path: root/Fundamentals_of_Turbomachinery_by_W_W_Peng/7-Axial_Flow_Compressors.ipynb

{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 7: Axial Flow Compressors"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.1: AFC.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('From (poe/poi)=[(Toi+ns*deltaTo)]^(k*eff*p/(k-1))')\n",
"disp('From the above mentioned equation we can find out the value of deltaTo')\n",
"delta_To=((8^(1/(3.5*0.87)))-1)*530/7\n",
"printf('deltaTo= %0.1f degrees Farenheit',delta_To)\n",
"\n",
"Q=450\n",
"r_m1=9\n",
"b=3\n",
"V_a1=(Q*144)/(2*%pi*r_m1*b)\n",
"printf('\n Thus Va1 = %0.0f fps',V_a1)\n",
"\n",
"N=12000\n",
"U=(N*%pi*9)/(30*12)\n",
"printf('\n N= %0.1f fps',U)\n",
"\n",
"disp('deltaho=U*Va*(tanß1-tanß2) and R=(Φ/2)*(tanß1+tanß2')\n",
"//let y= tanß1-tanß2\n",
"y=[(0.24*778*74.2*32.2)/(942.5*382)]\n",
"printf('\n Thus tanß1-tanß2= %0.2f',y)\n",
"\n",
"//let x=tanß1+tanß2\n",
"x=(0.5*2*942.5/382)\n",
"printf('\n Thus tanß1+tanß2= %0.3f',x)\n",
"\n",
"disp('Hence we get tanß1=1.853')\n",
"tanbeta1=1.853\n",
"beta_1=(atan(tanbeta1))*180/%pi\n",
"printf('\n The value of ß1= %0.1f degrees',beta_1)\n",
"disp('Also tanß2=0.613')\n",
"tanbeta2=0.613\n",
"beta_2=(atan(tanbeta2))*180/%pi\n",
"printf('\n The value of ß2=α1= %0.1f degrees',beta_2)\n",
"\n",
"disp('Ps=m*Cp*ns*deltaTo/etam')\n",
"P_s=(0.075*450*0.24*778*7*74.2)/(550*0.95)\n",
"printf('\n The total power required is Ps= %0.0f hp',P_s)\n",
"\n",
"disp('The adiabetic efficiency is given as ((poe/poi)^(((k-1)/k))-1/((Toe/Toi)-1)')\n",
"ETA_ad=0.811/0.979\n",
"printf('\n Thus adiabetic efficiency is %0.4f',ETA_ad)//answer given in the book is 0.827,but this is more accurate\n",
"\n",
"disp('ETAc=ETAm*ETAad')\n",
"ETA_m=0.95\n",
"ETA_ad=0.827\n",
"ETA_c=ETA_m*ETA_ad\n",
"printf(' Hence ETAc= %0.3f ',ETA_c)\n",
"disp('Thus ETAc=78.6%')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.2: AFC.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"r_h=15\n",
"r_t=24\n",
"r_m=((r_h^2+r_t^2)/2)^0.5\n",
"printf('rm= %0.0f in',r_m)\n",
"\n",
"N=6000\n",
"r_m=20\n",
"U_m=N*%pi*r_m/(12*30)\n",
"printf('\n U_m= %0.2f ft/s',U_m)\n",
"\n",
"disp('We have psia=lamda*psi=lambda*phi*(tanßm1-tanßm2)')\n",
"//let x=tanßm1-tanßm2\n",
"x=(0.24*778*32.2*35)/(0.92*1047.2*450)\n",
"printf('\n Hence we can find out tanßm1-tanßm2= %0.3f',x)\n",
"\n",
"disp('From equation 7.2B for Rm=0.5 we have αm1=ßm2')\n",
"disp('We get values of tanαm1+tanßm1=2.325 and tanαm2+tanßm2=2.325')\n",
"\n",
"disp('Hence we have tanßm1-tanαm1=0.485')\n",
"\n",
"tanalpham1=0.92\n",
"alpham_1=((atan(tanalpham1)))*180/%pi\n",
"printf('\n Thus αm1= %0.1f degrees',alpham_1)\n",
"\n",
"tanalpham2=1.405\n",
"alpham_2=((atan(tanalpham2)))*180/%pi\n",
"printf('\n Thus αm2= %0.2f degrees',alpham_2)\n",
"\n",
"disp('To determine the flow angles at the hub and tip,we use the free vortex condition of Vur=const, or rhtanαh=rttanαt=rmtanαm.Hence the flow angles can be determined.')\n",
"\n",
"tanalphah1=0.92*20/15\n",
"printf('\n tanαh1= %0.3f',tanalphah1)\n",
"alpha_h1=((atan(tanalphah1)))*180/%pi\n",
"printf('\nThus αh1= %0.1f degrees',alpha_h1)\n",
"\n",
"tanalphat1=0.92*20/24\n",
"printf('\n\n tanαt1= %0.3f',tanalphat1)\n",
"alpha_t1=((atan(tanalphat1)))*180/%pi\n",
"printf('\n Thus αt1= %0.1f degrees',alpha_t1)\n",
"\n",
"tanalphah2=1.405*20/15\n",
"printf('\n\n tanαh2= %0.3f',tanalphah2)\n",
"alpha_h2=((atan(tanalphah2)))*180/%pi\n",
"printf('\n Thus αh1= %0.1f degrees',alpha_h2)\n",
"\n",
"tanalphat2=1.405*20/24\n",
"printf('\n\n tanαt2= %0.3f',tanalphat2)\n",
"alpha_t2=((atan(tanalphat2)))*180/%pi\n",
"printf('\n Thus αh1= %0.1f degrees',alpha_t2)\n",
"\n",
"disp('The degree of reaction at the hub and tip can be determined.')\n",
"Rh=1-((1-0.5)/((15/20)^2))\n",
"Rt=1-((1-0.5)/((24/20)^2))\n",
"printf('\n Rh= %0.2f',Rh)\n",
"printf('\n Rt= %0.2f',Rt)\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",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.3: AFC.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('From the previous numerical wee have ßm1=αm2=54.5 degrees, tanßm1=1.405')\n",
"disp('ßm2=αm1=42.6 degrees. Thus tanßm2=0.92')\n",
"\n",
"disp('tanßm1-tanßm2=1.55(1+1.5*(s/c)),thus we can determine s/c')\n",
"//let x= s/c\n",
"x=[1.55/(1.405-0.92)-1]/1.5\n",
"printf('Thus (s/c)= %0.2f',x)\n",
"\n",
"disp('Also b/c=3,we have c=(rt-rh)/3')\n",
"rt=24\n",
"rh=15\n",
"c=(rt-rh)/3\n",
"printf('\n c= %0.0f in',c)\n",
"\n",
"s=1.47*c\n",
"printf('\n Hence we determine s to be equal to %0.1f in',s)\n",
"\n",
"rm=20\n",
"Zb=2*%pi*rm/s\n",
"printf('\n Zb= %0.0f',Zb)\n",
"\n",
"disp('The blade angles can be estimated from ßb1=ßm1-i and ßb2=ßm2-δ')\n",
"disp('Where i=3 degrees and δ=m*Θ*((s/c)^0.5)')\n",
"\n",
"//let n=a/c\n",
"n=0.5\n",
"disp('m=0.23*((2*(a/c))^2)+0.1*(ßm2/50)')\n",
"disp('Θ=ßm1-ßm2')\n",
"thita=11.9\n",
"m=0.23+(0.1*42.6/50)//for circular blade\n",
"printf('\n m= %0.3f',m)\n",
"m=0.315\n",
"x=1.47\n",
"delta=m*thita*(x^0.5)\n",
"printf('\n δ= %0.1f degrees',delta)\n",
"\n",
"beta_b1=54.5-3\n",
"printf('\n ßb1= %0.1f degrees',beta_b1)\n",
"\n",
"beta_b2=42.6-4.5\n",
"printf('\n ßb2= %0.1f degrees',beta_b2)\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.4: AFC.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('We have psia=lamda*psi=lambda*phi*(tanßm1-tanßm2) and R=0.5*φ(tanß1+tanß2) we can obtain the values of tanß1 and tanß2')\n",
"//let x=tanßm1-tanßm2\n",
"x=0.35/(0.92*0.5)\n",
"printf('\n tanßm1-tanßm2= %0.3f',x)\n",
"\n",
"//let y=tanß1+tanß2\n",
"y=2*0.5/0.5\n",
"printf('\nThus tanß1+tanß2= %0.3f',y)\n",
"\n",
"disp('Hence tanß1=1.38')\n",
"tanbeta1=1.38\n",
"beta_1=((atan(tanbeta1)))*180/%pi\n",
"printf('\nThus the value of ß1 is equal to %0.1f degrees',beta_1)\n",
"\n",
"disp('tanß2=0.619')\n",
"tanbeta2=0.619\n",
"beta_2=((atan(tanbeta2)))*180/%pi\n",
"printf('\nThus the value of ß1 is equal to %0.1f degrees',beta_2)\n",
"\n",
"disp('For each stage we have psi=Cp*deltaTos/((Um)^2)')\n",
"delta_T_os=0.35*(920^2)/6012\n",
"printf('\nHence deltaTos= %0.1f R',delta_T_os)\n",
"\n",
"Cp=0.24*778*32.2\n",
"printf('\nWhere Cp = %0.0f ft-lbf/slug',Cp)\n",
"\n",
"disp('For overall compressor form equation 47.4 we have (Poe/poi)=[(1+deltaToe/Toi)]^(k*eff*p/(k-1))')\n",
"delta_T_oe=530*[(4.5^(0.2857/0.9))-1]\n",
"printf('\nThus deltaToe= %0.0f R',delta_T_oe)\n",
"\n",
"disp('The number of stages can be calculated as  ns=deltaToe/deltaTos')\n",
"delta_T_oe=324\n",
"delta_T_os=49.3\n",
"ns=delta_T_oe/delta_T_os\n",
"printf('\nThus ns= %0.2f',ns)\n",
"disp('ns is approximately equal to 7')\n",
"\n",
"disp('Hence the actual values are: ')\n",
"delta_T_oe=7*49.3\n",
"printf('deltaToe= %0.1f R',delta_T_oe)\n",
"\n",
"//let f=poe/po1\n",
"f=(1+(345.1/530))^(0.9*3.5)\n",
"printf('\n poe/po1= %0.2f',f)\n",
"\n",
"disp('The adiabetic efficiency is given as ((poe/poi)^(((k-1)/k))-1/((Toe/Toi)-1)')\n",
"//let  k-1/k=d\n",
"d=0.2857\n",
"f=4.85\n",
"\n",
"ETA_ad=((f^(d))-1)/((345.1/530))\n",
"printf('\nETAad= %0.4f',ETA_ad)\n",
"//let r=ETAd*100\n",
"r=ETA_ad*100\n",
"printf('\n Thus ETAad= %0.2f percent',r)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.5: AFC.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('Assume constant axial flow velocity Va=500 ft/s , no prewhirl and hence V1=Va for the first stage and frre vortex conditions for all stages.')\n",
"\n",
"disp('1. First Stage')\n",
"\n",
"T1=530-((500^2)/(2*6012))\n",
"printf('T1= %0.1f R',T1)\n",
"\n",
"p1=14.7*((509.2/530)^3.5)\n",
"printf('\n p1= %0.2f psia',p1)\n",
"\n",
"rho_1=(12.78*144)/(53.33*509.2)\n",
"printf('\n rho_1= %0.5f lm/ft^3',rho_1)\n",
"\n",
"disp('With a selected nu=rh/rt=0.45, A=m/(rho_1*Va)')\n",
"m=65\n",
"rho_1=0.0677\n",
"Va=500\n",
"A=m/(rho_1*Va)\n",
"printf('\n A= %0.2f ft^2 = 276.5in^2',A)\n",
"\n",
"disp('A=pi*(r_t^2)*(1-nu^2)')\n",
"A=276.5\n",
"nu=0.45\n",
"r_t=sqrt(A/(%pi*(1-(nu^2))))\n",
"printf('\n r_t= %0.1f in',r_t)\n",
"\n",
"disp('rm=7.6in')\n",
"\n",
"disp('Ut=rt*omega')\n",
"N=15000\n",
"omega=N*%pi/30\n",
"rt=10.5/12//in feet\n",
"Ut=rt*omega\n",
"printf('\n Ut %0.1f ft/s',Ut)\n",
"\n",
"rh=4.7/12//in feet\n",
"Uh=rh*omega\n",
"printf('\n Uh= %0.1f ft/s',Uh)\n",
"\n",
"rm=7.6/12//in feet\n",
"Um=rm*omega\n",
"printf('\n Um= %0.1f ft/s',Um)//answer given in the book is 996.6,however 994.8 is more accurate\n",
"\n",
"disp('Without inlet whirl flow,we have Wt=(Ut^2+Va^2)^0.5')\n",
"Ut=1374.4\n",
"Va=500\n",
"Wt=sqrt(Ut^2+Va^2)\n",
"printf('\n Thus Wt = %0.0ft/s',Wt)\n",
"\n",
"Uh=615.2\n",
"Va=500\n",
"Wh=sqrt(Uh^2+Va^2)\n",
"printf('\n Thus Wh = %0.1ft/s',Wh)\n",
"\n",
"Um=996.6//This is the answer substituted in the book,although as mentioned earlier 994.8 is more accurate.\n",
"Va=500\n",
"Wm=sqrt(Um^2+Va^2)\n",
"printf('\n Thus Wm = %0.1ft/s',Wm)\n",
"\n",
"tanbeta1m=Um/Va\n",
"printf('\n tanß1m= %0.2f',tanbeta1m)\n",
"\n",
"beta1m=((atan(tanbeta1m)))*180/%pi\n",
"printf('\n ß1m= %0.1f degrees',beta1m)\n",
"\n",
"disp('With a1=(k*R*T1)^0.5=1309ft/s,the relative Mach numbers can be determined.')\n",
"Wt=1463\n",
"a1=1309\n",
"M_rt=Wt/a1\n",
"printf('\n M_rt= %0.2f',M_rt)\n",
"\n",
"Wh=792.8\n",
"M_rh=Wh/a1\n",
"printf('\nM_rh= %0.3f',M_rh)\n",
"\n",
"Wm=1115.0\n",
"M_rm=Wm/a1\n",
"printf('\n M_rm= %0.2f',M_rm)\n",
"\n",
"disp('Hence the relative flow at the leading edge is transonic,and a supersonic blade might be needed.')\n",
"disp('From the previous sum,to achieve a pressure ratio of 4.5,we have deltaTos=324/7=46.3R.')\n",
"disp('So deltaTos=45R is selected at the first stage.')\n",
"disp('From Cp*deltaTos=lambda*Um*Va*(tanß1m-tanß2m),with lambda=0.95,we obtain tanß2m=1.42. Thus ß2m=54.8 degrees')\n",
"\n",
"tanbeta1=1.99\n",
"tanbeta2=1.42\n",
"Va=500\n",
"Um=996.6\n",
"R=(Va/(2*Um))*(tanbeta1+tanbeta2)\n",
"printf('\n R=(Va/(2*Um))*(tanbeta1+tanbeta2)= %0.3f',R)\n",
"\n",
"disp('We have nu=(1-R)^0.5')\n",
"nu=sqrt(1-R)\n",
"printf('\n nu= %0.2f',nu)\n",
"\n",
"disp('This is less than the selected value and is acceptable.')\n",
"beta1=63.3\n",
"beta2=54.8\n",
"deHaller=cos(beta1*%pi/180)/cos(beta2*%pi/180)\n",
"printf('\n Also checking the deHaller number W2/W1=cosß1/cosß2=%0.2f >0.70,it is also acceptable',deHaller)\n",
"\n",
"disp('Before proceeding to the next stages,the following parameters have to be specified. The summation of deltaTos should add upto 324degrees.So the following arramgement is chosen')\n",
"Stage_number=[1 2 3 4 5 6 7 ];\n",
"deltaTos_R=[45 46 47 47 47 47 45];\n",
"lambda=[0.95 0.93 0.90 0.89 0.88 0.86 0.85];\n",
"R=[0.855 0.8 0.7 0.63 0.60 0.55 0.5];\n",
"table=[Stage_number' deltaTos_R' lambda' R']\n",
"disp('The table from left to right has values of Stagenumber deltaTos lamdbda and R')\n",
"disp(table)\n",
"\n",
"disp('2.Second Stage')\n",
"\n",
"To1=530+45\n",
"printf('To1= %0.0f R',To1)\n",
"\n",
"po1=14.7*((1+(0.9*45/530))^3.5)\n",
"printf('\n po1= %0.0f psia',po1)\n",
"\n",
"disp('With the specified values of deltaTos,lamda and R,equations 7.1 and 7.2A can be solved for tanß1 and tanß2.')\n",
"disp('Hence ß1=62.1 degrees and ß2=52.4 degrees,where W2/W1=0.76>0.70')\n",
"beta1=62.1*%pi/180//converting to radians \n",
"V1=[(Va^2)+(Um-(Va*tan(beta1)))^2]^0.5\n",
"printf('Also we have V1=[(Va^2)+(Um-(Va*tan(beta1)))^2]^0.5= %0.2f ft/s',V1)//answer in the book is 502.6,however the value found out here is more accurate\n",
"\n",
"Cp=6012\n",
"V1=502.6\n",
"T1=To1-((V1^2)/(2*Cp))\n",
"printf('\nT1= %0.0f R',T1)\n",
"\n",
"po1=19\n",
"T1=554\n",
"To1=575\n",
"p1=po1*((T1/To1)^3.5)\n",
"printf('\np1= %0.1f psia',p1)\n",
"\n",
"disp('rho1= p1/(R*T1)= 0.0813 lb_m/ft^3')\n",
"\n",
"\n",
"\n",
"disp('A=m/(rho1*Va)=230')\n",
"disp('A is also = 2*pi*r_m*b')\n",
"disp('Where b=rt-rh and rm is the same as that for the first stage. Hence we have rt=10 in,rh=5.2 in and nu=0.52,which is greater than nu_min(0.447) ')\n",
"\n",
"disp('3.Third to Seventh stage')\n",
"disp('The calculation procedure for these stages will be similar to the second stage. The calculations have not been repeated in the book,and hence the same is done here.')\n",
"disp('The results are as follows')\n",
"\n",
"Stagenumber=[1 2 3 4 5 6 7];\n",
"To1=[530 575 621 668 715 762 809];\n",
"po1=[14.7 19 24.3 30.5 37.9 46.3 55.9];\n",
"rt=[10.5 10 9.6 9.3 9.1 8.9 8.8];\n",
"rh=[4.7 5.2 5.6 5.9 6.1 6.3 6.5];\n",
"rm=[7.6 7.6 7.6 7.6 7.6 7.6 7.6 ];\n",
"beta1=[63.3 62.1 59.7 57.5 56.5 54.7 52.4];\n",
"beta2=[54.9 52.3 47.2 43.2 41.4 38 34.8]\n",
"table2=[Stagenumber' To1' po1' rt' rh' rm' beta1' beta2']\n",
"disp('Stageno. To1     po1     rt     rh     rm     beta1     beta2')\n",
"disp(table2)\n",
"\n",
"disp('The final discharge stagnation pressure can also be checked with the flow across the last stage such that poe=po3=po1(1+ETAs*(deltaTos/To1))^(k/(k-1))')\n",
"po1=55.9*(1+(0.9*45/809))^3.5\n",
"printf('po1= %0.1f psia',po1)\n",
"//let n=poe/poi\n",
"n=66.3/14.7\n",
"printf('\n This checks with poe/poi = %0.2f',n)\n",
"\n",
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"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 7.6: AFC.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('We have psi=lambda*[1-φ*(tanα1+tanß2)] and R=0.5+[(φ/2)*(tanß2-tanα1)]')\n",
"disp('We have tanα1+tanß2=(1-psi/lamda)=1.01')\n",
"disp('tanα1-tanß2=(1-2R)/φ=-0.545,or psi=lambda*(1-1.01*φ) and R=0.5+0.272*φ Hence φdash=0.495,assuming lamda to be constant we can determine the values of psi_dash and R_dash')\n",
"\n",
"lambda=0.9\n",
"//Let (tanalpha1+tanbeta2)=x\n",
"x=1.01\n",
"phi_dash=0.495\n",
"psi_dash=lambda*[1-(phi_dash*(x))]\n",
"printf('\n psi_dash= %0.2f',psi_dash)\n",
"\n",
"//Let= 0.5*(tanbeta2-tanalpha1=y\n",
"y=0.272\n",
"R_dash=0.5+[(phi_dash*0.272)]\n",
"printf('\n R_dash= %0.3f',R_dash)\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
}