{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 4: Second Law of Thermodynamics"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.10: Theoretical_problem.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.10\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"//The given example is a theoretical problem and it does not involve any numerical computation\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.1: EX4_1.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.1\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"Q1 = 250;//Heat absorbed in Kcal\n",
"T1 = (260+273);//Temperature at which engine absorbs heat\n",
"T0 = (40+273);//Temperature at which engine discards heat\n",
"//To Calculate work output, heat rejected, entropy change of system,surronding & total change in entropy and the efficiency of the heat engine\n",
"\n",
"//(i)Calculation of work output\n",
"W = (Q1*((T1-T0)/T1));//Work done using equations 4.7 & 4.9 given on page no 98\n",
"mprintf('(i)The work output of the heat engine is %f Kcal',W);\n",
"\n",
"//(ii)Calculation of heat rejected\n",
"Q2 = (Q1*T0)/T1;\n",
"mprintf('\n (ii)The heat rejected is %f Kcal',Q2);\n",
"\n",
"//(iii)Calculation of entropy\n",
"del_S1 = -(Q1/T1);//Change in the entropy of source in Kcal/Kg K\n",
"del_S2 = Q2/T0;//Change in the entropy of sink in Kcal/Kg K\n",
"del_St = del_S1+del_S2;//Total change in entropy in Kcal/Kg K\n",
"mprintf('\n (iii)Total change in entropy is %d confirming that the process is reversible',del_St);\n",
"\n",
"//(iv)Calculation of efficiency\n",
"n = (W/Q1)*100;\n",
"mprintf('\n (iv)The efficiency of the heat engine is %f percent',n);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.2: EX4_2.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.2\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"T1 = 373;//Temperature of the saturated steam in K\n",
"T2 = 298;//Temperature of the saturated water in K\n",
"//To calculate the total change in entropy and hence determine the reversibility of the process\n",
"\n",
"//del_H = del_Q+(V*del_P)\n",
"//del_H =del_Q; since it is a constant pressure process\n",
"\n",
"//From steam table,\n",
"//enthalpy of saturated steam at 373K is\n",
"H1 = 6348.5;// in Kcal/Kg\n",
"//enthalpy of saturated liquid water at 373K is\n",
"H2 = 99.15;//in Kcal/Kg\n",
"Q = H2-H1;//heat rejected in Kcal/Kg\n",
"del_S1 = Q/T1;//change in entropy of the system in Kcal/Kg K\n",
"del_S2 = Q/T2;//change in entropy of the surronding in Kcal/Kg K\n",
"del_St = del_S1+del_S2;//total change in the entropy in Kcal/Kg K\n",
"if(del_St == 0)\n",
"    mprintf('Process is reversible');\n",
"else\n",
"    mprintf('Process is irreversible');\n",
"end\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.3: To_find_out_entropy_change_of_block_air_and_total_entropy_change.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.3\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"Cp = 0.09;//specific heat of metal block in Kcal/Kg K\n",
"m = 10;//mass of metal block in Kg\n",
"T1 = 323;//initial temperature of the block in K\n",
"T2 = 298;//final temperature of the block in K\n",
"//constant pressure process\n",
"//To find out entropy change of block,air and total entropy change\n",
"\n",
"//(i)To calculate the entropy change of block\n",
"del_S1 = m*Cp*log(T2/T1);\n",
"mprintf('(i)Entropy change of block is %f Kcal/Kg K',del_S1);\n",
"\n",
"//(ii)To calculate the entropy change of air\n",
"Q = m*Cp*(T1-T2);//heat absorbed by air = heat rejected by block in Kcal\n",
"del_S2 = (Q/T2);\n",
"mprintf('\n (ii)Entropy change of air is %f Kcal/Kg K',del_S2);\n",
"\n",
"//(iii)To calculate the total entropy change\n",
"del_St = del_S1+del_S2;\n",
"mprintf('\n (iii)Total entropy change is %f Kcal/Kg K',del_St);\n",
"if(del_St == 0)\n",
"    mprintf('\n Process is reversible');\n",
"else\n",
"    mprintf('\n Process is irreversible');\n",
"end\n",
"//end "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.4: To_calculate_the_total_change_in_entropy.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.4\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"m1 = 10;//mass of metal block in Kg\n",
"m2 = 50;//mass of water in Kg\n",
"Cp1 = 0.09;//Specific heat of metal block in Kcal/Kg K\n",
"Cp2 = 1;//Specific heat of water in Kcal/Kg K\n",
"T1 = 50;//Initial temperature of block in deg celsius\n",
"T2 = 25;//Final temperature of block in deg celsius\n",
"\n",
"//To calculate the total change in entropy\n",
"//Heat lost by block = Heat gained by water\n",
"Tf = ((m1*Cp1*T1)+(m2*Cp2*T2))/((m1*Cp1)+(m2*Cp2));//final temperature of water in deg celsius\n",
"Tf1 = Tf+273.16;//final temperature in K\n",
"del_S1 = m1*Cp1*log(Tf1/(T1+273));//change in entropy of the block in Kcal/K\n",
"del_S2 = m2*Cp2*log(Tf1/(T2+273));//change in entropy of the block in Kcal/K\n",
"del_St = del_S1+del_S2;\n",
"mprintf('The total change entropy is %f Kcal/K',del_St);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.5: To_calculate_the_entropy_change.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.5\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"//Air at 20 deg celsius\n",
"//P1 = 250;initial pressure in atm\n",
"//P2 = 10;final pressure after throttling in atm\n",
"\n",
"//To calculate the entropy change\n",
"//According to the given conditions from figure4.5(page no 103)\n",
"S1 = -0.38;//initial entropy in Kcal/Kg K\n",
"S2 = -0.15;//final entroy in Kcal/Kg K\n",
"del_S = S2-S1;\n",
"mprintf('Change in entropy for the throttling process is %f Kcal/Kg K',del_S);\n",
"//From figure 4.6(page no 104), the final temperature is -10 deg celsius\n",
"//end "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.6: Theoretical_problem.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.6\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"//The given problem does not involve any numerical comptation\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.7: To_calculate_the_horse_power_of_the_compressor.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.7\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"//Basis: 1 hour\n",
"m = 10;//mass of air in Kg\n",
"T = 293;//Constant temperature throughout the process in K\n",
"//P1 = 1;//Initial pressure in atm\n",
"//P2 = 30;//Final pressure in atm\n",
"//According to the given data and using the graph  or figure A.2.7 given in page no 105\n",
"S1 = 0.02;//Initial entropy in Kcal/Kg\n",
"S2 = -0.23;//Final entropy in Kcal/Kg\n",
"H1 = 5;//Initial enthalpy in Kcal/Kg\n",
"H2 = 3;//Final enthalpy in Kcal/Kg\n",
"\n",
"W = -((H2-H1)+T*(S2-S1))*m*(427/(3600*75));\n",
"mprintf('The horse power of the compressor is %f hp',W);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.8: To_calculate_the_effectiveness_of_the_process.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.1\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"//Basis: 1 Kg of steam\n",
"//P1 = 30;Intial pressure in Kgf/cm^2\n",
"//P2 = 3;Final pressure in Kgf/cm^2\n",
"//T = 300;//Operating temperature\n",
"//From figure A.2.8, \n",
"H1 = 715;//Initial enthalpy of steam in Kcal/Kg\n",
"H2 = 625;//Final enthalpy of steam in Kcal/Kg\n",
"S1 = 1.56;//Initial entropy of steam in Kcal/Kg K\n",
"S2 = 1.61;//Final entropy of steam in Kcal/Kg K\n",
"Q = -1;//heat loss in Kcal/Kg\n",
"To = 298;//The lowest surronding temperature in K\n",
"\n",
"//To calculate the effectiveness of the process\n",
"W = (-(H2-H1)+Q);//Actual work output by the turbine in Kcal\n",
"//The maximum or available work can be calculated from equation 4.14\n",
"del_B = -((H2-H1)-(To*(S2-S1)));// Maximum work that can be obtained in Kcal\n",
"E = (W/del_B)*100;\n",
"mprintf('The effectiveness of the process is %f percent',E);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.9: To_Calculate_the_maximum_work_obtained_and_the_entropy_change.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 4\n",
"//Second Law of Thermodynamics\n",
"\n",
"//Example 4.9\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"m = 1;//mass of liquid water in Kg\n",
"T1 = 1350;//initial temperature in deg celsius\n",
"T2 = 400;//final temperature in deg celsius\n",
"Cp = 1;//Specific heat of water in Kcal/Kg K\n",
"Cpg = 0.2;//Specific heat of combustion gases in Kcal/Kg K\n",
"Hv = 468.35;//Heat of vapourisation at 14 Kgf/cm^2 and 194.16 deg celsius in Kcak/Kg\n",
"To = 298;//Surronding temperature\n",
"Tb = 194.16+273;//Boiling point of liquid water\n",
"\n",
"//To Calculate the maximum work obtained and the entropy change\n",
"//(i)Calculation of maximum work\n",
"//Q = del_H = m*Cp*(T2-T1); gas can be assumed to cool at constant pressure\n",
"//From equation 4.14 (page no 110)\n",
"del_B = -((m*Cpg*(T2-T1))-(To*m*Cp*log((T2+273)/(T1+273))));\n",
"mprintf('(i)The maximum work that can be obtained is %f Kcal/Kg of gas',del_B);\n",
"\n",
"//(ii)To Calculate the change in entropy\n",
"del_S =(m*Cp*log(Tb/To))+((m*Hv)/Tb);\n",
"mprintf('\n(ii)The entropy change per Kg of water is %f',del_S);\n",
"//end"
   ]
   }
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