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{
"metadata": {
"name": "ch 12"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 12 :Turbomachines"
]
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.2 Page no.697"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 12.2\n",
"#Find The ideal head rise\n",
"#and Tangential velocity component and The power transferred to the fluid.\n",
"\n",
"#Given\n",
"Q=1400.0 #gpm, water rate\n",
"N=1750.0 #rpm speed\n",
"b=2.0 #in height of blade\n",
"r1=1.9 #in inner radius\n",
"r2=7.0 #in outer radius\n",
"beta2=23.0 #degrees exit blade angle\n",
"beta2=beta2*3.14/180 #Radian\n",
"alpha1=90.0 #degrees water entering angle\n",
"\n",
"#Calculation\n",
"import math\n",
"U2=r2*2*math.pi*N/(60*12) #ft/sec\n",
"Vr2=(1400/(7.48*60*2*math.pi*(r2/12)*(b/12))) #ft/sec\n",
"V2tangential=round(U2,0)-(Vr2*1/(math.tan(beta2))) #ft/sec\n",
"hi=U2*V2tangential/32.2 #ft\n",
"print \"The ideal head rise=\",round(hi,2),\"ft\"\n",
"d=1.94 #slugs/(ft**3)\n",
"Wshaft=(d*Q*U2*V2tangential/(7.58*60))/550 #hp\n",
" \n",
"#result\n",
"print \"Tangential velocity component is \",round(V2tangential,0),\"ft/s\"\n",
"print \"The power transferred to the fluid=\",round(Wshaft,1),\"hp\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The ideal head rise= 315.27 ft\n",
"Tangential velocity component is 95.0 ft/s\n",
"The power transferred to the fluid= 110.226 hp\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.3 Page no.702\n"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 12.3\n",
"#Find the maximum height at which the pump can be located.\n",
"\n",
"#Given\n",
"Q=0.5 #(ft**3)/sec\n",
"NPSHr=15 #ft\n",
"T=80 #degree F\n",
"patm=14.7 #psi\n",
"KL=20\n",
"D=4.0 #in\n",
"\n",
"#Calculation\n",
"import math\n",
"V=Q/((math.pi/4)*(D/12.0)**2.0) #ft/sec\n",
"hL=KL*(V**2)/(2*32.2) #ft\n",
"#from value of T\n",
"pv=0.5069 #psi\n",
"sw=62.22 #lb/(ft**3) sw =specific weight\n",
"z1max=(patm*144/sw)-hL-(pv*144/sw)-NPSHr #ft\n",
"\n",
"#result\n",
"print \"The maximum height at which the pump can be located=\",round(z1max,2),\"feet\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The maximum height at which the pump can be located= 7.65 feet\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.5 Page no.708"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 12.5\n",
"#For peak efficiency predict the discharge actual head rise and shaft horsepower.\n",
"\n",
"#Given\n",
"D1=8.0 #in\n",
"N1=1200.0 #rpm\n",
"D2=12.0 #in\n",
"N2=1000.0 #rpm\n",
"T=60.0 #degree F\n",
"CQ=0.0625\n",
"\n",
"#Calculation\n",
"import math\n",
"Q1=CQ*(N1/60)*(2*math.pi)*(D1/12.0)**3.0 #(ft**3)/sec\n",
"print \"The discharge=\",round(Q1*7.48*60,0),\"gpm\"\n",
"CH=0.19\n",
"ha=CH*((N1*2*math.pi/60)**2.0)*((D1/12)**2.0)/32.2 #ft\n",
"print \"The actual headrise=\",round(ha,1),\"ft\"\n",
"CP=0.014\n",
"Wshaft=(CP*(1.94)*((N1*2*math.pi/60)**3.0)*((D1/12.0)**5.0))/550.0 #hp\n",
"\n",
"#Result\n",
"print \"The shaft horsepower=\",round(Wshaft,0),\"hp\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The discharge= 1044.0 gpm\n",
"The actual headrise= 41.4 ft\n",
"The shaft horsepower= 13.0 hp\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.6 Page no.719"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 12.6\n",
"#determine The nozzle diameter for maximum power outpu.\n",
"#The maximum power output and The angular velocity of the rotor.\n",
"\n",
"#Given\n",
"z0=200 #ft height of left section\n",
"l=1000 #ft length of pipe\n",
"f=0.02\n",
"D=8 #in. diameter of pipe\n",
"B=150 #degree \n",
"R=1.5 #ft radius of wheel\n",
"z1=0 #ft height of right section\n",
"#calculation\n",
"import math\n",
"#energy equation between a point on surface of lake and the nozzle outlet\n",
"#z0=(V1**2)/(2*32.2) + hL\n",
"#from continuity equation, V=(A1*V1/A)=((D1/D)**2)*V1\n",
"#neglecting minor losses, \n",
"#z0=(1+(f*l/D)*((D1/D)**4))*(V1**2)/(2*32.2)\n",
"#Wshaft=d*Q*u*(U-V1)*(1-cos(B))\n",
"#The maximum power occurs at U=V1/2 and dWshaft/dD1=0\n",
"a=(2*32.2*z0)**(0.5) \n",
"b=f*(l/(D/12.0))*(1/(D/12.0))**4.0 \n",
"c=a*math.pi*1.94*(1-math.cos(B*math.pi/180))/4.0 \n",
"d=(c*a*a/4) #1.04*(10**6)\n",
"#by the above conditions, and applying Q=(math.pi*(D1**2)*V1/4)\n",
"D1=(1/(2*b))**(0.25) #ft, nozzle diameter\n",
"\n",
"#result\n",
"print \"The nozzle diameter for maximum power output=\",round(D1,2),\"ft\"\n",
"Wshaft=-((d*D1**2)/(1+(b*D1**4))**(1.5))/550.0 #hp\n",
"print \"The maximum power output=\",round(Wshaft,0),\"hp\"\n",
"V1=a/(1+(b*(D1**4)))**0.5 #ft/sec\n",
"omega=(V1/(2*R))*60/(2*math.pi) #rpm\n",
"print \"The angular velocity of the rotor=\",round(omega,0),\"rpm\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The nozzle diameter for maximum power output= 0.24 ft\n",
"The maximum power output= -59.0 hp\n",
"The angular velocity of the rotor= 295.0 rpm\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.8 Page no.723"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 12.8\n",
"#find The shaft energy per unit mass of air.\n",
"#Given\n",
"ri=0.133 #in.\n",
"ro=0.168 #in.\n",
"N=300000 #rpm\n",
"\n",
"#calculation\n",
"import math\n",
"rm=0.5*(ro+ri)/12\n",
"U=(N*2*math.pi/60)*rm #ft/sec\n",
"wshaft=(-U)*(2*U)/32.174 #ft*lb/lbm\n",
"\n",
"#Result\n",
"print \"The shaft energy per unit mass of air=\",round(wshaft,0),\"ft*lb/lbm\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The shaft energy per unit mass of air= -9650.0 Ft*lb/lbm\n"
]
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 3,
"metadata": {},
"source": [
"Example 12.9 Page No.727"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#Example 12.9\n",
"#What type of turbine should we selected.\n",
"\n",
"#Given\n",
"w=6.0 #rev/s angular velocity\n",
"q=10.0 #ft**3/s , flow rate\n",
"h=20.0 #ft, head\n",
"gamma=62.4 #lb/ft**3, specific wt\n",
"n=94 # %, assumed efficiency\n",
"#calculation\n",
"import math\n",
"##rev/min\n",
"Wshaft=gamma*q*h*(n*10**-2)/(550.0) #shaft power\n",
"N=w1*math.sqrt(Wshaft)/(h)**(1.25)\n",
"print \"Accordint to information a mixed flow Francis Turbine must be used.\"\n",
"\n",
"g=32.2 #ft/s**2\n",
"V1=math.sqrt(2*g*h) #ft/s velocity\n",
"D=V1/(w*2*math.pi) #ft\n",
"d1=math.sqrt(4*q/(math.pi*V1))\n",
"print \"So a Pelton wheel with a diameter=\",round(D,3),\"ft\" \"supplied with water through nozzle of diameter=\",round(d1,3),\"ft\" \"is not a practical design.\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Accordint to information a mixed flow Francis Turbine must be used.\n",
"So a Pelton wheel with a diameter= 0.952 ftsupplied with water through nozzle of diameter= 0.596 ftis not a practical design.\n"
]
}
],
"prompt_number": 19
}
],
"metadata": {}
}
]
}
|
