Chapter 2 : Fluid Statics

Example 2.1 Page No : 24

In [1]:
import math 
	
#Initialization of variables
z = 20000.       	#ft
rate = -0.00356 	#F/ft
T = 59.          	#F
P = 14.7        	#psia
gamma = 0.076   	#lb/ft**3
	
#calculations
P2 = P*144 - gamma*(z)
P2f = P2/144
P3 = P*math.exp(-gamma*z/(P*144))
P4 = ((P*144)**0.285 -0.285*gamma*z*(P*144)**(-0.715))**(1/0.285)
P4f = P4/144.
P5 = P*((460+T)/(460+T+rate*z))**(gamma*(T+460)/(P*144*rate))
	
#Results
print ' Constant density'
print ' Final pressure  =  %.2f psia'%(P2f)
print ' \nIsothermal'
print ' Final pressure  =  %.2f psia'%(P3)
print ' \nIsentropic'
print ' Final pressure  =  %.1f psia'%(P4f)
print ' \nLinear decrease'
print ' Final pressure  =  %.1f psia'%(P5)

 Constant density
 Final pressure  =  4.14 psia
 
Isothermal
 Final pressure  =  7.17 psia
 
Isentropic
 Final pressure  =  6.6 psia
 
Linear decrease
 Final pressure  =  6.8 psia

Example 2.2 Page No : 37

In [4]:
import math 
	
#Initialization of variables
wA = 53.5 	#weight of A - lb/ft**3
wA2 = 8.4 	#kN/m**3
wB = 78.8 	#weight of B - lb/ft**3
wB2 = 12.3 	#kN/m**3
PB = 30. 	#pressure at B - psi
PB2 = 200. 	#kN/m**2
AB = 1.3 	#ft
AB2 = 40./100 	#m
BC = 6.5 	#ft
BC2 = 2. 	#m
CD = 10. 	#ft
CD2 = 3. 	#m
	
#calculations
PAbyGB = PB*144/wB - AB*13.55*62.4/wB - (BC+CD) + (AB+BC)*wA/wB
PA = PAbyGB*wB/144.
PAbyGB2 = PB2/wB2 - AB2*13.55*9.81/wB2 - (BC2+CD2) + (AB2+BC2)*wA2/wB2
PA2 = PAbyGB2*wB2
	
#Results
print ' English units'
print " Final pressure  =  %.1f psi"%(PA)
print ' \n SI Units'
print " Final pressure  =  %d kPa"%(PA2+1)

 English units
 Final pressure  =  16.2 psi
 
 SI Units
 Final pressure  =  106 kPa

Example 2.3 Page No : 41

In [3]:
import math 
from numpy import *

#Initialization of variables
W = 500. 	#weight of gate - lb
width = 2. 	#ft
len1 = 4. 	#ft
CGx = 1.2 	#ft
CGy = 0.9 	#ft
theta = 30. 	#degrees
gam = 62.4 	#lb/ft**3

#calculations
Fv = width*len1 	#multiply by gam*x
F = width/(2*math.cos(math.radians(theta))) 	#multiply by gam*x*x
vector = roots([F*gam*0.770/2,0, - Fv*gam*width,W*CGx])

#Result
print 'The gate will remain closed between %.2f ft and %.2f ft'%(vector[2],vector[1])

The gate will remain closed between 0.61 ft and 5.67 ft

Example 2.4 Page No : 43

In [5]:
import math 
	
#Initialization of variables
z1 = 1. 	#in
z2 = 2. 	#in
z3 = 2. 	#in
sOil = 0.8 
sWater = 1.
Pa = 3. 	#psi
	
#calculations
Pd = (Pa) + (z2+z1)*sOil*62.4/144 + 62.4*z3/144
Fa = Pa*144*math.pi*z3**2
Fb = sOil*62.4*(z2+z1-(z2+z3)*z2/((z2+z1)*math.pi))*(math.pi*z3**2 /2)
Fc = sOil*62.4*(z2+z1)*(math.pi*z3**2 /2)
Fd = 62.4*(z2+z3)*z2/((z2+z1)*math.pi)*(math.pi*z3**2 /2)
F = Fa+Fb+Fc+Fd
yPa = z2+z1
yCb = z2+z1-(z2+z3)*z2/((z2+z1)*math.pi) 
ICb = math.pi*(z2+z3)**4 /128 -0.5*math.pi*z2**2 *((z2+z3)*z2/((z2+z1)*math.pi))**2
yPb = yCb+ICb/(yCb*0.5*math.pi*z2**2)
yPc = z2+z1+ (z2+z3)*z2/((z2+z1)*math.pi) 
ICd = ICb
yPd = z2+z1 + (z2+z3)*z2/((z2+z1)*math.pi) + ICb/((z2+z3)*z2/((z2+z1)*math.pi)*0.5*math.pi*z3**2 )
yP = (Fa*yPa+Fb*yPb+Fc*yPc+Fd*yPd)/F
	
#Results
print ' case 1'
print ' Pressure at the bottom  =  %.1f psi'%(Pd)
print ' \n case 2'
print ' Net force  =  %d lb'%( F+3)
print ' Location of net force =  %.2f ft'%( yP)

 case 1
 Pressure at the bottom  =  4.9 psi
 
 case 2
 Net force  =  7380 lb
 Location of net force =  3.10 ft

Example 2.5 Page No : 46

In [6]:
import math 
	
#Initialization of variables
dia = 4. 	#m
P = 35. 	#kN/m**2
theta = 30. #degrees
	
#calculations
Fx = P*(dia-dia*(1-math.cos(math.radians(theta)))/2.)
Fz = P*dia*math.sin(math.radians(theta))/2
dist = (dia-dia*(1-math.cos(math.radians(theta)))/2.)
Fxb = 9.81*dist*dist/2 
Fzb = 9.81*((180+theta)*math.pi*(dia/2)**2/360 + math.sqrt(3) /2 + dia/2)
	
#Results
print ' part a'
print ' Horizontal force =  %.1f kN/m to the right'%( Fx)
print ' Vertical force  =  %.1f kN/m upward' %( Fz)
print ' \n part b'
print ' force by the fluid  =  %.1f kN/m to the right'%(Fxb)
print ' weight of the cross-hatched volume of liquid  = %.1f kN/m Upward'%(Fzb )

 part a
 Horizontal force =  130.6 kN/m to the right
 Vertical force  =  35.0 kN/m upward
 
 part b
 force by the fluid  =  68.3 kN/m to the right
 weight of the cross-hatched volume of liquid  = 100.0 kN/m Upward

Example 2.6 Page No : 49

In [11]:
import math 
from numpy.linalg import solve
from numpy import *

#Initialization of variables
d1 = 4. 	#diameter - in
h1 = 3.75 	#in
w1 = 0.85 	#weight of cylinder - lb
gamma = 52. #lb/ft**3
d2 = 5. 	#in
depth = 3. 	#in
	
#calculations
A = array([[(d1/2)*(d1/2), -(d2/2)*(d2/2)+(d1/2)*(d1/2)],[ 1,1]])
b = array([[0],[w1*12*(12*2/d1)**2 /(gamma*math.pi)]])
C = solve(A,b)
x = C[0]
y = C[1]
height = depth-x
	
#Results
print 'Bottom of the cylinder will be %.2f in above the bottom of hollow cylinder'%(height)

Bottom of the cylinder will be 2.19 in above the bottom of hollow cylinder

Example 2.7 Page No : 52

In [7]:
import math 
	
#Initialization of variables
v = 180. 	#velocity - m/s
angle = 40. #degrees
a = 4.   	#m/s**2
r = 2600. 	#radius - m
g = 9.81 	#m/s**2
	
#calculations
#Assume outward and right as positive
an = round(-v*v/r,1)
at = -a
ax = at*math.cos(math.radians(angle)) +an*math.sin(math.radians(angle))
az = at*math.sin(math.radians(angle)) -an*math.cos(math.radians(angle))
tangent = ax/(az+g)
theta = math.degrees(math.atan(tangent))


#Results
print 'Angle made by the free liquid  =  %.1f degrees'%(-theta)

Angle made by the free liquid  =  33.4 degrees