## Chapter 4 : Fluid Dynamics¶

### Example 4.1 Page no 159¶

In [1]:
# Difference in pressure at top and bottom

from math import *

from __future__ import division

# Given

d1 = 0.1                     # diameter in m

d2 = 0.05                     # diameter in m

Q = 0.1                      # discharge in m**3/s

A1 = pi*d1**2/4

A2 = pi*d2**2/4

gma =9810                   # specific weight

z= 6                        # difference in the height

g = 9.81

# Solution

V1 = Q/A1                    # velocity at section 1

V2 = Q/A2                    # velocity at section 2

dP = gma*((V2**2/(2*g))-(V1**2/(2*g))-z)/1000

print "Difference in pressure in section 1 and 2 = ",round(dP,1),"kN/m**2"

Difference in pressure in section 1 and 2 =  1157.0 kN/m**2


### Example 4.2 Page no 160¶

In [2]:
# Actual discharge

from math import *

from __future__ import division

# Given

d = 2.5                 # diameter in cm

h =200                        # head in cm

Cd = 0.65                   # coefficient of discharge

A  =pi*d**2/4

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

# Solution

Q = Cd*A*sqrt(2*g*h)/100

print "Actual discharge =",round(Q,2),"l/s"

Actual discharge = 2.0 l/s


### Example 4.3 Page no 162¶

In [4]:
# Discharge through the orifice

from __future__ import division

from math import *

from scipy import integrate

import numpy as np

# Given

H1 = 3               # height in m

H2 = 4               # height in m

b = 0.5              # width in m

Cd = 0.65            # co-efficient of discharge

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

# Solution

q = lambda h: h**(1/2)

Qt = Cd*b*sqrt(2*g)*Q

print "Discharge through the orifice =",round(Qt,2),"m**3/s"

Discharge through the orifice = 2.69 m**3/s


### Example 4.4 Page no 163¶

In [6]:
# discharge through orifice

from math import *

from scipy import integrate

from __future__ import division

import numpy as np

# Given

b = 1                  # bredth of the tank

d = 0.5                # depth of the tank

h1 = 0.2                 # height of the orifice in m

Cd = 0.6               # coefficient of discharge

H1 = 2                 # height in m

H2 = 2+h1               # height in m

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

A = 1*0.3              # area of submerged section in m**2

# Solution

q = lambda h: h**(1/2)

Q1 = Cd*b*sqrt(2*g)*(Q) # Flow through area 1

Q2 = Cd*sqrt(2*g*H2)*A

Td = Q1+Q2

print "Total Discharge =",round(Td,2),"m**3/s"

Total Discharge = 1.95 m**3/s


### Example 4.5 Page no 165¶

In [7]:
# Determine flow rate of water

from math import *

from __future__ import division

# Given

d1 = 2               # radius of pipe

d2 = 1               # radius of throat

D1 = 40

D2 = 20

A1 = pi*D1**2/4

A2 = pi*D2**2/4

Cd = 0.95

# Solution

V2 = sqrt(21582/0.9375)

Q = 1.52*pi*(d1/100)**2/4

Qa = Q*Cd

print "Actual discharge =",round(Qa,6),"m**3/s"

Actual discharge = 0.000454 m**3/s


### Example 4.6 Page no 166¶

In [8]:
# Velocity of stream point at the point of insertion

from math import *

from __future__ import division

# Given

dx = 0.5              # in ft

K = 1                 # constant

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

# solution

V = sqrt(2*g*dx)

print "velocity at the dept of 1 ft =",round(V,2),"ft/s"

velocity at the dept of 1 ft = 5.67 ft/s


### Example no 4.7 Page no 172¶

In [10]:
# Discharge throught the system

from math import *

from __future__ import division

gma= 0.8      # specific weight

V2 = 40       # velocity in m/s

z1 =25        # height at point 1

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

d = 15        # diameter of the nozzle in cm

# Solution

V2 = sqrt(2*g*z1/4.25)

A = pi*(d/100)**2/4

Q = A*V2*1000

print "Discharge throught the system =",round(Q,0),"l/s"

Discharge throught the system = 190.0 l/s


### Example 4.8 Page no 174¶

In [11]:
# Power input to the pump

from math import *

from __future__ import division

# Given

Eff = 0.8     # pump efficiency

Hl = 30       # head loss in m

D1 =6         # diameter in cm

D2 = 2        # diameter in cm

gma = 9810    # specific weight in N/m**3

V2 = 40       # velocity in m/s

P1 = -50      # pressure at point 1 in N/m**2

z2 = 100      # height at point 2

g = 9.8       # acceleration due to gravity in m/s**2

z1 = 30       # height in m

# Solution

V1=(2/6)**2*V2

Q = (pi*6**2/4)*V1*10**-4

Hs = z2 + (V2**2/(2*g)) + z1 + (50/gma) -(V1**2/(2*g))

P = gma*Q*Hs

Pi = (P/Eff)/1000

print "Power input = ",round(Pi,1),"kW"

Power input =  32.5 kW


### Example 4.9 Page no 176¶

In [12]:
# Pressure head at A and B

from math import *

# Given

Q = 0.2           # discharge in m**3/s

d1 = 0.25           # diameter of the pipe in m

A = pi*d1**2/4    # area of the pipe

za = 480          # height in m

z1 = 500          # height in m

z3 = 550          # elevation in m

gma =9810         # specific weight in N/m**2

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

# Solution

V=Q/A           # Velocity of in m/s

Hl1 = (0.02*100*V**2/(0.25*2*9.81))

Pa =(z1-za-(V**2/(2*g))-Hl1)

El = za+Pa

print "Elevation at height A =",round(El,2),"m"


Elevation at height A = 492.39 m