16 files changed, 316 insertions, 0 deletions
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