Chapter 12: Turbomachines

Example 12.1 Page no 443

In [1]:
# Example 12.1

from math import *

from __future__ import division

# Given

Q = 0.25                      # discharge from the pump in m**3/s

gma= 0.8*9810                 # specific weight in kg/m**3

H=25                          # elevation head in m

T = 350                       # Torque to drive the shaft in Nm

N = 1800                      # Speed in RPM

w = 2*pi*N/60                 # angular velocity

# Solution

Eff = gma*Q*H*100/(T*w)          # efficiency

print "Efficiency of th pump =",round(Eff,0),"%"
Efficiency of th pump = 74.0 %

Example 12.2 Page no 447

In [3]:
# Example 12.2

from math import *

from __future__ import division

# Given

d = 0.4                     # diameter of the pump in m

b = 0.03                    # width in m

theta = pi/3                # blade angle

N = 1500                    # speed in RPM

Q = 0.4                     # flow rate in m**3/s

g = 9.81                    # acceleration due to gravity in m/s**2

# Solution

w = 2*pi*N/60               # anggular velocity in rad/s

u2 = (d/2)*w                # blade velocity in m/s

V2r = Q/(2*pi*(d/2)*b)      # relative velocity in m/s

print "(a)"

print "Radial velocity at exit =",round(V2r,1),"m/s"

V2t = u2 - V2r*(cos(theta)/sin(theta))

print "Whirl velocity = ",round(V2t,1),"m/s"

v2 = V2r/sin(theta)

print "Relative velocity = ",round(v2,2),"m/s"

V2 = sqrt(V2t**2+V2r**2)

print "Actual velocity =",round(V2,2),"m/s"

print "(b)"

H = u2*V2t/g

print"Head added for no inlet whirl =",round(H,0),"m"

print "(c)"

P = g*Q*H

print "Power required =",round(P,1),"kW"
 (a)
Radial velocity at exit = 10.6 m/s
Whirl velocity =  25.3 m/s
Relative velocity =  12.25 m/s
Actual velocity = 27.43 m/s
(b)
Head added for no inlet whirl = 81.0 m
(c)
Power required = 317.8 kW

Example 12.3 Page no 450

In [4]:
# Example 12.3

from math import *

from __future__ import division

# Given

d = 0.36                       # diameter of the impeller of pump

N = 1500                       # Speed of impeller in RPM

# Solution

# For best efficiency

Q1 = 82                       # discharge in l/s

H1 = 17.5                     # Head in m

Eta = 0.8                     # efficiency 

Q2 = 100                      # discharge in l/s

H2 = 20                       # head in m

# Solving the simulataneous equation we get

D2 = 38.45

print "Impeller size =",round(D2,2),"cm"

N2 = 1500 

print "Speed of the pump =",round(N2,0),"RPM"
Impeller size = 38.45 cm
Speed of the pump = 1500.0 RPM

Example 12.4 Page no 454

In [5]:
# Example 12.4

from math import *

from __future__ import division

# Given

q = 500                         # discharge in cgm

Q = 500/449                     # discharge in ft**3/s

D = 0.667                       # diameter in ft

A = pi*D**2/4

V = Q/A                         # velocity in ft/s

g = 32.2                        # acceleration due to gravity in ft/s**2

N = 1800                        # speed in RPM

# Solution

# for water at 65 deg C

nu = 1.134*10**-5               # viscosity in ft**2/s

e = 0.00085                     # epssilon in ft

r = 0.001275                    

R = V*D/nu                      # reynolds no

f = 0.022                       # from moody's diagram

Hl = V**2*(12.1+(f*224.9))/64.4

hs = 119.4 + Hl

print "Discharge, Q = ",round(Q,2),"ft**3/s"
print "Dynamic head of the pump, H =",round(hs,1),"m"

Ns = N*sqrt(q)/(hs)**(3/4)

print "Specific speed of the pump, Ns =",round(Ns,0),"RPM"
Discharge, Q =  1.11 ft**3/s
Dynamic head of the pump, H = 122.1 m
Specific speed of the pump, Ns = 1096.0 RPM

Example 12.5 Page no 457

In [6]:
# Example 12.5 

from math import *

from __future__ import division

# Given

H = 60                        # height in m

Pb = 98*10**3                 # barometric pressure in N/m**2

Hl = 1                        # head in m

Pv = 1707                     # vapour pressure 

sigma = 0.08

w = 9810                      # specific weight

# Solution

Npsh_m = sigma*60             # minimum NPSH

Hsm = (Pb/w)-(Pv/w)-Npsh_m-Hl

print "Minimum value of static suction lift = ",round(Hsm,2),"m"
Minimum value of static suction lift =  4.02 m
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