path: root/Basic_Engineering_Thermodynamics_by_R_Joel/1-General_Introduction.ipynb

{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 1: General Introduction"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.10: Thermal_efficiency.sce"
		   ]
		  },
  {
"cell_type": "code",
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"clear;\n",
"clc;\n",
"disp('Example 1.10');\n",
"//  Given values\n",
"m_dot = 3.045; //  use of coal, [tonne/h]\n",
"c = 28; // calorific value of the coal, [MJ/kg]\n",
"P_out = 4.1; //  output of turbine, [MW]\n",
"//  solution\n",
"m_dot = m_dot*10^3/3600; // [kg/s]\n",
"P_in = m_dot*c; //  power input by coal, [MW]\n",
"n = P_out/P_in; //  thermal efficiency formula\n",
"mprintf('\n Thermal efficiency of the plant is  =  %f \n',n);\n",
"//End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.11: Power_output.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
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	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 1.11');\n",
"//  Given values\n",
"v = 50; //  speed, [km/h]\n",
"F = 900; //  Resistance to the motion of a car\n",
"//  solution\n",
"v = v*10^3/3600; //  [m/s]\n",
" Power = F*v; //  Power formula, [W]\n",
"mprintf('\n The power output of the engine is  =  %f  kW\n',Power*10^-3);\n",
" \n",
" //  End\n",
" "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.12: Power_output.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 1.12');\n",
"//  Given values\n",
"V = 230; //  volatage, [volts]\n",
"I = 60; //  current, [amps]\n",
"n_gen = .95; //  efficiency of generator\n",
"n_eng = .92; //  efficiency of engine\n",
"//  solution\n",
"P_gen = V*I; //  Power delivered by generator, [W]\n",
"P_gen=P_gen*10^-3; //  [kW]\n",
"P_in_eng=P_gen/n_gen;//Power input from engine,[kW]\n",
"P_out_eng=P_in_eng/n_eng;//Power output from engine,[kW]\n",
"mprintf('\n The power output from the engine is  =  %f  kW\n',P_out_eng);\n",
"//  End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.13: Current.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
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	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 1.13');\n",
"//  Given values\n",
"V = 230; // Voltage, [volts]\n",
"W = 4; //  Power of heater, [kW]\n",
"//  solution\n",
"//  using equation P=VI\n",
"I = W/V; //  current, [K amps]\n",
"mprintf('\n The current taken by heater is  =  %f  amps \n',I*10^3);\n",
"//  End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.14: Mass_of_coal_burnt.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 1.14');\n",
"//  Given values\n",
"P_out = 500; //  output of power station, [MW]\n",
"c = 29.5; //  calorific value of coal, [MJ/kg]\n",
"r=.28; \n",
"//  solution\n",
"//  since P represents only 28 percent of energy available from coal\n",
"P_coal = P_out/r; //  [MW]\n",
" \n",
"m_coal = P_coal/c; //  Mass of coal used, [kg/s]\n",
"m_coal = m_coal*3600; //  [kg/h]\n",
"//After one hour\n",
"m_coal = m_coal*1*10^-3; //  [tonne]\n",
"mprintf('\n Mass of coal burnt by the power station in 1 hour is  =  %f  tonne \n',m_coal);\n",
"//  End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.1: Work_done.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear ;\n",
"clc;\n",
"disp('Example 1.1');\n",
"// Given values\n",
"P = 700;   //pressure,[kN/m^2]\n",
"V1 = .28;   //initial volume,[m^3]\n",
"V2 = 1.68;   //final volume,[m^3]\n",
"//solution\n",
"W = P*(V2-V1);// // Formula for work done at constant pressure is, [kJ]\n",
"mprintf('\n The Work done is = %f MJ\n',W*10^-3);\n",
"//End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.2: Volume_of_the_gas.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
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	   "outputs": [],
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"clear;\n",
"clc;\n",
"disp('Example 1.2');\n",
"//Given values\n",
"P1 = 138;  // initial pressure,[kN/m^2]\n",
"V1 = .112;  //initial volume,[m^3]\n",
"P2 = 690; //  final pressure,[kN/m^2]\n",
"Gama=1.4; //  heat capacity ratio\n",
"//  solution\n",
"//  since gas is following, PV^1.4=constant,hence\n",
"V2 =V1*(P1/P2)^(1/Gama); //   final volume, [m^3] \n",
"mprintf('\n The new volume of the gas is = %f m^3\n',V2)\n",
"//End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.3: Work_done.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 1.3');\n",
"//  Given values\n",
"P1 = 2070;  //  initial pressure,  [kN/m^2]\n",
"V1 = .014;  //  initial volume,  [m^3]\n",
"P2 = 207;  //  final pressure, [kN/m^2]\n",
"n=1.35;  //  polytropic index\n",
"//  solution\n",
"//  since gas is following PV^n=constant\n",
"//  hence \n",
"V2 = V1*(P1/P2)^(1/n);  // final volume,  [m^3]\n",
"//  calculation of workdone\n",
"W=(P1*V1-P2*V2)/(1.35-1);  // using work done formula for polytropic process, [kJ]\n",
"mprintf('\n The Work done by gas during expansion is  =  %f kJ\n',W);\n",
"//End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.4: Final_Pressure_and_work_done.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 1.4');\n",
"//  Given values\n",
"P1 = 100;  //  initial pressure, [kN/m^2]\n",
"V1 = .056;  //  initial volume,  [m^3]\n",
"V2 = .007;  //  final volume,  [m^3]\n",
"//  To know  P2\n",
"//  since process is hyperbolic so, PV=constant\n",
"//  hence\n",
"P2 = P1*V1/V2;  //  final pressure, [kN/m^2]\n",
"mprintf('\n  The final pressure is  =  %f  kN/m^2\n',P2);\n",
"// calculation of workdone\n",
"W = P1*V1*log(V2/V1); //  formula for work done in this process, [kJ]\n",
"mprintf('\n  Work done on the gas is  =  %f  kJ\n',W);\n",
"//End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.5: Heat_transfer.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
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	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 1.5');\n",
"//  Given values\n",
"m = 5; //  mass,  [kg]\n",
"t1 = 15; //  inital temperature, [C]\n",
"t2 = 100; //  final temperature, [C]\n",
"c = 450; //  specific heat capacity, [J/kg K]\n",
"//  solution\n",
"//  using heat transfer equation,[1]\n",
"Q = m*c*(t2-t1); //  [J]\n",
"mprintf('\n The heat required is  =  %f  kJ\n',Q*10^-3);\n",
"//End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.6: Heat_transfer.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
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"clear;\n",
"clc;\n",
"disp('Example 1.6');\n",
"//  Given values\n",
"m_cop = 2; //  mass of copper vessel, [kg]\n",
"m_wat = 6; //  mass of water, [kg]\n",
"c_wat = 4.19; //  specific heat capacity of water,  [kJ/kg K]\n",
"t1 = 20; //  initial temperature, [C]\n",
"t2 = 90; //  final temperature, [C]\n",
"// From the table of average specific heat capacities\n",
"c_cop = .390;  // specific heat capacity of copper,[kJ/kg k]\n",
"// solution\n",
"Q_cop = m_cop*c_cop*(t2-t1); //  heat required by copper vessel, [kJ]\n",
"Q_wat = m_wat*c_wat*(t2-t1); //  heat required by water, [kJ]\n",
"// since there is no heat loss,so total heat transfer is sum of both\n",
"Q_total = Q_cop+Q_wat ; //  [kJ]\n",
"mprintf(' \n Required heat transfer to accomplish the change  =  %f  kJ\n',Q_total);\n",
"//End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.7: Temperature.sce"
		   ]
		  },
  {
"cell_type": "code",
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"clear;\n",
"clc;\n",
"disp('Example 1.7');\n",
"//  Given values\n",
"m = 10; //  mass of iron casting, [kg]\n",
"t1 = 200; //  initial temperature, [C]\n",
"Q = -715.5; //  [kJ], since heat is lost in this process\n",
"// From the table of average specific heat capacities\n",
"c = .50; //  specific heat capacity of casting iron, [kJ/kg K]\n",
"//  solution\n",
"//  using heat equation\n",
"//  Q = m*c*(t2-t1)\n",
"t2 = t1+Q/(m*c); // [C]\n",
"mprintf('\n The final temperature is  t2 =  %f C\n',t2);\n",
"// End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.8: Specific_heat_capacity.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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	   "outputs": [],
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"clear;\n",
"clc;\n",
"disp('Example 1.8');\n",
" \n",
"// Given values\n",
"m = 4; //  mass of the liquid, [kg]\n",
"t1 = 15; //  initial temperature, [C]\n",
"t2 = 100; //  final temperature, [C]\n",
"Q = 714; //  [kJ],required heat to accomplish this change\n",
"//  solution\n",
"//  using heat equation\n",
"//  Q=m*c*(t2-t1)\n",
"//  calculation of c\n",
"c=Q/(m*(t2-t1)); //  heat capacity, [kJ/kg K] \n",
"mprintf('\n The specific heat capacity of the liquid is  c =  %f  kJ/kg K\n',c);\n",
"//End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.9: Power_output_and_energy_rejected.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
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"clear;\n",
"clc;\n",
"disp('Example 1.9');\n",
"//  Given values\n",
"m_dot = 20.4; //  mass flowrate of petrol, [kg/h]\n",
"c = 43; //  calorific  value of petrol, [MJ/kg]\n",
"n = .2; //  Thermal efficiency of engine\n",
"//  solution\n",
"m_dot = 20.4/3600; // [kg/s]\n",
"c = 43*10^6; //  [J/kg]\n",
"//  power output\n",
"P_out = n*m_dot*c; //  [W]\n",
"mprintf('\n The power output of the engine is  =  %f  kJ\n',P_out*10^-3);\n",
" \n",
"//  power rejected\n",
"P_rej = m_dot*c*(1-n); //  [W]\n",
"P_rej = P_rej*60*10^-6; //  [MJ/min]\n",
"mprintf('\n The energy rejected by the engine is  =   %f  MJ/min \n',P_rej);\n",
"//End"
   ]
   }
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