{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 10: Hydraulic Turbines"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.1: HT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"//the values of omegas and energycoeeficient differ from the ones given in the book\n",
"//the reasons for the same are mentioned in the code below\n",
"H=85\n",
"Q=16\n",
"E=0.9//efficiency\n",
"g=9.8\n",
"rho=998\n",
"\n",
"P_o=E*rho*Q*g*H/1000\n",
"printf('The estimated power (Po) is equal to %0.0f kW',P_o)\n",
"\n",
"\n",
"disp('From figure 10.11,a Francis Turbine is selected. Then with the synchronous speed of 16 poles N is determined')\n",
"N=120*60/16\n",
"printf('N is equal to %grpm',N)\n",
"\n",
"\n",
"N_s=(N*Q^0.5)/H^0.75\n",
"printf('\nWe have value of Ns equal to %0.1f rpm(m^3/s)^0.5',N_s)\n",
"\n",
"\n",
"Ksigma=2.11\n",
"n=450/60\n",
"g=9.8\n",
"sigma=(Ksigma*n*Q^0.5)/((g*H)^0.75)\n",
"printf('\n Value of sigma is equal to %0.2f',sigma)\n",
"\n",
"\n",
"omega=(%pi*N)/30\n",
"V=16\n",
"omega_s=(omega*V^0.5)/((g*H)^0.75)//Answer given in the book is 1.33,this is because H has been wrongly substitued as 75. The correct substitution(H=85),gives the answer equal to 1.2157.\n",
"thita=1.9\n",
"K=1.054\n",
"printf('\n Value of omegas is equal to %g',omega_s)\n",
"\n",
"\n",
"disp('From figure 10.10 we have thita=1.9 for nq=Ns=64.3')\n",
"disp('Since K*D*(g*H)^0.25/Q^0.5=thita, we can determine D.')\n",
"\n",
"D=(thita*(Q^0.5))/(((g*H)^0.25)*K)\n",
"printf('\n Value of D is equal to %0.2f m ',D)\n",
"\n",
"\n",
"disp('From figure 10.9 we have efficiency=0.95,which is close to the original estimation')\n",
"D=1.34//value of D is approximately equal to 1.34\n",
"k_phi=(%pi^2)/4\n",
"k_psi=(%pi^2)/2\n",
"flowcoeff=Q/(k_phi*n*(D^3))\n",
"printf('The flow coefficient is equal to %0.3f',flowcoeff)\n",
"\n",
"\n",
"energycoeff=(g*H)/(k_psi*(n^2)*(D^2))\n",
"printf('\nThe energy coefficient is equal to %0.4f',energycoeff)//Answer given in the book is 1.47,this is because H has been wrongly substitued as 75. The correct substitution(H=85),gives the answer equal to 1.6713.\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.2: HT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"\n",
"disp('From psi=(g*H)/(omega^2*D^2)  and N=30*omega/pi we get N=172.7/(psi^0.5)')\n",
"disp('Also from phi=Q/(omega*D^3) and pi=Ps/(rho*omega^3*D^5) we get Q=0.353*phi*N and Ps=0.0087*N^3*pi')\n",
"disp('Pick the points along 80% gate opening curve,read the values for phi,psi, and efficiency from figure 10.14')\n",
"\n",
"phi=[0.158 0.151 0.14 0.127 0.108 0.092 0.076 0.066];\n",
"psi= [0.093 0.083 0.071 0.06 0.048 0.04 0.03 0.025];\n",
"E= [55 56.5 58 62.5 69 71.5 67.5 60];//efficiency\n",
"pai= [0.0078 0.0067 0.0058 0.0045 0.0034 0.0025 0.0015 0.001];\n",
"\n",
"N = zeros(1,length(phi));\n",
"Ps = zeros(1,length(phi));\n",
"Q = zeros(1,length(phi));\n",
"\n",
"for i = 1: length(phi)\n",
"   \n",
"\n",
"    N(i) = 172.7/sqrt(psi(i));\n",
"    Ps(i) = 0.0087*N(i)^3*pai(i)*10^-3;\n",
"    Q(i) = 0.353*phi(i)*N(i);\n",
"end\n",
"\n",
"disp('    phi     psi      eff(%)   pai        N(rpm)      Ps(mw)       Q(m^3/s)')\n",
"\n",
"table = [phi' psi' E' pai' N' Ps' Q'];\n",
"disp(table)\n",
"\n",
"plot(N,Ps,'o',N,Q,'d',N,E,'s')\n",
"legend('Ps (mw)','Q (m^3/s)','Eff (%)',-1)\n",
"xlabel('N (rpm)')\n",
"ylabel('Ps (mW), Q (m^3/s) , eff (%)')\n",
"set(gca(),'grid',[1 1])\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.3: HT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"D=3\n",
"dn=0.08\n",
"H=350\n",
"En=0.82\n",
"CVb=0.95\n",
"Em=0.90\n",
"Ev=0.96\n",
"g=9.8\n",
"\n",
"V2=(2*En*g*H)^0.5\n",
"printf(' The jet flow velocity is equal to %0.0f m/s',V2)\n",
"\n",
"Um=0.5*V2\n",
"printf('\n Optimum wheel tangential velocity is Um is equal to %0.1f m/s',Um)\n",
"\n",
"N=(60*Um)/(%pi*D)\n",
"printf('\n The rotating speed N is equal to %0.1f rpm',N)\n",
"\n",
"disp('Under the maximum utilization factor condition,we have beta3=90 degrees')\n",
"disp(' Since delta Emax=(1+CVb)*U(V2-U),we get the equation delta Emax=1.95*(U^2)')\n",
"delta_E_max=(1+CVb)*Um*(V2-Um)\n",
"printf(' The value of deltaEmax is equal to %g N-m/kg',delta_E_max)\n",
"\n",
"An=(%pi/4)*(dn^2)\n",
"Q=V2*An\n",
"printf('\n The flow rate is %0.3f m^3/s',Q)\n",
"\n",
"m=998*Q\n",
"Ps=Em*Ev*m*delta_E_max/1000\n",
"printf('\n The total shaft power output is %0.1f kW',Ps)\n",
"\n",
"Ns=(N*(Q^0.5))/(H^0.75)\n",
"printf('\n The specific speed can be calculated as %0.2f rpm((m^3/s)^0.5)/(m^0.75)',Ns)\n",
"\n",
"omega=%pi*N/30\n",
"omega_s=omega*(Q^0.5)/((g*H)^0.75)\n",
"printf('\n In non dimensional form , omegas is equal to %0.3f',omega_s)\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.4: HT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"H=80\n",
"Q=63\n",
"Es=0.97\n",
"N=400\n",
"V3=25\n",
"Dh3=2\n",
"rh3=1/12\n",
"g=32.2\n",
"\n",
"Ksigma=2.11\n",
"n=N/60\n",
"sigma=(Ksigma*n*(Q^0.5))/((g*H)^0.75)\n",
"printf(' The value of sigma is equal to %0.2f ', sigma)\n",
"\n",
"disp('We have thita=2.4. Thita is also equal to Kt*D2*((g*H)^0.25)/(Q^0.5)')\n",
"thita=2.4\n",
"Kt=1.054\n",
"D2=(thita*(Q^0.5))/(Kt*((g*H)^0.25))\n",
"printf(' Thus the value of D2 is %0.1f ft',D2)\n",
"\n",
"D2r=2.5//rounded off D2\n",
"U2=(D2r*N*%pi)/60\n",
"printf('\n U2 is equal to %0.2f ft/s',U2)\n",
"\n",
"V2=(2*g*H*Es)^0.5\n",
"printf('\n The inlet flow velocity V2 is equal to %0.2f ft/s',V2)\n",
"\n",
"disp('From the inlet velocity diagram for alpha2=20 degrees we have Vr2=V2*sinalpha2')\n",
"alpha2=20\n",
"Vr2=V2*(sin(alpha2*%pi/180))\n",
"printf(' Vr2 is equal to %0.2f ft/s',Vr2)\n",
"\n",
"tanbeta2=Vr2/(V2*(cos(alpha2*%pi/180))-U2)\n",
"printf('\n The value of tanbeta2 is %0.2f',tanbeta2)\n",
"beta2=(atan(tanbeta2))*180/%pi\n",
"printf('\n Thus value of beta2 is %0.1f degrees',beta2)\n",
"\n",
"disp('Selecting the incidence i=2.2 degrees we have betab2=62 degrees')\n",
"\n",
"disp('A2=Q/Vr2=(%pi*D2*b2) From this equation we can determine the value of b2.')\n",
"A2=Q/Vr2\n",
"printf(' A2 is equal to %0.2f ft^2',A2)\n",
"b2=A2/(%pi*D2)\n",
"printf('\n b2 is equal to %0.2f ft',b2)\n",
"disp('Thus b2= 4 inches')\n",
"\n",
"disp('At the outlet with rh3=1 inch, setting gamma=15 degrees and V3=25')\n",
"gamma1=15\n",
" A3=Q/V3\n",
"printf(' The value of A3 is equal to %g ft^2',A3)\n",
" rt3=((A3*(cos(gamma1*%pi/180)))/%pi+(rh3^2))^0.5\n",
" printf('\n The value of rt3 is %0.2f ft',rt3)\n",
" disp('On converting the value of rt3 from feet to inches we get rt3=10.6inches')\n",
" \n",
" rt3c=10.6//converted value of rt3\n",
" rh3c=1//converted value in inches\n",
" rm3=((((rt3c^2+ rh3c^2)/2)^0.5)/12)\n",
" printf(' The mean radius rm3 is equal to %0.3f ft',rm3)\n",
" Um3=26.3 \n",
" tanbetam3=V3/Um3\n",
" printf('\n The value of tanbetam3 %0.2f',tanbetam3)\n",
" betam3=atan(tanbetam3)*180/%pi\n",
" printf('\n The value of betam3 whih is equal to betabm3 if no deviation is assumed is equal to %0.2f degrees',betam3)\n",
" disp('On rounding off we get the value og betam3=43.6 degrees')\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 10.5: HT.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"clear all; clc;\n",
"disp('To use figure 10.21 we need the dimensional power specific speed. So the shaft power has to be estimated from figure 10.9 where the non dimensional omegas is needed.')\n",
"\n",
"Ve=5\n",
"Hl=0.7\n",
"\n",
"H=80\n",
"Q=63\n",
"Es=0.97\n",
"N=400\n",
"V3=25\n",
"Dh3=2\n",
"rh3=1/12\n",
"g=32.2\n",
"\n",
"omega=N*%pi/30\n",
"omega_s=omega*(Q^0.5)/((g*H)^0.75)\n",
"printf(' The value of omegas is %0.2f',omega_s)\n",
"\n",
"disp('We have efficiency=0.95')\n",
"E=0.95\n",
"rho=1.9378\n",
"Ps=E*rho*g*Q*H/550//conversion factor =1/550\n",
"printf(' The value of Ps %0.2f hp',Ps)\n",
"\n",
"Nsp=N*(Ps^0.5)/(80^1.25)\n",
"printf('\nThe value of Nsp is equal to %0.2f rpm(hp^0.5)/ft^1.25',Nsp)\n",
"\n",
"disp('From figure 10.21, we obtain sigma approximately equal to 0.1 or NSPHavail/H is greater than or equal to 0.1')\n",
"disp('NSPHavail =Ha-Z+Hl+Ve^2/(2*g)) and NSPH avail is greater than or equal to 8 ft.')\n",
"disp('At T=70 degrees farenheit we have the value of Ha equal to')\n",
"Ha=14.7*144/62.4\n",
"printf(' %0.2f ft.',Ha)\n",
"Hv=0.363*144/62.4\n",
"printf('\n The value of Hv is equal to %0.2f ft',Hv)\n",
"K=(Ve^2)/(2*g)\n",
"NPSHavail=8\n",
"printf('\n The value of (Ve^2)/(2*g) is equal to %0.2f ft',K)\n",
"//In the book the value of Zmax is directly stated\n",
"//I have used the given formualae and substiuted the values in it\n",
"//let NPSHavail=8\n",
"//then from the given formula we can find out the value of Zmax\n",
"\n",
"NPSH_avail=8\n",
"H_vr=0.84//rounded off value\n",
"Kr=0.39//rounded off value\n",
"H_ar=33.9//rounded off value\n",
"Z =H_ar-NPSH_avail+Hl+Kr-H_vr//Kr= rounded off value of (Ve^2)/(2*g)\n",
"printf('\n The value of Zmax is equal to %0.1f ft',Z)\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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