# Chapter 3 : Flow Through Orifices Mouthpieces Nozzles¶

### Example 3.1 page no : 70¶

In [12]:
#initialisation of variables

import math
Q = 16. 			#gpm
w = 62.4 			#lb/ft**3
d = 1. 			#in
h = 2+(5./12) 			#ft
g = 32.2 			#ft/sec**2
x = 11.5 			#ft
h1 = 1.2 			#in
#CALCULATIONS
Cd = Q*10/(60*w*(math.pi*(d/12)**2/4)*math.sqrt(2*g*h))
Cv = math.sqrt(x**2/(4*(h1/12)*h*12**2))
Cc = Cd/Cv
Cr = (1-Cv**2)/Cv**2
#RESULTS
print  'Cc  = %.3f  '%(Cc)
print  'Cv  = %.3f  '%(Cv)
print  'Cd  = %.3f  '%(Cd)
print  'Cr  = %.3f  '%(Cr)

# note : answers are slightly different because of rounding error.

Cc  = 0.644
Cv  = 0.975
Cd  = 0.628
Cr  = 0.053


### Example 3.2 page no : 71¶

In [1]:
import math

#initialisation of variables
Ww = 261. 			#lb/min
a = 1. 			#in**2
h = 4. 			#ft
y =  5. 			#ft
W1 = 10.65 			#lb
l = 1. 			#ft
Q = 261. 			#lb/min
w = 62.4 			#lb/ft**3
g = 32.2 			#ft/sec**2
#CALCULATIONS
v = Q*144./(w*60)
F = W1*l/y
v = F*g*60./Q
vth = math.sqrt(2*g*h)
Cv = v/vth
Q1 = Ww/w
Qth = vth*60./144
Cd = Q1/Qth
Cc = Cd/Cv
#RESULTS
print  'Cv  = %.3f  '%(Cv)
print  'Cd  = %.3f  '%(Cd)
print  'Cc  = %.3f  '%(Cc)

Cv  = 0.982
Cd  = 0.625
Cc  = 0.637


### Example 3.3 page no : 74¶

In [16]:
#initialisation of variables

import math
Q = 10. 			#ft**3/sec
a1 = 1. 			#ft**2
a2 = 4. 			#ft**2
g = 32.2 			#ft/sec**2
p1 = 12. 			#lb/in**2
v1 = 10. 			#ft/sec
w = 62.4 			#lb/ft**3
#RESULTS
v2 = v1*a1/a2
Hl = (v1-v2)**2/(2*g)
p2 = ((p1*144/w)+(v1**2/(2*g))-(v2**2/(2*g))-Hl)*(w/144)
W = Hl*v1*w/550.

#RESULTS
print  'Head lost  = %.3f ft of water '%(Hl)
print  'Pressure in larger part of pipe  = %.2f lb/in**2 '%(p2)
print  'Work done  = %.3f HP '%(W)

Head lost  = 0.873 ft of water
Pressure in larger part of pipe  = 12.25 lb/in**2
Work done  = 0.991 HP


### Example 3.4 page no : 78¶

In [18]:
#initialisation of variables

import math
Cc = 1.
Cv = 0.833
d = 2. 			#in
g = 32.2 			#ft/sec**2
H = 12. 			#ft
Pa = 34. 			#lb/in**2
#/CALCULATIONS
Q = Cc*Cv*math.pi*(d/12)**2*math.sqrt(2*g*H)/4
Cd = Cc*Cv
Pc = Pa-0.92*H
#RESULTS
#RESULTS
print  'Discharge  = %.3f cu ft/sec '%(Q)
print  'Coefficient of discharge  = %.3f  '%(Cd)
print  'Pressure at Vent-contraction = %.2f ft of water '%(Pc)

Discharge  = 0.505 cu ft/sec
Coefficient of discharge  = 0.833
Pressure at Vent-contraction = 22.96 ft of water


### Example 3.5 page no : 81¶

In [19]:
import math
H = 4. 			#ft
d = 1. 			#in
g = 32.2 			#ft/sec**2
Cc = 0.5
#CALCULATIONS
Q = Cc*math.pi*(d/12)**2*math.sqrt(2*g*H)/4
#RESULTS
print  'Actual Discharge  = %.4f cu ft/sec '%(Q)

Actual Discharge  = 0.0438 cu ft/sec


### Example 3.6 page no : 82¶

In [22]:
import math

#initialisation of variables

D = 4. 			#ft
d = 2. 			#in
H1 = 6. 			#ft
H2 = 2. 			#ft
t = 4. 			#min
g = 32.2 			#ft/sec**2
w = 62.4 			#lb/ft**3
H = 5. 			#ft
#CALCULATIONS
Cd = (2.*(math.pi/4.)*D**2*(math.sqrt(H1)-math.sqrt(H2)))/(t*60*(math.pi/4)*(d/12)**2*math.sqrt(2*g))
Q = Cd*(math.pi/4)*(d/12)**2*math.sqrt(2*g*H)*w*60/10
#RESULTS
print  'Cd  = %.3f '%(Cd)
print  'Discharge  = %.1f gpm'%(Q)

# note : answers are slightly different because of rounding error.

Cd  = 0.619
Discharge  = 90.8 gpm


### Example 3.7 page no : 84¶

In [26]:
import math

#initialisation of variables
H1 = 10. 			#ft
H2 = 2. 			#ft
Cd = 0.61
d1 = 8. 			#ft
g = 32.2 			#ft/sec**2
d2 = 3. 			#ft
#CALCULATIONS
a = d2**2./144
H0 = H1*d2/(d1-d2)
t = math.pi*(d1/2)**2*((2/5.)*(H1**(5./2)-H2**(5./2))+2*H0**2*(math.sqrt(H1)- \
math.sqrt(H2))+(4./3)*H0*(H1**(3./2)-H2**(3./2)))/(60*Cd*a*math.sqrt(2*g)*(H1+H0)**2)

#RESULTS
print  'time required to lower the water level  = %.2f min'%(t)

# Note : answer is different because of rounding error.

time required to lower the water level  = 5.14 min


### Example 3.8 page no : 85¶

In [27]:
import math
#initialisation of variables
D = 10. 			#ft
H1 = 17. 			#ft
H2 = 5. 			#ft
d = 3. 			#in
Cd = 0.62
g =32.2 			#ft/s**2
#CALCULATIONS
t1 = (2*math.pi*D**2/4)*(math.sqrt(H1)-math.sqrt(H2))/(Cd*math.sqrt(2*g)*math.pi*(d/12)**2/4)
t2 = math.pi*(14./15)*H2**(5./2)*4/(Cd*math.pi*(d/12.)**2*math.sqrt(2*g))
t = t1+t2
#RESULTS
print  'time required to empty the vessel  = %.f sec'%(t)

time required to empty the vessel  = 1885 sec


### Example 3.9 page no : 86¶

In [33]:
import math

#initialisation of variables
Cd = 0.8
g = 32.2 			#ft/sec**2
d = 3. 			#in
#CALCULATIONS
t = (60*2/(math.pi*(d/12)**2*math.sqrt(2*g)/4*Cd))*(6-d)**(3./2)/(3*60./2)
#RESULTS
print  'time to emptify biler  = %.2f min'%(t)

# note : answer is different because of rounding error.

time to emptify biler  = 21.98 min


### Example 3.10 page no : 87¶

In [45]:
import math

#initialisation of variables
A = 100 * 27      # sq ft
dif = 8 - 3.      # ft
a = 2.            # sq ft
Cd = 0.8          # Co-efficient

#CALCULATIONS
t1 = round(A*((a/3 * 22.7) - a/3 * 5.19 - (a/3*11.2))/(Cd*2*8.02*dif))
t2 = round(A*(2./3)*11.2/(Cd*a*8.02*dif))
t = t1 + t2

#RESULTS
print  'Total time to empty the tank = %d sec'%t

Total time to empty the tank = 491 sec


### Example 3.11 page no : 89¶

In [26]:
import math
#initialisation of variables
H1 = 9. 			#ft
H2 = 4. 			#ft
Cd = 0.6
a = 4. 			#in**2
A1 = 72. 			#ft**2
A2 = 24. 			#ft**2
g =32.2 			#ft/s**2
#CALCULATIONS
t = (2*A1*A2/(A1+A2))*(math.sqrt(H1)-math.sqrt(H2))*144/(Cd*60*a*math.sqrt(2*g))
#RESULTS
print  'time required to reduce the water level difference  = %.1f min'%(t)

time required to reduce the water level difference  = 4.5 min


### Example 3.12 pageno : 91¶

In [46]:
#initialisation of variables
import math
l = 80. 			#ft
w = 12. 			#ft
t = 3. 			#min
Hl = 12. 			#ft
g = 32.2 			#ft/sec**2
Cd = 0.6
#CALCULATIONS
s = math.sqrt(2*l*w*Hl**(1./2)/(Cd*math.sqrt(2*g)*t*60.))
#RESULTS
print  'side of the square orifice  = %.2f ft'%(s)

side of the square orifice  = 2.77 ft


### Example 3.13 page no : 92¶

In [10]:
#initialisation of variables
import math
g = 32.2 			#ft/sec**2
Cd = 0.6
d = 2. 			#in
H1 = 5. 			#ft

#CALCULATIONS
K = round(Cd * math.pi/4 * (d/12)**2 * math.sqrt(g*2),3)
t = d*math.pi*(0.5*math.log(1.89) - 0.235)/K**2
v = round(math.sqrt(2*g*H1)/2.)
q = v*Cd*math.pi*(d/12)**2./4

#RESULTS
print "Time required to raise the level is : %.2f sec"%t
print  'Total discharge  = %.3f cfs'%(q)

# Note : answers may vary because of rounding error. Please calculate manually.

Time required to raise the level is : 47.47 sec
Total discharge  = 0.118 cfs


### Example 3.14 page no : 95¶

In [54]:
import math
#initialisation of variables
Cd = 0.62
H = 9. 			#in
l = 3. 			#ft
g = 32.2 			#t/sec**2
#CALCULATIONS
Q1 = Cd*(H*l/12)*math.sqrt(2*g*3*H/24.)
Q2 = Cd*2*l*math.sqrt(2*g)*((H/6)**(3./2)-(H/12)**(3./2))/3
#RESULTS
print  'Discharge by appropriate formula  = %.2f cfs'%(Q1)
print  ' Discharge by exact formula  = %.2f cfs'%(Q2)

Discharge by appropriate formula  = 11.87 cfs
Discharge by exact formula  = 11.82 cfs


### Example 3.15 pageno : 95¶

In [60]:
import math
#initialisation of variables

Cd = 0.62
B = 2.5 			#ft
H2 = 8. 			#ft
H1 = 7. 			#ft
g = 32.2 			#ft/sec**2
h = 4. 			#ft
#CALCULATIONS
Q1 = round(2*Cd*B*math.sqrt(2*g)*(H2**(3./2)-H1**(3./2))/3)
Q2 = Cd*math.sqrt(2*g)*math.sqrt(H2)*B*(h-1)
Q = Q1+Q2
#RESULTS
print  'Total discharge  = %d cfs'%(Q)

Total discharge  = 139 cfs

In [ ]: