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+{
+"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"
+   ]
+   }
+,
+{
+		   "cell_type": "markdown",
+		   "metadata": {},
+		   "source": [
+			"## Example 10.9: power_and_thermal_efficiency.sce"
+		   ]
+		  },
+  {
+"cell_type": "code",
+	   "execution_count": null,
+	   "metadata": {
+	    "collapsed": true
+	   },
+	   "outputs": [],
+"source": [
+"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"
+   ]
+   }
+],
+"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
+}