Chapter 12 :Turbomachines

Example 12.2 Page no.697

In [10]:
Q=1400.0                                    #gpm, water rate
N=1750.0                                    #rpm  speed
b=2.0                                       #in  height of blade
r1=1.9                                      #in  inner radius
r2=7.0                                      #in  outer radius
beta2=23.0                                  #degrees  exit blade angle
beta2=beta2*3.14/180                        #Radian
alpha1=90.0                                 #degrees  water entering angle

#Calculation
import math
U2=r2*2*math.pi*N/(60*12)                            #ft/sec
Vr2=(1400/(7.48*60*2*math.pi*(r2/12)*(b/12)))        #ft/sec
V2tangential=round(U2,0)-(Vr2*1/(math.tan(beta2)))   #ft/sec
hi=U2*V2tangential/32.2                              #ft
print "The ideal head rise=",round(hi,2),"ft"
d=1.94                                               #slugs/(ft**3)
Wshaft=(d*Q*U2*V2tangential/(7.58*60))/550           #hp
  
#result
print "Tangential velocity component is ",round(V2tangential,0),"ft/s"
print "The power transferred to the fluid=",round(Wshaft,1),"hp"

The ideal head rise= 315.27 ft
Tangential velocity component is  95.0 ft/s
The power transferred to the fluid= 110.226 hp

Example 12.3 Page no.702

In [2]:
Q=0.5                 #(ft**3)/sec
NPSHr=15         #ft
T=80                 #degree F
patm=14.7          #psi
KL=20
D=4.0                #in

#Calculation
import math
V=Q/((math.pi/4)*(D/12.0)**2.0)         #ft/sec
hL=KL*(V**2)/(2*32.2)                #ft
#from value of T
pv=0.5069                                    #psi
sw=62.22            #lb/(ft**3) sw =specific weight
z1max=(patm*144/sw)-hL-(pv*144/sw)-NPSHr   #ft

#result
print "The maximum height at which the pump can be located=",round(z1max,2),"feet"

The maximum height at which the pump can be located= 7.65 feet

Example 12.5 Page no.708

In [4]:
D1=8.0                         #in
N1=1200.0                    #rpm
D2=12.0                        #in
N2=1000.0                    #rpm
T=60.0                         #degree F
CQ=0.0625

#Calculation
import math
Q1=CQ*(N1/60)*(2*math.pi)*(D1/12.0)**3.0    #(ft**3)/sec
print "The discharge=",round(Q1*7.48*60,0),"gpm"
CH=0.19
ha=CH*((N1*2*math.pi/60)**2.0)*((D1/12)**2.0)/32.2  #ft
print "The actual  headrise=",round(ha,1),"ft"
CP=0.014
Wshaft=(CP*(1.94)*((N1*2*math.pi/60)**3.0)*((D1/12.0)**5.0))/550.0  #hp

#Result
print "The shaft horsepower=",round(Wshaft,0),"hp"

The discharge= 1044.0 gpm
The actual  headrise= 41.4 ft
The shaft horsepower= 13.0 hp

Example 12.6 Page no.719

In [5]:
z0=200                     #ft  height of left section
l=1000                     #ft length of pipe
f=0.02
D=8                        #in. diameter of pipe
B=150                      #degree 
R=1.5                      #ft  radius of wheel
z1=0                       #ft    height of right section
#calculation
import math
#energy equation between a point on surface of lake and the nozzle outlet
#z0=(V1**2)/(2*32.2) + hL
#from continuity equation, V=(A1*V1/A)=((D1/D)**2)*V1
#neglecting minor losses, 
#z0=(1+(f*l/D)*((D1/D)**4))*(V1**2)/(2*32.2)
#Wshaft=d*Q*u*(U-V1)*(1-cos(B))
#The maximum power occurs at U=V1/2 and dWshaft/dD1=0
a=(2*32.2*z0)**(0.5)                                         
b=f*(l/(D/12.0))*(1/(D/12.0))**4.0                           
c=a*math.pi*1.94*(1-math.cos(B*math.pi/180))/4.0     
d=(c*a*a/4)                                          #1.04*(10**6)
#by the above conditions, and applying Q=(math.pi*(D1**2)*V1/4)
D1=(1/(2*b))**(0.25)                                 #ft, nozzle diameter

#result
print "The nozzle diameter for maximum power output=",round(D1,2),"ft"
Wshaft=-((d*D1**2)/(1+(b*D1**4))**(1.5))/550.0       #hp
print "The maximum power output=",round(Wshaft,0),"hp"
V1=a/(1+(b*(D1**4)))**0.5                            #ft/sec
omega=(V1/(2*R))*60/(2*math.pi)                      #rpm
print "The angular velocity of the rotor=",round(omega,0),"rpm"

The nozzle diameter for maximum power output= 0.24 ft
The maximum power output= -59.0 hp
The angular velocity of the rotor= 295.0 rpm

Example 12.8 Page no.723

In [7]:
ri=0.133                      #in.
ro=0.168                     #in.
N=300000                  #rpm

#calculation
import math
rm=0.5*(ro+ri)/12
U=(N*2*math.pi/60)*rm          #ft/sec
wshaft=(-U)*(2*U)/32.174       #ft*lb/lbm

#Result
print "The shaft energy per unit mass of air=",round(wshaft,0),"ft*lb/lbm"

The shaft energy per unit mass of air= -9650.0 Ft*lb/lbm

Example 12.9 Page No.727

In [19]:
w=6.0             #rev/s angular velocity
q=10.0            #ft**3/s , flow rate
h=20.0          #ft, head
gamma=62.4      #lb/ft**3, specific wt
n=94            # %, assumed efficiency
#calculation
import math
##rev/min
Wshaft=gamma*q*h*(n*10**-2)/(550.0) #shaft power
N=w1*math.sqrt(Wshaft)/(h)**(1.25)
print "Accordint to information a mixed flow Francis Turbine must be used."

g=32.2          #ft/s**2
V1=math.sqrt(2*g*h)   #ft/s  velocity
D=V1/(w*2*math.pi)    #ft
d1=math.sqrt(4*q/(math.pi*V1))
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."

Accordint to information a mixed flow Francis Turbine must be used.
So a Pelton wheel with a diameter= 0.952 ftsupplied with water through nozzle of diameter= 0.596 ftis not a practical design.