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/CH12/EX12.1/Example_12_1.sce | 25 +++++++++++++++++++++++
 479/CH12/EX12.2/Example_12_2.sce | 23 +++++++++++++++++++++
 479/CH12/EX12.2/Example_12_2.txt |  2 ++
 479/CH12/EX12.3/Example_12_3.sce | 44 ++++++++++++++++++++++++++++++++++++++++
 479/CH12/EX12.4/Example_12_4.sce | 22 ++++++++++++++++++++
 5 files changed, 116 insertions(+)
 create mode 100755 479/CH12/EX12.1/Example_12_1.sce
 create mode 100755 479/CH12/EX12.2/Example_12_2.sce
 create mode 100755 479/CH12/EX12.2/Example_12_2.txt
 create mode 100755 479/CH12/EX12.3/Example_12_3.sce
 create mode 100755 479/CH12/EX12.4/Example_12_4.sce

(limited to '479/CH12')

diff --git a/479/CH12/EX12.1/Example_12_1.sce b/479/CH12/EX12.1/Example_12_1.sce
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+++ b/479/CH12/EX12.1/Example_12_1.sce
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+//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
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+++ 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
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+++ b/479/CH12/EX12.4/Example_12_4.sce
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+//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
-- 
cgit