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+{
+"cells": [
+ {
+		   "cell_type": "markdown",
+	   "metadata": {},
+	   "source": [
+       "# Chapter 5: Entropy"
+	   ]
+	},
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.10: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_10.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 10')\n",
+"p1=3;//initial pressure in Mpa\n",
+"v1=0.05;//initial volume in m^3\n",
+"v2=0.3;//final volume in m^3\n",
+"disp('for reversible adiabatic process governing equation for expansion,')\n",
+"disp('P*V^1.4=constant')\n",
+"disp('also,for such process entropy change=0')\n",
+"disp('using p2/p1=(v1/v2)^1.4 or v=(p1*(v1^1.4)/p)^(1/1.4)')\n",
+"disp('final pressure(p2)in Mpa')\n",
+"disp('p2=p1*(v1/v2)^1.4')\n",
+"p2=p1*(v1/v2)^1.4\n",
+"disp('from first law,second law and definition of enthalpy;')\n",
+"disp('dH=T*dS+v*dP')\n",
+"disp('for adiabatic process of reversible type,dS=0')\n",
+"dS=0;//for adiabatic process of reversible type\n",
+"disp('so dH=v*dP')\n",
+"disp('integrating both side H2-H1=deltaH=v*dP in KJ')\n",
+"p1=3*1000;//initial pressure in Kpa\n",
+"p2=244;//final pressure in Kpa\n",
+"disp('so enthalpy change(deltaH)in KJ')\n",
+"function y = f(p), y =(p1*(v1^1.4)/p)^(1/1.4), endfunction\n",
+"deltaH = intg(p2, p1, f) \n",
+"disp('and entropy change=0')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.11: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_11.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 11')\n",
+"m=2;//mass of air in kg\n",
+"v1=1;//initial volume of air in m^3\n",
+"v2=10;//final volume of air in m^3\n",
+"R=287;//gas constant in J/kg K\n",
+"disp('during free expansion temperature remains same and it is an irreversible process.for getting change in entropy let us approximate this expansion process as a reversible isothermal expansion')\n",
+"disp('a> change in entropy of air(deltaS_air)in J/K')\n",
+"disp('deltaS_air=m*R*log(v2/v1)')\n",
+"deltaS_air=m*R*log(v2/v1)\n",
+"disp('b> during free expansion on heat is gained or lost to surrounding so,')\n",
+"disp('deltaS_surrounding=0')\n",
+"disp('entropy change of surroundings=0')\n",
+"deltaS_surrounding=0;//entropy change of surroundings\n",
+"disp('c> entropy change of universe(deltaS_universe)in J/K')\n",
+"disp('deltaS_universe=deltaS_air+deltaS_surrounding')\n",
+"deltaS_universe=deltaS_air+deltaS_surrounding"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.12: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_12.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 12')\n",
+"m=0.5;//mass of air in kg\n",
+"p1=1.013*10^5;//initial pressure of air in pa\n",
+"p2=0.8*10^6;//final pressure of air in pa\n",
+"T1=800;//initial temperature of air in K\n",
+"n=1.2;//polytropic expansion constant\n",
+"y=1.4;//expansion constant for air\n",
+"Cv=0.71;//specific heat at constant volume in KJ/kg K\n",
+"disp('let initial and final states be denoted by 1 and 2')\n",
+"disp('for poly tropic process pressure and temperature can be related as')\n",
+"disp('(p2/p1)^((n-1)/n)=T2/T1')\n",
+"disp('so temperature after compression(T2)=T1*(p2/p1)^((n-1)/n) in K')\n",
+"T2=T1*(p2/p1)^((n-1)/n)\n",
+"disp('substituting in entropy change expression for polytropic process,') \n",
+"disp('entropy change(deltaS)inKJ/kg K')\n",
+"disp('deltaS=Cv*((n-y)/(n-1))*log(T2/T1)')\n",
+"deltaS=Cv*((n-y)/(n-1))*log(T2/T1)\n",
+"disp('NOTE=>answer given in book i.e -244.54 KJ/kg K is incorrect,correct answer is -.24454 KJ/kg K')\n",
+"disp('total entropy change(deltaS)=m*deltaS*1000 in J/K')\n",
+"deltaS=m*deltaS*1000"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.13: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_13.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 13')\n",
+"disp('NOTE=>In question no. 13,formula for maximum work is derived which cannot be solve using scilab software')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.14: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_14.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 14')\n",
+"Q1=500;//heat supplied by source in kcal/s\n",
+"T1=600;//temperature of source in K\n",
+"T2=300;//temperature of sink in K\n",
+"disp('clausius inequality can be used for cyclic process as given below;consider 1 for source and 2 for sink')\n",
+"disp('K=dQ/T=Q1/T1-Q2/T2')\n",
+"disp('i> for Q2=200 kcal/s')\n",
+"Q2=200;//heat rejected by sink in kcal/s\n",
+"disp('K=Q1/T1-Q2/T2 in kcal/s K')\n",
+"K=Q1/T1-Q2/T2\n",
+"disp('as K is not greater than 0,therefore under these conditions engine is not possible')\n",
+"disp('ii> for Q2=400 kcal/s')\n",
+"Q2=400;//heat rejected by sink in kcal/s\n",
+"disp('K=Q1/T1-Q2/T2 in kcal/s K')\n",
+"K=Q1/T1-Q2/T2\n",
+"disp('as K is less than 0,so engine is feasible and cycle is reversible')\n",
+"disp('iii> for Q2=250 kcal/s')\n",
+"Q2=250;//heat rejected by sink in kcal/s\n",
+"disp('K=Q1/T1-Q2/T2 in kcal/s K')\n",
+"K=Q1/T1-Q2/T2\n",
+"disp('as K=0,so engine is feasible and cycle is reversible')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.15: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_15.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 15')\n",
+"p1=0.5;//initial pressure of air in Mpa\n",
+"T1=400;//initial temperature of air in K\n",
+"p2=0.3;//final pressure of air in Mpa\n",
+"T2=350;//initial temperature of air in K\n",
+"R=0.287;//gas constant in KJ/kg K\n",
+"Cp=1.004;//specific heat at constant pressure in KJ/kg K\n",
+"disp('let the two points be given as states 1 and 2,')\n",
+"disp('let us assume flow to be from 1 to 2')\n",
+"disp('so entropy change(deltaS1_2)=s1-s2=Cp*log(T1/T2)-R*log(p1/p2)in KJ/kg K')\n",
+"deltaS1_2=Cp*log(T1/T2)-R*log(p1/p2)\n",
+"disp('deltaS1_2=s1-s2=0.01254 KJ/kg K')\n",
+"disp('it means s2 > s1 hence the assumption that flow is from 1 to 2 is correct as from second law of thermodynamics the entropy increases in a process i.e s2 is greater than or equal to s1')\n",
+"disp('hence flow occurs from 1 to 2 i.e from 0.5 MPa,400K to 0.3 Mpa & 350 K')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.16: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_16.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 16')\n",
+"disp('NOTE=>In question no. 16,value of n is derived which cannot be solve using scilab software.')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.17: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_17.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 17')\n",
+"Q12=1000;//heat added during process 1-2 in KJ\n",
+"Q34=800;//heat added during process 3-4 in KJ\n",
+"T1=500;//operating temperature for process 1-2\n",
+"T3=400;//operating temperature for process 3-4\n",
+"T5=300;//operating temperature for process 5-6\n",
+"T2=T1;//isothermal process\n",
+"T4=T3;//isothermal process\n",
+"T6=T5;//isothermal process\n",
+"disp('total heat added(Q)in KJ')\n",
+"disp('Q=Q12+Q34')\n",
+"Q=Q12+Q34\n",
+"disp('for heat addition process 1-2')\n",
+"disp('Q12=T1*(s2-s1)')\n",
+"disp('deltaS=s2-s1=Q12/T1 in KJ/K')\n",
+"deltaS=Q12/T1\n",
+"disp('or heat addition process 3-4')\n",
+"disp('Q34=T3*(s4-s3)')\n",
+"disp('deltaS=s4-s3=Q34/T3 in KJ/K')\n",
+"deltaS=Q34/T3\n",
+"disp('or heat rejected in process 5-6(Q56)in KJ')\n",
+"disp('Q56=T5*(s5-s6)=T5*((s2-s1)+(s4-s3))=T5*(deltaS+deltaS)')\n",
+"Q56=T5*(deltaS+deltaS)\n",
+"disp('net work done=net heat(W_net)in KJ')\n",
+"disp('W_net=(Q12+Q34)-Q56')\n",
+"W_net=(Q12+Q34)-Q56\n",
+"disp('thermal efficiency of cycle(n)=W_net/Q')\n",
+"n=W_net/Q\n",
+"disp('or n=n*100 %')\n",
+"n=n*100 \n",
+"disp('so work done=600 KJ and thermal efficiency=33.33 %')\n",
+"\n",
+"\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.18: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_18.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 18')\n",
+"T1_a=800;//temperature of reservoir a in K\n",
+"T1_b=700;//temperature of reservoir b in K\n",
+"T1_c=600;//temperature of reservoir c in K\n",
+"T2=320;//temperature of sink in K\n",
+"W=20;//work done in KW\n",
+"Q2=10;//heat rejected to sink in KW\n",
+"disp('let heat supplied by reservoir at 800 K,700 K,600 K be Q1_a , Q1_b , Q1_c')\n",
+"disp('here Q1-Q2=W')\n",
+"disp('so heat supplied by source(Q1)=W+Q2 in KW')\n",
+"Q1=W+Q2\n",
+"disp('also given that,Q1_a=0.7*Q1_b.......eq 1')\n",
+"disp('Q1_c=Q1-(0.7*Q1_b+Q1_b)')\n",
+"disp('Q1_c=Q1-1.7*Q1_b........eq 2')\n",
+"disp('for reversible engine')\n",
+"disp('Q1_a/T1_a+Q1_b/T1_b+Q1_c/T1_c-Q2/T2=0......eq 3')\n",
+"disp('substitute eq 1 and eq 2 in eq 3 we get, ')\n",
+"disp('heat supplied by reservoir of 700 K(Q1_b)in KJ/s')\n",
+"disp('Q1_b=((Q2/T2)-(Q1/T1_c))/((0.7/T1_a)+(1/T1_b)-(1.7/T1_c))')\n",
+"Q1_b=((Q2/T2)-(Q1/T1_c))/((0.7/T1_a)+(1/T1_b)-(1.7/T1_c))\n",
+"disp('so heat supplied by reservoir of 800 K(Q1_a)in KJ/s')\n",
+"disp('Q1_a=0.7*Q1_b')\n",
+"Q1_a=0.7*Q1_b\n",
+"disp('and heat supplied by reservoir of 600 K(Q1_c)in KJ/s')\n",
+"disp('Q1_c=Q1-1.7*Q1_b')\n",
+"Q1_c=Q1-1.7*Q1_b\n",
+"disp('so heat supplied by reservoir at 800 K(Q1_a)')\n",
+"Q1_a\n",
+"disp('so heat supplied by reservoir at 700 K(Q1_b)')\n",
+"Q1_b\n",
+"disp('so heat supplied by reservoir at 600 K(Q1_c)')\n",
+"Q1_c=-Q1_c\n",
+"disp('NOTE=>answer given in book for heat supplied by reservoir at 800 K,700 K,600 K i.e Q1_a=61.94 KJ/s,Q1_b=88.48 KJ/s,Q1_c=120.42 KJ/s is wrong hence correct answer is calculated above.')\n",
+"\n",
+"\n",
+"\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.1: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_1.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 1')\n",
+"p1=5;//initial pressure of air\n",
+"T1=(27+273);//temperature of air in K\n",
+"p2=2;//final pressure of air in K\n",
+"R=0.287;//gas constant in KJ/kg K\n",
+"Cp_air=1.004;//specific heat of air at constant pressure in KJ/kg K\n",
+"disp('entropy change may be given as,')\n",
+"disp('s2-s1=((Cp_air*log(T2/T1)-(R*log(p2/p1))')\n",
+"disp('here for throttling process h1=h2=>Cp_air*T1=Cp_air*T2=>T1=T2')\n",
+"disp('so change in entropy(deltaS)in KJ/kg K')\n",
+"disp('deltaS=(Cp_air*log(1))-(R*log(p2/p1))')\n",
+"deltaS=(Cp_air*log(1))-(R*log(p2/p1))"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.2: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_2.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 2')\n",
+"T1=(27+273);//temperature of water in K\n",
+"T2=(100+273);//steam temperature of water in K\n",
+"m=5;//mass of water in kg\n",
+"q=2260;//heat of vaporisation at 100 degree celcius in KJ/kg\n",
+"Cp=4.2;//specific heat of water at constant pressure in KJ/kg K\n",
+"M=18;//molar mass for water/steam \n",
+"R1=8.314;//gas constant in KJ/kg K\n",
+"disp('total entropy change=entropy change during water temperature rise(deltaS1)+entropy change during water to steam change(deltaS2)+entropy change during steam temperature rise(deltaS3)')\n",
+"disp('deltaS1=Q1/T1,where Q1=m*Cp*deltaT')\n",
+"disp('heat added for increasing water temperature from 27 to 100 degree celcius(Q1)in KJ')\n",
+"disp('Q1=m*Cp*(T2-T1)')\n",
+"Q1=m*Cp*(T2-T1)\n",
+"disp('deltaS1=Q1/T1 in KJ/K')\n",
+"deltaS1=Q1/T1\n",
+"disp('now heat of vaporisation(Q2)=m*q in KJ')\n",
+"Q2=m*q\n",
+"disp('entropy change during phase transformation(deltaS2)in KJ/K')\n",
+"disp('deltaS2=Q2/T2')\n",
+"deltaS2=Q2/T2\n",
+"disp('entropy change during steam temperature rise(deltaS3)in KJ/K')\n",
+"disp('deltaS3=m*Cp_steam*dT/T')\n",
+"disp('here Cp_steam=R*(3.5+1.2*T+0.14*T^2)*10^-3 in KJ/kg K')\n",
+"disp('R=R1/M in KJ/kg K')\n",
+"R=R1/M\n",
+"T2=(100+273.15);//steam temperature of water in K\n",
+"T3=(400+273.15);//temperature of steam in K\n",
+"disp('now deltaS3=(m*R*(3.5+1.2*T+0.14*T^2)*10^-3)*dT/T in KJ/K')\n",
+"function y = f(T), y =(m*R*(3.5+1.2*T+0.14*T^2)*10^-3)/T , endfunction\n",
+"deltaS3 = intg(T2, T3, f) \n",
+"disp('total entropy change(deltaS)=deltaS1+deltaS2+deltaS3 in KJ/K')\n",
+"deltaS3=51.84;//approximately\n",
+"deltaS=deltaS1+deltaS2+deltaS3\n",
+"\n",
+""
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.3: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_3.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 3')\n",
+"R1=8.314;//gas constant in KJ/kg K\n",
+"M=32;//molar mass for O2 \n",
+"T1=(27+273);//initial temperature of O2 in K\n",
+"p1=125;//initial pressure of O2 in Kpa\n",
+"p2=375;//final pressure of O2 in Kpa\n",
+"Cp=1.004;//specific heat of air at constant pressure in KJ/kg K\n",
+"disp('gas constant for oxygen(R)in KJ/kg K')\n",
+"disp('R=R1/M')\n",
+"R=R1/M\n",
+"disp('for reversible process the change in entropy may be given as')\n",
+"disp('deltaS=(Cp*log(T2/T1))-(R*log(p2/p1))in KJ/kg K')\n",
+"T2=T1;//isothermal process\n",
+"deltaS=(Cp*log(T2/T1))-(R*log(p2/p1))\n",
+"disp('so entropy change=deltaS in KJ/kg K')\n",
+"deltaS"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.4: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_4.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 4')\n",
+"T1=(150+273.15);//temperature of copper block in K\n",
+"T2=(25+273.15);//temperature of sea water in K\n",
+"m=1;//mass of copper block in kg\n",
+"C=0.393;//heat capacity of copper in KJ/kg K\n",
+"disp('entropy change in universe(deltaS_universe)=deltaS_block+deltaS_water')\n",
+"disp('where deltaS_block=m*C*log(T2/T1)')\n",
+"disp('here hot block is put into sea water,so block shall cool down upto sea water at 25 degree celcius as sea may be treated as sink')\n",
+"disp('therefore deltaS_block=m*C*log(T2/T1)in KJ/K')\n",
+"deltaS_block=m*C*log(T2/T1)\n",
+"disp('heat loss by block =heat gained by water(Q)in KJ')\n",
+"disp('Q=-m*C*(T1-T2)')\n",
+"Q=-m*C*(T1-T2)\n",
+"disp('therefore deltaS_water=-Q/T2 in KJ/K')\n",
+"deltaS_water=-Q/T2\n",
+"disp('thus deltaS_universe=(deltaS_block+deltaS_water)*1000 in J/K')\n",
+"deltaS_universe=(deltaS_block+deltaS_water)*1000"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.5: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_5.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 5')\n",
+"m=1;//mass of copper block in kg\n",
+"T=(27+273);//temperature of copper block in K\n",
+"h=200;//height from which copper block dropped in sea water in m\n",
+"C=0.393;//heat capacity for copper in KJ/kg K\n",
+"g=9.81;//acceleration due to gravity in m/s^2\n",
+"disp('deltaS_universe=(deltaS_block+deltaS_seawater)')\n",
+"disp('since block and sea water both are at same temperature so,')\n",
+"disp('deltaS_universe=deltaS_seawater')\n",
+"disp('conservation of energy equation yields,')\n",
+"disp('Q-W=deltaU+deltaP.E+deltaK.E')\n",
+"disp('since in this case,W=0,deltaK.E=0,deltaU=0')\n",
+"disp('Q=deltaP.E')\n",
+"disp('change in potential energy=deltaP.E=m*g*h in J')\n",
+"deltaPE=m*g*h\n",
+"Q=deltaPE\n",
+"disp('deltaS_universe=deltaS_seawater=Q/T in J/kg K')\n",
+"deltaS_universe=Q/T\n",
+"disp('entropy change of universe(deltaS_universe)in J/kg K')\n",
+"deltaS_universe"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.6: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_6.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 6')\n",
+"m1=1;//mass of first copper block in kg\n",
+"m2=0.5;//mass of second copper block in kg\n",
+"T1=(150+273.15);//temperature of first copper block in K\n",
+"T2=(0+273.15);//temperature of second copper block in K\n",
+"Cp_1=0.393;//heat capacity for copper block 1 in KJ/kg K\n",
+"Cp_2=0.381;//heat capacity for copper block 2 in KJ/kg K\n",
+"disp('here deltaS_universe=deltaS_block1+deltaS_block2')\n",
+"disp('two blocks at different temperatures shall first attain equilibrium temperature.let equilibrium temperature be Tf')\n",
+"disp('then from energy conservation')\n",
+"disp('m1*Cp_1*(T1-Tf)=m2*Cp_2*(Tf-T2)')\n",
+"disp('Tf=((m1*Cp_1*T1)+(m2*Cp_2*T2))/(m1*Cp_1+m2*Cp_2)in K')\n",
+"Tf=((m1*Cp_1*T1)+(m2*Cp_2*T2))/(m1*Cp_1+m2*Cp_2)\n",
+"disp('hence,entropy change in block 1(deltaS1),due to temperature changing from Tf to T1')\n",
+"disp('deltaS1=m1*Cp_1*log(Tf/T1)in KJ/K')\n",
+"deltaS1=m1*Cp_1*log(Tf/T1)\n",
+"disp('entropy change in block 2(deltaS2)in KJ/K')\n",
+"disp('deltaS2=m2*Cp_2*log(Tf/T2)')\n",
+"deltaS2=m2*Cp_2*log(Tf/T2)\n",
+"disp('entropy change of universe(deltaS)=deltaS1+deltaS2 in KJ/K')\n",
+"deltaS=deltaS1+deltaS2"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.7: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_7.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 7')\n",
+"disp('NOTE=>in this question formula is derived which cannot be solve using scilab software')"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.8: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_8.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 8')\n",
+"T1=1800;//temperature of high temperature reservoir in K\n",
+"T2=300;//temperature of low temperature reservoir in K\n",
+"Q1=5;//heat addition in MW\n",
+"W=2;//work done in MW\n",
+"disp('for irreversible operation of engine,')\n",
+"disp('rate of entropy generation=Q1/T1+Q2/T2')\n",
+"disp('W=Q1-Q2=>Q2=Q1-W in MW')\n",
+"Q2=Q1-W\n",
+"disp('entropy generated(deltaS_gen)in MW')\n",
+"disp('deltaS_gen=Q1/T1+Q2/T2')\n",
+"Q1=-5;//heat addition in MW\n",
+"deltaS_gen=Q1/T1+Q2/T2\n",
+"disp('work lost(W_lost)in MW')\n",
+"disp('W_lost=T2*deltaS_gen')\n",
+"W_lost=T2*deltaS_gen"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 5.9: Engineering_Thermodynamics_by_Onkar_Singh_Chapter_5_Example_9.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"// Display mode\n",
+"mode(0);\n",
+"// Display warning for floating point exception\n",
+"ieee(1);\n",
+"clear;\n",
+"clc;\n",
+"disp('Engineering Thermodynamics by Onkar Singh Chapter 5 Example 9')\n",
+"T1=500;//temperature of system in K\n",
+"T2=300;//temperature of reservoir in K\n",
+"disp('system and reservoir can be treated as source and sink.device thought of can be a carnot engine operating between these two limits.maximum heat available from system shall be the heat rejected till its temperature drops from 500 K to 300 K')\n",
+"disp('therefore,maximum heat(Q1)=(C*dT)in J')\n",
+"disp('here C=0.05*T^2+0.10*T+0.085 in J/K')\n",
+"disp('so Q1=(0.05*T^2+0.10*T+0.085)*dT')\n",
+"function y = f(T), y = (0.05*T^2+0.10*T+0.085), endfunction\n",
+"Q1 = intg(T1, T2, f)\n",
+"Q1=-Q1\n",
+"disp('entropy change of system,deltaS_system=C*dT/T in J/K')\n",
+"disp('so deltaS_system=(0.05*T^2+0.10*T+0.085)*dT/T')\n",
+"function y = k(T), y = (0.05*T^2+0.10*T+0.085)/T, endfunction\n",
+"deltaS_system = intg(T1, T2, k)\n",
+"disp('deltaS_reservoir=Q2/T2=(Q1-W)/T2')\n",
+"disp('also,we know from entropy principle,deltaS_universe is greater than equal to 0')\n",
+"disp('deltaS_universe=deltaS_system+deltaS_reservoir')\n",
+"disp('thus,upon substituting,deltaS_system+deltaS_reservoir is greater than equal to 0')\n",
+"disp('W is less than or equal to(Q1+deltaS_system*T2)/1000 in KJ')\n",
+"W=(Q1+deltaS_system*T2)/1000\n",
+"disp('hence maximum work=W in KJ')\n",
+"W"
+   ]
+   }
+],
+"metadata": {
+		  "kernelspec": {
+		   "display_name": "Scilab",
+		   "language": "scilab",
+		   "name": "scilab"
+		  },
+		  "language_info": {
+		   "file_extension": ".sce",
+		   "help_links": [
+			{
+			 "text": "MetaKernel Magics",
+			 "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+			}
+		   ],
+		   "mimetype": "text/x-octave",
+		   "name": "scilab",
+		   "version": "0.7.1"
+		  }
+		 },
+		 "nbformat": 4,
+		 "nbformat_minor": 0
+}