_2.sce b/479/CH1/EX1.2/Example_1_2.sce
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+//Chemical Engineering Thermodynamics

+//Chapter

+//Introduction

+

+//Example 1.2

+clear;

+clc;

+

+//Given

+//The given example is a theoretical problem and it does not involve any numerical computation

+//end
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diff --git a/479/CH1/EX1.3/Example_1_3.sce b/479/CH1/EX1.3/Example_1_3.sce
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+//Chemical Engineering Thermodynamics

+//Chapter

+//Introduction

+

+//Example 1.3

+clear;

+clc;

+

+//Given

+//The given example is a theoretical problem and it does not involve any numerical computation

+//end
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diff --git a/479/CH1/EX1.4/Example_1_4.sce b/479/CH1/EX1.4/Example_1_4.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 1

+//Introduction

+

+//Example 1.4

+clear;

+clc;

+

+//Given

+h = 100;//h is the height of the water fall in m

+n = .855;//n is the efficiency of the turbine

+g = 9.8;//g is the acceleration due to gravity in m/(s^2)

+E = 100*10*3600;//E is the potential enery of water available to the bulb for 10 hours in watt or J/s

+

+//To determine the mass of water required

+m = (E/(g*h*n));//m is the mass of water required for lighting the bulb for 10 hours in Kg

+mprintf('Mass of water required for lighting the bulb for 10 hours in Kg= %f Kg',m);

+mprintf('\n Mass of water required for lighting the bulb for 10 hours in tonnes= %f Kg',m/907.2);

+//end

diff --git a/479/CH1/EX1.5/Example_1_5.sce b/479/CH1/EX1.5/Example_1_5.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 1

+//Introduction

+

+//Example 1.5

+clear;

+clc;

+

+//Given

+n = 1;//n is the Kg mole of an ideal gas

+P = 700*(10^4);//P is the pressure of the system in N/(m^2)

+W = 45;//W is the weight of the mass in Kg

+M = 20;//M is the weight of the piston and piston rod together in Kg

+T = 300;//T is the constant temperature of the bath in K

+h = .4;//h is the height difference of the piston after expansion in m

+

+//To calculate the work obtained

+a = (10^-4);//a is the cross sectional area of the cylinder in m^2

+V = h*a;//V is the volume changed as gas expands in m^3

+

+//(i). If gas alone is the system

+//1Kgf = 9.8065Nm

+P1 = ((W+M)*9.8065)/(10^-4);//P1 is the resisting pressure when the gas confined in the cylinder taken as a system

+W1 = P1*V;//W1 is the work done if the gas confined in the cylinder us taken as system

+mprintf('Work done by the system if the gas confined in the cylinder is taken as a system is %f Nm',W1);

+

+//(ii). If gas + piston + piston rod is a system

+P2 = ((W*9.8065)/(10^-4));//P2 is the resisting pressure when the gas plus piston plus piston rod is taken as a system

+W2 = P2*V;//W2 is the Work done by the system if the gas plus piston plus piston rod is taken as a system

+mprintf('\n Work done by the system if the gas plus piston plus piston rod is taken as system is %f Nm',W2);

+

+//(iii). If gas + piston + piston rod +weight is system

+P3 = 0;//P3 is the resisting pressure when the gas plus piston plus piston rod plus weight is taken as a system

+W3 = P3*V;//W3 is the work done by the system if the gas plus piston plus piston rod plus weight is taken as a system

+mprintf('\n Work done by the system if the gas plus piston plus piston rod plus weight is taken as a system is %f',W3);

+//end
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diff --git a/479/CH1/EX1.6/Example_1_6.sce b/479/CH1/EX1.6/Example_1_6.sce
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+//Chemical Engineering Thermodynamics

+//Chapter 1

+//Introduction

+

+//Example 1.6

+clear;

+clc;

+

+//Given

+n = 1;//n is the Kg mole of ideal gas.

+P1 = 700*(10^4);//P1 is the initial pressure of the system in N/(m^2)

+P2 = 638*(10^4);//P2 is the final pressure of the system in N/(m^2)

+T = 300;//T is temperature of the system in K

+R = 8314.4;//R is gas constant in Nm/Kgmole deg K

+

+//To calculate the work done

+W = n*R*T*log(P1/P2);//W is the work done by the system in Nm

+mprintf('Work done by the system is %4.2e Nm',W);

+//end
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