path: root/Fundamentals_of_Turbomachinery_by_W_W_Peng/8-Gas_Turbines.ipynb

{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 8: Gas Turbines"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.1: GT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('From ETAad=(1-((p0e/p0i)^(ETAs*(k-1)/k)))/(1-((p0e/p0i)^(k-1)/k)))')\n",
"ETA_ad=(1-((14.7/200)^(0.85/3.5)))/(1-(14.7/200)^(0.2857))\n",
"printf('\n ETAad= %0.3f',ETA_ad)\n",
"\n",
"disp('T_soe/T_0i=((p0e/p0i)^((k-1)/k))')\n",
"//let T_soe/T_0i=w\n",
"w=(14.7/200)^0.2857\n",
"printf('\n T_soe/T_0i= %0.4f',w)\n",
"\n",
"T_soe=0.4743*(1800+460)\n",
"printf('\n T_soe= %0.0f R',T_soe)\n",
"\n",
"disp('ETAad=(T_0i-T_0e)/(T_0i-T_soe)=0.893')\n",
"T_0i=T_soe/w//since T_soe/T_0i=w\n",
"T_0e=T_0i-(ETA_ad*(T_0i-T_soe))\n",
"printf(' Thus T_0e= %0.0f R=739 degrees Farenheit',T_0e)\n",
"\n",
"disp('Also for impulse turbine, we have ß2=ß3.Hence from delta_H0=Um*(V_u2-V_u3)=Um*[Um+Va*tanß2-(Um-Va*tan(ß3))]=2*Um*Va*tanß')\n",
"delta_h0=2*750*400*tan(50*%pi/180)\n",
"printf('\n delta_h0= %g is approximately equal to 7.15*10^5 ((ft/s)^2)',delta_h0)\n",
"//Let 0.24*778*32.2*delta_T0=u\n",
"u=7.15*10^5\n",
"delta_T0=u/(0.24*778*32.2)\n",
"printf('\n\n Or 0.24*778*32.2*delta_T0=7.15*10^5,we have temperature rise per stage = %0.0f degrees Farenheit',delta_T0)\n",
"\n",
"n_s=(1800-739)/119\n",
"printf('\n Hence the number of stages n_s= %0.2f is approximately equal to 9',n_s)"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.2: GT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('Referring to figure 8.3,we have T_s2=T_01*((p2/p01)^((k-1)/k))')\n",
"T_01=2860\n",
"p2=180\n",
"p01=250\n",
"k=1.4//k=(Cp/Cv)\n",
"T_s2=T_01*((p2/p01)^((k-1)/k))\n",
"printf(' T_s2= %0.0f R',T_s2)\n",
"\n",
"disp(' From ε_s=(h2-hs2)/((V_2^2)/2),we have Cp*(T2-Ts2=ε_s*(V_2^2)/2.Combining with Cp*(T_02-T_2)=((V_2^2)/2),where T02=T01, we have Cp*(T_02-Ts2)=(1+εs)*((V_2^2)/2)')\n",
"disp('V2=[2*Cp*(T_02-T_s2)/(1+εs)]^0.5')\n",
"T_02=T_01\n",
"Cp=0.24*778*32.2\n",
"epsilon_s=0.07\n",
"T_s2=2604\n",
"V2=[2*Cp*(T_02-T_s2)/(1+epsilon_s)]^0.5\n",
"printf(' V2= %0.0f ft/s',V2)\n",
"\n",
"V2=1696\n",
"alpha2=65*%pi/180//converting to radians\n",
"V_u2=V2*sin(alpha2)\n",
"printf('\n Hence we have V_u2=V2*sin(α2) = %0.0f ft/s',V_u2)\n",
"\n",
"Va=V2*cos(alpha2)\n",
"printf('\n Va=V2*cos(alpha2)= %0.1f ft/s',Va)\n",
"\n",
"disp('T2=T_02-(V_2^2)/(2*Cp)')\n",
"T_02=2860\n",
"V2=1696\n",
"Cp=0.24*778*32.2\n",
"T2=T_02-(V2^2)/(2*Cp)\n",
"printf(' Hence we have T2= %0.0f R ',T2)\n",
"\n",
"disp('Since V1=Va,we have T1=T_01-((V1^2)/(2*Cp))')\n",
"V1=716.8\n",
"T_01=2860//2860R\n",
"T1=T_01-((V1^2)/(2*Cp))\n",
"printf('T1= %0.0f R',T1)\n",
"\n",
"disp('From delta_E=Cp*delta_T0s=U*(V_u2+V_u3)=U*V_u2, we have U=Ps/(m*V_u2)')\n",
"Ps=375*550*32.2//converting unit of Ps\n",
"m=3\n",
"V_u2=1537\n",
"U=Ps/(m*V_u2)\n",
"printf('U=%0.0f ft/s',U)\n",
"\n",
"disp('φ=Va/U')\n",
"Va=716.8\n",
"U=1440\n",
"phi=Va/U\n",
"printf('φ=%0.3f',phi)\n",
"\n",
"tanbeta3=U/Va\n",
"printf('\n tanß3= %0.2f',tanbeta3)\n",
"\n",
"beta3=((atan(tanbeta3))*180/%pi)\n",
"printf('\n ß3= %0.1f degrees',beta3)\n",
"\n",
"alpha3=0\n",
"phi=0.498\n",
"alpha2=65*%pi/180\n",
"R=1+((phi/2)*(tan(alpha3)-tan(alpha2)))\n",
"printf('\n R= %0.3f',R)\n",
"\n",
"disp('Also from the velocity diagram in figure 8.4,we have tanß2=tanα2-(1/φ)=0.136,so ß2=7.8 degrees')\n",
"disp('Similarly we have W3=Va/cosß3=1606 ft/s and W2=Va/cosß2=723.5 ft/s')\n",
"disp('Across the rotor we have h2+(W2^2)/2=h3+(W3^2)/2. Hence T3=T2+(W2^2)-(W3^2)/(2*Cp)=2450R')\n",
"disp('We have Ts3=T3-εr*(W3^2)/(2*Cp)=2424R')\n",
"disp('Also p3=p2*(Ts3/T2)^(k/(k-1))=136.9 psia')\n",
"ETAs=(1+(0.12*(1606^2)+0.07*(1696^2)*(2450/2621))/(2*0.24*778*32.2*(2817-2450)))^-1\n",
"printf(' From equation 8.2 we have ETAs= %0.4f',ETAs)\n",
"\n",
"//Let j=0.498/2\n",
"j=0.498/2\n",
"//Let k=0.12*[(sec(63.5*%pi/180))^2]\n",
"k=0.12*[(sec(63.5*%pi/180))^2]\n",
"//Let l=0.07*(2450/2621)*[(sec(65*%pi/180))^2]\n",
"l=0.07*(2450/2621)*[(sec(65*%pi/180))^2]\n",
"//let m=tan(63.5*%pi/180)+tan(7.8*%pi/180)\n",
"m=tan(63.5*%pi/180)+tan(7.8*%pi/180)\n",
"\n",
"ETAs=[1+((j*(k+l))/m)]^-1\n",
"printf('\n From equation 8.3 we have ETAs= %0.4f',ETAs)\n",
"\n",
"disp('Also ETAs can be calculated from ETAs=(T_01-T_03)/(T_01-T_ss03)')\n",
"disp('We have T_03=T3+V3^2/(2*Cp)')\n",
"disp('p03=p3*(T_03/T3)^(k/(k-1))')\n",
"disp('T_ss03=T_01*(p_03/(p_01)*((k-1)/k))')\n",
"ETAs=(2860-2493)/(2860-2450)\n",
"printf(' Hence we have ETAs= %0.3f',ETAs)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
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"\n",
"\n",
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"\n",
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"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
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"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.3: GT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('Use the velocity diagram shown in figure 8.2 or 8.4')\n",
"disp('We have Vatanß2=Vatanα2-Um')\n",
"disp('Or tanß2=tanα2-Um/Va')\n",
"\n",
"tanbeta2=tan(75*%pi/180)-1200/500\n",
"printf(' Thus tanß2= %0.2f',tanbeta2)\n",
"beta2=((atan(tanbeta2))*180/%pi)\n",
"printf('\n Thus ß2= %0.2f',beta2)\n",
"\n",
"disp('Also Vatanß3=Vatanα3+Um')\n",
"disp('tanß3=tanα3+Um/Va')\n",
"tanbeta3=tan(10*%pi/180)+1200/500\n",
"printf(' tanß3= %0.2f',tanbeta3)\n",
"beta3=((atan(tanbeta3))*180/%pi)\n",
"printf('\n ß3= %0.2f',beta3)\n",
"\n",
"disp('From Cp*deltaT0s=deltaE=Um(Vu2+Vu3)=UmVa(tanα2+tanα3)')\n",
"deltaE=1200*500*(tan(75*%pi/180)+tan(10*%pi/180))\n",
"printf(' Thus Cp*deltaT0s=deltaE= %g=2.34*10^6 ((ft/s)^2)',deltaE)\n",
"\n",
"deltaT0s=(2.34*(10^6))/(0.24*778*32.2)\n",
"disp('deltaT0s=(2.34*(10^6))/(0.24*778*32.2)')\n",
"printf(' Thus deltaT0s= %0.2f R which is rounded off to 890R',deltaT0s)\n",
"\n",
"disp('Hence neglecting leakage and mechanical losses , we have shaft power output Ps=mCpdeltaT0s')\n",
"Ps=50*2.34*(10^6)/(32.2*550)\n",
"printf('\n Hence we have Ps= %0.2f hp wich is rounded off to 6607hp',Ps)\n",
"\n",
"disp('The degree of reaction at the mean radius can be determined from equation 8.5A')\n",
"R=(500/(2*1200))*(tan(68.8*%pi/180)-tan(53.1*%pi/180))\n",
"printf('\n Thus R = %0.3f',R)\n",
"\n",
"disp('To determine the radii the flow area A2 can be determined from m=rho2*A2*Va. The density rho2 can be determined from p2 and T2 which can be caalculated as follows.')\n",
"disp('From Cp*T02=CpT2+V2^2/2 and V2= VA/cosα2')\n",
"V2=500/(cos(75*%pi/180))\n",
"printf('\n V2= %0.0f ft/s',V2)\n",
"\n",
"T2=2000-(1932^2)/(2*0.24*778*32.2)\n",
"printf('\n Thus we have T2= %0.0f degrees Farenheit = 2150R',T2)\n",
"\n",
"disp('From the definition of loss coefficient εe we have Ts2=T2-εsV2^2/(2*Cp)')\n",
"Ts2=1690-(0.08*(1932^2))/(2*.24*778*32.2)\n",
"printf('\n Ts2= %0.2f degrees Farenheit which is equal to 2125.2R',Ts2)\n",
"\n",
"//Let x= P2/p01\n",
"x=(2125.2/2460)^(1.4/0.4)\n",
"printf('\n and P2/p01= %0.2f',x)\n",
"\n",
"\n",
"P2=200*0.60\n",
"printf('\n P2= %0.0f psia',P2)\n",
"\n",
"disp('Hence the density can be calculated as rho2=p2/(R*T2)')\n",
"\n",
"rho2=120*144/(53.3*2150)\n",
"printf('\n Thus ro2=%0.3f lbm/ft^3',rho2)\n",
"\n",
"A2=50/(0.151*500)\n",
"printf('\n A2=m/(rho2*Va)=%0.3f ft^2',A2)\n",
"\n",
"//let y=rt^2-rh^2\n",
"y=0.662/%pi\n",
"printf('\n rt^2-rh^2=A2/pi=%0.3f',y)\n",
"\n",
"rm=30*1200/(8000*%pi)\n",
"printf(' and rm= %0.2f ft',rm)\n",
"\n",
"disp('rt^2+rh^2=2*rm^2=4.09ft^2')\n",
"disp('and 1.466ft,rh=1.393ft')\n",
"rt=1.466\n",
"rh=1.393\n",
"b=rt-rh\n",
"printf(' b= %0.3f ft=0.88in',b)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
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"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.4: GT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('For N/(T01^1/2)=40,the following data of p02/p01,τ/p01 and Eta can be obtained from figure 8.9a')\n",
"//let x=p02/p01\n",
"x=[0.70 0.75 0.8];\n",
"//let y=τ/p01\n",
"y=[8.7 5.3 2.2];\n",
"Eta=[0.81 0.64 0.41];\n",
"//let z=Ps/(p01*((T01)^1/2))\n",
"z=[0.066 0.040 0.017];\n",
"//let i=m*((T01)^1/2)/p01\n",
"i=[2.48 2.34 2.0];\n",
"table=[x' y'Eta' z' i'];\n",
"disp(' The columns of the table are in the order p02/p01 τ/p01 Eta Ps/(p01*((T01)^1/2)) and m*((T01)^1/2)/p01')\n",
"disp(table)\n",
"\n",
"disp('The power and mass flow rate have to be obtained with the following manipulations. ')\n",
"disp('Frpm Ps=τ*omega, wee obtain:')\n",
"disp('Ps/(p01*((T01)^0.5))=τ*N*pi/(30*550*p01*((T01)^0.5))')\n",
"disp('Also from Ps/m=Eta*Cp*T01[1-(p02/p01)^((k-1)/k)] we obtain')\n",
"disp('m*(T01^0.5)/p01={[Ps/(p01*T01^0.5)]/(Eta*Cp)}*[1-(p02/p01)^((k-1)/k)]^-1')\n",
"disp('Where (k-1)/k=0.40/1.4')\n",
"\n",
"//Let j=(k-1)/k=0.40/1.4\n",
"j=0.40/1.4\n",
"printf('Thus=(k-1)/k %0.4f',j)\n",
"\n",
"disp('And Cp= 0.24*Btu/(1bm*R)')\n",
"Cp=0.24*778/(550)\n",
"printf('Thus Cp= %0.4f hp-s/(lbm*R)',Cp)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.5: GT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('From Um=rm*omega')\n",
"disp('We have rm=30*Um/(N*pi)')\n",
"rm=30*850/(8000*%pi)\n",
"printf(' Thus rm= %0.2f ft=12.1 in',rm)\n",
"\n",
"disp('rm=(rt^2+rh^2/2)^0.5 or rt^2+rh^2=293.8')\n",
"disp('Combined with b=rt-rh=4 in,we have rt^2-4rt-138.9=0,thus rt=13.95 in')\n",
"\n",
"disp('and rh=9.95 in')\n",
"\n",
"disp('To find the number of stages required the exhaust air temperature can be estimated as T0e=T0i*(p0e/p0i)^((k-1)/k)')\n",
"T0e=1400*(14.7/60)^0.2857\n",
"printf(' Thus T0e= %0.1f R',T0e)\n",
"\n",
"disp('The maximum energy available per unit mass of air is delta_Hs=Cp*(T0i-T0e)')\n",
"delta_Hs=0.24*(1400-936.7)\n",
"printf(' delta_Hs= %0.1f Btu/lbm',delta_Hs)\n",
"\n",
"disp('The maximum energy transfer per stage with an impulse turbine is deltaEi=2*Um^2')\n",
"delta_Ei=2*(850^2)/(32.2*778)\n",
"printf(' delta_Ei= %0.2f Btu/lbm',delta_Ei)\n",
"\n",
"disp('Hence the required number of stages is ETAsi=delta_Hs/delta_Ei')\n",
"ETAsi=111.2/57.68\n",
"printf(' ETAsi= %0.2f which is approximately equal to 2',ETAsi)\n",
"\n",
"disp('With the reaction turbine stages,it will be delta_Er=Um^2')\n",
"delta_Er=850^2/(32.2*778)\n",
"printf(' delta_Er= %0.2f Btu/lbm',delta_Er)\n",
"disp('And ETAse=3.85 is approximatly equal to 4')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.6: GT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('From figure 8.14c we have Pl=620hp at N=18400rpm. Pick a point on the curve of N/(T01^0.5)=18400/(530^0.5)=800')\n",
"disp('In figure 8.14a,say p02/p01=5')\n",
"disp('So we have')\n",
"\n",
"disp('m(T01^0.5)/p01   ETAc   p02(psia)   m(lbm/s)   p03(psia)   p03/p04')\n",
"disp('   5.7            0.85     73.5       3.64       71.5        4.86')\n",
"\n",
"disp('where p02=5*14.7')\n",
" p02=5*14.7\n",
" printf(' Thus p02= %0.2f',p02)\n",
" \n",
"disp('m=5.7*14.7/(530^0.5)')\n",
" m=5.7*14.7/(530^0.5)\n",
" printf(' m= %0.2f',m)\n",
" \n",
"disp('73.5-2')\n",
" p03=73.5-2\n",
" printf(' m= %0.2f psia',p03)\n",
" \n",
"//Let i=p03/p04\n",
"i=71.5/14.7\n",
"printf('\n p03/p04= %0.2f',i)\n",
"\n",
"disp('Then from figure 8.14b,with p03/p04 and m3=m2 we have')\n",
"disp('m(T01^0.5)/p01     T03(R)   N/(T03^0.5)      ETAt')\n",
"disp('2.56                2528        366           0.87')\n",
"\n",
"disp('where T03=(2.56*71.5/3.64)^2')\n",
"T03=(2.56*71.5/3.64)^2\n",
"printf(' T03= %0.2f',T03)\n",
"\n",
"disp('N/T03=18400/(2528^0.5)')\n",
"//let k=N/T03\n",
"k=18400/(sqrt(2528))\n",
"printf(' Thus T03= %0.2f',k)\n",
"\n",
"disp('So from equations (8.1),(7.4) and(8.11) we have:')\n",
"disp('delta_T034=ETAt*T03*[1-(p04/p03)^((k-1)/k)]')\n",
"delta_T034=0.87*2528*[1-(4.86)^(-0.248)]\n",
"printf(' delta_T034 = %0.0f R',delta_T034)\n",
"\n",
"disp('delta_T012=(T01/ETAc)*[(p02/p01)^((k-1)/k)-1]')\n",
"delta_T012=(530/0.85)*[(5^0.2857)-1]\n",
"printf(' delta_T012= %0.0fR',delta_T012)\n",
"\n",
"P0=3.64*(0.28*713-0.24*(364/0.95))\n",
"printf(' \n and P0=3.64*(0.28*713-0.24*(364/0.95))= %0.0f Btu/s=554hp, which is less than Pl',P0)\n",
"\n",
"disp('So we pick another point on the same curve , say p02/p01=5.2, and repeat the calculations ')\n",
"\n",
"disp('m(T01^0.5)/p01  ETAc  p02(psia)  m(lbm/s)  p03(psia)  p03/p04  m(T01^0.5)/p01  T03(R)  N/(T03^0.5)  ETAt')\n",
"disp('5.6             0.88   76.4       3.57       74.4       5.06     2.55            2824    346         0.85')\n",
"\n",
"delta_T034=0.85*2824*[1-(5.06^(-0.248))]\n",
"printf('\n The new delta_T034= %0.2fR',delta_T034)//the book has rounded off the value to 794R,the value calculated in this code is more accurate\n",
"\n",
"delta_T012=(530/0.88)*[(5.2^0.2857)-1]\n",
"printf('\n delta_T012= %0.0fR',delta_T012)\n",
"\n",
"P0=3.57*(0.28*794-0.24*362/0.95)\n",
"printf('\n Net output power P0= %0.0f Btu/s=660hp, which is much greater than Pl \n\n',P0)\n",
"\n",
"disp('Pick another point say p02/p01=5.15')\n",
"\n",
"disp('m(T01^0.5)/p01  ETAc  p02(psia)  m(lbm/s)  p03(psia)  p03/p04  m(T01^0.5)/p01  T03(R)  N/(T03^0.5)  ETAt')\n",
"disp('5.65            0.87   75.7       3.61       73.7       5.01     2.55            2710    353         0.86')\n",
"\n",
"delta_T034=0.86*2710*[1-(5.01^(-0.248))]\n",
"printf('\n\n From new values delta_T034= %0.0f R',delta_T034)\n",
"\n",
"delta_T012=(530/0.87)*[(5.15^(0.2857))-1]\n",
"printf('\n and delta_T012= %0.0f R',delta_T012)\n",
"\n",
"P0=3.61*(0.28*768-0.24*(364/0.95))\n",
"printf('also P0= %0.1f Btu/s =628hp',P0)\n",
"\n",
"disp('P0 is close to Pl')\n",
"disp('So the running point is around p02/p01=5.15, m(T01^0.5)/p01=5.65')\n",
"disp('and N/(T01^0.5)=800 on the compressor characteristics')\n",
"disp('It is not too close to the surge line and hence is safe.')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
" \n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 8.7: GT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('Velocity diagrams at the rotor inlet and outlet are given.')\n",
"\n",
"disp('Velocities at the rotor inlet can be calculated .')\n",
"\n",
"r2=5/12\n",
"N=15000\n",
"U2=r2*N*%pi/30\n",
"printf(' U2=r2*N*pi/30 = %0.1f ft/s',U2)\n",
"\n",
"alpha2=85*%pi/180//converting to radians\n",
"V2=U2/sin(alpha2)\n",
"printf('\n V2= %0.0f ft/s',V2)\n",
"\n",
"Vr2=U2/(tan(alpha2))\n",
"printf('\n Vr2=W2= %0.1f ft/s',Vr2)\n",
"\n",
"disp('Hence from Cp(T02-T2)=V2^2/2, where T02=T01, we have')\n",
"T02=2000\n",
"V2=657\n",
"Cp=0.24*778*32.2\n",
"T2=T02-(V2^2)/(2*Cp)\n",
"printf('\n T2=T02-(V2^2)/(2*Cp) =%0.1fR',T2)\n",
"\n",
"disp('From εn=(T2-T2_dash)/((V2^2)/(2*Cp)), we have')\n",
"T2=1964.1\n",
"epsilon_n=0.08\n",
"V2=657\n",
"T2_dash=T2-epsilon_n*V2^2/(2*Cp)\n",
"printf(' T2_dash=T2-epsilon_n*V2^2/(2*Cp)= %0.1fR',T2_dash)\n",
"\n",
"p01=50\n",
"T2_dash=1961.2\n",
"T01=2000\n",
"//let i=(k/(k-1))\n",
"i=3.5\n",
"p2=p01*(T2_dash/T01)^i\n",
"printf('\n p2= %0.1f psia',p2)\n",
"\n",
"p2=46.7\n",
"R=53.3\n",
"T2=1964.1\n",
"rho2=p2*144/(R*T2)//conversion factor=144\n",
"printf('\n rho2= %0.3f lb,/ft^3',rho2)\n",
"\n",
"rho2=0.064\n",
"Vr2=57.3\n",
"A2=(2*%pi*5*2/144)\n",
"m=rho2*Vr2*A2\n",
"printf('\n So the mass flow rate m=rho2*Vr2*A2= %0.2f lbm/s',m)\n",
"\n",
"disp('Assuming whirl-free flow at the rotor outlet under the design condition, we have')\n",
"U2=654.5\n",
"delta_E=(U2)^2\n",
"printf('\n delta_E=U2*Vu2=U2^2 %g ((ft/s)^2)=428370/(32.2*550)=24.2hp/(lbm/s)',delta_E)\n",
"\n",
"m=1.60\n",
"delta_E=24.2//after converting to new units\n",
"Ps=m*delta_E\n",
"printf('\n Ps= %0.1f hp',Ps)\n",
"\n",
"rm3=2.06/12\n",
"Um3=rm3*N*%pi/30\n",
"printf('\n Um3= %0.2f ft/s',Um3)\n",
"\n",
"beta3=30*%pi/180//converting to radians\n",
"V3=Um3/(tan(beta3))\n",
"printf('\n V3 = %0.0f ft/s',V3)\n",
"\n",
"W3=Um3/sin(beta3)\n",
"printf('\n W3 = %0.2f ft/s',W3)//the value has been rounded off to 539.2 in the book,however the value found here is more accurate\n",
"\n",
"disp('The turbine efficiency can be determined from equations 8.12 and 8.13. Without detailed calculations wthe result is given as ETAt=0.691')\n",
"\n",
"disp('The exhaust pressure/temperature can be determined from te following calculations with the help of figure 8.21')\n",
"delta_E=428370/(32.2*778)\n",
"printf(' From Cp(T01-T03)=delta_E= %0.1fBtu/lbm',delta_E)\n",
"\n",
"disp('(T01-T03)/(T01-T3dash)=Etat=0.691')\n",
"\n",
"T03=2000-(17.1/0.24)\n",
"printf(' T03= %0.0f R',T03)\n",
"\n",
"T01=2000\n",
"T03=1929\n",
"ETAt=0.691\n",
"T3_dash=T01-(T01-T03)/ETAt\n",
"printf('\n T3_dash=T01-(T01-T03)/ETAt %0.0fR =',T3_dash)\n",
"\n",
"//let i=k/(k-1)\n",
"i=3.5\n",
"p01=50\n",
"T3_dash=1897\n",
"T01=2000\n",
"p3=p01*(T3_dash/T01)^i\n",
"printf('\n p3=p01*(T3_dash/T01)^i= %0.1f psia',p3)\n",
"\n",
"T2=1964.4\n",
"p3=41.6\n",
"p2=46.7\n",
"// Let l=(k-1)/k\n",
"l=0.2857\n",
"T3_dbldash=T2*(p3/p2)^(l)\n",
"printf('\n T3_dbldash=T2*(p3/p2)^(1/i)=%0.2fR',T3_dbldash)//answer given in the book is 1900.3 R,however the value tabulated here is more accurate\n",
"\n",
"T3_dbldash=1900.3\n",
"epsilon_r=0.45\n",
"W3=539.2\n",
"Cp=0.24*32.2*778\n",
"T3=T3_dbldash+epsilon_r*(W3^2)/(2*Cp)\n",
"printf('\n T3= %0.1f R',T3)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
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