Chapter 11 : Flow and Power Transmission through Pipe-lines

Example 11.2.1 page no : 307

In [2]:
import math 
#initialisation of variables
Q= 0.5 			#ft**3/sec
d= 3. 			#in
d1= 4. 			#in
d2= 2. 			#in
h= 12.7 			#in
g= 32.2 			#ft/sec**2
s= 13.6 			#kg/m**3
			
#CALCULATIONS
k  = (s-1)*(h/2.)*2.*g*(math.pi/(Q*d**2*4))**2.+((d1/d)**4.-1)
Cc = (d1/d2)**2./(math.sqrt(k)+1.)
			
#CALCULATIONS
print  ' contraction coefficient= %.3f '%(Cc)
print  ' ANSWER GIVEN IN THE TEXTBOOK IS WRONG'

 contraction coefficient= 0.294 
 ANSWER GIVEN IN THE TEXTBOOK IS WRONG

Example 11.3.1 page no : 310

In [5]:
    
import math 
#initialisation of variables
Q= 400. 			#gallons
d= 4. 			#in
d1=6. 			#in
C= 0.66
g= 32.2 			#ft/sec**2
w=62.4			#lbf/ft**3
			
#CALCULATIONS
u1= (Q/60.)*d1**2./(math.pi*6.23)
u2= (d/d1)**2.*u1
h= (u1-u2)**2./(2*g)
w=62.4			#lbf/ft**3
p= (((u1**2-u2**2)/(2*g))-h)*w
h1= ((1/C)-1.)**2*(u1**2/(2*g))
p1= (((u1**2.-u2**2)/(2*g))+h1)*w
p2= (u1**2.-u2**2)
			
#RESULTS
print  ' Loss of head due to the sudden enlargement= %.3f ft'%(h)
print  '  difference in pressure = %.1f lbf/ft**2'%(p)
print  '  difference in pressure = %.f lbf/ft**2'%round(p1,-1)
print  '  difference in pressure = %d lbf/ft**2'%(p2)

 Loss of head due to the sudden enlargement= 0.721 ft
  difference in pressure = 71.9 lbf/ft**2
  difference in pressure = 160 lbf/ft**2
  difference in pressure = 120 lbf/ft**2

Example 11.8.1 page no : 323

In [3]:
# variables
f = .0075
H = 96           # ft
u1 = 25.         # ft
g = 64.4        
l1 = 5280        # ft


# calculations
d = (4*f*u1*(3*l1+(4*l1)/4))/(H*g)

# results
print "Diameter of Pipe = %.2f ft"%d

Diameter of Pipe = 2.56 ft

Example 11.8.2 page no : 324

In [6]:
import math
#initialisation of variables
l= 9. 			#ft
g= 32.2 			#ft/sec**2
Q= 160.
A= 21.*math.pi
			
#CALCULATIONS
d= 6./(l*2*g*(A/Q)**2-1.5)
			
#RESULTS
print  'diameter = %.2f ft'%(d)

#ANSWER GIVEN IN THE TEXTBOOKIS WRONG

diameter = 0.06 ft

Example 11.8.3 pageno :325

In [11]:
import math 
#initialisation of variables
a= 2.493
b= 6.8
c= -393./(4*2.493)
d= 0.75 			#ft
			
#CALCULATIONS
u2= (-b+math.sqrt(b**2-4*a*c))/(2*a)
Q= math.pi*d**2*u2/4.
			
#RESULTS
print  ' Rate of flow = %.2f ft**3/sec'%(Q)

 Rate of flow = 1.25 ft**3/sec

Example 11.10.2 page no : 332

In [12]:
    
import math 
#initialisation of variables
g= 32.2 			#ft/sec**2
H= 100. 			#ft
L= 1000. 			#ft
h1= 0.03
h= 0.05
h2= 0.4
d= 6. 			#in
le= 1021. 			#ft
			
#CALCULATIONS
u1= math.sqrt((2*g*H)/(1+h+h2+(h1*L/0.5)))
Q= math.pi*(d/12.)**2*u1/4.
u2= math.sqrt((H*2*g)/(1+h+(1./16)*(1+h+h2+(h1*L/0.5))))
Q1= math.pi*(d/24)**2*u2/4
r= math.sqrt((d/12)/(2*h1*le))
			
#RESULTS
print  ' rate of discharge without a nozzle= %.2f ft**3/sec'%(Q)
print  '  rate of discharge= %.2f ft**3/sec'%(Q1)
print  '  diameter of nozzle= %.2f in'%(r)

 rate of discharge without a nozzle= 2.01 ft**3/sec
  rate of discharge= 1.78 ft**3/sec
  diameter of nozzle= 0.09 in

Example 11.10.3 pageno : 332

In [15]:
import math
#initialisation of variables

L= 1730. 			#ft
l= 104. 			#ft
hl= 234. 			#ft
u= 2.3 			#ft/sec
g= 32.2 			#ft/sec**2
p1 = 750*144.
w = 62.3
			
#CALCULATIONS
Q = round(3*math.pi/4.*1./4 *u,3)       # ft**3/sec
H1 = round(p1/w + (u**2)/(2*g),-1)      # ft
Pi = round(w*Q*H1/550)                  # h.p.
H2= L-l
R= u**2/(2*g)
H1= round(H2+R+hl)
z = H2*100./H1
			
#RESULTS
print  ' efficiency of tramsmission = %.1f per cent'%(z)

 efficiency of tramsmission = 87.4 per cent

Example 11.11.1 page no : 335

In [14]:
import math 
#initialisation of variables
g= 32.2 			#ft/sec**2
w= 62.3 			#lbf/ft**3
p= 40. 			#lbf/in**2
k= 0.44
d= 2. 			#in
			
#CALCULATIONS
Cu = math.sqrt(1/1.44)
Q= (math.pi*(d/2)**2./144.)*math.sqrt(2*g*p*144/(w*0.981))
P= w*Q**3*(144./math.pi)**2/(2*g*550)
			
#RESULTS
print ' Coefficient of velocity = %.3f'%Cu
print  ' discharge rate = %.2f ft**3/min'%(Q)
print  ' Power of jet = %.2f h.p'%(P)

 Coefficient of velocity = 0.833
 discharge rate = 1.70 ft**3/min
 Power of jet = 18.14 h.p

Example 11.11.2 page no : 336

In [15]:
    
import math 
#initialisation of variables
u= 80. 			#ft/sec
d= 1.5 			#in
Cu= 0.97
f= 0.007
l= 150. 			#ft
g= 32.2 			#ft/sec**2
n= 70. 			#percent
Ho= -10. 			#ft
w= 62.3 			#lb/ft**3
			
#CALCULATIONS
Q=math.pi*d**2*u/(4*144.)
ut= u/Cu
H2= ut**2/(2*g)
H1= 5*H2/4
hf= H1/5
D= ((4/math.pi)**2*4*f*l*Q**2*0.00237/(hf*2*g))**(1/5.)*12*(3.95/1.18)
Ps= w*2*Q*(H1-Ho)*100/(n*550.)
			
#RESULTS
print  ' Diameter = %.2f in'%(D)
print  '  shaft power of the pump = %.1f h.p'%(Ps)

 Diameter = 3.95 in
  shaft power of the pump = 45.1 h.p

Example 11.13.1 pageno : 342

In [16]:
#initialisation of variables

v= 10. 			#ft/sec
g= 32. 			#ft/sec**2
w= 62.3 			#lbf/ft**3
l= 200. 			#ft
t= 0.5 			#sec
			
#CALCULATIONS
dp= w*l*v/(g*t*144)
			
#RESULTS
print  ' Rise in pressure = %.1f lbf/in**2'%(dp)

# Answer may vary because of rounding error.

 Rise in pressure = 54.1 lbf/in**2

Example 11.14.1 page no : 343

In [17]:
import math 
#initialisation of variables
w= 62.3 			#lb/ft**3
g= 32.2			#ft/se**2
k= 3.*10.**5 			#lbf/in**2
u= 10.			#ft/sec
			
#CALCULATIONS
P= u*math.sqrt(w*k/g)/12.
			
#RESULTS
print  ' Rise in pressure = %.f lbf/in**2'%(P)

 Rise in pressure = 635 lbf/in**2

Example 11.14.2 page no: 344

In [18]:
import math 
#initialisation of variables
g= 32.2 			#ft/sec**2
w= 62.3 			#lb/ft**3
k= 3.*10.**5 			#lbf/in**2
			
#CALCULATIONS
v= math.sqrt(k*g*144/w)
			
#RESULTS
print  ' velocity of sound in the fluid = %.f ft/sec'%(v)

# answer may vary because of rounding error

 velocity of sound in the fluid = 4725 ft/sec

Example 11.14.4 pageno : 347

In [19]:
import math 
#initialisation of variables
w= 62.3 			#lb/ft**3
d= 6. 			#in
t= 5./8 			#in
k= 3.*10**5 			#lbf/in**2
E= 18.*10**6 			#lbf/in**2
M= 3. 			#tonf
			
#CALCULATIONS
u= math.sqrt(((M*2240)**2/w)*(t*2/d)*32.2*114*((t*2/(d*k))+(2/E)))
Q= (math.pi*(d/2)**2/144)*u
			
#RESULTS
print  ' maximum permissible flow = %.2f ft**3/sec'%(Q)

# answer may vary because of rounding error.

 maximum permissible flow = 4.15 ft**3/sec