{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 10: Steam plant"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.10: mass_and_thermal_efficiency.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 10.10');\n",
"\n",
"// aim : To determine\n",
"// (a) the mass of steam bled to each feed heater in kg/kg of supply steam\n",
"// (b) the thermal efficiency of the arrangement\n",
"\n",
"// given values\n",
"P1 = 7;// steam initial pressure, [MN/m^2]\n",
"T1 = 273+500;// steam initil temperature, [K]\n",
"P2 = 2;// pressure at stage 1, [MN/m^2]\n",
"P3 = .5;// pressure at stage 2, [MN/m^2]\n",
"P4 = .05;// condenser pressure,[MN/m^2]\n",
"SE = .82;// stage efficiency of turbine\n",
"\n",
"// solution\n",
"// from the enthalpy-entropy chart(Fig10.23) values of specific enthalpies are\n",
"h1 = 3410;// [kJ/kg]\n",
"h2_prim = 3045;// [kJ/kg]\n",
"// h1-h2=SE*(h1-h2_prim), so\n",
"h2 = h1-SE*(h1-h2_prim);// [kJ/kg]\n",
"\n",
"h3_prim = 2790;// [kJ/kg]\n",
"// h2-h3=SE*(h2-h3_prim), so\n",
"h3 = h2-SE*(h2-h3_prim);// [kJ/kg]\n",
"\n",
"h4_prim = 2450;// [kJ/kg]\n",
"// h3-h4 = SE*(h3-h4_prim), so\n",
"h4 = h3-SE*(h3-h4_prim);// [kJ/kg]\n",
"\n",
"// from steam table\n",
"// @ 2 MN/m^2\n",
"hf2 = 908.6;// [kJ/kg]\n",
"// @ .5 MN/m^2\n",
"hf3 = 640.1;// [kJ/kg] \n",
"// @ .05 MN/m^2\n",
"hf4 = 340.6;// [kJ/kg]\n",
"\n",
"// (a) \n",
"// for feed heater1\n",
"m1 = (hf2-hf3)/(h2-hf3);// mass of bled steam, [kg/kg supplied steam]\n",
"// for feed heater2\n",
"m2 = (1-m1)*(hf3-hf4)/(h3-hf4);// \n",
"mprintf('\n (a) The mass of steam bled in feed heater 1 is  =  %f  kg/kg supply steam\n',m1);\n",
"mprintf('\n      The mass of steam bled in feed heater 2 is  =  %f kg/kg supply steam\n',m2);\n",
"\n",
"// (b)\n",
"W = (h1-h2)+(1-m1)*(h2-h3)+(1-m1-m2)*(h3-h4);// theoretical work done, [kJ/kg]\n",
"Eb = h1-hf2;// energy input in the boiler, [kJ/kg]\n",
"TE1 = W/Eb;// thermal efficiency\n",
"mprintf('\n (b) The thermal efficiency of the arrangement is  =  %f percent\n',TE1*100);\n",
"\n",
"// If there is no feed heating\n",
"hf5 = hf4;\n",
"h5_prim = 2370;// [kJ/kg]\n",
"// h1-h5 = SE*(h1-h5_prim), so\n",
"h5 = h1-SE*(h1-h5_prim);// [kJ/kg]\n",
"Ei = h1-hf5;//energy input, [kJ/kg]\n",
"W = h1-h5;//  theoretical work, [kJ/kg]\n",
"TE2 = W/Ei;// thermal efficiency\n",
"mprintf('\n      The thermal efficiency if there is no feed heating is  =  %f percent\n',TE2*100);\n",
"\n",
"//  End "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.1: equivalent_evaporation.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 10.1');\n",
"\n",
"//  aim : To determine\n",
"//   the equivalent evaporation\n",
"\n",
"//  Given\n",
"P = 1.4;//  [MN/m^2]\n",
"m = 8;//  mass of water,[kg]\n",
"T1 = 39;//  entering temperature,[C]\n",
"T2 = 100;//  [C]\n",
"x = .95;//dryness fraction \n",
"\n",
"//  solution\n",
"hf = 830.1;//  [kJ/kg]\n",
"hfg = 1957.7;//  [kJ/kg]\n",
"//  steam is wet so specific enthalpy of steam is\n",
"h = hf+x*hfg;//  [kJ/kg]\n",
"\n",
"//  at 39 C\n",
"h1 = 163.4;//  [kJ/kg]\n",
"//  hence\n",
"q = h-h1;//  [kJ/kg]\n",
"Q = m*q;//  [kJ]\n",
"\n",
"evap = Q/2256.9;//  equivalent evaporation[kg steam/(kg coal)]\n",
"\n",
"mprintf('\n The equivalent evaporation, from and at 100 C is  =  %f  kg steam/kg coal\n ',evap);\n",
"\n",
"// End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.2: mass_fraction_of_enthalpy_drop_and_heat_transfer.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 10.2');\n",
"\n",
"// aim : To determine \n",
"//  the mass of oil used per hour and the fraction of enthalpy drop through the turbine\n",
"//  heat transfer available per kilogram of exhaust steam\n",
"\n",
"//  Given values\n",
"ms_dot = 5000;// generation of steam, [kg/h]\n",
"P1 = 1.8;// generated steam pressure, [MN/m^2]\n",
"T1 = 273+325;// generated steam temperature, [K]\n",
"Tf = 273+49.4;// feed temperature, [K]\n",
"neta = .8;// efficiency of boiler plant \n",
"c = 45500;// calorific value, [kJ/kg]\n",
"P = 500;// turbine generated power, [kW]\n",
"Pt = .18;// turbine exhaust pressure, [MN/m^2]\n",
"x = .98;// dryness farction of steam\n",
"\n",
"//  solution\n",
"//  using steam table at 1.8 MN/m^2\n",
"hf1 = 3106;// [kJ/kg]\n",
"hg1 = 3080;// [kJ/kg]\n",
"//  so\n",
"h1 = hf1-neta*(hf1-hg1);// [kJ/kg]\n",
"//  again using steam table specific enthalpy of feed water is\n",
"hwf = 206.9;// [kJ/kg]\n",
"h_rais = ms_dot*(h1-hwf);// energy to raise steam, [kJ]\n",
"\n",
"h_fue = h_rais/neta;// energy from fuel per hour, [kJ]\n",
"m_oil = h_fue/c;// mass of fuel per hour, [kg]\n",
"\n",
"//  from steam table at exhaust\n",
"hf = 490.7;// [kJ/kg]\n",
"hfg = 2210.8;//  [kJ/kg]\n",
"//  hence\n",
"h = hf+x*hfg;// [kJ/kg]\n",
"//  now\n",
"h_drop = (h1-h)*ms_dot/3600;// specific enthalpy drop in turbine [kJ]\n",
"f = P/h_drop;// fraction ofenthalpy drop converted into work\n",
"//  heat transfer available in exhaust is\n",
"Q = h-hwf;// [kJ/kg]\n",
"mprintf('\n The mass of oil used per hour is  =  %f  kg\n',m_oil);\n",
"mprintf('\n The fraction of the enthalpy drop through the turbine that is converted into useful work is  =  %f\n',f);\n",
"mprintf('\n The heat transfer available in exhaust steam above 49.4 C is  =  %f  kJ/kg\n',Q);\n",
"\n",
"//  End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.3: efficiency_equivalent_evaporation_and_coal_consumption.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 10.3');\n",
"\n",
"//  aim : To determine\n",
"//  (a) the thermal efficiency of the boiler\n",
"//  (b) the equivalent evaporation of the boiler\n",
"//  (c) the new coal consumption \n",
"\n",
"//  given values\n",
"ms_dot = 5400;// steam feed rate, [kg/h]\n",
"P = 750;// steam pressure, [kN/m^2]\n",
"x = .98;// steam dryness fraction\n",
"Tf1 = 41.5;// feed water temperature, [C]\n",
"CV = 31000;// calorific value of coal used in the boiler, [kJ/kg]\n",
"mc1 = 670;// rate of burning of coal/h, [kg]\n",
"Tf2 = 100;// increased water temperature, [C]\n",
"\n",
"// solution\n",
"//   (a)\n",
"SRC = ms_dot/mc1;// steam raised/kg coal, [kg]\n",
"hf = 709.3;// [kJ/kg]\n",
"hfg = 2055.5;// [kJ/kg]\n",
"h1 = hf+x*hfg;// specific enthalpy of steam raised, [kJ/kg]\n",
"//  from steam table \n",
"hfw = 173.9;// specific enthalpy of feed water, [kJ/kg]\n",
"EOB = SRC*(h1-hfw)/CV;// efficiency of boiler\n",
"mprintf('\n (a) The thermal efficiency of the boiler is  =  %f percent\n',EOB*100);\n",
"\n",
"// (b)\n",
"he = 2256.9;// specific enthalpy of evaporation, [kJ/kg]\n",
"Ee = SRC*(h1-hfw)/he;// equivalent evaporation[kg/kg coal]\n",
"mprintf('\n (b) The equivalent evaporation of boiler is  =  %f  kg/kg coal\n',Ee);\n",
"\n",
"// (c)\n",
"hw = 419.1;// specific enthalpy of feed water at 100 C, [kJ/kg]\n",
"Eos = ms_dot*(h1-hw);// energy of steam under new condition, [kJ/h]\n",
"neb = EOB+.05;// given condition new efficiency of boiler if 5%more than previous\n",
"Ec = Eos/neb;// energy from coal, [kJ/h]\n",
"mc2 = Ec/CV;// mass of coal used per hour in new condition, [kg]\n",
"mprintf('\n (c) Mass of coal used in new condition is  =  %f kg\n',mc2);\n",
"mprintf('\n      The saving in coal per hour is  =  %f  kg\n',mc1-mc2);\n",
"\n",
"//  End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.4: heat_transfer_and_volume.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 10.4');\n",
"\n",
"//  aim : To determine the\n",
"//  (a) Heat transfer in the boiler\n",
"//   (b) Heat transfer in the superheater\n",
"//  (c) Gas used\n",
"\n",
"//  given values\n",
"P = 100;// boiler operating pressure, [bar]\n",
"Tf = 256;// feed water temperature, [C]\n",
"x = .9;// steam dryness fraction.\n",
"Th = 450;// superheater exit temperature, [C]\n",
"m = 1200;// steam generation/h, [tonne]\n",
"TE = .92;// thermal efficiency\n",
"CV = 38;// calorific value of fuel, [MJ/m^3]\n",
"\n",
"// solution\n",
"// (a)\n",
"// from steam table\n",
"hw = 1115.4;// specific enthalpy of feed water, [kJ/kg]\n",
"// for wet steam\n",
"hf = 1408;// specific enthalpy, [kJ/kg]\n",
"hg = 2727.7;// specific enthalpy, [kJ/kg]\n",
"//  so\n",
"h = hf+x*(hg-hf);// total specific enthalpy of wet steam, [kJ/kg]\n",
"//  hence\n",
"Qb = m*(h-hw);// heat transfer/h for wet steam, [MJ]\n",
"mprintf('\n (a) The heat transfer/h in producing wet steam in the boiler is  =  %f  MJ\n',Qb);\n",
"\n",
"// (b)\n",
"// again from steam table\n",
"// specific enthalpy of superheated stem at given condition is,\n",
"hs = 3244;// [kJ/kg]\n",
"\n",
"Qs = m*(hs-h);// heat transfer/h in superheater, [MJ]\n",
"mprintf('\n (b) The heat transfer/h in superheater is  =  %f  MJ\n',Qs);\n",
"\n",
"// (c)\n",
"V = (Qb+Qs)/(TE*CV);// volume of gs used/h, [m^3]\n",
"mprintf('\n (c) The volume of gas used/h is  =  %f m^3\n',V);\n",
"\n",
"//  There is calculation mistake in the book so our answer is not matching\n",
"\n",
"//  End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.5: flow_rate.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 10.5');\n",
"\n",
"//aim : To determine \n",
"// the flow rate of cooling water\n",
"\n",
"//Given values\n",
"P=24;//pressure, [kN/m^2]\n",
"ms_dot=1.8;//steam condense rate,[tonne/h]\n",
"x=.98;//dryness fraction\n",
"T1=21;//entrance temperature of cooling water,[C]\n",
"T2=57;//outlet temperature of cooling water,[C]\n",
"\n",
"//solution\n",
"//at 24 kN/m^2, for steam\n",
"hfg=2616.8;//[kJ/kg]\n",
"hf1=268.2;//[kJ/kg]\n",
"//hence\n",
"h1=hf1+x*(hfg-hf1);//[kJ/kg]\n",
"\n",
"//for cooling water\n",
"hf3=238.6;//[kJ/kg]\n",
"hf2=88.1;//[kJ/kg]\n",
"\n",
"//using equation [3]\n",
"//ms_dot*(hf3-hf2)=mw_dot*(h1-hf1),so\n",
"mw_dot=ms_dot*(h1-hf1)/(hf3-hf2);//[tonne/h]\n",
"disp('tonne/h',mw_dot,'The flow rate of the cooling water is =')\n",
"\n",
"//End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.6: energy_supplied_dryness_fraction_and_Rankine_efficiency.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 10.6');\n",
"\n",
"//  aim : To determine\n",
"//  (a) the energy supplied in the boiler\n",
"//  (b) the dryness fraction of the steam entering the condenser\n",
"//  (c) the rankine efficiency\n",
"\n",
"//  given values\n",
"P1 = 3.5;// steam entering pressure, [MN/m^2]\n",
"T1 = 273+350;// entering temperature, [K]\n",
"P2 = 10;//steam exhaust pressure, [kN/m^2]\n",
"\n",
"// solution\n",
"//  (a)\n",
"//  from steam table, at P1 is,\n",
"hf1 = 3139;// [kJ/kg]\n",
"hg1 = 3095;// [kJ/kg]\n",
"h1 = hf1-1.5/2*(hf1-hg1);\n",
"// at Point 3\n",
"h3 = 191.8;// [kJ/kg]\n",
"Es = h1-h3;// energy supplied, [kJ/kg]\n",
"mprintf('\n (a) The energy supplied in boiler/kg steam is  =  %f kJ/kg\n',Es);\n",
"\n",
"// (b)\n",
"// at P1\n",
"sf1 = 6.960;// [kJ/kg K]\n",
"sg1 = 6.587;// [kJ/kg K]\n",
"s1 = sf1-1.5/2*(sf1-sg1);// [kJ/kg K]\n",
"// at P2\n",
"sf2 = .649;// [kJ/kg K] \n",
" sg2 = 8.151;// [kJ/kg K]\n",
" // s2=sf2+x2(sg2-sf2)\n",
" // theoretically expansion through turbine is isentropic so s1=s2\n",
" // hence\n",
" s2 = s1;\n",
" x2 = (s2-sf2)/(sg2-sf2);// dryness fraction\n",
" mprintf('\n (b) The dryness fraction of steam entering the condenser is =  %f \n',x2);\n",
" \n",
" // (c)\n",
" // at point 2\n",
" hf2 = 191.8;// [kJ/kg]\n",
" hfg2 = 2392.9;// [kJ/kg]\n",
" h2 = hf2+x2*hfg2;// [kJ/kg]\n",
" Re = (h1-h2)/(h1-h3);// rankine efficiency\n",
" mprintf('\n (c) The Rankine efficiency is  =  %f percent\n',Re*100);\n",
" \n",
" //  End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.7: Rankine_efficiency_and_specific_work_done.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 10.7');\n",
"\n",
"// aim : To determine\n",
"//  the specific work done and compare this with that obtained when determining the rankine effficiency\n",
"\n",
"// given values\n",
"P1 = 1000;// steam entering pressure, [kN/m^2]\n",
"x1 = .97;// steam entering dryness fraction\n",
"P2 = 15;//steam exhaust pressure, [kN/m^2]\n",
"n = 1.135;// polytropic index\n",
"\n",
"// solution\n",
"// (a)\n",
"// from steam table, at P1 is\n",
"hf1 = 762.6;// [kJ/kg]\n",
"hfg1 = 2013.6;// [kJ/kg]\n",
"h1 = hf1+hfg1; // [kJ/kg]\n",
"\n",
"sf1 = 2.138;// [kJ/kg K]\n",
"sg1 = 6.583;// [kJ/kg K]\n",
"s1 = sf1+x1*(sg1-sf1);// [kJ/kg K]\n",
"\n",
"// at P2\n",
"sf2 = .755;// [kJ/kg K] \n",
" sg2 = 8.009;// [kJ/kg K]\n",
"// s2 = sf2+x2(sg2-sf2)\n",
"// since expansion through turbine is isentropic so s1=s2\n",
" // hence\n",
" s2 = s1;\n",
" x2 = (s2-sf2)/(sg2-sf2);// dryness fraction\n",
" \n",
"  // at point 2\n",
" hf2 = 226.0;// [kJ/kg]\n",
" hfg2 = 2373.2;// [kJ/kg]\n",
" h2 = hf2+x2*hfg2;// [kJ/kg]\n",
" \n",
"// at Point 3\n",
"h3 = 226.0;// [kJ/kg]\n",
"\n",
"// (a)\n",
" Re = (h1-h2)/(h1-h3);// rankine efficiency\n",
" mprintf('\n (a) The Rankine efficiency is  =  %f percent\n',Re*100);\n",
" \n",
"// (b)\n",
"vg1 = .1943;// specific volume at P1, [m^3/kg]\n",
"vg2 = 10.02;// specific volume at P2, [m^3/kg]\n",
"V1 = x1*vg1;// [m^3/kg]\n",
"V2 = x2*vg2;// [m^3/kg]\n",
"\n",
"W1 = n/(n-1)*(P1*V1-P2*V2);// specific work done, [kJ/kg]\n",
"\n",
"//  from rankine cycle\n",
"W2 = h1-h2;// [kJ/kg]\n",
"mprintf('\n (b) The specific work done is  =  %f kJ/kg\n',W1);\n",
"mprintf('\n     The specific work done (from rankine) is  =  %f kJ/kg\n',W2);\n",
"\n",
"// there is calculation mistake in the book so our answer is not matching\n",
"\n",
"//  End"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.8: Rankine_efficiency_steam_consumption_and_Carnot_efficiency.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear;\n",
"clc;\n",
"disp('Example 10.8');\n",
"\n",
"//  aim : To determine\n",
"//  (a) the rankine fficiency\n",
"//  (b) the specific steam consumption\n",
"//  (c) the carnot efficiency of the cycle\n",
"\n",
"// given values\n",
"P1 = 1100;// steam entering pressure, [kN/m^2]\n",
"T1 = 273+250;// steam entering temperature, [K]\n",
"P2 = 280;// pressure at point 2, [kN/m^2]\n",
"P3 = 35;// pressure at point 3, [kN/m^2]\n",
"\n",
"// solution\n",
"// (a)\n",
"// from steam table, at P1  and T1 is\n",
"hf1 = 2943;// [kJ/kg]\n",
"hg1 = 2902;// [kJ/kg]\n",
"h1 = hf1-.1*(hf1-hg1); // [kJ/kg]\n",
"\n",
"sf1 = 6.926;// [kJ/kg K]\n",
"sg1 = 6.545;// [kJ/kg K]\n",
"s1 = sf1-.1*(sf1-sg1);// [kJ/kg K]\n",
"\n",
"// at P2\n",
"sf2 = 1.647;// [kJ/kg K] \n",
" sg2 = 7.014;// [kJ/kg K]\n",
"// s2=sf2+x2(sg2-sf2)\n",
"// since expansion through turbine is isentropic so s1=s2\n",
" // hence\n",
" s2  = s1;\n",
" x2 = (s2-sf2)/(sg2-sf2);// dryness fraction\n",
" \n",
" // at point 2\n",
" hf2 = 551.4;// [kJ/kg]\n",
" hfg2 = 2170.1;// [kJ/kg]\n",
" h2 = hf2+x2*hfg2;// [kJ/kg]\n",
" vg2 = .646;// [m^3/kg]\n",
" v2 = x2*vg2;// [m^3/kg]\n",
" \n",
" // by Fig10.20.\n",
" A6125 = h1-h2;// area of 6125, [kJ/kg]\n",
" A5234 = v2*(P2-P3);// area 5234, [kJ/kg]\n",
" W = A6125+A5234;// work done \n",
" hf = 304.3;// specific enthalpy of water at condenser pressuer, [kJ/kg]\n",
" ER = h1-hf;// energy received, [kJ/kg]\n",
" Re = W/ER;// rankine efficiency\n",
" mprintf('\n (a) The rankine efficiency is  =  %f percent\n',Re*100);\n",
" \n",
" // (b)\n",
" kWh = 3600;// [kJ]\n",
" SSC = kWh/W;// specific steam consumption, [kJ/kWh]\n",
" mprintf('\n (b) The specific steam consumption is = %f kJ/kWh\n',SSC);\n",
" \n",
" // (c)\n",
" // from steam table \n",
"T3 = 273+72.7;// temperature at point 3\n",
"CE = (T1-T3)/T1;// carnot efficiency\n",
"mprintf('\n (c) The carnot efficiency of the cycle is  =  %f percent\n',CE*100);\n",
"\n",
"//  End"
   ]
   }
,
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		   "source": [
			"## Example 10.9: power_and_thermal_efficiency.sce"
		   ]
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"clear;\n",
"clc;\n",
"disp('Example 10.9');\n",
"\n",
"// aim : To determine\n",
"// (a) the theoretical power of steam passing through the turbine\n",
"// (b) the thermal efficiency of the cycle\n",
"// (c) the thermal efficiency of the cycle assuming there is no reheat\n",
"\n",
"// given values\n",
"P1 = 6;// initial pressure, [MN/m^2]\n",
"T1 = 450;// initial temperature, [C]\n",
"P2 = 1;// pressure at stage 1, [MN/m^2]\n",
"P3 = 1;// pressure at stage 2, [MN/m^2]\n",
"T3 = 370;// temperature, [C]\n",
"P4 = .02;// pressure at stage 3, [MN/m^2]\n",
"P5 = .02;// pressure at stage 4, [MN/m^2]\n",
"T5 = 320;// temperature, [C]\n",
"P6 = .02;// pressure at stage 5, [MN/m^2]\n",
"P7 = .02;// final pressure , [MN/m^2]\n",
"\n",
"// solution\n",
"// (a) \n",
"// using Fig 10.21\n",
"h1 = 3305;// specific enthalpy, [kJ/kg]\n",
"h2 = 2850;// specific enthalpy, [kJ/kg]\n",
"h3 = 3202;// specific enthalpy, [kJ/kg]\n",
"h4 = 2810;// specific enthalpy, [kJ/kg]\n",
"h5 = 3115;// specific enthalpy, [kJ/kg]\n",
"h6 = 2630;// specific enthalpy, [kJ/kg]\n",
"h7 = 2215;// specific enthalpy, [kJ/kg]\n",
"W = (h1-h2)+(h3-h4)+(h5-h6);// specific work through the turbine, [kJ/kg]\n",
"mprintf('\n (a) The theoretical power/kg steam/s is  =  %f kW\n',W);\n",
"\n",
"// (b)\n",
"// from steam table\n",
"hf6 = 251.5;// [kJ/kg]\n",
"\n",
"TE1 = ((h1-h2)+(h3-h4)+(h5-h6))/((h1-hf6)+(h3-h2)+(h5-h4));// thermal efficiency\n",
"mprintf('\n (b) The thermal efficiency of the cycle is  =  %f percent\n',TE1*100);\n",
"\n",
"// (c)\n",
"// if there is no heat\n",
"hf7 = hf6;\n",
"TE2 = (h1-h7)/(h1-hf7);// thermal efficiency\n",
"mprintf('\n (c) The thermal efficiency of the cycle if there is no heat is  =  %f percent\n',TE2*100);\n",
"\n",
"//  End"
   ]
   }
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