initial commit / add all books

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diff --git a/1919/CH2/EX2.1/Ex2_1.sce b/1919/CH2/EX2.1/Ex2_1.sce
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+// Theory and Problems of Thermodynamics

+// Chapter 2

+// First Law of Thermodynamics

+// Example 1

+

+clear ;clc;

+

+//Given data

+// a system is taken from state 1 to state 2 along a path A

+Q_A_21  = 200   //heat energy received in kJ

+W_A_21  = 100   //work done  in kJ

+// if the system is taken different path from state 2 to state 1 

+W_A_12  = -150   //spend work on system 

+// Path C

+Q_C_12 = 0;                     // for adiabatic Q = 0

+

+

+//Path A

+U_A_21 = Q_A_21 - W_A_21;     // U_A_21 = U2 - U1 = Q - W

+

+// determine heat interaction along path B

+U_B_12 = - U_A_21;               // U_B_12 = U1 - U2 = -(U2 - U1)

+Q_A_12 = U_B_12 + W_A_12        // U1 - U2 = Q - W

+

+// determine work done on system in surroundings in adiabatic path C

+U_C_12 = U_B_12;            

+W_C_12 = Q_C_12 - U_C_12;       // work done

+

+// Results

+mprintf('Heat interaction along path B = %3.0f kJ', Q_A_12)

+mprintf('\n The work done on surroundings in adiabatic path B = %3.0f kJ', W_C_12)

diff --git a/1919/CH2/EX2.2/Ex2_2.sce b/1919/CH2/EX2.2/Ex2_2.sce
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+// Theory and Problems of Thermodynamics

+// Chapter 2

+// First Law of Thermodynamics

+// Example 2

+

+clear ;clc;

+

+//Given data

+r = 1.4;         // r = CP/CV

+P1 = 0.5;        // pressure at state 1 in MPa

+T1 = 300;        // temperature at state 1 in K

+V_21 = 2;        // ratio of V2 and V1

+T2 = T1 * V_21;  // temperature at state 2 in K

+T3 = T1;       // temperature at state 3 in K

+R = 8.314;       // gas constant in J/K

+

+//For reversible adiabatic path

+T_23 = T2 / T3                // T_23 ratio of T2 and T3

+V_32 = T_23 ^ (1/(r-1));      // ratio of V3 and V2

+

+/// Q = Q12 = Q23 = Cp*T1

+Q_A = R * r * T1 / (r -1)     // heat interactions for adiabatic path

+Q_A = Q_A / 1000;             // units conversion J to kJ

+

+// W = W12 + W23 = R*T1 + Cv*T1 = Cp*T1

+W_A = R * r * T1 / (r -1)     // work done for adiabatic path

+W_A = W_A / 1000;             // units conversion J to kJ

+

+// for reversible isothermal process

+// W_I = integrate('P','V',3,1) = R*T1*log(V1/V3) = R*T1*ln(V1/V2)*(V2/V3)

+W_I = R * T1 * log(1 / (V_21 * V_32) );     // work for reversible isothermal process

+W_I = W_I / 1000;             // units conversion J to kJ

+

+// Results

+mprintf('heat interaction for reversible adiabatic path = %4.2f kJ', Q_A)

+mprintf('\n work done for adiabatic path = %4.2f kJ', W_A)

+mprintf('\n work done for reversible isothermal process = %5.3f kJ', W_I)

+

diff --git a/1919/CH2/EX2.3/Ex2_3.sce b/1919/CH2/EX2.3/Ex2_3.sce
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+// Theory and Problems of Thermodynamics

+// Chapter 2

+// First Law of Thermodynamics

+// Example 3

+

+clear ;clc;

+

+//Given data

+T1 = 300        // initial temperature in K

+P1 = 100        // initial pressure in kPa

+I = 3           // current in amperes

+V = 200         // voltage in volts

+t = 60          // time period for flow through resistor in s

+r = 1.4;        // CP/CV

+R = 8.314       // gas constant in J/K

+

+// determine work done by the gas and temperature

+// U2-U1 = Q-W = Q-(integrate('P','V') - W_e) = W_e-(integrate('P','V'))

+// W = I * V * t - P*(V2-V1) = I * V * t - R*(T2-T1)

+

+W_e = I * V * t     // electrical work done on system

+T2 = T1 + W_e * (r-1)/(R * r);  // final temperature

+W = R * (T2 - T1);     // work done by gas

+W = W/1000;            // units conversion J to kJ 

+

+// Results

+mprintf('Final temperature = %5.2f K', T2)

+mprintf('\n Work done by the gas = %5.3f kJ', W)

+

+

+

diff --git a/1919/CH2/EX2.4/Ex2_4.sce b/1919/CH2/EX2.4/Ex2_4.sce
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index 000000000..a06fd2009
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+// Theory and Problems of Thermodynamics

+// Chapter 2

+// First Law of Thermodynamics

+// Example 4

+

+clear ;clc;

+

+//Given data

+T = 303          // initial temperature in K

+P = 100          // initial pressure in kPa

+V_w = 0.001004   // specific volume  for water in m3/kg

+V_v = 1.694910   // specific volume  for vapor in m3/kg

+H_w = 125.6      // enthalpy of water in kJ/kg

+H_v = 2675.47    // enthalphy of vapor in kJ/kg

+

+// determine heat and work interactions

+W = P * (V_v - V_w);    // work interaction

+del_H = H_v - H_w       // enthalpy difference

+Q = del_H               //heat interaction

+

+// Results

+mprintf('Work interaction = %7.2f kJ', W)

+mprintf('\n Heat interaction = %5.2f kJ', Q)

+

+

diff --git a/1919/CH2/EX2.5/Ex2_5.sce b/1919/CH2/EX2.5/Ex2_5.sce
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+// Theory and Problems of Thermodynamics

+// Chapter 2

+// First Law of Thermodynamics

+// Example 5

+

+clear ;clc;

+

+//Given data

+

+//the given problem is theoritical and does not involve any numerical computation

diff --git a/1919/CH2/EX2.6/Ex2_6.sce b/1919/CH2/EX2.6/Ex2_6.sce
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+// Theory and Problems of Thermodynamics

+// Chapter 2

+// First Law of Thermodynamics

+// Example 6

+

+clear ;clc;

+

+//Given data

+

+//the given problem is theoritical and does not involve any numerical computation

diff --git a/1919/CH2/EX2.7/Ex2_7.sce b/1919/CH2/EX2.7/Ex2_7.sce
new file mode 100755
index 000000000..0f57b2e59
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+++ b/1919/CH2/EX2.7/Ex2_7.sce
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+// Theory and Problems of Thermodynamics

+// Chapter 2

+// First Law of Thermodynamics

+// Example 7

+

+clear ;clc;

+

+//Given data

+T1 = 300          // initial temperature in K

+P1 = 100          // initial pressure in kPa

+P2 = 1.5          // initial pressure in MPa

+r = 1.4           // r = CP/CV

+n = 1.2           //P(V^1.2) = constant

+R = 8.314         // gas constant in J/K

+

+// determine work done and energy transferred as heat

+P2 = P2 * 1000      // units conversion MPa to kPa

+V_12 = (P2/P1)^(1/1.2)  //ratio of V1 and V2

+T2 = P2 * T1 / (P1 * V_12)  // final temperature in K

+

+W = R * (T1 - T2)/(n-1);       // work done on gas

+W = W/1000;                    // units conversion J to kJ

+Q = (R * (T1-T2)/(r-1))/1000 + W      // energy transferred as heat

+

+// Results

+mprintf('Work done on gas = %5.4f kJ', W)

+mprintf('\n Energy transfered as heat = %5.3f kJ', Q)

diff --git a/1919/CH2/EX2.8/Ex2_8.sce b/1919/CH2/EX2.8/Ex2_8.sce
new file mode 100755
index 000000000..d0205c58d
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+// Theory and Problems of Thermodynamics

+// Chapter 2

+// First Law of Thermodynamics

+// Example 8

+

+clear ;clc;

+

+//Given data

+V1 = 1            // one compartment volume in m^3 

+T1 = 300          // first compartment temperature in K

+P1 = 0.1          // first compartment pressure in MPa

+V2 = 2            // second compartment volume in m^3 

+T2 = 1000         // second compartment temperature in K

+P2 = 1            // second compartment pressure in MPa

+R = 8.314         // gas constant in J/K

+del_U = 0;        // tank is rigid and insulated W = 0 & Q = 0

+

+// determine final temperature and pressure of gas if the partion between 

+//               the two compartment is ruptured

+P1 = P1 * 1e6      // units conversion MPa to Pa

+P2 = P2 * 1e6      // units conversion MPa to Pa

+

+N1 = P1 * V1 / (R * T1);        // quantity of gas in first compartment 

+N2 = P2 * V2 / (R * T2);        // quantity of gas in second compartment

+

+T_f = (T1 * N1 + T2 * N2)/ (N1 + N2)        //Final state Temperature

+

+P_f = (N1+N2)*R*T_f/(V1+V2);                //Final stste Pressure

+P_f = P_f/1000000                           // Units conversion Pa to MPa

+

+// Results

+mprintf('Final state Temperature = %3.0f K', T_f)

+mprintf('\n Final state pressure = %2.1f MPa', P_f)

+

+