# Chapter 2 : Hydrostatic Forces and Centre of Pressure¶

### Example 2.3.1 page no : 31¶

In [1]:
#initialisation of variables
sw= 62.3 			#lbf/ft**3
d= 288. 			#ft
p= 1. 			#lbf/in**2

#CALCULATIONS
P= sw*d/144.
D= p*144./sw

#RESULTS
print  'pressure at a depth of 288 ft= %.1f lbf/in**2'%(P)
print  'depth= %.2f ft'%(D)

pressure at a depth of 288 ft= 124.6 lbf/in**2
depth= 2.31 ft


### Example 2.3.2 page no : 31¶

In [3]:
#initialisation of variables

w= 62.3 			#lbf/ft**3
d= 11.5 			#ft

#CALCULATIONS
p= w*d/144.

#RESULTS
print  'pressure required to bubble air slowly through the tank= %.f lbf/in**2 guage'%(p)

pressure required to bubble air slowly through the tank= 5 lbf/in**2 guage


### Example 2.3.3 page no : 32¶

In [6]:

#initialisation of variables
w= 62.3 			#lbf/ft**3
d= 23.1 			#ft

#CALCULATIONS
dp=  w*d/144.

#RESULTS
print  ' pressure guage= %.f lbf/in**2'%(dp)

 pressure guage= 10 lbf/in**2


### Example 2.3.4 page no : 32¶

In [8]:
#initialisation of variables
import math
d= 1. 			#ft
s= 0.8
h= 2. 			#ft
w= 62.3 			#lbf/ft**3
d1= 5. 			#ft

#CALCULATIONS
F= (math.pi/4)*d**2*s*w*(d/2)
F1= (math.pi/4)*d**2*s*w*(d1/2)

#RESULTS
print  'Force= %.2f lbf'%(F)
print  'Force= %.f lbf'%(F1)

Force= 19.57 lbf
Force= 98 lbf


### Example 2.3.5 page no : 32¶

In [9]:
#initialisation of variables
F= 100. 			#tonf
p= 2000. 			#lbf/in**2
x= 12. 			#in
x1= 48. 			#in
p1= 40. 			#lbf/in**2

#CALCULATIONS
A3= (F/p)*2240.
A2= A3*x/x1
A1= A2*p/p1

#RESULTS
print  'Piston area= %.f in**2'%(A2)
print  'Ram area= %.f in**2'%(A1)

Piston area= 28 in**2
Ram area= 1400 in**2


### Example 2.4.1 page no : 34¶

In [11]:
#initialisation of variables

Va= 100. 			#ft**3
h= 10. 			#ft
V1= 60. 			#ft**3
Pabyw= 34. 			#ft
h1= 4. 			#ft

#CALCULATIONS
H= Pabyw*((Va/V1)-1)-(h-h1)
h2= H+h
Va1= (1+(h2/Pabyw))*Va
V= Va1-Va

#RESULTS
print  'depth of the river= %.1f ft'%(h2)
print  'volume to be pumped= %.f ft**3'%(V)

depth of the river= 26.7 ft
volume to be pumped= 78 ft**3


### Example 2.6.2 page no : 38¶

In [12]:
#initialisation of variables
import math
w= 62.3 			#lbf/ft**3
dg= 4. 			#ft
d= 1. 			#ft
h= 2. 			#ft
HP= 1.0156 			#ft

#CALCULATIONS
F= w*dg*d**2*(math.pi/4.)
F1= F*HP/h

#RESULTS
print  ' Minimum force= %.1f lbf'%(F1)

 Minimum force= 99.4 lbf


### Example 2.6.3 page no : 39¶

In [3]:
#initialisation of variables
import math
s= 1.03
w= 64.3 		#lbf/ft**3
dg= 14. 		#ft
A= 40. 			#ft**2
b= 5. 			#ft
d= 8. 			#ft
b1= 2.5 		#ft
y= 10. 			#ft
x= 3. 			#ft
z= 4. 			#ft

#CALCULATIONS
F= w*dg*A*16.05/36000
r= ((b*d**3./12)/(A*dg))
F3= F*b1/b
F1= (F*(x-r)-(F/2)*x)/(y-z)
F2= F-(F1+F3)

#RESULTS
print  ' force F3 = %.3f tonf'%(F3)
print  ' force F2 = %.2f tonf'%(F2)

 force F3 = 8.027 tonf
force F2 = 5.03 tonf


### Example 2.6.5 page no : 42¶

In [1]:
#initialisation of variables
import math
w= 62.4 			#lb/ft**3
H1= 15. 			#ft
B= 10. 			#ft
H2= 5. 			#ft
r= math.sqrt(19.)/10.
l= 16. 			#ft

#CALCULATIONS
F= 0.5*w*(H1**2-H2**2)*B/2240
N= F*0.5/r
h= H2*(1-(H2/H1)**3)/(1-(H2/H1)**2)
R1= N*h/l
R2= N-R1

#RESULTS
print 'Resultant water force on one gate : %.1f tonf'%F
print  ' Normal reaction = %.2f tonf'%(N)
print  '  reaction forces on the finges= %.1f tonf'%(R2)

Resultant water force on one gate : 27.9 tonf
Normal reaction = 31.95 tonf
reaction forces on the finges= 21.1 tonf


### Example 2.7.1 page no : 44¶

In [4]:
#initialisation of variables
import math
w= 62.3 			#lb/ft**3
b= 1. 			#ft
s= 42. 			#ft
d= 170. 			#ft
l= 15.75
a= 170/3.

#CALCULATIONS
W= 2*w*b*s*d/3.
F= w*b*d**2/2.
L= l+a*(F/W)

#RESULTS
print  ' distance= %.f ft from O'%(L)

 distance= 188 ft from O


### Example 2.7.2 page no : 45¶

In [17]:
#initialisation of variables
import math
d= 2. 			#ft
a= 30. 			#degrees
p= 200. 			#ft
w= 62.3 			#lbf/ft**3

#CALCULATIONS

 Resultant static thrust= 9.05 tonf