From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001
From: priyanka
Date: Wed, 24 Jun 2015 15:03:17 +0530
Subject: initial commit / add all books

---
 479/CH4/EX4.1/Example_4_1.sce   | 32 ++++++++++++++++++++++++++++++++
 479/CH4/EX4.10/Example_4_10.sce | 11 +++++++++++
 479/CH4/EX4.2/Example_4_2.sce   | 31 +++++++++++++++++++++++++++++++
 479/CH4/EX4.3/Example_4_3.sce   | 34 ++++++++++++++++++++++++++++++++++
 479/CH4/EX4.4/Example__4_4.sce  | 25 +++++++++++++++++++++++++
 479/CH4/EX4.5/Example_4_5.sce   | 21 +++++++++++++++++++++
 479/CH4/EX4.6/Example_4_6.sce   | 11 +++++++++++
 479/CH4/EX4.7/Example_4_7.sce   | 23 +++++++++++++++++++++++
 479/CH4/EX4.8/Example_4_8.sce   | 28 ++++++++++++++++++++++++++++
 479/CH4/EX4.9/Example_4_9.sce   | 29 +++++++++++++++++++++++++++++
 10 files changed, 245 insertions(+)
 create mode 100755 479/CH4/EX4.1/Example_4_1.sce
 create mode 100755 479/CH4/EX4.10/Example_4_10.sce
 create mode 100755 479/CH4/EX4.2/Example_4_2.sce
 create mode 100755 479/CH4/EX4.3/Example_4_3.sce
 create mode 100755 479/CH4/EX4.4/Example__4_4.sce
 create mode 100755 479/CH4/EX4.5/Example_4_5.sce
 create mode 100755 479/CH4/EX4.6/Example_4_6.sce
 create mode 100755 479/CH4/EX4.7/Example_4_7.sce
 create mode 100755 479/CH4/EX4.8/Example_4_8.sce
 create mode 100755 479/CH4/EX4.9/Example_4_9.sce

(limited to '479/CH4')

diff --git a/479/CH4/EX4.1/Example_4_1.sce b/479/CH4/EX4.1/Example_4_1.sce
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+//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
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diff --git a/479/CH4/EX4.10/Example_4_10.sce b/479/CH4/EX4.10/Example_4_10.sce
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+//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
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+//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
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+//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 
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diff --git a/479/CH4/EX4.4/Example__4_4.sce b/479/CH4/EX4.4/Example__4_4.sce
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+//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
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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
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+//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 
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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
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+++ 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
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+//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
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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
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+//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
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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
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+++ 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
-- 
cgit