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+{
+"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",
+"\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 8.3: 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",
+"\n",
+"\n",
+"\n",
+"\n",
+"\n",
+"\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",
+""
+ ]
+ }
+],
+"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
+}
generated by cgit (git 2.34.1) at 2024-12-19 17:53:31 +0000