Chapter 15 : Impulse Turbines

Example 15.1 Page No : 486

In [3]:
import math 
from numpy import *
	
#Initialization of variables
z2 = 500.	#ft
z1 = 300.	#ft
D = array([1, 1.5, 2 ,2.5, 3, 4, 6])
g = 32.2
gam = 62.4
	
#calculations
Dj = D/12
Vj = sqrt((z2-z1)*2*g/(1.04 + 640.*Dj**4))
Aj = math.pi/4 *Dj**2
Q = Aj*Vj
Pjet = gam*Q*Vj**2 /(2*g) /550
Pj = max(Pjet)
for i in range(0,len(Pjet)):
    if(Pjet[i] == Pj):
        break
    
diameter = D[i]
	
#Results
print "Dj,in     Dj,ft     Vj,fps     Aj,ft**2     Q=AjVj,cfs     Pjet,hp"
for i in range(len(D)):
    print "%5.1f     %5.3f     %5.f     %7.4f        %5.2f        %5.1f"%(D[i],Dj[i],Vj[i],Aj[i],Q[i],Pjet[i])
print "Thus a pipe of %d in will be the optimum"%(diameter)

# answer are slightly different because of rounding off error

Dj,in     Dj,ft     Vj,fps     Aj,ft**2     Q=AjVj,cfs     Pjet,hp
  1.0     0.083       110      0.0055         0.60         12.7
  1.5     0.125       104      0.0123         1.27         24.2
  2.0     0.167        92      0.0218         2.00         29.6
  2.5     0.208        76      0.0341         2.58         26.1
  3.0     0.250        60      0.0491         2.96         19.0
  4.0     0.333        38      0.0873         3.31          8.4
  6.0     0.500        18      0.1963         3.48          1.9
Thus a pipe of 2 in will be the optimum

Example 15.2 Page No : 498

In [14]:
import math 
from sympy.functions.elementary.trigonometric import acot
	
#Initialization of variables
phi = 0.46
g = 32.2
k = 0.44
cv = 0.98
d = 10.      	#in
A = 0.545 	    #ft**2
beta = 160.  	#degrees
	
#calculations
u = phi*math.sqrt(2*g)
V1 = cv*math.sqrt(2*g)
gQ = 62.4*A*V1
T = d/2 *gQ/g *(1 - math.cos(math.radians(beta)) /math.sqrt(1+k) )*math.sqrt(2*g)*(cv-phi)
Power = T*2*u/d
	
#Results
print "Torque required  =  %d ft lb"%(T)
print " Power transferred  =  %d ft lb/s"%(Power)
Pi = gQ
He = Power/Pi
print " Hydraulic efficiency  =  %.2f"%(He)
v1 = V1-u
v2 = v1/(math.sqrt(1+k))
hl = k*v2**2 /(2*g)
print "Head loss in bucket friction  =  %.1f %%"%(hl*100)
Hn = (1/cv**2 -1)*V1**2 /(2*g)
print " Head loss in  nozzle  =  %.4f"%(Hn*100)
V2cos = u+v2*math.cos(math.radians(beta))
V2sin = v2*math.sin(math.radians(beta))
#alpha = math.degrees(1/math.atan(V2cos/V2sin))
alpha = math.degrees(acot(V2cos/V2sin))
V2 = V2sin/math.sin(math.radians(alpha))
Hd = V2**2/(2*g)
print " Head loss at discharge  =  %.1f %%"%(Hd*100)
Htotal = Hd+Hn+hl
print " Total head loss  =  %.2f %%"%(Htotal*100)

# rounding off error

Torque required  =  309 ft lb
 Power transferred  =  228 ft lb/s
 Hydraulic efficiency  =  0.85
Head loss in bucket friction  =  8.3 %
 Head loss in  nozzle  =  3.9600
 Head loss at discharge  =  2.5 %
 Total head loss  =  14.70 %

Example 15.3 Page No : 501

In [1]:
import math 
	
#Initialization of variables
cv = 0.98
g = 32.2
h = 1320. 	#ft
A = 0.196 	#ft**2
eta = 0.85
ne = 400.
phi = 0.45
	
#calculations
V = cv*math.sqrt(2*g*h)
Q = A*V
bhp = eta*62.4*Q*h/550
ns = ne*math.sqrt(bhp) /h**(5./4)
u = phi*math.sqrt(2*g*h)
D = u*60/math.pi/ne
	
#Results
print "Pitch diameter  =  %.2f ft"%(D)


# part b
#Initialization of variables
cv = 0.98
g = 32.2
h = 1320. 	#ft
A = 0.196 	#ft**2
eta = 0.85
ne = 400.
phi = 0.45
	
#calculations
V = cv*math.sqrt(2*g*h)
Q = A*V/3
bhp = eta*62.4*Q*h/550
ne2 = 600.
ns1 = ne2*math.sqrt(bhp) /h**(5./4)
D = 2500./ne2
Dj = math.sqrt(Q*4/V/math.pi)
	
#Results
print " Jet diameter  =  %.3f ft"%(Dj)
print " Specific speed  =  %.2f "%(ns1)
print " Pitch Diameter  =  %.2f ft"%(D)
print " Operating speed  =  %d rpm"%(ne2)

# rounding off error

Pitch diameter  =  6.26 ft
 Jet diameter  =  0.288 ft
 Specific speed  =  3.68 
 Pitch Diameter  =  4.17 ft
 Operating speed  =  600 rpm