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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /479 | |
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initial commit / add all books
Diffstat (limited to '479')
108 files changed, 3073 insertions, 0 deletions
diff --git a/479/CH1/EX1.1/Example_1_1.sce b/479/CH1/EX1.1/Example_1_1.sce new file mode 100755 index 000000000..915204987 --- /dev/null +++ b/479/CH1/EX1.1/Example_1_1.sce @@ -0,0 +1,18 @@ +//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
\ No newline at end of file diff --git a/479/CH1/EX1.2/Example_1_2.sce b/479/CH1/EX1.2/Example_1_2.sce new file mode 100755 index 000000000..6a0bf072a --- /dev/null +++ b/479/CH1/EX1.2/Example_1_2.sce @@ -0,0 +1,11 @@ +//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
\ No newline at end of file diff --git a/479/CH1/EX1.3/Example_1_3.sce b/479/CH1/EX1.3/Example_1_3.sce new file mode 100755 index 000000000..40c1957c5 --- /dev/null +++ b/479/CH1/EX1.3/Example_1_3.sce @@ -0,0 +1,11 @@ +//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
\ No newline at end of file diff --git a/479/CH1/EX1.4/Example_1_4.sce b/479/CH1/EX1.4/Example_1_4.sce new file mode 100755 index 000000000..18d9ab207 --- /dev/null +++ b/479/CH1/EX1.4/Example_1_4.sce @@ -0,0 +1,19 @@ +//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 new file mode 100755 index 000000000..e2f416143 --- /dev/null +++ b/479/CH1/EX1.5/Example_1_5.sce @@ -0,0 +1,36 @@ +//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
\ No newline at end of file diff --git a/479/CH1/EX1.6/Example_1_6.sce b/479/CH1/EX1.6/Example_1_6.sce new file mode 100755 index 000000000..f64b60019 --- /dev/null +++ b/479/CH1/EX1.6/Example_1_6.sce @@ -0,0 +1,19 @@ +//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
\ No newline at end of file diff --git a/479/CH10/EX10.1/Example_10_1.sce b/479/CH10/EX10.1/Example_10_1.sce new file mode 100755 index 000000000..c18726e16 --- /dev/null +++ b/479/CH10/EX10.1/Example_10_1.sce @@ -0,0 +1,33 @@ +//Chemical Engineering Thermodynamics
+//Chapter 10
+//Compressor
+//Example 10.1
+clear;
+clc;
+
+//Given
+V1 = 2.7;//flow rate of CO2 in cubic meter/min
+T1 = 273-51;//temperature in K
+P1 = 1;//initial pressure in Kgf/sq cm
+P2 = 10;//final pressure in Kgf/sq cm
+y = 1.3;//gamma
+v1 = 0.41;//specific volume in cubic meter/Kg
+H1 = 158.7;// initial enthalpy in Kcal/Kg
+H2 = 188.7;//final enthalpy in Kcal/Kg
+
+//process is isentropic
+//To calculate the horsepower required
+
+//(i)Assuming ideal gas behaviour
+//From equation 10.3 (page no 189)
+W = (y/(y-1))*(P1*1.03*10^4*V1)*(1-(P2/P1)^((y-1)/y));//work in m Kgf/min
+W1 = W/4500;
+mprintf('(i)The horsepower required is %f hp',W1);
+
+//(ii)Using the given data for CO2
+//From equation 10.2 (page no 189)
+W = -(H2 - H1);//work in Kcal/Kg
+M = V1/v1;//Mass rate of gas in Kg/min
+W1 = W*M*(427/4500);
+mprintf('\n (ii)Compressor work is %f hp',W1);
+//end
\ No newline at end of file diff --git a/479/CH10/EX10.2/Example_10_2.sce b/479/CH10/EX10.2/Example_10_2.sce new file mode 100755 index 000000000..54e541a61 --- /dev/null +++ b/479/CH10/EX10.2/Example_10_2.sce @@ -0,0 +1,24 @@ +//Chemical Engineering Thermodynamics
+//Chapter 10
+//Compressor
+//Example 10.2
+clear;
+clc;
+
+//Given
+P1 = 1;//Initial pressure in atm
+P2 = 29;//Final pressure in atm
+C = 0.05;//Clearance
+y = 1.4;//gamma of air
+
+//To calculate the volumetric efficiency and the maximum possible pressure that can be attained in a single stage
+//(i)Calulation of volumetric efficiency
+//From equation 10.11 (page no 194)
+V_E = 1+C-C*(P2/P1)^(1/y);
+mprintf('(i)Volumetric efficiency is %f percent',V_E*100);
+
+//(ii)Calculation of maximum pressure
+V_E = 0;//Minimum efficiency
+P2 = P1*(((1+C-V_E)/C)^y);
+mprintf('\n (ii)The maximum possible pressure attained is %f atm',P2);
+//end
\ No newline at end of file diff --git a/479/CH10/EX10.2/Example_10_2.txt b/479/CH10/EX10.2/Example_10_2.txt new file mode 100755 index 000000000..bac31c4c3 --- /dev/null +++ b/479/CH10/EX10.2/Example_10_2.txt @@ -0,0 +1,3 @@ +For example 10.2 (ii),
+The answer given in the book is 79 atm but i am getting 70.97 atm.
+There might be some calculation mistake in the book.
diff --git a/479/CH10/EX10.3/Example_10_3.sce b/479/CH10/EX10.3/Example_10_3.sce new file mode 100755 index 000000000..ebcc43c8b --- /dev/null +++ b/479/CH10/EX10.3/Example_10_3.sce @@ -0,0 +1,25 @@ +//Chemical Engineering Thermodynamics
+//Chapter 10
+//Compressor
+
+//Example 10.3
+clear;
+clc;
+
+//Given
+V_d = 5.15;//displacement volume in cubic meter/min
+P1 = 1;//initial pressure in Kgf/sq cm
+P2 = 8.5;//final pressure in Kgf/sq cm
+C = 0.06;//Clearance
+M_E = 0.8;//Mechenical efficiency
+y = 1.31;//gamma
+
+//To calculate the capacity and the actual horse power of the compressor
+v1 = V_d*(1+C-(C*((P2/P1)^(1/y))));
+mprintf('The capacity of the copressor is %f cubic meter/min',v1);
+//From equation 10.6 (page no 192)
+W = (y/(y-1))*(P1*1*10^4*v1)*(1-(P2/P1)^((y-1)/y));//work in Kgf/min
+W1 = W/4500;//work in hp
+W2 = W1/M_E;
+mprintf('\n The actual horse power of the compressor is %f hp',W2);
+//end
\ No newline at end of file diff --git a/479/CH10/EX10.4/Example_10_4.sce b/479/CH10/EX10.4/Example_10_4.sce new file mode 100755 index 000000000..644f2d286 --- /dev/null +++ b/479/CH10/EX10.4/Example_10_4.sce @@ -0,0 +1,22 @@ +//Chemical Engineering Thermodynamics
+//Chapter 10
+//Compressor
+
+//Example 10.4
+clear;
+clc;
+
+//Given
+P1 = 1;//Initial pressure in Kgf/sq cm
+Pn = 13;//Final pressure in Kgf/sq cm
+V1 =27;//flow rate of gas in cubic meter/min
+y = 1.6;//gamma of the gas
+n = [1 2 3 4 7 10];//number of stages
+mprintf('No of stages Horse power in hp');
+//To Calculate the theoretical horse power required
+for i = 1:6
+ W(i) = n(i)*(y/(y-1))*((P1*10^4)/4500)*V1*(1-(Pn/P1)^((y-1)/(n(i)*y)));
+ mprintf('\n %d',n(i));
+ mprintf(' %f',-W(i))
+end
+//end
diff --git a/479/CH10/EX10.5/Example_10_5.sce b/479/CH10/EX10.5/Example_10_5.sce new file mode 100755 index 000000000..3bc48b05a --- /dev/null +++ b/479/CH10/EX10.5/Example_10_5.sce @@ -0,0 +1,22 @@ +//Chemical Engineering Thermodynamics
+//Chapter 10
+//Compressor
+
+//Example 10.5
+clear;
+clc;
+
+//Given
+P1 = 1;//Initial pressure in Kgf/sq cm
+P4 = 200;//Final pressure in Kgf/sq cm
+n = 4;//no of stages
+
+//To find out the presure between stages
+r = (P4/P1)^(1/n);//Compression ratio
+P2 = r*P1;
+mprintf('The pressure after 1st stage is %f Kgf/sq cm',P2);
+P3 = r*P2;
+mprintf('\n The pressure after 2nd stage is %f Kgf/sq cm',P3);
+P4 = r*P3;
+mprintf('\n The pressure after 3rd stage is %f Kgf/sq cm',P4);
+//end
\ No newline at end of file diff --git a/479/CH11/EX11.1/Example_11_1.sce b/479/CH11/EX11.1/Example_11_1.sce new file mode 100755 index 000000000..463ff2230 --- /dev/null +++ b/479/CH11/EX11.1/Example_11_1.sce @@ -0,0 +1,33 @@ +//Chemical Engineering Thermodynamics
+//Chapter 11
+//Liquefaction of Gases
+
+//Example 11.1
+clear;
+clc;
+
+//Given
+P1 = 8.74;//Initial pressure in Kgf/sq cm
+P2 = 2.41;//Final pressure in Kgf/sq cm
+H1 = 327.13;//Enthalpy of inlet stream in Kcal/Kg
+H2 = H1;//Enthalpy of exit stream in Kcal/Kg ,since throttling is isenthalpic
+Hl = 26.8;//Enthalpy of liquid at the final condition in Kcal/Kg
+Hg = 340.3;//Enthalpy of gas at the final condition in Kcal/Kg
+vl = 152*10^-5;//Specific volume of liquid at the final condition in cubic meter/Kg
+vg = 0.509;//Specific volume of gas at the final condition in cubic meter/Kg
+v1 = 0.1494;//Initial specific volume in cubic meter/Kg
+
+//To Calculate the dryness fraction of exit stream and the ratio of upstream to downstream diameters
+//(i)Calculation of the dryness fraction of exit stream
+//From equation 3.13(a) (page no 82)
+x = (H2- Hl)/(Hg-Hl);
+mprintf('(i)The dryness fraction of the exit stream is %f',x);
+
+//(ii)Calculation of the ratio of upstream to downstream pipe diameters
+//From equation 3.13(b) (page no 82)
+v2 = (vl*(1-x))+(x*vg);//Total specific volume at the final condition in cubic meter/Kg
+//u1 = u2; since KE changes are negligible
+//From continuity equation: A2/A1 = D2^2/D1^2 = v2/v1 ; let required ratio,r = D2/D1;
+r = (v2/v1)^(1/2);
+mprintf('\n (ii)The ratio of upstream to downstream diameters is %f',r);
+//end
\ No newline at end of file diff --git a/479/CH11/EX11.2/Example_11_2.sce b/479/CH11/EX11.2/Example_11_2.sce new file mode 100755 index 000000000..bb4594cc3 --- /dev/null +++ b/479/CH11/EX11.2/Example_11_2.sce @@ -0,0 +1,20 @@ +//Chemical Engineering Thermodynamics
+//Chapter 11
+//Liquefaction of Gases
+
+//Example 11.2
+clear;
+clc;
+
+//Given
+P1 = 1000*1.033*10^4;//Initial pressure in Kgf/sq m
+P2 = 1*1.033*10^4;//Final pressure in Kgf/sq m
+T1 = 300;//Inital temperature in K
+Cp = 7;//Specific heat of the gas in Kcal/Kgmole K
+//Gas obeys the relation: v = (R*T)/P+(b*(T^2))
+b = 5.4392*10^-8;//in cubic meter/Kgmole K^2
+
+//To Calculate the temperature of the throttled gas
+//From equation (a) (page no 212);which we got after integration
+T2 = 1/((1/T1)-((b/Cp)*((P2-P1)/427)));
+mprintf('The throttled gas is cooled to %f K',T2);
\ No newline at end of file diff --git a/479/CH11/EX11.3/Example_11_3.sce b/479/CH11/EX11.3/Example_11_3.sce new file mode 100755 index 000000000..31f8f6804 --- /dev/null +++ b/479/CH11/EX11.3/Example_11_3.sce @@ -0,0 +1,27 @@ +//Chemical Engineering Thermodynamics
+//Chapter 11
+//Liquefaction of Gases
+
+//Example 11.3
+clear;
+clc;
+
+//Given
+//From the figure 11.8 (page no 216) & from figure A.2.7
+H3 = 0;
+H7 = -47;//in Kcal/Kg
+H6 = -93;//in Kcal/Kg
+H8 = 7;//in Kcal/Kg
+
+//To Calculate the fraction of air liquified at steady state and temperature of air before throttling
+//(i)Calculation of fraction of air liquified
+//From equation 11.3 (page no 215)
+x = (H8-H3)/(H8-H6);
+mprintf('The fraction of air liquified is %f',x);
+
+//(ii)Calculation of temperature
+H4 = H3+(H7*(1-x))-(H8*(1-x));//enthalpy of the gas before throttling
+//From figure A.2.7 temperature corresponds to pressure 160 atm and the enthalpy H4 is
+T = -112;
+mprintf('\n The temperature of air before throttling is %d deg celsius',T);
+//end
\ No newline at end of file diff --git a/479/CH12/EX12.1/Example_12_1.sce b/479/CH12/EX12.1/Example_12_1.sce new file mode 100755 index 000000000..197da396b --- /dev/null +++ b/479/CH12/EX12.1/Example_12_1.sce @@ -0,0 +1,25 @@ +//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
\ No newline at end of file diff --git a/479/CH12/EX12.2/Example_12_2.sce b/479/CH12/EX12.2/Example_12_2.sce new file mode 100755 index 000000000..40e68b232 --- /dev/null +++ b/479/CH12/EX12.2/Example_12_2.sce @@ -0,0 +1,23 @@ +//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
\ No newline at end of file diff --git a/479/CH12/EX12.2/Example_12_2.txt b/479/CH12/EX12.2/Example_12_2.txt new file mode 100755 index 000000000..9a81d99a9 --- /dev/null +++ b/479/CH12/EX12.2/Example_12_2.txt @@ -0,0 +1,2 @@ +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.
\ No newline at end of file diff --git a/479/CH12/EX12.3/Example_12_3.sce b/479/CH12/EX12.3/Example_12_3.sce new file mode 100755 index 000000000..cac9b868b --- /dev/null +++ b/479/CH12/EX12.3/Example_12_3.sce @@ -0,0 +1,44 @@ +//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
\ No newline at end of file diff --git a/479/CH12/EX12.4/Example_12_4.sce b/479/CH12/EX12.4/Example_12_4.sce new file mode 100755 index 000000000..cb7e915e7 --- /dev/null +++ b/479/CH12/EX12.4/Example_12_4.sce @@ -0,0 +1,22 @@ +//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
\ No newline at end of file diff --git a/479/CH13/EX13.1/Example_13_1.sce b/479/CH13/EX13.1/Example_13_1.sce new file mode 100755 index 000000000..5f1481c3f --- /dev/null +++ b/479/CH13/EX13.1/Example_13_1.sce @@ -0,0 +1,30 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.1
+clear;
+clc;
+
+//Given
+//N2 obeys the relation : Z = 1+(2.11*10^-4*P)
+Tc = 126;//Critical temperature in K
+Pc = 33.5;//Critical pressure in atm
+T = 373;//in K
+P = 100;//in atm
+
+//To Calculate the fugacity of N2 at 373K and 100 atm
+//(i)Using the Z relation given above
+//From equation 13.12 (page no 239)
+phi = %e^(2.11*10^-4*(P-0));//fugacity coefficient
+f = phi*P;
+mprintf('(i)The fugacity of N2 using the given Z relation is %f atm',f);
+
+//(ii)Using the fugacity chart given in figure A.2.9
+Pr = P/Pc;//Reduced pressure in atm
+Tr = T/Tc;//Reduced temperature in K
+//From figure A.2.9,
+phi = 1.04
+f = phi*P;
+mprintf('\n (ii)The fugacity of N2 using the fugacity chart is %f atm',f);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.10/Exampe_13_10.sce b/479/CH13/EX13.10/Exampe_13_10.sce new file mode 100755 index 000000000..e6a74d9ba --- /dev/null +++ b/479/CH13/EX13.10/Exampe_13_10.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.10
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.11/Example_13_11.sce b/479/CH13/EX13.11/Example_13_11.sce new file mode 100755 index 000000000..dbd0fb26e --- /dev/null +++ b/479/CH13/EX13.11/Example_13_11.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.11
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.12/Example_13_12.sce b/479/CH13/EX13.12/Example_13_12.sce new file mode 100755 index 000000000..c31b3fdf3 --- /dev/null +++ b/479/CH13/EX13.12/Example_13_12.sce @@ -0,0 +1,31 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.12
+clear;
+clc;
+
+//Given
+x_A = [0 0.0435 0.0942 0.1711 0.2403 0.3380 0.5981];//mole fraction of acetic acid
+p_A = [0 17.2 30.5 46.5 57.8 69.3 95.7];//partial pressure of acetic acid in mmHg
+P_T1 = 202;//vapour pressure of toulene in mmHg
+P_T2_ex = 167.3;//experimental partial pressure in mmmHg
+
+//To Calculate the partial pressure of toulene in the solution and check with the experimental value
+//From the equation 13.95,
+//ln(P_T2/P_T1) = -intg(x_A/((1-x_A)*p_A))
+for i = 1:7
+ if (p_A(i) ~= 0)
+ x(i) = (x_A(i)/((1-x_A(i))*p_A(i)))*10^4
+ end
+end
+plot(x,p_A);
+xtitle(" ","(x_A/((1-x_A)*p_A))*10^4", "p_A");
+//Area of the graph drawn is
+A = -0.138;
+P_T2 = (%e^A)*P_T1;
+e = ((P_T2-P_T2_ex)*100)/P_T2_ex;
+mprintf('The partial pressure of toulene is %f mmHg',P_T2);
+mprintf('\n This deviates %i percent from the reported value',e);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.13/Example_13_13.sce b/479/CH13/EX13.13/Example_13_13.sce new file mode 100755 index 000000000..8360364b0 --- /dev/null +++ b/479/CH13/EX13.13/Example_13_13.sce @@ -0,0 +1,43 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.13
+clear;
+clc;
+
+//Given
+P = 760;//pressure at maximum boiling azeotrope of A and B in mmHg
+x_A = 0.6;//mole fraction of A in liquid phase
+x_B = 0.4;//mole fraction of B in liquid phase
+p_A = 600;//vapour pressure of A at 90 deg cel
+p_B = 300;//vapour pressure of B at 90 deg cel
+
+//To Check whether the activity coefficient of the solution can be represented by the Margules equation
+y_A = P/p_A;//Activity coefficient of A
+y_B = P/p_B;//Activity coefficient of B
+//From the Margules equation or equation (a) & (b)
+U = [((x_B^2)-(2*(x_B^2)*x_A)) (2*(x_B^2)*x_A); (2*(x_A^2)*x_B) ((x_A^2)-(2*(x_A^2)*x_B))];
+V = [log(y_A); log(y_B)];
+W = U\V;
+//Now the value of constants A and B in equations(a)&(b) are given as
+A = W(1);
+B = W(2);
+//let us assume
+x_A = [0.0 0.2 0.4 0.6 0.8 1.0];
+x_B = [1.0 0.8 0.6 0.4 0.2 0.0];
+//C = lny_A; D = lny_B; E = ln(y_A/y_B)
+for i = 1:6
+ C(i) = (x_B(i)^2)*(2*(B-A)*x_A(i)+A);
+ D(i) = (x_A(i)^2)*(2*(A-B)*x_B(i)+B);
+ E(i) = C(i)-D(i);
+end
+clf;
+plot(x_A,E);
+xtitle(" ","x_A","ln(y_A/y_B)");
+a = get("current_axes");
+set(a,"x_location","origin");
+//Since the graph drawn is approximately symmetrical.Thus it satisfies the Redlich-Kister Test
+mprintf('The actvity coefficients of the system can be represented by Margules equation');
+//end
+
diff --git a/479/CH13/EX13.14/Example_13_14.sce b/479/CH13/EX13.14/Example_13_14.sce new file mode 100755 index 000000000..5c812b361 --- /dev/null +++ b/479/CH13/EX13.14/Example_13_14.sce @@ -0,0 +1,28 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.14
+clear;
+clc;
+
+//Given
+P = 760;//Total pressure of the mixture in mmHg
+T = [80 90 95 100];//Temperature in deg celsius
+P1 = [87.4 129.0 162.0 187.0];//vapour pressure of 1,1,2,2-tetrachloroethane in mmHg
+P2 = [356 526 648 760];//Vapour pressure of water in mmHg
+
+//To Calculate the composition of the vapour evolved
+clf;
+plot2d(T,P1,style=3);
+plot2d(T,P2,style=5);
+xtitle(" ","Temp in deg cel","Vapour pressure in mmHg");
+legend("1,1,2,2-tetrachloroethane","Water");
+//From the graph we conclude that at 93.8 deg cel
+P1 = 155;//in mm Hg
+P2 = 605;//in mm Hg
+y_1 = P1/P;
+y_2 = P2/P;
+mprintf('Mole fraction of 1,1,2,2-tetrachloroethane in vapour is %f',y_1);
+mprintf('\n Mole fraction of water in vapour is %f',y_2);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.15/Example_13_15.sce b/479/CH13/EX13.15/Example_13_15.sce new file mode 100755 index 000000000..bcf443b19 --- /dev/null +++ b/479/CH13/EX13.15/Example_13_15.sce @@ -0,0 +1,30 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.15
+clear;
+clc;
+
+//Given
+T = [146.2 142.3 126.1 115.9 95.0 98.0 100];//Temperature in deg cel
+P1 = [760.0 685.0 450.3 313.0];//Vapour pressure of 1,1,2,2-tetrachloroethane at the coressponding temperature in mm Hg
+P2_5 = 648.0;//Vapour pressure of water at 95 deg cel in mm Hg
+P2_6 = 711.0;//Vapour pressure of water at 98 deg cel in mm Hg
+P = 760;//Total pressure of mixture in mm Hg
+
+x1 = [0 0 0 0 0 0 0];
+//To plot a graph between temperature and vapour phase composition
+for i = 1:4
+ x1(i) = P1(i)/P;//mole fraction of 1,1,2,2-tetrachloroethane
+end
+x2_5 = P2_5/P;//mole fraction of water at 95 deg cel
+x2_6 = P2_6/P;//mole fraction of water at 98 deg cel
+x1(5) = 1-x2_5;
+x1(6) = 1-x2_6;
+
+clf;
+plot(x1,T);
+xtitle("","mole fraction of 1,1,2,2-tetrachloroethane","Temperature in deg cel");
+mprintf('The required graph has been ploted in the graphic window');
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.16/Example_13_16.sce b/479/CH13/EX13.16/Example_13_16.sce new file mode 100755 index 000000000..fd786013e --- /dev/null +++ b/479/CH13/EX13.16/Example_13_16.sce @@ -0,0 +1,48 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.16
+clear;
+clc;
+
+//Given
+//B = -(1.203*10^10)*(T^2.7); second virial coefficient, T is in K
+//log P = 6.95464-(1344.8/(219.482+t))...(a);Vapour pressure of toulene
+t = 107.2;//Temperature in deg cel
+T = t+273.16;//in K
+H_ex = 7964;//experimental value of heat of vapourisation in Kcal/Kgmole
+d = 800;//density of liquid toulene in Kg/cubic meter
+R = 1.98;//gas constant in Kcal/Kgmole K
+M = 92.14;//molecular weight of toulene
+
+//To Calculate the heat of vapourization of toulene by using ideal gas law, second virial coefficient but neglecting vl and including vl
+//From equation (a), let K = dlogP/dT
+K = 1344.8/(219.482+t)^2;
+//(i)Using ideal gas behaviour
+//From equation 13.112(page no 286)
+H_c = (2.303*R*(T^2))*K;
+mprintf('(i)The heat of vapourization using ideal gas behaviour is %f Kcal/Kgmole',H_c);
+D = ((H_c-H_ex)/H_c)*100;
+mprintf('\n The deviation is %f percent',D);
+
+//(ii)Using second virial coeff but neglecting vl
+//From equation(a)
+P = 10^(6.95464-1344.8/(219.482+t));//in mm Hg
+P1 = P*1.033*10^4/760;//in Kgf/sq m
+B = -((1.203*10^10)/(T^2.7))*10^-3;//in cubic meter/Kgmole
+//From equation 13.111 (page no 286) neglecting vl,
+l = (R*T)+((B*P1)/427);//in Kcal/Kgmole
+H_c = K*2.303*T*l;
+mprintf('\n\n(ii)The heat of vapourisation using second virial coefficient but neglecting vl is %f Kcal/Kgmole',H_c);
+D = ((H_c-H_ex)/H_c)*100;
+mprintf('\n The deviation in this case is %f percent',D);
+
+//(iii)Using second virial coeff including vl
+vl = M/d;//Liquid specific volume in cubic meter/Kgmole
+n = P1*vl/427;//in Kcal/Kgmole
+H_c = K*2.303*T*(l-n);
+mprintf('\n\n(iii)The heat of vapourisation using second virial coefficient including vl is %f Kcal/Kgmole',H_c);
+D = ((H_c-H_ex)/H_c)*100;
+mprintf('\n The deviation in this case is %f',D);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.17/Example_13_17.sce b/479/CH13/EX13.17/Example_13_17.sce new file mode 100755 index 000000000..02cb59dac --- /dev/null +++ b/479/CH13/EX13.17/Example_13_17.sce @@ -0,0 +1,40 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.17
+clear;
+clc;
+
+//Given
+H_ex = 539;//Heat of vapoization of water in Kcal/Kg
+Tc = 647;//Critical temperature in K
+Pc = 218;//Critical pressure in atm
+Tb = 373;//Boiling point of water in K
+t = 100;//temperature in deg cel
+M = 18;//Molecular weight of water
+P = 1;//pressure at boiling point in atm
+P1 = 1.033*10^4;//pressure in Kgf/sq m
+
+//To Calculate the heat of vapourisation of water by Vishwanath and Kuloor method and by Riedel's method and compare with the experimental value
+//(i) Using Vishwanath and Kuloor method
+H_c = (4.7*Tc*((1-(P/Pc))^0.69)*log(P/Pc))/((1-(Tc/Tb))*18);
+mprintf('(i)The heat of vapourisation of water using Vishwanath and Kuloor method is %f Kcal/Kg',H_c);
+D = (H_c-H_ex)*100/H_c;
+mprintf('\n The deviation occurs using this method is %f percent',D);
+
+//(ii)Using Riedel's method
+H_c = (Tb*2.17*(log(218)-1))/((0.93-(Tb/Tc))*18);
+mprintf('\n\n(ii)The heat of vapourisation of water using Riedel method is %f Kcal/Kg',H_c);
+D = (H_c-H_ex)*100/H_c;
+mprintf('\n The deviation occurs using this method is %f percent',D);
+
+//(iii)By using given vapour equation; logP = 8.2157-(2218.8537/(273.16+t)), t is in deg cel
+//From steam table,
+Vv = 1.673;//in cubic meter/Kg
+Vl = 0.001;//in cubic meter/Kg
+H_c = (2218.8/(273.16+t)^2)*(2.3*Tb*P1*(Vv-Vl)/427);
+mprintf('\n\n(iii)The heat of vapourisation using the given vapour equation is %f Kcal/Kg',H_c);
+D = (H_c-H_ex)*100/H_c;
+mprintf('\n The deviation occurs using this method is %f percent',D);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.17/Example_13_17.txt b/479/CH13/EX13.17/Example_13_17.txt new file mode 100755 index 000000000..59ddec64e --- /dev/null +++ b/479/CH13/EX13.17/Example_13_17.txt @@ -0,0 +1,3 @@ +For example 13.17(i), the answer given in the book, heat of vapourisation H_c = 532Kcal/Kg and the deviation as 1.3 %
+ but i am getting H_c = 1234.397 Kcal/Kg,using the Vishwanath and Kuloor method
+ and the deviation is 56.335 %.There might be some calculation mistake in the book.
\ No newline at end of file diff --git a/479/CH13/EX13.18/Example_13_18.sce b/479/CH13/EX13.18/Example_13_18.sce new file mode 100755 index 000000000..748a8434e --- /dev/null +++ b/479/CH13/EX13.18/Example_13_18.sce @@ -0,0 +1,33 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.18
+clear;
+clc;
+
+//Given
+T1 = 273-87;//temp in K
+T2 = 273;//temp in K
+H1 = 115;//Latent heat of saturated ethane at 1 atm and -87 deg cel in Kcal/Kg
+H2_ex = 72.44;//Experimental value of latent heat at 0 deg cel in Kcal/Kg
+Tc = 306;//Critical temperature in K
+M = 30;//Molecular weight of ethane
+
+//To Calculate the latent heat of saturated ethane at 0 deg cel
+Tr1 = T1/Tc;//reduced temp in K
+Tr2 = T2/Tc;//reduced temp in K
+//(i)Using Waton's method:
+H2_c = H1*((1-Tr2)/(1-Tr1))^0.38;
+mprintf('(i)The latent heat of saturated ethane at 0 deg cel using Waton method is %f Kcal/Kg',H2_c);
+D = (H2_ex-H2_c)*100/H2_ex;
+mprintf('\n The deviation occurs using this method is %f percent',D);
+
+//(ii)Using Vishwanath and Kuloor method
+//From equation 13.117 (page no 289)
+n = (0.00133*(H1*M/T1)+0.8794)^(1/0.1);
+H2_c = H1*((1-Tr2)/(1-Tr1))^n;
+mprintf('\n\n(ii)The latent heat of saturated ethane at 0 deg cel using Vishwanath and Kuloor method is %f Kcal/Kg',H2_c);
+D = (H2_ex-H2_c)*100/H2_ex;
+mprintf('\n The deviation occurs using this method is %f percent',D);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.19/Example_13_19.sce b/479/CH13/EX13.19/Example_13_19.sce new file mode 100755 index 000000000..09dfa1409 --- /dev/null +++ b/479/CH13/EX13.19/Example_13_19.sce @@ -0,0 +1,36 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.19
+clear;
+clc;
+
+//Given
+H_s_ex = 32.7;//experimental value of latent heat of the solution in KJ/mole
+x1 = 0.536;//mole percent of toulene in the solution
+x2 = 1-0.536;//mole percent of 1,1,1-trichloroethane in the solution
+H1 = 33.34;//Latent heat of toulene in KJ/gmole
+H2 = 29.72;//Latent heat of 1,1,1-trichloroethane in KJ/gmole
+He = 0;//excess enthalpy is neglected
+Cp1 = 39.55;//Specific heat of toulene in cal/gmole deg cel
+Cp2 = 24.62;//Specific heat of 1,1,1-trichloroethane in cal/gmole deg cel
+T_D = 100;//dew point temperature in deg cel
+T_B = 92.6;//bubble point temperature in deg cel
+
+//To calculate the latent heat of the solution and compare it with the one which calculated from the given vapour pressure equation
+//(i)Calculation of latent heat of the solution
+//From equation 13.118 (page no 291)
+H_s = H1*x1+H2*x2+He+(Cp1*x1+Cp2*x2)*10^-3*4.17*(T_D-T_B);
+mprintf('(i)The latent heat of the solution is %f KJ/gmole',H_s);
+D = ((H_s_ex-H_s)*100)/H_s_ex;
+mprintf('\n The deviation occurs using this method is %f percent',D);
+
+//(ii)Calculation of latent heat from the vapour pressure equation
+//From equation (a) (page no 291)
+K = 1657.599/((273.16+5)^2);
+H_s = (K*2.303*8.314*(273.16+5)^2)*10^-3;
+mprintf('\n\n(ii)The latent heat of the solution is %f KJ/gmole',H_s);
+D = ((H_s_ex-H_s)*100)/H_s_ex;
+mprintf('\n The deviation occurs using this method is %f percent',D);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.19/Example_13_19.txt b/479/CH13/EX13.19/Example_13_19.txt new file mode 100755 index 000000000..aeffb16d5 --- /dev/null +++ b/479/CH13/EX13.19/Example_13_19.txt @@ -0,0 +1,4 @@ +For example 13.19, the answer given in the book in part(i), deviation as 0.91% but i am getting deviation as 0.1%.
+ and for part(ii) i am getting heat of vapourization as 31.74 KJ/gmole and deviation as 2.94%.
+ but in the book it is given as heat of vapourization as 32.1 KJ/gmole and deviation as 1.8%.
+ There might be some calculation mistake in the book.
\ No newline at end of file diff --git a/479/CH13/EX13.2/Example_13_2.sce b/479/CH13/EX13.2/Example_13_2.sce new file mode 100755 index 000000000..a770036e1 --- /dev/null +++ b/479/CH13/EX13.2/Example_13_2.sce @@ -0,0 +1,22 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.2
+clear;
+clc;
+
+//Given
+P1 = 50*1.03*10^4;//Initial pressure in Kgf/sq m
+T = 373;//Temperature in K
+P2 = 1.03*10^4;//Final pressure in Kgf/sq m
+V = 0.001*18;//Volume in cubic meter
+R = 848;//gas constant in m Kgf/Kgmole K
+
+//To Calculate the fugacity of liquid water
+//From equation 13.13(page no 240)
+del_u = (V/(R*T))*(P2-P1);//del_u = ln(f2/f1); Change in chemical potential
+f1 = P2;//in Kgf/sq m
+f2 = f1*(%e^del_u);
+mprintf('The fugacity of the liquid water at 50 atm is %4.2e Kgf/sq m',f2);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.3/Example_13_3.sce b/479/CH13/EX13.3/Example_13_3.sce new file mode 100755 index 000000000..cfe3f9ae3 --- /dev/null +++ b/479/CH13/EX13.3/Example_13_3.sce @@ -0,0 +1,27 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.3
+clear;
+clc;
+
+//Given
+x1 = 0.1;//mole fraction of methane
+x2 = 0.9;//mole fraction of propane
+P = [28.1 31.6 35.1];//Pressure in Kgf/sq cm are
+K1 = [5.8 5.10 4.36];//Vapourisation constants of methane at the corresponding presssures
+K2 = [0.61 0.58 0.56];//Vapourisation constants of propane at the correspondig pressures
+
+//To Calculate the bubble point pressure of the solution
+//From equation 13.27 (page no 245)
+for i = 1:3
+ y1(i) = K1(i)*x1;//mole fraction of methane in the vapour phase
+ y2(i)= K2(i)*x2;//mole fraction of propane in the vapour phase
+ y(i) = y1(i)+y2(i);//sum of the mole fraction in the vapour phase
+end
+plot(P,y);
+xtitle("y vs pressure","P","y");
+P1 = interpln([y';P],1);// in Kgf/sq cm
+mprintf('The bubble point pressure of the solution is %f Kgf/sq cm',P1);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.4/Example_13_4.sce b/479/CH13/EX13.4/Example_13_4.sce new file mode 100755 index 000000000..5c1116c47 --- /dev/null +++ b/479/CH13/EX13.4/Example_13_4.sce @@ -0,0 +1,37 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.4
+clear;
+clc;
+
+//Given
+T = [80.6 79.0 77.3 61.4];//Various temperature in deg cel
+x1 = [0.0 15.0 29.0 100.0];//mole fraction of CHCl3 in liquid phase
+y1 = [0.0 20.0 40.0 100.0];//mole fraction of CHCl3 in vapour phase
+P1 = [1370 1310 1230 700];//Vapour pressure of CHCl3 in mm Hg
+P = 760;//Total pressure in mm Hg
+
+//To Calculate the equilibrium data i.e y/x and compare with the experimental values
+//From equation 13.27 (page no 245);K = y1/x1 = Pi/P
+mprintf('Temperature Experimental Calculated');
+
+for i = 1:4
+ mprintf('\n %f',T(i));
+ if x1(i) == 0
+ mprintf(' Not defined');
+ else
+ K_ex(i) = y1(i)/x1(i);
+mprintf(' %f',K_ex(i));
+ end
+K_c(i) = P1(i)/P;
+mprintf(' %f',K_c(i));
+end
+
+if K_ex(i) == K_c(i)
+ then mprintf('\n\n The liquid solution is perfect');
+else
+ mprintf('\n\n The liquid solution is imperfect');
+end
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.5/Example_13_5.sce b/479/CH13/EX13.5/Example_13_5.sce new file mode 100755 index 000000000..72f3a8d68 --- /dev/null +++ b/479/CH13/EX13.5/Example_13_5.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.5
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and does not contain any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.6/Example_13_6.sce b/479/CH13/EX13.6/Example_13_6.sce new file mode 100755 index 000000000..990304a97 --- /dev/null +++ b/479/CH13/EX13.6/Example_13_6.sce @@ -0,0 +1,24 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.6
+clear;
+clc;
+
+//Given
+x1 = 0.1;//Mole fraction of dichloromethane (CCl2H2)
+x2 = 0.9;//Mole fraction of methyl acetate (C3H6O2)
+M1 = 85;//Molecular weight of CCl2H2
+M2 = 74;//Molecular weight of C3H602
+D1 = 1.3163;//Density of CCl2H2 in gm/cc
+D2 = 0.9279;//Density of C3H6O2 in gm/cc
+
+//To Calculate the volume of 10% dichloromethane solution
+V1 = M1/D1;//Specific volume of pure CCL2H2 in cc/gmole
+V2 = M2/D2;//Specific volume of C3H6O2 in cc/gmole
+//From equation 13.62(page no 256)& 13.78 (page no 257)
+V_e = x1*x2*(1.2672-0.771*x1);//excess volume in cc/gmole
+V = V1*x1+V2*x2+V_e;
+mprintf('The volume of 10 percent dichloromethane is %f cc/gmole',V);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.7/Example_13_7.sce b/479/CH13/EX13.7/Example_13_7.sce new file mode 100755 index 000000000..695af3a9d --- /dev/null +++ b/479/CH13/EX13.7/Example_13_7.sce @@ -0,0 +1,52 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.7
+clear;
+clc;
+
+//Given
+x_T = 0.957;//mole fraction of Toluene
+x_D = 0.043;//mole fraction of 1,2-dichloroethane
+t = [90; 100; 110];//temperature in deg cel
+R = 1.98;//gas constant in Kcal/Kgmole K
+
+//To Calculate the vapour pressure of the solution, bubble point at 686 mm Hg and the vapour composition at equilibrium,
+//compare the experimental value of 91.2% toluene in vapour with the calculated value & calculate the free energy of mixing
+//(1)Calculation of vapour pressure
+mprintf('(1)Temp(deg cel) P_T(mmHg) P_D(mmHg) P_s(mmHg)');
+for i = 1:3
+ P_T(i) = 10^(6.95464-(1344.8/(219.482+t(i))));//Given as equation(a)(page no 260)
+ P_D(i) = 10^(7.03993-(1274.079/(223+t(i))));//Given as equation(b)(page no 260)
+ P_s(i) = x_T*P_T(i)+x_D*P_D(i);//pressure of the solution in mm Hg
+ mprintf('\n %f',t(i));
+ mprintf(' %f',P_T(i));
+ mprintf(' %f',P_D(i));
+ mprintf(' %f',P_s(i));
+end
+
+//(2)Calculation of bubble point and comparison of values
+clf;
+plot(t,P_s);
+xtitle("t vs P_s","t","P_s");
+T = interpln([P_s';t'],686);
+P = 686;//pressure of solution in mm Hg
+y_T_e = 0.912;//experimental value of mole fraction of toluene
+//From the graph we found that the temperature at P = 686 mm Hg is
+//t = 105.3;//in deg cel
+mprintf('\n\n(2)The bubble point is %f deg cel',T);
+//From equation (a)(page no 260)
+P_T = 10^(6.95464-(1344.8/(219.482+T)));//vapour pressure of Toluene in mmHg
+//From equation 13.27 (page no 245)
+y_T_c = (x_T*P_T)/P;
+y_D_c = 1-y_T_c;
+mprintf('\n The vapour composition of toluene is %f',y_T_c);
+mprintf('\n The vapour composition of 1,2-dichloroethane is %f',y_D_c);
+e = ((y_T_e-y_T_c)/y_T_e)*100;
+mprintf('\n The percentage error is %f percent',e);
+
+//(3)Calculation of free energy
+del_F = R*(T+273)*((x_T*log(x_T))+(x_D*log(x_D)));
+mprintf('\n\n(3)The free energy of mixing is %f Kcal/Kgmole',del_F);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.8/Example_13_8.sce b/479/CH13/EX13.8/Example_13_8.sce new file mode 100755 index 000000000..2699ffb34 --- /dev/null +++ b/479/CH13/EX13.8/Example_13_8.sce @@ -0,0 +1,26 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.8
+clear;
+clc;
+
+//Given
+//Consider the diagram shown in page no 263
+w1 = 100;//weight of LiBr entered as feed in the evaporator per hour in Kg
+x1 = 0.45;//weight fraction of LiBr entered as feed
+x2 = 0;//weight fraction of steam in the LiBr soln
+x3 = 0.65;//weight fraction of LiBr formed as product
+H1 = -39;//Enthalpy of 45% solution at 25 deg cel in Kcal/Kg
+H3 = -4.15;//Enthalpy of 65% solution at 114.4 deg cel in Kcal/Kg
+H2 = 649;//Enthalpy of superheated steam at 100 mmHg and 114.4 deg cel in Kcal/Kg
+
+//To Calculate the heating load required for the process
+//According to material balance
+w3 = (w1*x1)/x3;//weight of LiBr solution formed after evaporation per hour in Kg
+w2 = w1-w3;// weight of steam formed in Kg/hr
+//According to energy balance
+Q = (w2*H2)+(w3*H3)-(w1*H1);
+mprintf('The heat that has to be supplied for this concentration process is %f Kcal/hr',Q);
+//end
\ No newline at end of file diff --git a/479/CH13/EX13.9/Example_13_9.sce b/479/CH13/EX13.9/Example_13_9.sce new file mode 100755 index 000000000..b260581f6 --- /dev/null +++ b/479/CH13/EX13.9/Example_13_9.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 13
+//Thermodynamics in Phase Equilibria
+
+//Example 13.9
+clear;
+clc;
+
+//Given
+// In the given example, all the values were determined from the graph given as figure 13.9 and it does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.1/Example_14_1.sce b/479/CH14/EX14.1/Example_14_1.sce new file mode 100755 index 000000000..fa2d933eb --- /dev/null +++ b/479/CH14/EX14.1/Example_14_1.sce @@ -0,0 +1,17 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.1
+clear;
+clc;
+
+//Given
+H_f_C2H4 = 12500;//Standard heat of formation of ethylene at 25 deg cel in Kcal/Kgmole
+H_f_C2H4O = -11667;//Standard heat of formation of ethylene oxide at 25 deg cel in Kcal/Kgmole
+
+//To Calculate the standard heats of reaction at 25 deg celsius
+//The reaction is: C2H4 + (1/2)O2 - C2H4O
+del_H_rxn = H_f_C2H4O-H_f_C2H4;//Since, Standard heat of formation of O2 is zero
+mprintf('The standard heats of reaction at 25 deg celsius is %d Kcal/Kgmole',del_H_rxn);
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.10/Example_14_10.sce b/479/CH14/EX14.10/Example_14_10.sce new file mode 100755 index 000000000..f2d9716eb --- /dev/null +++ b/479/CH14/EX14.10/Example_14_10.sce @@ -0,0 +1,49 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.10
+clear;
+clc;
+
+//Given
+//SO2(A) + (1/2)O2 (B) - SO3(C)
+//Basis: 1 Kgmole SO2
+n_A = 1;//Kgmole of SO2 fed
+n_B = n_A;//Kgmole of O2 fed
+T1 = 273+400;//Temperature in K at which reactants enter
+To = 298;//room temperature in K
+del_H = -23490;//Standar heat of reaction at 25 deg cel from example 14.6 in Kcal/Kgmole
+
+//At T1,
+C_A_T1 = 11.0; C_B_T1 = 7.4;//in Kcal/Kgmole
+//Assume the various temperature be
+T = [913 1073 1373 1573];//in K
+//Mean specific heats of the components A,B & C at various temperature are given below in Kcal/Kgmole K
+C_A = [11.6 11.8 12.3 12.5];
+C_B = [7.7 7.8 8.0 8.2];
+C_C = [ 16.6 17.2 18.2 18.6];
+
+//To Calculate the final temperature for various conversions and the maximum conversion that can be attained in a single reactor operating adiabatically
+//In equation 14.18 (page no 307), H2-Hr = K & (Hp-H3)= L(say)
+K = ((n_A*C_A_T1)+(n_B*C_B_T1))*(To-T1);//in Kcal/Kgmole
+mprintf('Adiabatic reaction temp in K pecentage conversion of SO2');
+for i = 1:4
+ n_C(i) = (-K-(C_A(i)*(T(i)-To))-(C_B(i)*(T(i)-To)))/((T(i)-To)*(C_C(i)-C_A(i)-0.5*C_B(i))+del_H);
+mprintf('\n %d',T(i));
+mprintf(' %f',n_C(i)*100);
+end
+
+clf;
+//plot(T,n_C*100);
+plot2d(T,n_C*100,style =2);
+//Now equilibrium conversion at various temperature taken from figure 14.7 (page no 325) are given as
+Ta = [850 900 1000 1100 1200 1300 1400];
+n_C1 = [93.5 88.2 69.0 49.0 37.0 21.5 6.25];
+plot2d(Ta,n_C1,style=3);
+xtitle("Temperature vs Percentage Conversion","Temperature in K","% Conversion");
+//From the graph,it can be seen that the curve cut each other approximately at the temp
+T1 = 1140;//in Kelvin
+C = interpln([Ta;n_C1],T1);
+mprintf('\n\n The maximum conversion that can be attained is %d percent',C);
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.11/Example_14_11.sce b/479/CH14/EX14.11/Example_14_11.sce new file mode 100755 index 000000000..f16386146 --- /dev/null +++ b/479/CH14/EX14.11/Example_14_11.sce @@ -0,0 +1,30 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.11
+clear;
+clc;
+
+//Given
+//Basis: 1 Kgmole of benzene
+//C6H6 (A) + HNO3 (B) - C6H5NO2 (C) + H2O (D)
+T = 298;//Temperature in K
+R = 1.98;//gas constant in Kcal/Kgmole K
+//Standard enthalpy in Kcal/Kgmole at 25 deg celsius of the above components are given as
+H_A = 11718;
+H_B = -41404;
+H_C = -68371;
+H_D = 3800;
+//Standard entropy in Kcal/Kgmole K at 25 deg celsius of the above components are given as
+S_A = 41.30;
+S_B = 37.19;
+S_C = 16.72;
+S_D = 53.60;
+
+//To Calculate the conversion of benzene at 25 degree celsius and 1 atm
+del_F = (H_C+H_D-(T*(S_C+S_D)))-(H_A+H_B-(T*(S_A+S_B)));
+Ka = %e^(-del_F/(R*T));//Equilibrium constant
+x = (Ka^(1/2)/(1+(Ka^(1/2))));
+mprintf('The conversion is almost %f percent for this reaction.',x*100);
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.12/Example_14_12.sce b/479/CH14/EX14.12/Example_14_12.sce new file mode 100755 index 000000000..da72bbb33 --- /dev/null +++ b/479/CH14/EX14.12/Example_14_12.sce @@ -0,0 +1,37 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.12
+clear;
+clc;
+
+//Given
+//C + 2H2 - CH4
+//Basis: 1 Kgmole of C fed
+T = 1000;//Temperature in K
+P1 = 2;//Pressure in atm
+del_F = 4580;//Standard free energy in Kcal/Kgmole
+
+
+//To Calculate the maximum CH4 concentration under the condition of 2 atm and the quantity of methane obtained if pressure is 1 atm
+Ka = %e^(-del_F/(R*T));//Equilibrium constant
+//In relation (d) (page no 339) p_H2 = p (say)
+p = poly(0,'p');
+q = Ka*(p^2)+p-P1;
+r = roots(q);
+p_H2 = r(2);//partial pressure of H2
+p_CH4 = P1-p_H2;//partial pressure of CH4
+X_H2 = p_H2*100/P1;//mole percent of H2
+X_CH4 = p_CH4*100/P1;//mole percent of CH4
+mprintf('Under the conditions of 2 atm and 1000 K,the maximum CH4 concentration is %f percent and further increase is not pssible',X_CH4);
+//Now.pressure has become
+P2 = 1;//in atm
+q = Ka*(p^2)+p-P2;
+r = roots(q);
+p_H2 = r(2);//partial pressure of H2
+p_CH4 = P2-p_H2;//partial pressure of CH4
+X_H2 = p_H2*100/P2;//mole percent of H2
+X_CH4 = p_CH4*100/P2;//mole percent of CH4
+mprintf('\n\n Under the conditions of 1 atm and 1000 K,Methane = %f percent and Hydrogen = %f percent',X_CH4,X_H2);
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.13/Example_14_13.sce b/479/CH14/EX14.13/Example_14_13.sce new file mode 100755 index 000000000..f728cb876 --- /dev/null +++ b/479/CH14/EX14.13/Example_14_13.sce @@ -0,0 +1,47 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.13
+clear;
+clc;
+
+//Given
+T1 = 273+110;//Temperature in K
+T = 298;//Room temperature in K
+P = 1;//Pressure in atm
+R = 1.98;//gas constant in Kcal/Kgmole
+//Ag2CO3(s) (A) = Ag2O(s) (B) + CO2(g) (C)
+a = 1;//Stoichiometry coefficient of A
+b = 1;//Stoichiometry coefficient of B
+c = 1;//Stoichiometry coefficient of C
+//Standard entropy of the above components in Kcal/Kgmole K at 25 deg cel are given as
+S_A = 40.17;
+S_B = 29.09;
+S_C = 51.08;
+//Standard enthalpy of the above components in Kcal/Kgmole at 25 deg cel are given as
+H_A = -119900;
+H_B = -6950;
+H_C = -94036;
+//Specific heat (assumed constant) of the above components in Kcal/Kgmole K are given as
+C_A = 26.1;
+C_B = 16.5;
+C_C = 9.6;
+
+//To Calculate the partial pressure of CO2 required for decomposition and thus determine whether Ag2CO3 will decomposes or not at the given pressure and temperature
+del_H = H_C+H_B-H_A;//Standard heat of reaction at 25 deg cel in Kcal/Kgmole
+del_C = b*C_B+c*C_C-a*C_A;
+//From equation 14.15 (page no 340)
+del_Ht = del_H +del_C*(T1-T);
+del_F = del_H-(T*(S_B+S_C-S_A));//Standard free energy in Kcal/Kgmole
+Ka1 = %e^(-del_F/(R*T));//Equilibrium constant at temperature T
+//Since del_Ht is constant(as del_C = 0),Ka2 can be calculated by equation 14.43 (page no 316)
+Ka2 = Ka1* %e^((del_Ht/R)*(1/T1-1/T));
+p_CO2 = Ka2;
+mprintf('The partial pressure of CO2 required for decomposition is %4.3e atm',p_CO2);
+if p_CO2 < P
+ mprintf('\n Silver carbonate will not decompose at 110 deg celsius as the pressure given is 1 atm');
+else
+ mprintf('\n Silver carbonate will decompose at 110 deg celsius');
+end
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.13/Example_14_13.txt b/479/CH14/EX14.13/Example_14_13.txt new file mode 100755 index 000000000..b6ffe1ee8 --- /dev/null +++ b/479/CH14/EX14.13/Example_14_13.txt @@ -0,0 +1,4 @@ +For example 14.13,
+the answer given in the book is partial pressure of CO2 as 0.0012
+But,I am getting partial pressure of CO2 as 5.787e-009 atm.
+There might be some calculation mistake in the book.
\ No newline at end of file diff --git a/479/CH14/EX14.14/Example_14_14.sce b/479/CH14/EX14.14/Example_14_14.sce new file mode 100755 index 000000000..7aa02282a --- /dev/null +++ b/479/CH14/EX14.14/Example_14_14.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.14
+clear;
+clc;
+
+//Given
+//The given example is an theoretical problem and it does not involve any numerical computation.
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.15/Example_14_15.sce b/479/CH14/EX14.15/Example_14_15.sce new file mode 100755 index 000000000..7335a10be --- /dev/null +++ b/479/CH14/EX14.15/Example_14_15.sce @@ -0,0 +1,27 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.15
+clear;
+clc;
+
+//Given
+//C3H8 = C3H6 + H2... (i)
+//C3H8 = C2H4 +CH4...(ii)
+Kp1 = 7.88;//Equilibrium constant of equation (i)
+Kp2 = 775;//Equilibrium constant of equation (ii)
+T = 760+273;//Temperature in K
+
+//To Calculate the equilibrium composition of the mixture
+//Basis: 1 mole of C3H8 in feed
+//From the equations (a) &(b) (page no 343); y/x = z (say)
+z = (Kp2/Kp1)^(1/2);
+//Substituting y = z*x in equation(a), we got the equation p:
+x = poly(0,'x')
+p = (1+Kp1+Kp1*(z^2)+2*Kp1)*x^2 - Kp1;
+q = roots(p);
+x = q(1);
+y = z*x;
+mprintf('Moles of H2, C3H6, C2H4 and CH4 formed at equilibrium are %f %f %f %f respectively.',x,x,y,y);
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.2/Example_14_2.sce b/479/CH14/EX14.2/Example_14_2.sce new file mode 100755 index 000000000..3dd420bff --- /dev/null +++ b/479/CH14/EX14.2/Example_14_2.sce @@ -0,0 +1,19 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.2
+clear;
+clc;
+
+//Given
+T1 = 25+273;//Initial temperature in K
+T2 = 450+273;//Final temperature in K
+//Specific heat of sulphur dioxide is given by the relation:
+//Cp = 7.116+9.512*10^-3*T+(3.511*10^-6)*T^2
+
+//To Calculate the sensible heat required
+//Basis: 1 Kgmole of sulphur dioxide
+Q = 7.116*(T2-T1)+(9.512*10^-3*(T2^2-T1^2)/2)+((3.51*10^-6)*(T2^3-T1^3)/3);
+mprintf('Sensible heat required is %d Kcal/Kgmole',Q);
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.3/Example_14_3.sce b/479/CH14/EX14.3/Example_14_3.sce new file mode 100755 index 000000000..71bb32ae9 --- /dev/null +++ b/479/CH14/EX14.3/Example_14_3.sce @@ -0,0 +1,51 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.3
+clear;
+clc;
+
+//Given
+H_C2H4 = 12500;//Standard heat of formation of ethylene at 25 de cel
+H_C2H4O = -12190;//Standard heat of formation of ethylene oxide at 25 deg cel
+H_CO2 = -94052;//Standard heat of formation of CO2 at 25 deg cel
+H_H2O = -57798;//Standard heat of formation of H2O at 25 deg cel
+T1 = 200;//temperature at which mixture entered in deg cel
+Ti = 25;//intermediate temperature in deg cel
+T2 = 260;//product temperature in deg cel
+Cp_air_a = 7;//Mean specific heat for 25 to 200 deg cel in Kcal/Kgmole
+Cp_C2H4_a = 18;//Mean specific heat for 25 to 200 deg cel in Kcal/Kgmole
+//Mean specific heat for 25 to 260 deg cel in Kcal/Kgmole are given as:
+Cp_C2H4_b = 19;
+Cp_C2H4O_b = 21;
+Cp_O2_b = 7.30;
+Cp_N2_b = 7.00;
+Cp_CO2_b = 10.00;
+Cp_H2O_b = 8.25;
+//Basis: 1 Kgmole of ethylene as feed
+n_air = 9;//Kgmoles
+n_C2H4 =1;//Kgmoles
+n_C2H4_1 = 0.6//ethylene consumed while converting in C2H4O in Kgmoles (Rxn 1)
+n_C2H4_2 = 0.3;//ethylene burnt completely to CO2 in Kgmoles (Rxn 2)
+
+//To Calculate the heat must be removed if the product temperature should not exceed 260 deg cel
+n_O2 = n_air*0.21;//Kgmoles of O2 fed
+n_N2 = n_air-n_O2;//Kgmoles of N2 fed
+//Rxn 1: (0.6)C2H4 + (0.3)O2 - (0.6)C2H4O;...(i) 60% of C2H4 is converted to C2H4O
+n_O2_1 = n_C2H4_1/2;//Kgmoles of oxygen consumed
+n_C2H4O_1 = n_C2H4_1;//C2H4O formed in Kgmoles
+//Rxn 2: (0.3)C2H4 + (0.9)O2 - (0.6)CO2+ (0.6)H20;...(ii) 30%conversion
+n_O2_2 = 3*n_C2H4_2;//Kgmoles of O2 reacted
+n_CO2_2 = 2*n_C2H4_2;//Kgmoles of CO2 formed
+n_H2O_2 = 2*n_C2H4_2;//Kgmoles of H2O formed
+n_C2H4_r = n_C2H4 - n_C2H4_1 - n_C2H4_2;//unreacted ethylene in Kgmoles
+n_O2_r = n_O2 - n_O2_1-n_O2_2;//unreacted O2 in Kgmoles
+//The overall rxn is given by (i)&(ii),
+//(0.9)C2H4 + (1.2)O2 - (0.6)C2H4O + (0.6)CO2 + (0.6)H2O...(iii)
+del_H = (0.6*H_C2H4O)+(0.6*H_CO2)+(0.6*H_H2O)-(0.9*H_C2H4);//since,standard enthalpy of O2 is zero; Standard heat of rxn in Kcal/Kgmole
+Q1 = (n_C2H4*Cp_C2H4_a + n_air*Cp_air_a)*(Ti-T1);//Sensible heat in feed in Kcal
+Q2 = (n_C2H4_r*Cp_C2H4_b + n_C2H4O_1*Cp_C2H4O_b + n_CO2_2*Cp_CO2_b + n_H2O_2*Cp_H2O_b + n_O2_r*Cp_O2_b + n_N2*Cp_N2_b)*(T2-Ti);//Sensible heat in product in Kcal
+Q = Q1+Q2+del_H;
+mprintf('The heat to be removed is %f Kcal so that the product temperature is 260 deg celsius',-Q);
+//end
diff --git a/479/CH14/EX14.4/Example_14_4.sce b/479/CH14/EX14.4/Example_14_4.sce new file mode 100755 index 000000000..883ee4d45 --- /dev/null +++ b/479/CH14/EX14.4/Example_14_4.sce @@ -0,0 +1,47 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.4
+clear;
+clc;
+
+//Given
+//Basis: 1Kgmole of C0
+//CO + (1/2)O2 - CO2
+//Whem mixture entered,their compositions are given as:
+n_CO = 1;//Kgmole
+del_H = -67636;//Standard heat of reaction in Kcal/Kgmole at 25 deg cel
+T1 = 95;//Temperature at which mixture entered in deg celsius
+T2 = 25;//Intermediate temperature in deg celsius
+//Mean specific heat values for the temperature between 25 and 95 deg celsius in Kcal/Kgmole are given as (from figure 14.4)(page no 303)
+Cpm_CO = 6.95;
+Cpm_O2 = 7.1;
+Cpm_N2 = 6.95;
+
+//To Calculate the theoretical flame temperature when both air and CO2 enter at 95 deg celsius
+n_O2 = 1;//Kgmole, as 100% excess air is given
+n_N2 = n_O2*(0.79/0.21);//Kgmole
+//After the rxn:
+n_CO2 = n_CO;//Kgmole
+n_O2_r = n_O2-(n_CO/2);//remaining Kgmole of O2
+//In equation 14.18 (page no 307) say: H_2-H_R = Ha, H_P-H_3 = Hb, Hc = del_H+Ha & Ht = Hc+Hb
+Ha = (n_CO*Cpm_CO + n_O2*Cpm_O2 + n_N2*Cpm_N2)*(T2-T1);//in Kcal/Kgmole
+Hc = del_H+Ha;//in Kcal/Kgmole
+//For calculating Hb let us assume the temperature as
+T = [530 1000 1650];//in deg celsius
+Cpm_CO2 = [10.85 12 12.75];//Mean specific heat of CO2 at the coresspondig temperature (from figure 14.4)
+Cpm_O2 = [7.55 7.8 8.3];//Mean specific heat of O2 at the coresspondig temperature (from figure 14.4)
+Cpm_N2 = [7.15 7.5 7.85];//Mean specific heat of N2 at the coresspondig temperature (from figure 14.4)
+for i = 1:3
+ Hb(i) = (n_CO2*Cpm_CO2(i)+n_O2_r*Cpm_O2(i)+n_N2*Cpm_N2(i))*(T(i)-T2);//in Kcal/Kgmole
+ Ht(i) = Hc+Hb(i);//in Kcal/Kgmole
+end
+clf;
+plot(T,Ht);
+xtitle(" ","Temperature, deg celsius","Ht in Kcal/Kgmole");
+a = get("current_axes");
+set(a,"x_location","origin");
+Tf = interpln([Ht';T],0);
+mprintf('The adiabatic temperature is read as %d degree celsius',Tf);
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.4/Example_14_4.txt b/479/CH14/EX14.4/Example_14_4.txt new file mode 100755 index 000000000..c590523d9 --- /dev/null +++ b/479/CH14/EX14.4/Example_14_4.txt @@ -0,0 +1,3 @@ +For example 14.4,the answer given in the book is temperature as 1610 degree celsius.
+ But i am getting temperature as 1560 degree celsius.
+ There might be some calculation mistake in the book.
\ No newline at end of file diff --git a/479/CH14/EX14.5/Example_14_5.sce b/479/CH14/EX14.5/Example_14_5.sce new file mode 100755 index 000000000..3a2d45ea5 --- /dev/null +++ b/479/CH14/EX14.5/Example_14_5.sce @@ -0,0 +1,29 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.5
+clear;
+clc;
+
+//Given
+T = 298;//Temperature in Kelvin
+//Standard enthalpy in Kcal/Kgmole of the following components at 298 K are given as
+H_SO2 = -70960;
+H_SO3 = -94450;
+H_O2 = 0;
+//Standard entropy in Kcal/Kgmole K of the following components at 298 K are given as
+S_SO2 = 2.48;
+S_SO3 = -19.7
+S_O2 = 0;
+//Basis: 1 Kgmole of SO2
+//SO2 +(1/2)O2 - SO3
+
+//To Calculate the standard free energy for the reaction
+n_SO2 = 1;//Kgmole of SO2 fed
+n_O2 = (1/2)*2;//Kgmole of O2 fed as 100% excess O2 is given
+n_SO3 = n_SO2;//Kgmole of SO3 formed
+//From equation 14.38 (page no 312)
+del_F = (H_SO3-(T*S_SO3))-(H_SO2-(T*S_SO2))-(H_O2-(T*S_O2));
+mprintf('The standard free energy for the reaction at 25 degree celsius is %f Kcal/Kgmole',del_F);
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.6/Example_14_6.sce b/479/CH14/EX14.6/Example_14_6.sce new file mode 100755 index 000000000..e81c94b47 --- /dev/null +++ b/479/CH14/EX14.6/Example_14_6.sce @@ -0,0 +1,63 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.6
+clear;
+clc;
+
+//Given
+//SO2 + (1/2)O2 - SO3
+//Basis: 1 Kgmole of SO2
+n_SO2 = 1;// SO2 fed in Kgmole
+//From table 14.1 (page no 301)
+//alpha values for the following components are given as
+a_SO2 = 7.116;
+a_O2 = 6.148;
+a_SO3 = 6.077;
+//beta values for the following components are given as
+b_SO2 = 9.512*10^-3;
+b_O2 = 3.102*10^-3;
+b_SO3 = 25.537*10^-3;
+//Standard enthalpy of the following components at 25 deg cel in Kcal/Kgmole are given as
+H_SO2 -70960;
+H_O2 = 0;
+H_SO3 = -94450;
+//Standard free energy of the following components at 25 deg cel in Kcal/Kgmole K are given as
+F_SO2 = -71680;
+F_O2 = 0;
+F_SO3 = -88590;
+n_O2 = n_SO2;//O2 fed in Kgmole; since 50 mole percent mixture of SO2 & O2 is fed
+n_SO3 = n_SO2;//SO3 formed in Kgmole
+n_O2_e = n_O2-(n_O2/2);//Kgmoles of O2 in exit gas
+n_O2_r = n_O2/2;//Kgmoles of O2 reacted
+R = 1.98;//gas constant in Kcal/Kgmole K
+
+//To show the variation of the standard heats of reaction with temperature and the equilibrium constant with temperature graphically in the given temperature range
+//(i)Variation of the standard heats of reaction with temperature
+del_H = (n_SO3*H_SO3)-(n_O2_r*H_O2)-(n_SO2*H_SO2);// in Kcal/Kgmole
+del_F = (n_SO3*F_SO3)-(n_O2_r*F_O2)-(n_SO2*F_SO2);// in Kcal/Kgmole
+//From equation 14.10 (page no 301)
+del_a = (n_SO3*a_SO3)-(n_O2_r*a_O2)-(n_SO2*a_SO2);
+del_b = (n_SO3*b_SO3)-(n_O2_r*b_O2)-(n_SO2*b_SO2);
+//In equation 14.11 (page no 302), substituting del_H at
+T = 298;//in deg cel
+I = del_H - del_a*T - (del_b*(T^2)/2);// integrating constant
+mprintf('(i)The standard heat of reaction at any tempperature can be calculated by the relation:');
+mprintf('\n del_Ht = %fT + %fT^2 %f',del_a,del_b/2,I);
+
+//(ii)Variation of the equilibrium constant with temperature
+//K1 = lnKa (say)
+K1 = -del_F/(R*T);
+//From equation 14.42 (page no 316); M1 = M/R (say)
+M1 = K1-(del_a/R)*log(T)-(del_b/(2*R))*T+(I/(T*R));
+//Let us assume the temperature in the range 800K to 1500K as
+Ta = [700 800 825 850 900 1000 1100 1300 1500];
+for i = 1:9
+ Ka(i) = %e^((del_a/R)*log(Ta(i))+(del_b*Ta(i)/(2*R))-(I/(Ta(i)*R))+M1);
+end
+clf;
+plot(Ta,Ka);
+xtitle(" ","Temperature in K","equilibrium constant K");
+mprintf('\n\n(ii)From the graph it can be seen that as temperature increases Ka decreases exponentially,so the reaction is exothermic.');
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.7/Example_14_7.sce b/479/CH14/EX14.7/Example_14_7.sce new file mode 100755 index 000000000..f7ce60dcd --- /dev/null +++ b/479/CH14/EX14.7/Example_14_7.sce @@ -0,0 +1,140 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.7
+clear;
+clc;
+
+//Given
+//SO2 + (1/2)O2 - SO3
+//Basis: 1 Kgmole of SO2
+n_SO2 = 1;// SO2 fed in Kgmole
+n_O2 = n_SO2;//O2 fed in kgmole
+
+//To Calculate the conversion of SO2 to SO3 at 1atm and at various temperature
+//(1)Calculate the conversion of SO2 to SO3
+P = 1;//Pressure in atm
+T = 850;//Temperature in K
+m = 1-1-(1/2);
+//From example 14.6
+Ta = [700 800 825 850 900 1000 1100 1300 1500];
+Ka = [395.40 52.51 34.60 23.44 11.59 3.527 1.48 0.398 0.0016];
+clf;
+xset('window',2);
+plot2d(Ta,Ka,style=3);
+xtitle("Equilibrium constant vs Temperature","Temperature in K","Ka");
+Ka1 = interpln([Ta;Ka],850);
+//Let Nc be the moles of SO3 at equilibrium
+Nc = [0.1 0.2 0.3 0.4 0.5 0.7 0.8 0.9 0.930 0.95 0.98 0.988 0.989 0.9895 0.9897 0.9899 0.9900];
+//From equation 14.49 (page no 320) and using the given data ,we got equation (b) (page no 323)
+for i = 1:17
+ Ka(i) = (((n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i)));
+end
+xset('window',1);
+plot2d(Nc,Ka,style=2);
+xtitle("Equilibrium constant vs Kgmoles of SO3","Kg moles of SO3","Ka");
+Nc1 = interpln([Ka;Nc],Ka1);
+C = Nc1*100/n_SO2;
+mprintf('(1)The conversion of SO2 to SO3 at 1atm and 850K is %f percent',C);
+
+//(2)Calculation of conversion at 1 atm and 850 K under the following conditions
+//(i) Given:
+n_N2 = 3.75;//Kgmoles of N2 fed
+//Let Nc be the moles of SO3 at equilibrium
+Nc = [0.85 0.87 0.90];
+//From equation 14.49 (page no 320) and using the given data ,we got equation (c) (page no 324)
+for i = 1:3
+ Ka2(i) = (((+n_N2+n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i)));
+end
+xset('window',1);
+plot2d(Nc,Ka2,style=5);
+Nc2 = interpln([Ka2';Nc],Ka1);
+C2 = Nc2*100/n_SO2;
+mprintf('\n\n (2)(i)The conversion of SO2 to SO3 at 1 atm and 850 K when inert gas is also added is %f percent',C2);
+
+//(ii)SO3 is also sent along the original feed
+n_SO3 = 1;//Kgmoles of SO3 fed
+//Let Nc be the moles of SO3 at equilibrium
+Nc = [0.80 0.86 0.92];
+//From equation 14.49 (page no 320) and using the given data ,we got equation (d) (page no 326)
+for i = 1:3
+ Ka3(i) = (((+n_SO3+n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*((n_SO3+Nc(i))/(n_SO2-Nc(i)));
+end
+xset('window',1);
+plot2d(Nc,Ka3,style=6);
+Nc3 = interpln([Ka3';Nc],Ka1);
+C3 = Nc3*100/n_SO2;
+mprintf('\n (ii)The conversion of SO2 to SO3 at 1 atm and 850 K when SO3 is also added along the original feed is %f percent',C3);
+
+//(iii)Variation of SO2 to O2 ratio:
+//(a)SO2:O2 = 1:1 ; This has been worked out in part 1
+mprintf('\n (iii)(a)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1:1 is %f percent',C);
+Xc = Nc1/(n_SO2+n_O2-0.5*Nc1);
+
+//(b)SO2:O2 = 1.1:0.5,Now
+n_SO2 = 1.1;//Kgmoles of SO2 fed
+n_O2 = 0.5;//Kgmoles of O2 fed
+//Let Nc be the moles of SO3 at equilibrium
+Nc = [0.9 0.91 0.92];
+//From equation 14.49 (page no 320) and using the given data ,we got equation (e) (page no 327)
+for i = 1:3
+ Ka4(i) = (((n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i)));
+end
+xset('window',1);
+plot2d(Nc,Ka4,style=1);
+Nc4 = interpln([Ka4';Nc],Ka1);
+C4 = Nc4*100/n_SO2;
+mprintf('\n (iii)(b)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1.1:0.5 is %f percent',C4);
+Xc1 = Nc4/(n_SO2+n_O2-0.5*Nc4);
+
+//(c)SO2:O2 = 1:0.5
+n_SO2 = 1;//Kgmoles of SO2 fed
+n_O2 = 0.5;//Kgmoles of O2 fed
+//Let Nc be the moles of SO3 at equilibrium
+Nc = [0.8 0.85 0.86 0.87];
+//From equation (a)
+for i = 1:4
+ Ka5(i) = (((n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i)));
+end
+xset('window',1);
+plot2d(Nc,Ka5,style=4);
+Nc5 = interpln([Ka5';Nc],Ka1);
+C5 = Nc5*100/n_SO2;
+mprintf('\n (iii)(c)The conversion of SO2 to SO3 at 1atm and 850K when SO2:O2 = 1:0.5 is %f percent',C5);
+Xc2 = Nc5/(n_SO2+n_O2-0.5*Nc5);
+
+if(Xc2>Xc) and (Xc2>Xc1)
+ mprintf('\n SO2:O2 = 1:0.5 gives the maximum concentration of SO3 at equilibrium.');
+else
+ if(Xc1>Xc) and (Xc1>Xc2)
+ mprintf('\n SO2:O2 = 1.1:0.5 gives the maximum concentration of SO3 at equilibrium');
+ else
+ if(Xc>Xc1) and (Xc>Xc2)
+ mprintf('\n SO2:O2 = 1:1 gives the maximum concentration of SO3 at equilibrium');
+ end
+ end
+end
+
+//(3)Conversion of SO2 to SO3 at 50 atm and 850 K when SO2:O2 = 1:1
+n_SO2 = 1;//Kgmole of SO2 fed
+n_O2 = 1;//Kgmoles of O2 fed
+P = 50;//Pressure in atm
+//From figure A.2.9
+phi_SO2 = 0.99;
+phi_SO3 = 0.972;
+phi_O2 = 1;
+//From equation 14.48 (page no320), Ka = Ky*(P^m)*K_phi
+K_phi = phi_SO3/(phi_SO2*(phi_O2^2));
+//Let Nc be the moles of SO3 at equilibrium
+Nc = [0.99 0.985 0.97 0.96];
+for i = 1:4
+ Ka6(i) = K_phi*(P^m)*((((n_SO2+n_O2-0.5*Nc(i))/(n_O2-0.5*Nc(i)))^(1/2))*(Nc(i)/(n_SO2-Nc(i))));
+end
+xset('window',1);
+plot2d(Nc,Ka6,style=3);
+Nc6 = interpln([Ka6';Nc],Ka1);
+C = Nc6*100/n_SO2;
+mprintf('\n\n (3)The conversion of SO2 to SO3 at 50atm and 850K when SO2:O2 = 1:1 is %f percent',C);
+legend("1 part","2.(i) part","2.(ii)part","2.(iii).(b)part","2.(iii).(c)part","3 part");
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.8/Example_14_8.sce b/479/CH14/EX14.8/Example_14_8.sce new file mode 100755 index 000000000..b162663c5 --- /dev/null +++ b/479/CH14/EX14.8/Example_14_8.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.8
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and it does not involve any numerical computation.
+//end
\ No newline at end of file diff --git a/479/CH14/EX14.9/Example_14_9.sce b/479/CH14/EX14.9/Example_14_9.sce new file mode 100755 index 000000000..5021dade5 --- /dev/null +++ b/479/CH14/EX14.9/Example_14_9.sce @@ -0,0 +1,57 @@ +//Chemical Engineering Thermodynamics
+//Chapter 14
+//Thermodynamics of Chemical Reactions
+
+//Example 14.9
+clear;
+clc;
+
+//Given
+//2A(g) - B(g)+C(g)
+T = 400;//Temperature in Kelvin
+P = 1;//Pressure in atm
+R = 1.98;//gas constant in Kcal/Kgmole K
+del_Fo = 3000;//Standard free energy of the reaction in Kcal/Kgmole
+n_A = 1;//Kgmoles of A
+n_B = 0.1;//Kgmoles of B
+n_C = 0.1;//Kgmoles of C
+
+//To find out the equilibrium has been attained or not. If not then calculate the equilibrium composition and also find out whether the equilibrium composition will change or not if pressure were increased to 3 atm.
+//(1)To find out the equilibrium has been attained or not. If not then calculate the equilibrium composition
+//P_B/P_C = 1 (Given)
+n_T = n_A+n_B+n_C;//Total Kgmoles of the component
+p_A = (n_A/n_T)*P;//Partial pressure of A
+p_B = (n_B/n_T)*P;//Partial pressure of B
+p_C = (n_C/n_T)*P;//Partial pressure of C
+//Using the relation 14.36 (page no 312)
+del_F = del_Fo + (R*T)*log((p_B*p_C)/(p_A^2));
+if del_F == 0;
+ then mprintf('1.Equilibrium has been attained.');
+else
+ mprintf('1.Equilibrium has not been attained.');
+end
+//Equilibrium composition
+//At equilibrium del_F = 0
+//From equations 14.35(page no 312) and 14.49(page no 320), we got the relations (a),(b)(page no 331) &(c)(page no 332) and
+//ln(p_C^2/(P-(2*p_C^2)))= -del_Fo/(R*T);
+Kp = %e^(-del_Fo/(R*T))//equilibrium constant in terms of pressure
+p_C = (Kp^(1/2))/(1+2*(Kp^(1/2)));//Partial pressure of C at equilibrium in atm
+p_B = p_C;//as p_B/p_C = 1
+p_A = P-(2*p_C);//Partial pressure of A at equilibrium in atm
+N_A = p_A*P*100;
+N_B = p_B*P*100;
+N_C = p_C*P*100;
+mprintf('\n Equilibrium composition of A,B and C are %f,%f and %f mole percent respectively.',N_A,N_B,N_C);
+
+//2.Calculation of composition at 400 K and 3 atm and vapour pressure of C at 400 K is 0.3 atm
+P = 3;//Pressure in atm
+P_C = 0.3;//Vapour pressure of C in atm
+//Since m=0,pressure will not have an effect on equilibrium compositions.
+mprintf('\n\n 2.The compositions will be the same as above on increasing the pressure.');
+p_B1 = p_B*P;//Partial pressure of B in atm
+p_C1 = p_B1;//Partial pressure of C in atm
+p_A1 = p_A*P;//Partial pressure of A in atm
+if P_C < p_C1
+p_A2 = ((p_B1*P_C)/Kp)^(1/2);//decreased partial pressure of A in atm
+mprintf('\n The partial pressure of A drops from %f to %f to compensate for the amount of C condensed.',p_A1,p_A2);
+//end
diff --git a/479/CH15/EX15.1/Example_15_1.sce b/479/CH15/EX15.1/Example_15_1.sce new file mode 100755 index 000000000..d0704423f --- /dev/null +++ b/479/CH15/EX15.1/Example_15_1.sce @@ -0,0 +1,24 @@ +//Chemical Engineering Thermodynamics
+//Chapter 15
+//Fuel Cells
+
+//Example 15.1
+clear;
+clc;
+
+//Given
+del_F = -56.29;//Standard free energy change in Kcal/Kgmole
+del_H = -68.317;//Standard heat of reaction in Kcal/kgmole
+F = 23.06;//Electro-chemical equivalent in Kcal/volt
+J = 2;//Valance for H2
+
+//To Calculate the emf of the cell, cell efficiency and heat to be removed to maintain isothermal conditions
+//Basis: 1 Kgmole of H2
+//From equation 15.4 (page no 355)
+E = -del_F/(F*J);
+mprintf('1.The emf of the cell is %f volt.',E);
+n = del_F/del_H*100;
+mprintf('\n 2.The cell efficiency is %f percent.',n);
+Q = del_H-del_F;
+mprintf('\n 3.The heat to be removed is %f Kcal to maintain the temperature at 25 degree celsius.',Q);
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.1/Example_2_1.sce b/479/CH2/EX2.1/Example_2_1.sce new file mode 100755 index 000000000..2f010bdf9 --- /dev/null +++ b/479/CH2/EX2.1/Example_2_1.sce @@ -0,0 +1,18 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.1
+clear;
+clc;
+
+//Given
+m = 140;//m is the mass of N2 in Kg
+P = 4.052*(10^5);//P is the pressure of the system in Pa
+V = 30;//V is the volume of the system in m^3
+R = 8314.4;// R is the gas constant
+
+//To determine temperature required
+T = P*V/((m/28)*R);//T is the temperature of the system in K
+mprintf('Temperature of the system is %f K',T);
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.10/Example_2_10.sce b/479/CH2/EX2.10/Example_2_10.sce new file mode 100755 index 000000000..a135dcfee --- /dev/null +++ b/479/CH2/EX2.10/Example_2_10.sce @@ -0,0 +1,57 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.10
+clear;
+clc;
+
+//Given
+yN2 = 1/4;//mole faction of N2 in the mixture
+yH2 = 3/4;//mole fraction of H2 in the mixture
+V = 5.7;//V is the rate at which mixture enters in m^3 in 1 hour
+P = 600;//P is in atm
+T = 298;//T is in K
+TcN2 = 126;//critical temp of N2 in K
+TcH2 = 33.3;//critical temp of H2 in K
+TcNh3 = 406.0;//critical temp of NH3 in K
+PcN2 = 33.5;//critical pressure of N2 in atm
+PcH2 = 12.8;//critical pressure of H2 in atm
+PcNH3 = 111.0;//critical pressure of NH3 in atm
+R = 0.082;//gas constant
+
+//To calculate the amount of ammonia leaving the reactor and the velocity of gaseous product leaving the reactor
+//(i)Calculation of amount of NH3 leaving the reactor
+Tcm = (TcN2*yN2)+(TcH2*yH2);//critical temperature of the mixture
+Pcm = (PcN2*yN2)+(PcH2*yH2);//critical pressure of the mixture
+Trm = T/Tcm;
+Prm = P/Pcm;
+//From figure A.2.3
+Zm = 1.57;//compressibility factor of the mixture
+N = (P*V)/(Zm*R*T);//Kg mole of the mixture
+N1 = 0.25*N;//Kg mole of N2 in feed
+//N2+3H2 - 2NH3
+W = 2*0.15*N1*17;
+mprintf('(i)Ammonia formed per hour is %f Kg',W);
+
+//(ii)Calculation of velocity
+N1 = 0.25*N-(0.25*N*0.15);//Kg mole of N2 after reactor
+N2 = 0.75*N-(0.75*N*0.15);//Kg mole of H2 after reactor
+N3 = 0.25*N*2*0.15;//Kg mole of NH3 after reactor
+Nt = N1+N2+N3;//total Kg moles after reactor
+y1NH3 = N3/Nt;//mole fraction of NH3 after reactor
+y1N2 = N1/Nt;//mole fraction of N2 after reactor
+y1H2 = N2/Nt;//mole fraction of H2 after reactor
+T1cm = (TcN2*y1N2)+(TcH2*y1H2);
+P1cm = (PcN2*y1N2)+(PcH2*y1H2);
+T1 = 448;//in K
+P1 = 550;//in atm
+T1rm = T1/T1cm;
+P1rm = P1/P1cm;
+//From Figure A.2.2
+Zm1 = 1.38;
+V1 = (Zm1*Nt*R*T1)/P1;
+d = 5*(10^-2);//diameter of pipe
+v = V1/((%pi/4)*(d^2)*3600);
+mprintf('\n (ii)Velocity in pipe is %f m/sec',v);
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.2/Example_2_2.sce b/479/CH2/EX2.2/Example_2_2.sce new file mode 100755 index 000000000..e01570f74 --- /dev/null +++ b/479/CH2/EX2.2/Example_2_2.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.2
+clear;
+clc;
+
+//Given
+//This example is a theoretical problem and does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.3/Example_2_3.sce b/479/CH2/EX2.3/Example_2_3.sce new file mode 100755 index 000000000..10f075406 --- /dev/null +++ b/479/CH2/EX2.3/Example_2_3.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.3
+clear;
+clc;
+
+//Given
+//This example is a theoretical problem and does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.4/Example_2_4.sce b/479/CH2/EX2.4/Example_2_4.sce new file mode 100755 index 000000000..b94ffe17f --- /dev/null +++ b/479/CH2/EX2.4/Example_2_4.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.4
+clear;
+clc;
+
+//Given
+//This example is a theoretical problem and does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.5/Example_2_5.sce b/479/CH2/EX2.5/Example_2_5.sce new file mode 100755 index 000000000..728348458 --- /dev/null +++ b/479/CH2/EX2.5/Example_2_5.sce @@ -0,0 +1,53 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.5
+clear;
+clc;
+
+//Given
+n = 1;//n is Kg moles of methane
+T = 423;//T is the temperatue of the system in kelvin
+P = 100;//P is the pressure of the system in atm
+Tc = 191;//Tc is the critical temperature of the system in K
+Pc = 45.8;//Pc is the critical pressure of the system in atm
+R = 0.08206;//R is the gas constant in (m^3 atm/Kg mole K)
+
+//To calculate the volume of methane
+//(i)Using ideal gas equation
+V1 = (n*R*T)/P;//V1 is the volume of the gas in m^3
+mprintf('(i)Volume of the gas using ideal gas equation is %f cubic meter',V1);
+
+//(ii)Using Vander Waals' equation
+a = (27*(R^2)*(Tc^2))/(64*Pc);//Vander Waais constant
+b = (R*Tc)/(8*Pc);//Vander Waais constant
+v = poly(0, 'v');
+q = -((a*b)/P)+(a/P)*v-(((R*T)+(b*P))/P)*v^2+(v^3);//According to Vander Waals equation
+r = roots(q);
+mprintf('\n (ii)Volume of the gas using Vander Waals equation is %f cubic meter',r(1));
+
+//(iii)Using generalized Z chart
+Tr = T/Tc;//Tr is the reduced temperatue
+Pr = P/Pc;//Pr is the reduced pressure
+//From the figure A.2.2,
+Z = 0.97;//Z is the compressibility factor
+V = (Z*R*T)/P;
+mprintf('\n (iii)Volume of the gas using Z chart is %f cubic meter',V);
+
+//(iv)Using molar polarisation method
+//From Table 2.2
+Pmc = 6.82;//Pmc is the molar polarisation for methane
+//From figure A.2.4
+Z0 = .965;
+Z1 = 14.8*(10^-4);
+Z = Z0+(Z1*Pmc);
+V = (Z*R*T)/P;
+mprintf('\n (iv)Volume of the gas using molar polarisation method is %f cubic meter',V);
+
+//(v)From experiment
+//Given
+Z = 0.9848;
+V = (0.9848*n*R*T)/P;
+mprintf('\n (v)Volume of the gas calculated by experimental Z value is %f cubic meter',V);
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.6/Example_2_6.sce b/479/CH2/EX2.6/Example_2_6.sce new file mode 100755 index 000000000..6de8f8e07 --- /dev/null +++ b/479/CH2/EX2.6/Example_2_6.sce @@ -0,0 +1,49 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.6
+clear;
+clc;
+
+//Given
+P1 = 266;
+T1 = 473.16;//Initial temperature in Kelvin
+T2 = 273.16;//Final temperature in Kelvin
+V1 = 80; V2 = 80;//Initial & final volume in litres
+N1 = (14.28/28); N2 = (14.28/28);//Initial and final Kg moles are equal
+Tc = 126;//Critical temperature of N2 in K
+Pc = 33.5;//Critical pressure of N2 in atm
+
+//To calculate the final pressure achieved
+//(i)Using ideal gas law
+p2 = (P1*V1*N2*T2)/(V2*N1*T1);
+mprintf('(i)Final pressure of N2 using ideal gas law is %f atm',p2);
+
+//(ii)Using generalized Z chart
+Tr1 = T1/Tc;//reduced initial temp in k
+Pr1 = P1/Pc;//reduced initial press in K
+//From the Z-chart compressibility factor coressponding to the above Tr1 &Pr1 is
+Z1 = 1.07;
+P2 = [125,135,150];
+Z2 = [0.95, 0.96, 0.98];
+F = [0,0,0];
+for i = 1:3
+ F(i) = (P2(i)/(Z2(i)*T2))-(P1/(Z1*T1));
+end
+clf;
+plot(P2,F);
+xtitle("P2 vs F","P2","F");
+P3 = interpln([F;P2],0);
+mprintf('\n (ii)Final pressure of N2 from Z chart is %f atm',P3);
+
+//(iii)Using Pseudo reduced density chart
+R = 0.082;//gas constant
+v = V1/N1;//Volume per moles of nitrogen in m^3/Kg mole
+Dr = (R*Tc)/(Pc*v);
+Tr2 = T2/Tc;//final reduced temp in K
+//From figure A.2.1, reduced pressure coressponding to this Dr and Tr2 is
+Pr2 = 4.1//final reduced pressure in atm
+p2_ = Pr2*Pc;
+mprintf('\n (iii)Final pressure achieved using Dr chart is %f atm',p2_);
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.7/Example_2_7.sce b/479/CH2/EX2.7/Example_2_7.sce new file mode 100755 index 000000000..fd7d89932 --- /dev/null +++ b/479/CH2/EX2.7/Example_2_7.sce @@ -0,0 +1,31 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.7
+clear;
+clc;
+
+//Given
+n = 1;//n is the Kg mole of methane gas
+T = 298;//T is the constant temperature in K
+P1 = 1;//P1 is the initial pressure of the system
+P2 = 100;//P2 is the final pressure of the system
+R = 8314.4;//R is the gas constant in Nm/Kgmole deg K
+
+//To compute the work required
+//(i)Using ideal gas law
+W = R*T*log(P1/P2);
+mprintf('(i)Work done by the system if the gas obeys ideal gas law is %4.2e Nm',W);
+
+//(ii)Using Vander Waals' equation
+//Given
+//For methane
+a = 2.32*(10^5);//Vander Wals' constant a in N/m^2
+b = 0.0428;//Vanderwaals' constant b in m^3
+//V1 and V2 are evaluated by trial and error using Vanderwaals' equation as P1 and P2 are known
+V1 = 11.1;//initial volume of the gas in m^3
+V2 = 0.089;//final volume of the gas in m^3
+W = (R*T*log((V2-b)/(V1-b)))+(a*((1/V2)-(1/V1)))
+mprintf('\n (ii)Work done by the system if the gas obeys Vander Waals equation is %4.2e Nm',W);
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.8/Example_2_8.sce b/479/CH2/EX2.8/Example_2_8.sce new file mode 100755 index 000000000..22019a4ec --- /dev/null +++ b/479/CH2/EX2.8/Example_2_8.sce @@ -0,0 +1,63 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.8
+clear;
+clc;
+
+//Given
+V = 27*(10^-3);//Volume of the container in m^3
+n = (15/70.91);//n is the Kg moles of chlorine
+T = 293;//T is the temperature in K
+R = 0.08206;
+P = 10^(4.39-(1045/293));//P is the vapour pressure of chlorine
+Pc = 76.1;//Critical pressure of Chlorine
+Tc = 417;//Critical temperature of Chlorine
+Pr = P/Pc;//Reduced pressure of Chlorine
+Tr = T/Tc;//Critical temperature of Chlorine
+M = 70.91;//Molecular weight of the Chlorine
+
+//To determine the vapour pressure of chlorine, amount of liquid Cl2 and temperature required
+//(i)Specific volume of liquid Chlorine
+//From figure A.2.2
+Zg = 0.93;
+//From figure A.2.6
+Zl = 0.013;
+vl = ((Zl*R*T)/P);
+mprintf('(i)Specific volume of liquid Chlorine from compressibility chart is %f cubic meter /Kgmole',vl);
+
+//From Francis relation, taking the constants from Table 2.3
+D = (1.606-(216*(10^-5)*20)-(28/(200-20)))*10^3;//Density of liq Cl2 in Kg/m^3
+Vl = M/D;
+mprintf('\n Specific volume of liquid Chlorine from Francis relation is %f cubic meter /Kgmole',Vl);
+
+//(ii)Amount of liquid Cl2 present in the cylinder
+vg = ((Zg*R*T)/P);
+V1 = V-vg;//V1 is the volume of liquid Chlorine
+Vct = 0.027;//volume of the container
+Vg = (0.212-(Vct/vl))/((1/vg)-(1/vl));//By material balance
+W = ((V-Vg)*70.9)/vl;
+mprintf('\n\n (ii)Weight of Chlorine at 20deg cel is %f Kg',W);
+
+//(iii)Calculation of temperature required to evaporate all the liquid chlorine
+//log P' = 4.39 - 1045/T (given)
+//Assume the various temperature
+Ng = 0.212;//total Kg moles of gas
+Ta = [413 415 417];
+N = [0,0,0];
+for i = 1:3
+ Tr(i) = Ta(i)/Tc;//reduced temperature in K
+ P(i) = 10^(4.39-(1045/Ta(i)));
+ Pr(i) = P(i)/Pc;//reduced pressure in K
+//From the compressibility factor chart,Z values coressponding to the above Tr &Pr are given as
+Z = [0.4 0.328 0.208];
+N(i) = (P(i)*Vct)/(Z(i)*R*Ta(i));
+end
+
+clf;
+plot(N,Ta);
+xtitle("Ta vs N","N","Ta");
+T1 = interpln([N;Ta],0.212);//in K
+mprintf('\n (iii)The temperature required to evaporate all the liquid chlorine is %f deg celsius',T1-273);
+//end
\ No newline at end of file diff --git a/479/CH2/EX2.9/Example_2_9.sce b/479/CH2/EX2.9/Example_2_9.sce new file mode 100755 index 000000000..45519a7c2 --- /dev/null +++ b/479/CH2/EX2.9/Example_2_9.sce @@ -0,0 +1,81 @@ +//Chemical Engineering Thermodynamics
+//Chapter 2
+//P-V-T Relations
+
+//Example 2.9
+clear;
+clc;
+
+//Given
+N1 = 0.7;//Kg mole of CH4
+N2 = 0.3;//Kg mole of N2
+R = 0.08206;//Gas constant
+T = 323;//Temperature in Kelvin
+V = 0.04;//Volume in m^3
+a1 = 2.280; b1 = 0.0428;//Vanderwaals constants for CH4
+a2 = 1.345;b2 = 0.0386;//Vanderwaals constants for N2
+Tc1 = 191; Pc1 = 45.8;//Critical temperature in K and pressure of CH4 in atm
+Tc2 = 126;Pc2 = 33.5;//Critical temperature in K and pressure of N2 in atm
+
+//To find Approx Value
+function[A]=approx(V,n)
+ A=round(V*10^n)/10^n;//V-Value n-To what place
+ funcprot(0)
+endfunction
+
+//To calculate the pressure exerted by the gas mixture
+//(i)Using ideal gas law
+P = (N1+N2)*((R*T)/V);
+mprintf('(i) Pressure exerted by the gas mixture using ideal gas law is %d atm',P);
+
+//(ii)Using Vander waal equation
+P1 = ((N1*R*T)/(V-(N1*b1)))-((a1*(N1^2))/(V^2));//Partial pressure of CH4
+P2 = ((N2*R*T)/(V-(N2*b2)))-((a2*(N2^2))/(V^2));//Partial pressure of N2
+Pt = P1+P2;
+mprintf('\n(ii) Pressure exerted by the gas mixture using Vander waal equation is %f atm', Pt);
+
+//(iii)Using Zchart and Dalton's law
+Tra = T/Tc1;//reduced temperature of CH4
+Trb = T/Tc2;//reduced temperature of N2
+//Asssume the pressure
+P = [660 732 793 815 831];
+for i = 1:5
+ Pa(i) = N1*P(i);// partial pressure of CH4 for the ith total pressure
+ Pb(i) = N2*P(i);// partial pressure of N2 for the ith total pressure
+ Pra(i) = Pa(i)/Pc1;//reduced pressure of CH4 for the ith total pressure
+ Prb(i) = Pb(i)/Pc2;//reduced pressure of N2 for the ith total pressure
+end
+
+//For the above Pr and Tr values compressibility factors from the figure A.2.3 are given as
+Za = [1.154 1.280 1.331 1.370 1.390];//Z values of CH4
+Zb = [1 1 1 1 1];//Z values of N2
+V3 = 0.0421;
+for i = 1:5
+ Pa(i) = Za(i)*N1*((R*T)/V);//partial pressure of CH4 coressponding to the ith total presure
+ Pb(i) = Zb(i)*N2*((R*T)/V);//partial pressure of N2 coressponding to the ith total pressure
+ Pt(i) = Pa(i)+Pb(i);//total pressure of the gas mixture
+ if Pt(i)-P(i) < 15
+ mprintf('\n(iii) pressure exerted by the gas mixture using Z chart and Dalton Law is %d atm',Pt(i));
+ else
+ end
+end
+
+//(iv)Using Amagat's law and Z chart
+P = [1000 1200 1500 1700];
+for i=1:4
+ Pra(i) = P(i)/Pc1;
+ Prb(i) = P(i)/Pc2;
+end
+//For the above Pr and Tr values compressibility factors from the figure A.2.3 are given as
+Za = [1.87 2.14 2.52 2.77];
+Zb = [1.80 2.10 2.37 2.54];
+for i = 1:4
+ Va(i) = approx((N1*Za(i)*((R*T)/P(i))),4);
+ Vb(i) = approx((N2*Zb(i)*((R*T)/P(i))),4);
+ V1(i) = approx((Va(i)+Vb(i)),4);
+ if V1(i)-V <= 0.003
+ mprintf('\n(iv) Pressure exerted by the gas mixture using Amagat law and Zchart is %d atm ',P(i));
+ else
+end
+end
+//end
diff --git a/479/CH2/EX2.9/Example_2_9.txt b/479/CH2/EX2.9/Example_2_9.txt new file mode 100755 index 000000000..791f69feb --- /dev/null +++ b/479/CH2/EX2.9/Example_2_9.txt @@ -0,0 +1,4 @@ +For example 2.9,
+In the (iii) part, i am getting answer as P = 843 atm.
+But the answer given in the book is 833 atm.
+There might be some calculation mistke in the book.
\ No newline at end of file diff --git a/479/CH3/EX3.1/Example_3_1.sce b/479/CH3/EX3.1/Example_3_1.sce new file mode 100755 index 000000000..e9b3d2382 --- /dev/null +++ b/479/CH3/EX3.1/Example_3_1.sce @@ -0,0 +1,17 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.1
+clear;
+clc;
+
+//Given
+W = -((2*745.6*(10^-3)/4.18)*3600);//work added to the system in Kcal/hr
+m = 10;//Amount of fluid in tank in Kg
+Q = -378;//Heat losses from the system in Kcal/hr
+
+//To calculate the change in internal energy
+delE=(Q-W)/m;// Change in internal energy in Kcal/hr kg
+mprintf('Change in Internal energy is %f Kcal/hr Kg',delE);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.10/Example_3_10.sce b/479/CH3/EX3.10/Example_3_10.sce new file mode 100755 index 000000000..c998d9d00 --- /dev/null +++ b/479/CH3/EX3.10/Example_3_10.sce @@ -0,0 +1,26 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.10
+clear;
+clc;
+
+//Given
+V = 0.3;//Volume of the tank in m^3
+P1 = 1;//Initial pressure of the tank in atm
+P2 = 0;//Final pressure of the tank in atm
+T = 298;//Temperature of the tank in K
+t = 10;//evacuation time in min
+
+//delN=(V/(R*T)*delP)..(a) change in moles as V and T are constant
+//delW=delN*R*T*lnP..(b)pump work required
+//From (a)&(b),delW=V*delP*lnP
+
+//To calculate the pump work required
+//On doing integration of dW we will get
+
+W = V*(P1-P2);//pump work done in J/sec
+W1=(W*(1.033*10^4))/(75*600);
+mprintf('The pump work required is %f hp',W1);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.11/Example_3_11.sce b/479/CH3/EX3.11/Example_3_11.sce new file mode 100755 index 000000000..3192a9b75 --- /dev/null +++ b/479/CH3/EX3.11/Example_3_11.sce @@ -0,0 +1,20 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.11
+clear;
+clc;
+
+//Given
+H1 = 680.6;//Enthalpy of entering steam at 6Kgf/cm^2 &200 deg cel in Kcal/Kg
+u1 = 60;//velocity at which steam entered the nozzle in m/sec
+u2 = 600;//velocity at which steam left the nozzle in m/sec
+g = 9.8;
+Hg = 642.8; Hlq = 110.2;//Enthalpy of saturated vapour & saturated liquid at 1.46 Kgf/cm^2 respectively
+
+//To calculate the quality of exit steam
+H2 = H1+((u1^2)-(u2^2))/(2*g*427);//enthalpy of leaving steam in Kcal/Kg
+x = (H2-Hlq)/(Hg-Hlq);
+mprintf('The quality of exit steam is %f percent',x*100);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.12/Example_3_12.sce b/479/CH3/EX3.12/Example_3_12.sce new file mode 100755 index 000000000..a3de9a7fb --- /dev/null +++ b/479/CH3/EX3.12/Example_3_12.sce @@ -0,0 +1,21 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.12
+clear;
+clc;
+
+//Given
+W = 0;//pump work
+Mi = 0;//chamber is initially evacuated
+M2 = 0;//no exist stream
+H1 = 684.2;//enthalpy of steam at 200 deg cel & 3 Kgf/cm^2
+
+//To calculate the internal energy of the steam in the chamber
+//Q=150*m1;.. (a) heat lost from the chamber in Kcal/Kg
+//m1=mf;..(b) mass of steam added from large pipe is equal to steam in chamber
+ //H1*M1-Q=Mf*Ef; using (a)&(b)
+ Ef = H1-150;
+ mprintf('The internal energy of steam in chamber is %f Kcal',Ef);
+ //end
\ No newline at end of file diff --git a/479/CH3/EX3.13/Example_3_13.sce b/479/CH3/EX3.13/Example_3_13.sce new file mode 100755 index 000000000..ff466298f --- /dev/null +++ b/479/CH3/EX3.13/Example_3_13.sce @@ -0,0 +1,35 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.13
+clear;
+clc;
+
+//Given
+//Q=W=delPE=delKE=0;
+//M2=0; no exit stream
+Ti = 288;//initial temperature in K
+H = 7*Ti;//enthalpy of air in Kcal/Kgmole
+Ei = 5*Ti;// initial internal energy of air in Kcal/Kgmole
+//Ef=5*Tf;Final internal energy of air in Kcal/Kgmole
+Pi = 0.3;//initial pressure in atm
+V = 0.57;//volume of the tank in m^3
+R = 848;//gas constant in mKgf/Kg mole K
+Pf = 1;//final prssure in atm
+
+//To calculate the final weight and the final temperature of the air in the tank
+Mi = (Pi*V*1.03*10^4)/(R*Ti);//initial quantity of air in tank in Kg mole
+//Tf=(Pf*V*1.033*10^4)/(Mf*R)..(a) final temperature,Mf=final quantity of air in tank in Kg mole
+//M1=Mf-Mi..(b) M1 is mass of steam added in Kg mole
+//H*M1=(Ef*Mf)-(Ei*Mi)
+//H*M1=((5*Pf*V*1.033*10^4)/(Tf*R))*Tf-(Ei*Mi)...(c)
+A = [1 -1;0 -H];
+B = [Mi;((Ei*Mi)-((5*Pf*V*1.03*10^4)/R))];
+x = A\B;
+Mf = x(1);
+mprintf('The final weight of air in the tank is%f Kg',Mf);
+
+Tf = (Pf*V*1.03*10^4)/(Mf*R);
+mprintf('\n The final temperature of air in the tank is %f K',Tf);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.13/Example_3_13.txt b/479/CH3/EX3.13/Example_3_13.txt new file mode 100755 index 000000000..46ad81a5f --- /dev/null +++ b/479/CH3/EX3.13/Example_3_13.txt @@ -0,0 +1,4 @@ +For example 3.13,
+ the answers are given in the book, Mf = 0.017 Kgmole & Tf = 406K.
+ But i am getting Mf = 0.019232 Kgmole & Tf = 360K.
+ There might be some calculation mistake in the book.
diff --git a/479/CH3/EX3.14/Example_3_14.sce b/479/CH3/EX3.14/Example_3_14.sce new file mode 100755 index 000000000..9c1f0928a --- /dev/null +++ b/479/CH3/EX3.14/Example_3_14.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.14
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and it does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.2/Example_3_2.sce b/479/CH3/EX3.2/Example_3_2.sce new file mode 100755 index 000000000..2e7c14f6b --- /dev/null +++ b/479/CH3/EX3.2/Example_3_2.sce @@ -0,0 +1,19 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.2
+clear;
+clc;
+
+//Given
+n = 1;//kg moles of a gas
+Cv = 5;//specific heat in Kcal/Kgmole
+delT = 15;//increase in temperature in deg celsius
+
+//To calculate the change in internal energy
+Q = n*Cv*delT;//heat given to the system in Kcal
+W = 0;//work done
+delE = Q-W;//Change in internal energy
+mprintf('Change in internal energy is %f Kcal',delE);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.3/Example_3_3.sce b/479/CH3/EX3.3/Example_3_3.sce new file mode 100755 index 000000000..39b2fd9a6 --- /dev/null +++ b/479/CH3/EX3.3/Example_3_3.sce @@ -0,0 +1,27 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.3
+clear;
+clc;
+
+//Given
+P = 1;//constant pressure throughout the process in atm
+T1 = 273;//Initial temperature in K
+T2 = 373;//Final temperature in K
+V1 = 0;//Volume of liquid water or initial volume
+V0 = 22.4;//volume of vapour at standard condition in cubic meter
+Q = 9.7//Heat of vapourisation in Kcal
+
+//To calculate the work done by the expanding gas and increase in internal energy
+//(i)Calculation of work done
+V2 = 22.4*(T2/T1)*(P)*(10^-3);//Volume of final vapour in cubic meter
+w = P*(V2-V1);//Work done in atm cubic meter
+W = w*(1.03*10^4)/427;//Work done in Kcal
+mprintf('(i)Work done by the expanding gas is %f Kcal',W);
+
+//(ii)Calculation of change in internal energy
+delE = Q-W;
+mprintf('\n\n (ii)Increase in internal energy is %f Kcal',delE);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.4/Example_3_4.sce b/479/CH3/EX3.4/Example_3_4.sce new file mode 100755 index 000000000..497c963a7 --- /dev/null +++ b/479/CH3/EX3.4/Example_3_4.sce @@ -0,0 +1,25 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.4
+clear;
+clc;
+
+//Given
+W = 0;//work done during the process
+P1 = 1;//Initial pressure in atm
+P2 = 10;//Final pressure in atm
+V2 = V1;//Initial & final volume are equal
+Cv = 0.23//specific heat at constant volume in Kcal/Kg deg K
+//(delQ/delT)=Q
+Q = 1.3;//Rate of heat addition in Kcal/min
+m = 2.5//Weight of an ideal gas in Kg
+T1 = 298//Initial temperature in Kelvin
+
+//To calculate the time taken for the gas to attain 10 atm
+//Q = m*Cv*(delT/delt)=1.3
+T2 = (P2*T1)/(P1);//Final temperature in Kelvin
+t = ((m*Cv)/1.3)*(T2-T1);//time taken in minutes
+mprintf('The time taken to attain a pressure of 10 atm is %f hrs',t/60);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.5/Example_3_5.sce b/479/CH3/EX3.5/Example_3_5.sce new file mode 100755 index 000000000..c7d74e359 --- /dev/null +++ b/479/CH3/EX3.5/Example_3_5.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.5
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and it does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.6/Example_3_6.sce b/479/CH3/EX3.6/Example_3_6.sce new file mode 100755 index 000000000..f95d8e5f9 --- /dev/null +++ b/479/CH3/EX3.6/Example_3_6.sce @@ -0,0 +1,19 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.6
+clear;
+clc;
+
+//Given
+R = 1.98;//gas constant in kcal/Kgmole deg K
+T = 293;//Temperature in K
+M = 29;//Molecular weight of air
+
+//To calculate the flow work per kg of air
+//W=(P*V)=(R*T)
+W = R*T;//Flow work in Kcal/Kg mole
+W1 = W/M;
+mprintf('Flow work is %f Kcal/Kg',W1);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.7/Example_3_7.sce b/479/CH3/EX3.7/Example_3_7.sce new file mode 100755 index 000000000..81f9e705d --- /dev/null +++ b/479/CH3/EX3.7/Example_3_7.sce @@ -0,0 +1,27 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.7
+clear;
+clc;
+
+//Given
+m = 5000;//Amount of steam recived per hour in Kg
+H1 = 666;//Specific enthalpy when steam entered in the turbine in Kcal/Kg
+H2 = 540;//Specific enthalpy when steam left the turbine in Kcal/Kg
+u1 = 3000/60;//velocity at which steam entered in m/sec
+u2 = 600/60;//velocity at which steam left in m/sec
+Z1 = 5;//height at which steam entered in m
+Z2 = 1;//height at which steam left in m
+Q = -4000;//heat lost in Kcal
+g = 9.81;
+
+//To calculate the horsepuwer output of the turbine
+delH = H2-H1;//change in enthalpy in Kcal
+delKE = ((u2^2)-(u1^2)/(2*g))/(9.8065*427);//change in kinetic energy in Kcal; 1kgf = 9.8065 N
+delPE = ((Z2-Z1)*g)/(9.8065*427);//change in potential energy in Kcal
+W = -(m*(delH+delKE+delPE))+Q;//work delivered in Kcal/hr
+W1 = W*(427/(3600*75));//work delivered by turbine in hp
+mprintf('Work delivered by turbine is %f hp',W1);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.7/Example_3_7.txt b/479/CH3/EX3.7/Example_3_7.txt new file mode 100755 index 000000000..62da72829 --- /dev/null +++ b/479/CH3/EX3.7/Example_3_7.txt @@ -0,0 +1,3 @@ +For Example 3.7,
+the answer given in the book is 999 hp but i am getting 990.13329 hp.
+There might be some calculation mistake in the book.
\ No newline at end of file diff --git a/479/CH3/EX3.8/Example_3_8.sce b/479/CH3/EX3.8/Example_3_8.sce new file mode 100755 index 000000000..87e846a6a --- /dev/null +++ b/479/CH3/EX3.8/Example_3_8.sce @@ -0,0 +1,27 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.8
+clear;
+clc;
+
+//Given
+m = 183;//rate of water flow in Kg/min
+H1 = 95;//enthalpy of storage tank 1 in Kcal/Kg
+h = 15;//height difference between two storage tanks in m
+Q = -10100;//extraced heat from storage tank 1 in a heat exchanger in Kcal/min
+W = -2;//work delivered by motor in hp
+
+// To find out the enthalpy of water tank2 and the temperature of water in the second tank
+delPE = h/427;//change in potential energy in Kcal/Kg
+delKE = 0;//change in kinetic energy
+W1 = W*(75/427);//work delivered by motor in Kcal/sec
+W2 = W1*60;//work delivered by motor in Kcal/min
+H2 = ((Q+W2)/m)-delKE-delPE+H1;//enthalpy of storage tank 2 in Kcal/Kg
+mprintf('The enthalpy of storage tank 2 is %f Kcal/Kg',H2);
+
+//The enthalpy H2=39.66 corresponds to the temperature T according to steam table
+T=40;//Temperature is in deg celsius
+mprintf('\n The temperature of water in the second tank is %d deg celsius',T);
+//end
\ No newline at end of file diff --git a/479/CH3/EX3.9/Example_3_9.sce b/479/CH3/EX3.9/Example_3_9.sce new file mode 100755 index 000000000..009a49e57 --- /dev/null +++ b/479/CH3/EX3.9/Example_3_9.sce @@ -0,0 +1,24 @@ +//Chemical Engineering Thermodynamics
+//Chapter 3
+//First Law of Thermodynamics
+
+//Example 3.9
+clear;
+clc;
+
+//To calculate the mass of steam required
+//Given
+m2 = 100;//mass of water to be heated
+//From diagram,
+//m3 = m1+m2;..(a)
+//Hs = H1;..(b) since throttling is a constant enthalpy process
+//m3*H3-(m1*H1+m2*H2)=0;..(c) since delH=0
+
+//From steam tables,
+Hs = 681.7//enthalpy of steam at 200 deg cel bleeded at the rate of 5Kgf/(cm^2) in Kcal/Kg
+H2 = 5.03;//enthalpy of liquid water at 5 deg cel
+H3 = 64.98;//enthalpy of liquid water at 65 deg cel
+//from equn (a),(b)&(c);(page no 80)
+m1 = ((H3-H2)/(Hs-H3))*m2;//mass of steam required in Kg (page no 80)
+mprintf('The mass of steam required to heat 100 Kg of water is %f Kg',m1);
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.1/Example_4_1.sce b/479/CH4/EX4.1/Example_4_1.sce new file mode 100755 index 000000000..a33d16f7d --- /dev/null +++ b/479/CH4/EX4.1/Example_4_1.sce @@ -0,0 +1,32 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.1
+clear;
+clc;
+
+//Given
+Q1 = 250;//Heat absorbed in Kcal
+T1 = (260+273);//Temperature at which engine absorbs heat
+T0 = (40+273);//Temperature at which engine discards heat
+//To Calculate work output, heat rejected, entropy change of system,surronding & total change in entropy and the efficiency of the heat engine
+
+//(i)Calculation of work output
+W = (Q1*((T1-T0)/T1));//Work done using equations 4.7 & 4.9 given on page no 98
+mprintf('(i)The work output of the heat engine is %f Kcal',W);
+
+//(ii)Calculation of heat rejected
+Q2 = (Q1*T0)/T1;
+mprintf('\n (ii)The heat rejected is %f Kcal',Q2);
+
+//(iii)Calculation of entropy
+del_S1 = -(Q1/T1);//Change in the entropy of source in Kcal/Kg K
+del_S2 = Q2/T0;//Change in the entropy of sink in Kcal/Kg K
+del_St = del_S1+del_S2;//Total change in entropy in Kcal/Kg K
+mprintf('\n (iii)Total change in entropy is %d confirming that the process is reversible',del_St);
+
+//(iv)Calculation of efficiency
+n = (W/Q1)*100;
+mprintf('\n (iv)The efficiency of the heat engine is %f percent',n);
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.10/Example_4_10.sce b/479/CH4/EX4.10/Example_4_10.sce new file mode 100755 index 000000000..65432306e --- /dev/null +++ b/479/CH4/EX4.10/Example_4_10.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.10
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and it does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.2/Example_4_2.sce b/479/CH4/EX4.2/Example_4_2.sce new file mode 100755 index 000000000..013071aa1 --- /dev/null +++ b/479/CH4/EX4.2/Example_4_2.sce @@ -0,0 +1,31 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.2
+clear;
+clc;
+
+//Given
+T1 = 373;//Temperature of the saturated steam in K
+T2 = 298;//Temperature of the saturated water in K
+//To calculate the total change in entropy and hence determine the reversibility of the process
+
+//del_H = del_Q+(V*del_P)
+//del_H =del_Q; since it is a constant pressure process
+
+//From steam table,
+//enthalpy of saturated steam at 373K is
+H1 = 6348.5;// in Kcal/Kg
+//enthalpy of saturated liquid water at 373K is
+H2 = 99.15;//in Kcal/Kg
+Q = H2-H1;//heat rejected in Kcal/Kg
+del_S1 = Q/T1;//change in entropy of the system in Kcal/Kg K
+del_S2 = Q/T2;//change in entropy of the surronding in Kcal/Kg K
+del_St = del_S1+del_S2;//total change in the entropy in Kcal/Kg K
+if(del_St == 0)
+ mprintf('Process is reversible');
+else
+ mprintf('Process is irreversible');
+end
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.3/Example_4_3.sce b/479/CH4/EX4.3/Example_4_3.sce new file mode 100755 index 000000000..f72a688e8 --- /dev/null +++ b/479/CH4/EX4.3/Example_4_3.sce @@ -0,0 +1,34 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.3
+clear;
+clc;
+
+//Given
+Cp = 0.09;//specific heat of metal block in Kcal/Kg K
+m = 10;//mass of metal block in Kg
+T1 = 323;//initial temperature of the block in K
+T2 = 298;//final temperature of the block in K
+//constant pressure process
+//To find out entropy change of block,air and total entropy change
+
+//(i)To calculate the entropy change of block
+del_S1 = m*Cp*log(T2/T1);
+mprintf('(i)Entropy change of block is %f Kcal/Kg K',del_S1);
+
+//(ii)To calculate the entropy change of air
+Q = m*Cp*(T1-T2);//heat absorbed by air = heat rejected by block in Kcal
+del_S2 = (Q/T2);
+mprintf('\n (ii)Entropy change of air is %f Kcal/Kg K',del_S2);
+
+//(iii)To calculate the total entropy change
+del_St = del_S1+del_S2;
+mprintf('\n (iii)Total entropy change is %f Kcal/Kg K',del_St);
+if(del_St == 0)
+ mprintf('\n Process is reversible');
+else
+ mprintf('\n Process is irreversible');
+end
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.4/Example__4_4.sce b/479/CH4/EX4.4/Example__4_4.sce new file mode 100755 index 000000000..8ad2ad919 --- /dev/null +++ b/479/CH4/EX4.4/Example__4_4.sce @@ -0,0 +1,25 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.4
+clear;
+clc;
+
+//Given
+m1 = 10;//mass of metal block in Kg
+m2 = 50;//mass of water in Kg
+Cp1 = 0.09;//Specific heat of metal block in Kcal/Kg K
+Cp2 = 1;//Specific heat of water in Kcal/Kg K
+T1 = 50;//Initial temperature of block in deg celsius
+T2 = 25;//Final temperature of block in deg celsius
+
+//To calculate the total change in entropy
+//Heat lost by block = Heat gained by water
+Tf = ((m1*Cp1*T1)+(m2*Cp2*T2))/((m1*Cp1)+(m2*Cp2));//final temperature of water in deg celsius
+Tf1 = Tf+273.16;//final temperature in K
+del_S1 = m1*Cp1*log(Tf1/(T1+273));//change in entropy of the block in Kcal/K
+del_S2 = m2*Cp2*log(Tf1/(T2+273));//change in entropy of the block in Kcal/K
+del_St = del_S1+del_S2;
+mprintf('The total change entropy is %f Kcal/K',del_St);
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.5/Example_4_5.sce b/479/CH4/EX4.5/Example_4_5.sce new file mode 100755 index 000000000..c5cd1b51b --- /dev/null +++ b/479/CH4/EX4.5/Example_4_5.sce @@ -0,0 +1,21 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.5
+clear;
+clc;
+
+//Given
+//Air at 20 deg celsius
+//P1 = 250;initial pressure in atm
+//P2 = 10;final pressure after throttling in atm
+
+//To calculate the entropy change
+//According to the given conditions from figure4.5(page no 103)
+S1 = -0.38;//initial entropy in Kcal/Kg K
+S2 = -0.15;//final entroy in Kcal/Kg K
+del_S = S2-S1;
+mprintf('Change in entropy for the throttling process is %f Kcal/Kg K',del_S);
+//From figure 4.6(page no 104), the final temperature is -10 deg celsius
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.6/Example_4_6.sce b/479/CH4/EX4.6/Example_4_6.sce new file mode 100755 index 000000000..38c7e19c9 --- /dev/null +++ b/479/CH4/EX4.6/Example_4_6.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.6
+clear;
+clc;
+
+//Given
+//The given problem does not involve any numerical comptation
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.7/Example_4_7.sce b/479/CH4/EX4.7/Example_4_7.sce new file mode 100755 index 000000000..c3fe7aebb --- /dev/null +++ b/479/CH4/EX4.7/Example_4_7.sce @@ -0,0 +1,23 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.7
+clear;
+clc;
+
+//Given
+//Basis: 1 hour
+m = 10;//mass of air in Kg
+T = 293;//Constant temperature throughout the process in K
+//P1 = 1;//Initial pressure in atm
+//P2 = 30;//Final pressure in atm
+//According to the given data and using the graph or figure A.2.7 given in page no 105
+S1 = 0.02;//Initial entropy in Kcal/Kg
+S2 = -0.23;//Final entropy in Kcal/Kg
+H1 = 5;//Initial enthalpy in Kcal/Kg
+H2 = 3;//Final enthalpy in Kcal/Kg
+
+W = -((H2-H1)+T*(S2-S1))*m*(427/(3600*75));
+mprintf('The horse power of the compressor is %f hp',W);
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.8/Example_4_8.sce b/479/CH4/EX4.8/Example_4_8.sce new file mode 100755 index 000000000..8164a0694 --- /dev/null +++ b/479/CH4/EX4.8/Example_4_8.sce @@ -0,0 +1,28 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.1
+clear;
+clc;
+
+//Given
+//Basis: 1 Kg of steam
+//P1 = 30;Intial pressure in Kgf/cm^2
+//P2 = 3;Final pressure in Kgf/cm^2
+//T = 300;//Operating temperature
+//From figure A.2.8,
+H1 = 715;//Initial enthalpy of steam in Kcal/Kg
+H2 = 625;//Final enthalpy of steam in Kcal/Kg
+S1 = 1.56;//Initial entropy of steam in Kcal/Kg K
+S2 = 1.61;//Final entropy of steam in Kcal/Kg K
+Q = -1;//heat loss in Kcal/Kg
+To = 298;//The lowest surronding temperature in K
+
+//To calculate the effectiveness of the process
+W = (-(H2-H1)+Q);//Actual work output by the turbine in Kcal
+//The maximum or available work can be calculated from equation 4.14
+del_B = -((H2-H1)-(To*(S2-S1)));// Maximum work that can be obtained in Kcal
+E = (W/del_B)*100;
+mprintf('The effectiveness of the process is %f percent',E);
+//end
\ No newline at end of file diff --git a/479/CH4/EX4.9/Example_4_9.sce b/479/CH4/EX4.9/Example_4_9.sce new file mode 100755 index 000000000..9f32b8e39 --- /dev/null +++ b/479/CH4/EX4.9/Example_4_9.sce @@ -0,0 +1,29 @@ +//Chemical Engineering Thermodynamics
+//Chapter 4
+//Second Law of Thermodynamics
+
+//Example 4.9
+clear;
+clc;
+
+//Given
+m = 1;//mass of liquid water in Kg
+T1 = 1350;//initial temperature in deg celsius
+T2 = 400;//final temperature in deg celsius
+Cp = 1;//Specific heat of water in Kcal/Kg K
+Cpg = 0.2;//Specific heat of combustion gases in Kcal/Kg K
+Hv = 468.35;//Heat of vapourisation at 14 Kgf/cm^2 and 194.16 deg celsius in Kcak/Kg
+To = 298;//Surronding temperature
+Tb = 194.16+273;//Boiling point of liquid water
+
+//To Calculate the maximum work obtained and the entropy change
+//(i)Calculation of maximum work
+//Q = del_H = m*Cp*(T2-T1); gas can be assumed to cool at constant pressure
+//From equation 4.14 (page no 110)
+del_B = -((m*Cpg*(T2-T1))-(To*m*Cp*log((T2+273)/(T1+273))));
+mprintf('(i)The maximum work that can be obtained is %f Kcal/Kg of gas',del_B);
+
+//(ii)To Calculate the change in entropy
+del_S =(m*Cp*log(Tb/To))+((m*Hv)/Tb);
+mprintf('\n(ii)The entropy change per Kg of water is %f',del_S);
+//end
\ No newline at end of file diff --git a/479/CH5/EX5.1/Example_5_1.sce b/479/CH5/EX5.1/Example_5_1.sce new file mode 100755 index 000000000..605ae1e8d --- /dev/null +++ b/479/CH5/EX5.1/Example_5_1.sce @@ -0,0 +1,39 @@ +//Chemical Engineering Thermodynamics
+//Chapter 5
+//Thermodynamic Potentials and Maxwell's Relation
+
+//Example 5.1
+clear;
+clc;
+
+//Given
+T = 293;//Constant temperature in K
+w_NH3 = 20/100;//weight of NH3 in an aqueous solution in Kg
+w_H2O = 80/100;//weight of H2O in an aqueous solution in Kg
+V_f = 40;//feed rate in Kg/min
+M_NH3 = 17;//Molecular weight of NH3
+M_H2O = 18;//Molecular weight of H20
+R = 1.98;//gas constant in Kcl/Kg mole K
+V_s = 62;//Rate of heating steam in Kg/min
+P1_H2O = 11.6;//Vapour pressure of water in feed in mm Hg
+P2_H2O = 17.5;//Vapour pressure of pure water in mm Hg
+P1_NH3 = 227;//Vapour pressure of NH3 in feed in mm Hg
+P2_NH3 = 6350;//Vapor pressure of pure NH3 in mm Hg
+//From steam tables:
+Hs = 666.4;//Enthalpy of steam at 160 deg celsius & 2Kgf/cm^2 in Kcal /Kg
+Ss = 1.75;//Entropy of steam at 160 deg celsius & 2Kgf/cm^2V in Kcal/Kg K
+Hl = 20.03;//Enthalpy of liquid water at 20 deg celsius in Kcal/Kg
+Sl = 0.0612;//Entropy of liquid water at 20 deg celsius in Kcal/Kg K
+
+//To Calculate the efficiency of the separation process
+//Material Balance:
+n_NH3 = (V*w_NH3)/M_NH3;//Kg moles of NH3 in feed(tops)
+n_H2O = (V*w_H2O)/M_H2O;//Kg moles of H20 in feed(bottoms)
+//del_F = del_F_NH3 +del_F_H2O;
+del_F = (R*T*n_NH3*log(P2_NH3/P1_NH3))+(R*T*n_H2O*log(P2_H2O/P1_H2O));//Theoretical minimum work done in Kcal
+//The available energy of the steam can be calculated from equation 4.14(page no 110)
+del_B = -V_s*((Hl-Hs)-T*(Sl-Ss));//Available energy of the steam in Kcal
+E = (del_F/del_B)*100;
+mprintf('The efficiency of the separation process is %f percent',E);
+//end
+
diff --git a/479/CH6/EX6.0/Chapter_6.sce b/479/CH6/EX6.0/Chapter_6.sce new file mode 100755 index 000000000..18c0234ec --- /dev/null +++ b/479/CH6/EX6.0/Chapter_6.sce @@ -0,0 +1,10 @@ +//Chemical Engineering Thermodynamics
+//Chapter 6
+//Thermodynamic Relations
+
+clear;
+clc;
+
+//Given
+//The chapter does not contain any example which involve numerical computation
+//end
\ No newline at end of file diff --git a/479/CH7/EX7.1/Example_7_1.sce b/479/CH7/EX7.1/Example_7_1.sce new file mode 100755 index 000000000..4e8c6b092 --- /dev/null +++ b/479/CH7/EX7.1/Example_7_1.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 7
+//Ideal Gases
+
+//Example 7.1
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and it does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH7/EX7.2/Example_7_2.sce b/479/CH7/EX7.2/Example_7_2.sce new file mode 100755 index 000000000..ef0c05c86 --- /dev/null +++ b/479/CH7/EX7.2/Example_7_2.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 7
+//Ideal Gases
+
+//Example 7.2
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and it does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH7/EX7.3/Example_7_3.sce b/479/CH7/EX7.3/Example_7_3.sce new file mode 100755 index 000000000..c40244580 --- /dev/null +++ b/479/CH7/EX7.3/Example_7_3.sce @@ -0,0 +1,47 @@ +//Chemical Engineering Thermodynamics
+//Chapter 7
+//Ideal Gases
+
+//Example 7.3
+clear;
+clc;
+
+//Given
+P1 = 15;//initial pressure in Kgf/cm^2
+P2 = 1;//final pressure in Kgf/cm^2
+V1 = 0.012;//initial volume in m^3
+V2 = 0.06;//final volume in m^3
+T1 = 420;//initial temperature in K
+M = 28;//molecular weight of the gas
+Cp = 0.25;//specific heat at constant pressure in Kcal/Kg K
+R = 1.98;//gas constant in Kcal/Kg mole K
+R2 = 848;//gas constant in mKgf/Kgmole K
+//Cv = a+0.0005*T1; Specific heat at constant volume
+
+//To Calculate the final temperature of the ideal gas, work done in an open and closed system,internal energy change for the process
+//(a)Calculation of final temperature
+//Using ideal gas law:(P*V)/(R*T)
+T2 = (P2*V2*T1)/(P1*V1);
+mprintf('(a)The final temperature is %d K',T2);
+
+//(b)Calculation of work in an open and closed system
+//From equation 7.22(page no 147): P1*(V1^n)=P2*(V2^n)
+n = (log(P2/P1))/(log(V1/V2));
+//From equation 7.25(page no 149)
+W = ((P1*V1)-(P2*V2))/(n-1)*10^4;//work in mKgf
+W1 = W/427;//Work in Kcal
+mprintf('\n (b)The work in a closed system is %f Kcal',W1);
+Ws = n*W1;//from equation 7.28(page no 149)
+mprintf('\n The work in an open system is %f Kcal',Ws);
+
+//(c)Calculation of internal energy change
+R1 = R/M;//gas constant in Kcal/Kg
+Cv = Cp-R1;//specific heat at constant volume in Kcal/Kg K
+a = Cv-(0.0005*T1);
+m = (P1*10^4*V1*M)/(R2*T1);//mass of gas in Kg
+function y = f(T)
+ y = m*(a+(0.0005*T));
+endfunction
+del_E = intg(T1,T2,f);//internal energy change in Kcal/Kg
+del_E1 = M*del_E;//internal energy change in Kcal/Kgmole
+mprintf('\n (c)The internal energy change for the process is %f Kcal/Kgmole',del_E1);
\ No newline at end of file diff --git a/479/CH7/EX7.4/Example_7_4.sce b/479/CH7/EX7.4/Example_7_4.sce new file mode 100755 index 000000000..cbdffd2bf --- /dev/null +++ b/479/CH7/EX7.4/Example_7_4.sce @@ -0,0 +1,11 @@ +//Chemical Engineering Thermodynamics
+//Chapter 7
+//Ideal Gases
+
+//Example 7.4
+clear;
+clc;
+
+//Given
+//The given example is a theoretical problem and it does not involve any numerical computation
+//end
\ No newline at end of file diff --git a/479/CH7/EX7.5/Example_7_5.sce b/479/CH7/EX7.5/Example_7_5.sce new file mode 100755 index 000000000..6fc292e8c --- /dev/null +++ b/479/CH7/EX7.5/Example_7_5.sce @@ -0,0 +1,158 @@ +//Chemical Engineering Thermodynamics
+//Chapter 7
+//Ideal Gases
+
+//Example 7.5
+clear;
+clc;
+
+//Given
+P1 = 1;//Initial pressure of air in atm
+T1 = 15+273;//Initial temperature in K
+P2 = 5;//Final pressure of air in atm
+T2 = 15+273;//Final temperature in K
+Cv = 5;//specific heat of air at constant volume in Kcal/Kgmole K
+Cp = 7;//specific heat of air at constant pressure in Kcal/Kgmole K
+R = 0.082;//gas constant in atm-m^3/Kgmole K
+R1 = 2;//gas constant in Kcal/Kgmole K
+//From the P-V diagram given in page no 155:
+//Line 12 represents Isothermal process
+//Line b2,c2 & 1a represent Isometric process
+//Line a2 & 1c represent Isobaric process
+//Line 1b reprsent Adiabatic process
+
+//To find Approx Value
+function[A]=approx(V,n)
+ A=round(V*10^n)/10^n;//V-Value n-To what place
+ funcprot(0)
+endfunction
+
+//To Compute del_H, del_E, Q, W, del_S for the processes given above.
+//To indicate the quantities that are state functions
+//To verify that the work required in an isothermal process is less than that in an adiabatic process
+
+//Basis:1 Kgmole of air
+V1 = (R*T1)/P1;//Initial volume in cubic meter
+V2 = (R*T2)/P2;//Final volume in cubic meter
+
+//(i)Isothermal path 12
+//Equations 7.7 to 7.9 will be used (page no 145)
+del_E_12 = Cv*(T2-T1);
+del_H_12 = Cp*(T2-T1);
+W_12 = R1*T1*log(P1/P2);
+Q_12 = W_12;
+del_S_12 = approx((R1*log(P1/P2)),4);
+mprintf('\n(i)For isothermal process or path 12 change in internal energy is %f',del_E_12);
+mprintf('\n For isothermal process or path 12 change in enthalpy is %f',del_H_12);
+mprintf('\n For isothermal process or path 12 heat released is %f Kcal',Q_12);
+mprintf('\n For isothermal process or path 12 work is %f Kcal',W_12);
+mprintf('\n For isothermal process or path 12 change in entropy is %f Kcal/Kgmole K',del_S_12);
+
+//(ii)Path 1a2 = 1a(isometric)+a2(isobaric)
+//Equation 7.1 to 7.6 will be used (page no 144 & 145)
+Pa = P2;
+Ta = (Pa/P1)*T1;//in K
+Q_1a = Cv*(Ta-T1);
+del_E_1a = Q_1a;
+del_H_1a = Cp*(Ta-T1);
+W_1a = 0;// Constant volume process
+del_E_a2 = Cv*(T2-Ta);
+del_H_a2 = Cp*(T2-Ta);
+Q_a2 = del_H_a2;
+W_a2 = P2*(V2-V1)*((10^4*1.03)/427);
+del_H_1a2 = del_H_1a+del_H_a2;
+del_E_1a2 = del_E_1a+del_E_a2;
+Q_1a2 = Q_1a+Q_a2;
+W_1a2 = W_1a+W_a2;
+del_S_1a = Cv*log(Ta/T1);
+del_S_a2 = Cp*log(T2/Ta);
+del_S_1a2 = approx((del_S_1a+del_S_a2),4);
+mprintf('\n\n(ii)For path 1a2 change in internal energy is %f',del_E_1a2);
+mprintf('\n For path 1a2 change in enthalpy is %f',del_H_1a2);
+mprintf('\n For path 1a2 heat released is %f Kcal',Q_1a2);
+mprintf('\n For path 1a2 work is %f Kcal',W_1a2);
+mprintf('\n For path 1a2 change in entropy is %f Kcal/Kgmole K',del_S_1a2);
+
+//(iii)Path 1b2 = 1b(adiabatic)+b2(isometric)
+//From equation 7.11 (page no 146)
+y = Cp/Cv;
+Tb = T1*((V1/V2))^(y-1);
+//From equation 7.1 to 7.3,7.10 & 7.21,(page no 144,146,147)
+Q_1b = 0;//adiabatic process
+del_E_1b = Cv*(Tb-T1);
+del_H_1b = Cp*(Tb-T1);
+W_1b = -del_E_1b;
+Q_b2 = Cv*(T1-Tb);
+del_H_b2 = Cp*(T1-Tb);
+W_b2 = 0;//constant volume prcess
+del_E_b2 = Cv*(T2-Tb);
+del_H_1b2 = del_H_1b+del_H_b2;
+del_E_1b2 = del_E_1b+del_E_b2;
+W_1b2 = W_1b+W_b2;
+Q_1b2 = Q_1b+Q_b2;
+del_S_1b2 = approx((Cv*log(T1/Tb)),4);
+mprintf('\n\n(iii)For path 1b2 change in internal energy is %f',del_E_1b2);
+mprintf('\n For path 1b2 change in enthalpy is %f',del_H_1b2);
+mprintf('\n For path 1b2 heat released is %f Kcal',Q_1b2);
+mprintf('\n For path 1b2 work is %f Kcal',W_1b2);
+mprintf('\n For path 1b2 change in entropy is %f Kcal/Kgmole K',del_S_1b2);
+
+//(iv)Path 1c2 = 1c(isobaric)+c2(isometric);
+Pc = P1;
+Vc = V2;
+Tc = (Pc/P2)*T2;
+del_E_1c = Cv*(Tc-T1);
+Q_1c = Cp*(Tc-T1);
+del_H_1c = Q_1c;
+W_1c = P1*(Vc-V1)*((10^4*1.03)/427);
+del_E_c2 = Cv*(T2-Tc);
+Q_c2 = del_E_c2;
+del_H_c2 = Cp*(T2-Tc);
+W_c2 = 0;//constant volume process
+del_E_1c2 = del_E_1c+del_E_c2;
+del_H_1c2 = del_H_1c+del_H_c2;
+Q_1c2 = Q_1c+Q_c2;
+W_1c2 = W_1c+W_c2;
+del_S_1c = Cp*log(Tc/T1);
+del_S_c2 = Cv*log(T2/Tc);
+del_S_1c2 = approx((del_S_1c+del_S_c2),4);
+mprintf('\n\n(iv)For path 1c2 change in internal energy is %f',del_E_1c2);
+mprintf('\n For path 1c2 change in enthalpy is %f',del_H_1c2);
+mprintf('\n For path 1c2 heat released is %f Kcal',Q_1c2);
+mprintf('\n For path 1c2 work is %f Kcal',W_1c2);
+mprintf('\n For path 1c2 change in entropy is %f Kcal/Kgmole K',del_S_1c2);
+
+//Determination of state & path functions
+if((del_E_12 == del_E_1a2)&(del_E_12 == del_E_1b2)&(del_E_12 == del_E_1c2))
+ mprintf('\n\n del_E is a state function');
+else
+ mprintf('\n\n del_E is a path function');
+end
+if((del_H_12 == del_H_1a2)&(del_H_12 == del_H_1b2)&(del_H_12 == del_H_1c2))
+ mprintf('\n\n del_H is a state function');
+else
+ mprintf('\n\n del_H is a path function');
+end
+if(del_S_12 == del_S_1a2)&(del_S_12 == del_S_1b2)&(del_S_12 == del_S_1c2)
+ mprintf('\n\n del_S is a state function');
+else
+ mprintf('\n\n del_S is a path function');
+end
+if((Q_12 == Q_1a2)&(Q_12 == Q_1b2)&(Q_12 == Q_1c2))
+ mprintf('\n\n Q is a state function');
+else
+ mprintf('\n\n Q is a path function');
+end
+if((W_12 == W_1a2)&(W_12 == W_1b2)&(W_12 == W_1c2))
+ mprintf('\n\n W is a state function');
+else
+ mprintf('\n\n W is a path function');
+end
+
+//Comparison of work required by isothermal & adiabatic process
+if(-(W_12)<-(W_1b2))
+ mprintf('\n\n Work required by isothermal process is less than the work required by an adiabatic process');
+else
+ mprintf('\n\n Statement is incorrect');
+end
+//end
diff --git a/479/CH8/EX8.1/Example_8_1.sce b/479/CH8/EX8.1/Example_8_1.sce new file mode 100755 index 000000000..88a273668 --- /dev/null +++ b/479/CH8/EX8.1/Example_8_1.sce @@ -0,0 +1,28 @@ +//Chemical Engineering Thermodynamics
+//Chapter 8
+//Third Law of Thermodynamics
+
+//Example 8.1
+clear;
+clc;
+
+//Given
+C_ps = 0.1;//Molal heat capacity of copper at 20 K
+Ti = 0;//Initial temperature in K
+Tf = 20;//melting point in K
+Tb = 300;//boiling point in K
+
+//To calculate the absolute entropy of copper at 300 K
+//From equation 8.4(page no 164)
+a = C_ps/(Tf^3);// a is the charateristic constant
+C_p = [0.1 0.80 1.94 3.0 3.9 5.0];
+//T1 = log(T);
+T1 = [1.301 1.6021 1.7782 1.9031 2.000 2.1761];
+plot(T1,C_p);
+xtitle("C_p vs T1","T1","Cp");
+// Area under the curve is given as
+A = 7.82;
+//From equation 8.5(page no 164)
+S = (a*((Tf^3)/3))+A;
+mprintf('The absolute entropy of copper is %f Kcal/Kgmole',S);
+//end
\ No newline at end of file diff --git a/479/CH9/EX9.1/Example_9_1.sce b/479/CH9/EX9.1/Example_9_1.sce new file mode 100755 index 000000000..e17aa8354 --- /dev/null +++ b/479/CH9/EX9.1/Example_9_1.sce @@ -0,0 +1,37 @@ +//Chemical Engineering Thermodynamics
+//Chapter 9
+//Fluid Flow in Pipes and Nozzles
+//Example 9.1
+clear;
+clc;
+
+//Given
+R = 848;//gas constant in m Kgf/Kgmole K
+M = 29;//molecular weight of air
+g = 9.81;
+T1 = 90+273;//initial temperature in K
+y = 1.4;//gamma = Cp/Cv
+W = 800/3600;//Mass rate of air in Kg/sec
+P1 = 3.5;//initial pressure in atm
+d = 2.5;//diameter of the pipe in cm
+
+//To find out the pressure at the final point
+v1 = (R*T1)/(M*P1*1.033*10^4);//specific volume in cubic meter/Kg
+u1 = (W*v1)/(%pi*(d^2*(10^-4))/4);//inital velocity in m/sec
+//Assume final temperature as
+T2 = [300 310];
+//Assume specific heat capacity in J/KgK corresponding to the above temperature as
+Cp = [2987.56 2983.56];
+for i = 1:2
+ us(i) = (g*y*R*T2(i)/M)^(1/2);//sonic velocity attained in m/sec
+ u2(i) = ((u1^2)-((2*g*Cp(i)/M)*(T2(i)-T1)))^(1/2);//From equation 9.18 & 9.19 (page no 170)
+end
+if us(i)-u2(i) <= 1
+ u2 = u2(i);
+ T2 = T2(i);
+else
+end
+v2 = u2*(%pi/4)*(d^2/10^4)*(1/W);
+P2 = (P1*v1*T2)/(T1*v2);
+mprintf('The pressure at the final point is %f atm',P2);
+//end
\ No newline at end of file diff --git a/479/CH9/EX9.2/Exampe_9_2.sce b/479/CH9/EX9.2/Exampe_9_2.sce new file mode 100755 index 000000000..0138db882 --- /dev/null +++ b/479/CH9/EX9.2/Exampe_9_2.sce @@ -0,0 +1,19 @@ +//Chemical Engineering Thermodynamics
+//Chapter 9
+//Fluid Flow in Pipes and Nozzles
+//Example 9.2
+clear;
+clc;
+
+//Given
+A1 = 0.002;//inlet area in sq meter
+A2 = 0.0005;//throat area in sq meter
+P1 = 1.3*10^4;//inlet pressure in Kgf/sq m
+P2 = 0.7*10^4;//throat pressure in Kgf/sq m
+g = 9.81;
+v = 12*10^-4;//specific volume in cubic m /Kg
+
+//To find out the mass rate of alcohol
+u2 = ((v*2*g*(P1-P2))/(1-((A2/A1)^2)))^(1/2);//throat velocity in m/sec
+W = (u2*A2)/v;
+mprintf('The mass rate of alcohol is %f Kg/sec',W);
\ No newline at end of file diff --git a/479/CH9/EX9.3/Example_9_3.sce b/479/CH9/EX9.3/Example_9_3.sce new file mode 100755 index 000000000..031486d28 --- /dev/null +++ b/479/CH9/EX9.3/Example_9_3.sce @@ -0,0 +1,62 @@ +//Chemical Engineering Thermodynamics
+//Chapter 9
+//Fluid Flow in Pipes and Nozzles
+//Example 9.3
+clear;
+clc;
+
+//Given
+P1 = 50;//initial pressure in Kgf/sq m
+T1 = 45+273;//initial temperature in K
+g = 9.81;
+y = 1.35;//gamma
+R = 848;//gas constant in m Kgf/Kgmole K
+M = 29;//molecular weight of air
+d = 1;//pipe diameter in cm
+
+//(i)To plot velocity,specific volume,mass velocity against P2/P1
+//(ii)To calculate the critical pressure,critical mass velocity and mass rate of flow
+//(i)Plotting of graph
+V1 = (R*T1)/(M*P1*1.033*10^4);//initial volume of the gas in cubic m/Kg
+//P3 = P2/P1 (say)
+//Assume P3 values as
+P3 = [1.0 0.8 0.6 0.4 0.2 0.1 0];
+G = [0 0 0 0 0 0 0];
+for i = 1:7
+ u2(i) = (((2*g*y*R*T1)/((y-1)*M))*(1-(P3(i)^((y-1)/y))))^(1/2);//final velocity in m/sec
+end
+for i = 1:6
+ v2(i) = V1/(P3(i)^(1/y));//final specific volume in cubic meter/Kg
+end
+for i = 1:6
+ G(i) = u2(i)/v2(i);//Mass velocity in Kg/sq m sec
+end
+
+clf;
+xset('window',4);
+plot(P3,u2,"o-");
+xtitle("Velocity vs P2/P1","P2/P1","Velocity");
+xset('window',5);
+plot(P3,G,"+-");
+xtitle("Mass velocity vs P2/P1","P2/P1","Mass velocity");
+xset('window',6);
+P_3 = [1.0 0.8 0.6 0.4 0.2 0.1];
+plot(P_3,v2,"*-");
+xtitle("Sp. volume vs P2/P1","P2/P1","Specific volume");
+
+//(ii)Calculation of critical pressure,critical mass velocity and mass rate of flow
+//From equation 9.37(page no 181)
+P2 = P1*(2/(y+1))^(y/(y-1));
+mprintf('The critical pressure is %f atm',P2);
+//From equation a (page no 183)
+u2 = (((2*g*y*R*T1)/((y-1)*M))*(1-((P2/P1)^((y-1)/y))))^(1/2);
+mprintf('\n The critical velocity is %f m/sec',u2);
+//From equation b (page no 183)
+v2 = ((R*T1)/(M*P1*1.033*10^4))/((P2/P1)^(1/y));
+mprintf('\n The critical specific volume is %f cubic meter/Kg',v2);
+//From relation c (page no 183)
+G = u2/v2;
+mprintf('\n The critical mass velocity is %f Kg/sq meter sec',G);
+W = G*(%pi/4)*(d/(100))^2;
+mprintf('\n Mass rate of flow through nozzle is %f Kg/sec',W);
+//end
\ No newline at end of file diff --git a/479/CH9/EX9.4/Example_9_4.sce b/479/CH9/EX9.4/Example_9_4.sce new file mode 100755 index 000000000..ad48dd080 --- /dev/null +++ b/479/CH9/EX9.4/Example_9_4.sce @@ -0,0 +1,27 @@ +//Chemical Engineering Thermodynamics
+//Chapter 9
+//Fluid Flow in Pipes and Nozzles
+//Example 9.4
+clear;
+clc;
+
+//Given
+A1 = 0.1;//Inlet area in sq meter
+u1 = 60;//inlet velocity in m/sec
+v1 = 0.185;//inlet specific volume in cubic meter/Kg
+H1 = 715;//inlet enthalpy in Kcal/Kg
+H2 = 660;//exit enthalpy in Kcal/Kg
+v2 = 0.495;//exit specific volume in cubic meter/Kg
+g = 9.81
+
+//To calculate the area at exit of nozzle and hence decide the type of the nozzle
+//From the first law
+u2 = ((u1^2)-(2*g*(H2-H1)*427))^(1/2);
+W = (u1*A1)/v1;//Mass rate of gas in Kg/sec
+A2 = (W*v2)/u2;//Area at exit of nozzle
+if(A2 < A1)
+ mprintf('The nozzle is convergent');
+else
+ mprintf('The nozzle is divergent');
+end
+//end
\ No newline at end of file |