{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 3: First Law of Thermodynamics"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.10: To_calculate_the_pump_work_required.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.10\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"V = 0.3;//Volume of the tank in m^3\n",
"P1 = 1;//Initial pressure of the tank in atm\n",
"P2 = 0;//Final pressure of the tank in atm\n",
"T = 298;//Temperature of the tank in K\n",
"t = 10;//evacuation time in min\n",
"\n",
"//delN=(V/(R*T)*delP)..(a) change in moles as V and T are constant\n",
"//delW=delN*R*T*lnP..(b)pump work required\n",
"//From (a)&(b),delW=V*delP*lnP\n",
"\n",
"//To calculate the pump work required\n",
"//On doing integration of dW we will get\n",
"\n",
"W = V*(P1-P2);//pump work done in J/sec\n",
"W1=(W*(1.033*10^4))/(75*600);\n",
"mprintf('The pump work required is %f hp',W1);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.11: To_calculate_the_quality_of_exit_steam.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.11\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"H1 = 680.6;//Enthalpy of entering steam at 6Kgf/cm^2 &200 deg cel in Kcal/Kg\n",
"u1 = 60;//velocity at which steam entered the nozzle in m/sec\n",
"u2 = 600;//velocity at which steam left the nozzle in m/sec\n",
"g = 9.8;\n",
"Hg = 642.8; Hlq = 110.2;//Enthalpy of saturated vapour & saturated liquid at 1.46 Kgf/cm^2 respectively\n",
"\n",
"//To calculate the quality of exit steam\n",
"H2 = H1+((u1^2)-(u2^2))/(2*g*427);//enthalpy of leaving steam in Kcal/Kg\n",
"x = (H2-Hlq)/(Hg-Hlq);\n",
"mprintf('The quality of exit steam is %f percent',x*100);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.12: To_calculate_the_internal_energy_of_the_steam_in_the_chamber.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.12\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"W = 0;//pump work\n",
"Mi = 0;//chamber is initially evacuated\n",
"M2 = 0;//no exist stream\n",
"H1 = 684.2;//enthalpy of steam at 200 deg cel & 3 Kgf/cm^2\n",
"\n",
"//To calculate the internal energy of the steam in the chamber\n",
"//Q=150*m1;.. (a) heat lost from the chamber in Kcal/Kg\n",
"//m1=mf;..(b) mass of steam added from large pipe is equal to steam in chamber\n",
" //H1*M1-Q=Mf*Ef; using (a)&(b)\n",
" Ef = H1-150;\n",
" mprintf('The internal energy of steam in chamber is %f Kcal',Ef);\n",
" //end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.13: To_calculate_the_final_weight_and_the_final_temperature_of_the_air_in_the_tank.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.13\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"//Q=W=delPE=delKE=0;\n",
"//M2=0; no exit stream\n",
"Ti = 288;//initial temperature in K\n",
"H = 7*Ti;//enthalpy of air in Kcal/Kgmole\n",
"Ei = 5*Ti;// initial internal energy of air in Kcal/Kgmole\n",
"//Ef=5*Tf;Final internal energy of air in Kcal/Kgmole\n",
"Pi = 0.3;//initial pressure in atm\n",
"V = 0.57;//volume of the tank in m^3\n",
"R = 848;//gas constant in mKgf/Kg mole K\n",
"Pf = 1;//final prssure in atm\n",
"\n",
"//To calculate the final weight and the final temperature of the air in the tank\n",
"Mi = (Pi*V*1.03*10^4)/(R*Ti);//initial quantity of air in tank in Kg mole\n",
"//Tf=(Pf*V*1.033*10^4)/(Mf*R)..(a) final temperature,Mf=final quantity of air in tank in Kg mole\n",
"//M1=Mf-Mi..(b) M1 is mass of steam added in Kg mole\n",
"//H*M1=(Ef*Mf)-(Ei*Mi)\n",
"//H*M1=((5*Pf*V*1.033*10^4)/(Tf*R))*Tf-(Ei*Mi)...(c)\n",
"A = [1 -1;0 -H];\n",
"B = [Mi;((Ei*Mi)-((5*Pf*V*1.03*10^4)/R))];\n",
"x = A\B;\n",
"Mf = x(1);\n",
"mprintf('The final weight of air in the tank is%f Kg',Mf);\n",
"\n",
"Tf = (Pf*V*1.03*10^4)/(Mf*R);\n",
"mprintf('\n The final temperature of air in the tank is %f K',Tf);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.14: Theoretical_problem.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.14\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 3.1: To_calculate_the_change_in_internal_energy.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.1\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"W = -((2*745.6*(10^-3)/4.18)*3600);//work added to the system in Kcal/hr\n",
"m = 10;//Amount of fluid in tank in Kg\n",
"Q = -378;//Heat losses from the system in Kcal/hr\n",
"\n",
"//To calculate the change in internal energy\n",
"delE=(Q-W)/m;// Change in internal energy in Kcal/hr kg\n",
"mprintf('Change in Internal energy is %f Kcal/hr Kg',delE);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.2: To_calculate_the_change_in_internal_energy.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.2\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"n = 1;//kg moles of a gas\n",
"Cv = 5;//specific heat in Kcal/Kgmole\n",
"delT = 15;//increase in temperature in deg celsius\n",
"\n",
"//To calculate the change in internal energy\n",
"Q = n*Cv*delT;//heat given to the system in Kcal\n",
"W = 0;//work done\n",
"delE = Q-W;//Change in internal energy\n",
"mprintf('Change in internal energy is %f Kcal',delE);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.3: To_calculate_the_work_done_by_the_expanding_gas_and_increase_in_internal_energy.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.3\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"P = 1;//constant pressure throughout the process in atm\n",
"T1 = 273;//Initial temperature in K\n",
"T2 = 373;//Final temperature in K\n",
"V1 = 0;//Volume of liquid water or initial volume\n",
"V0 = 22.4;//volume of vapour at standard condition in cubic meter\n",
"Q = 9.7//Heat of vapourisation in Kcal\n",
"\n",
"//To calculate the work done by the expanding gas and increase in internal energy\n",
"//(i)Calculation of work done\n",
"V2 = 22.4*(T2/T1)*(P)*(10^-3);//Volume of final vapour in cubic meter\n",
"w = P*(V2-V1);//Work done in atm cubic meter\n",
"W = w*(1.03*10^4)/427;//Work done in Kcal\n",
"mprintf('(i)Work done by the expanding gas is %f Kcal',W);\n",
"\n",
"//(ii)Calculation of change in internal energy\n",
"delE = Q-W;\n",
"mprintf('\n\n (ii)Increase in internal energy is %f Kcal',delE);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.4: To_calculate_the_time_taken_for_the_gas_to_attain_10_atm.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.4\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"W = 0;//work done during the process\n",
"P1 = 1;//Initial pressure in atm\n",
"P2 = 10;//Final pressure in atm\n",
"V2 = V1;//Initial & final volume are equal\n",
"Cv = 0.23//specific heat at constant volume in Kcal/Kg deg K\n",
"//(delQ/delT)=Q\n",
"Q = 1.3;//Rate of heat addition in Kcal/min\n",
"m = 2.5//Weight of an ideal gas in Kg\n",
"T1 = 298//Initial temperature in Kelvin\n",
"\n",
"//To calculate the time taken for the gas to attain 10 atm\n",
"//Q = m*Cv*(delT/delt)=1.3\n",
"T2 = (P2*T1)/(P1);//Final temperature in Kelvin\n",
"t = ((m*Cv)/1.3)*(T2-T1);//time taken in minutes\n",
"mprintf('The time taken to attain a pressure of 10 atm is %f hrs',t/60);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.5: Theoretical_problem.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.5\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 3.6: To_calculate_the_flow_work_per_kg_of_air.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.6\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"R = 1.98;//gas constant in kcal/Kgmole deg K\n",
"T = 293;//Temperature in K\n",
"M = 29;//Molecular weight of air\n",
"\n",
"//To calculate the flow work per kg of air\n",
"//W=(P*V)=(R*T)\n",
"W = R*T;//Flow work in Kcal/Kg mole\n",
"W1 = W/M;\n",
"mprintf('Flow work is %f Kcal/Kg',W1);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.7: To_calculate_the_horse_power_output_of_the_turbine.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.7\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"m = 5000;//Amount of steam recived per hour in Kg\n",
"H1 = 666;//Specific enthalpy when steam entered in the turbine in Kcal/Kg\n",
"H2 = 540;//Specific enthalpy when steam left the turbine in Kcal/Kg\n",
"u1 = 3000/60;//velocity at which steam entered in m/sec\n",
"u2 = 600/60;//velocity at which steam left in m/sec\n",
"Z1 = 5;//height at which steam entered in m\n",
"Z2 = 1;//height at which steam left in m\n",
"Q = -4000;//heat lost in Kcal\n",
"g = 9.81;\n",
"\n",
"//To calculate the horsepuwer output of the turbine\n",
"delH = H2-H1;//change in enthalpy in Kcal\n",
"delKE = ((u2^2)-(u1^2)/(2*g))/(9.8065*427);//change in kinetic energy in Kcal; 1kgf = 9.8065 N\n",
"delPE = ((Z2-Z1)*g)/(9.8065*427);//change in potential energy in Kcal\n",
"W = -(m*(delH+delKE+delPE))+Q;//work delivered in Kcal/hr\n",
"W1 = W*(427/(3600*75));//work delivered by turbine in hp\n",
"mprintf('Work delivered by turbine is %f hp',W1);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.8: EX3_8.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.8\n",
"clear;\n",
"clc;\n",
"\n",
"//Given\n",
"m = 183;//rate of water flow in Kg/min\n",
"H1 = 95;//enthalpy of storage tank 1 in Kcal/Kg\n",
"h = 15;//height difference between two storage tanks in m\n",
"Q = -10100;//extraced heat from storage tank 1 in a heat exchanger in Kcal/min\n",
"W = -2;//work delivered by motor in hp\n",
"\n",
"// To find out the  enthalpy of water tank2 and the temperature of water in the second tank\n",
"delPE = h/427;//change in potential energy in Kcal/Kg\n",
"delKE = 0;//change in kinetic energy\n",
"W1 = W*(75/427);//work delivered by motor in Kcal/sec\n",
"W2 = W1*60;//work delivered by motor in Kcal/min\n",
"H2 = ((Q+W2)/m)-delKE-delPE+H1;//enthalpy of storage tank 2 in Kcal/Kg\n",
"mprintf('The enthalpy of storage tank 2 is %f Kcal/Kg',H2);\n",
"\n",
"//The enthalpy H2=39.66 corresponds to the temperature T according to steam table\n",
"T=40;//Temperature is in deg celsius\n",
"mprintf('\n The temperature of water in the second tank is %d deg celsius',T);\n",
"//end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 3.9: To_calculate_the_mass_of_steam_required.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Chemical Engineering Thermodynamics\n",
"//Chapter 3\n",
"//First Law of Thermodynamics\n",
"\n",
"//Example 3.9\n",
"clear;\n",
"clc;\n",
"\n",
"//To calculate the mass of steam required\n",
"//Given\n",
"m2 = 100;//mass of water to be heated\n",
"//From diagram, \n",
"//m3 = m1+m2;..(a)\n",
"//Hs = H1;..(b) since throttling is a constant enthalpy process\n",
"//m3*H3-(m1*H1+m2*H2)=0;..(c) since delH=0\n",
"\n",
"//From steam tables, \n",
"Hs = 681.7//enthalpy of steam at 200 deg cel bleeded at the rate of 5Kgf/(cm^2) in Kcal/Kg\n",
"H2 = 5.03;//enthalpy of liquid water at 5 deg cel\n",
"H3 = 64.98;//enthalpy of liquid water at 65 deg cel\n",
"//from equn (a),(b)&(c);(page no 80)\n",
"m1 = ((H3-H2)/(Hs-H3))*m2;//mass of steam required in Kg  (page no 80)\n",
"mprintf('The mass of steam required to heat 100 Kg of water is %f Kg',m1);\n",
"//end "
   ]
   }
],
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