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 {
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	   "source": [
       "# Chapter 1: Fundamentals Concepts and Definitions of Thermodynamics"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.11: Power_of_the_engine.sce"
		   ]
		  },
  {
"cell_type": "code",
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	   "metadata": {
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	   },
	   "outputs": [],
"source": [
"// Exa 1.11\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"V =  50;// km per hr\n",
"V = V * (1000/3600);// in m per sec\n",
"F = 900;// in N\n",
"P = F * V;// in watt\n",
"P = P *10^-3;// in kW\n",
"disp(P,'Power of the engine of a car in kW is : ');\n",
"H = P * 60;// in kJ \n",
"disp(H,'Heat equivalent of work per minute in kJ is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.12: Power_required.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 1.12\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"E_air = 200-100;// in kJ/kg\n",
"E_lost = 40;// in kJ/kg\n",
"E_total = E_air + E_lost;// in kJ per kg\n",
"M = 0.5;// mass flow rate in kg per s\n",
"P = M * E_total;// in kJ/s\n",
"disp(P,'Power required for an air mass flowin kJ/s is :');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.13: Specific_heat.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 1.13\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"m_b = 1;// in kg\n",
"t_ib = 80;//in degree c\n",
"m_w = 10;// in kg\n",
"t_iw = 25;// in degree c\n",
"del_t = 5;// in degree c\n",
"S_w = 4.18;//  in kJ/kg\n",
"t_equ = (t_iw + del_t);// in degree c\n",
"// Heat loss by metal = Heat gained by water\n",
"S_b = m_w * S_w * (t_equ - t_iw)/(m_b * (t_ib - t_equ));// in kJ/kg-K\n",
"disp(S_b,'Specific heat of metal block in kJ/kg-K is');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.15: Convert_reading_in_kPa.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 1.15\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"P_gauge = 90;// in cm of hg\n",
"P_atm = 760;// in mm of hg\n",
"P_atm = 76;// in cm of hg\n",
"P_abs = P_gauge + P_atm;// in cm of hg\n",
"P_abs = P_abs * (101.32/76);// in kPa\n",
"disp(P_abs,'Reading of pressure in kPa');\n",
"// Part (b)\n",
"P_vacuum = 40;// in cm of hg\n",
"P_abs = P_atm - P_vacuum;// in cm of hg\n",
"P_abs = P_abs * (101.32/76);// in kpa\n",
"disp(P_abs,'Reading of pressure to kpa');\n",
"// Part (c)\n",
"Rho = 1000;// in kg per m^3\n",
"g = 9.81;// \n",
"h = 1.2;// in m \n",
"P_gauge = Rho * g * h;// in N m^2\n",
"P_gauge= P_gauge*10^-3;// in kPa\n",
"P_atm = 101.32;// in kPa\n",
"P_abs = P_gauge + P_atm;// in kpa\n",
"disp(P_abs,'Reading of pressure in kPa');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.16: Depth_of_atmosphere.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 1.16\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"g=9.81;// in m/s^2\n",
"P=1.0332*10^5;// in kN/m^2\n",
"v='(2.3*10^4/p)^1/1.4';// given expression\n",
"H= 1/g*(2.3*10^4)^(1/1.4)*integrate('(1/p)^(1/1.4)','p',0,P);// in m\n",
"disp(H*10^-3,'The value of H in km is : ')\n",
"\n",
"// Note: There is the calculation error in the book in this question, so the answer in the book is wrong."
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.2: Absolute_pressure.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 1.2\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"Pvacc = 700;// mm of hg\n",
"Patm = 760;// mm of hg\n",
"Pabs = Patm - Pvacc;// mm of hg\n",
"disp(Pabs/760,'Absolute pressure in in kg/cm^2 is :');// as 1kg/cm^2= 760 mm of Hg\n",
"disp(Pabs*1.01325/760,'Absolute pressure in bar is : ');// as 1.01325 bar = 760 mm of Hg\n",
"disp(Pabs*1.01325/760*10^2,'Absolute pressure in in kPa');// as 1 bar = 10^2 kPa"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.3: Tank_pressure.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//Exa 1.3\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"Patm = 101;// kpa\n",
"Rho = 13.6 * 10^3;// in kg per m^3\n",
"h = 250;// in cm\n",
"h = h*10^-2;// in m\n",
"g = 9.806;\n",
"p = Rho * g * h;// in N/m^2\n",
"p= p*10^-3;// in kPa\n",
"// Total pressure in tank\n",
"p = p + Patm;// in kpa\n",
"p = p*10^-3;// in Mpa\n",
"disp(p,'Total pressure in tank in Mpa');"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.4: Work_done_by_the_piston.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 1.4\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"m = 1.5;// in kg\n",
"pi = 0.1;// in MPa\n",
"pi= pi*10^6;// in Pa\n",
"pf = 0.7;// in MPa\n",
"pf= pf*10^6;// in Pa\n",
"rho_i = 1.16;// kg per m^3\n",
"vi = m/rho_i;// in m^3\n",
"WorkDone= pi*vi*log(pi/pf);// in J\n",
"disp(WorkDone*10^-3,'Work done in kJ is : ')\n",
"if WorkDone<0 then\n",
"disp('The -ve sign indicates work is done on the system, hence');\n",
"disp((WorkDone*10^-3),'The work done by the piston in kJ is : ')\n",
"end"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.5: Work_done.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exav1.5\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"p = 1.0;// in Mpa\n",
"p = p*10^6;// in N per m^2\n",
"del_v = 1.5;//m^3 per min\n",
"del_v = del_v*60;// m^3 per h\n",
"W = p * del_v;// W stands for work done in J\n",
"W = W*10^-6;// in MJ\n",
"disp(W,'Work done by the pump upon the water in an hour in MJ is : '); "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.6: Height_from_which_the_mass_should_fall.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 1.6\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"// W = 2*g*h\n",
"// due to stirring of water\n",
"g = 9.81;\n",
"J = 4.1868*10^3;\n",
"m = (0.2+10*10^-3)*10^3;// in gm\n",
"s= 1;// in cal per gm°C\n",
"del_T = 2;// in ° C\n",
"H = m * s * del_T;// in cal\n",
"H = H*10^-3;// in kcal\n",
"// W = JH and W= 2*g*h\n",
"h = J*H/(2 * g);// in m\n",
"disp(h,'The height from the mass should be fall in meter is :');\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.7: Power_of_feed_pump.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Exa 1.7\n",
"clc;\n",
"clear;\n",
"close;\n",
"// Given data\n",
"// mass of 1 litr of water is 1 kg. so\n",
"m = 5000;// in kg\n",
"h = 10-1;// in m\n",
"g= 9.81;//\n",
"PE = m * g * h;// in N m\n",
"PE = PE*10^-3;// in kj\n",
"Eta = 0.85;\n",
"// Eta = energy output/energy input\n",
"E_input = PE/Eta;// in Kj\n",
"E_input = E_input*10^3;// in J\n",
"t = 45;// time in min\n",
"t = t*60;// in sec\n",
"P = E_input/t;// in J/s\n",
"P = P*10^-3;// in kW\n",
"disp(P,'Power required for the feed pump in kW is :');\n",
""
   ]
   }
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