Chapter 2 : Fluid Statics

Example 2.1 Page No : 32

In [1]:
# variables		
T = 68;		#degreeF
p = 10;		# psi
d = 15;		# feet
rho = 1.59;		#specific gravity

# calculations 
gam = rho*62.4;		#lb/cuft
p1 = gam*d + p*144;		#psf

# results 
print 'p1 = %d psf = %.1f psi '%(p1,p1*0.00694);

#incorrect answer given in the textbook
p1 = 2928 psf = 20.3 psi 

Example 2.2 Page No : 32

In [1]:
# variables		
h = 35000;		# feet
p1 = 14.7;		# psia
T1 = 519;		# degreeR
gam1 = 0.0765;		# lb/cuft
p2 = 504;		# psfa

# calculations 
T2 = T1 - h*0.00356;		# degreeR
gam2 = p2/(53.3*T2);		# lb/cuft

# results 
print 'p2 = %d psfa = %.2f psia \nspecific weight = %.3f lb/cuft'%(p2,p2*0.00695,gam2);
p2 = 504 psfa = 3.50 psia 
specific weight = 0.024 lb/cuft

Example 2.3 Page No : 35

In [4]:
		
# variables
h1 = 12.5;		# inches
p1 = 14.50;		# psia

# calculations 
p = p1 - h1*(14.70/29.92);		#absolute pressure in psia

# results 
print 'Absolute pressure = %.2f psia'%(p);
Absolute pressure = 8.36 psia

Example 2.4 Page No : 37

In [5]:
# variables		
gam1 = 0.9*62.4;
gam2 = 13.55*62.4;
l1 = 10;		# feet
l2 = 15./12;		# feet

# calculations 
p_x = gam2*l2 - gam1*l1;		# psf

# results 
print 'The gauge reading = %d psf = %.2f psi'%(p_x,0.00694*p_x);
The gauge reading = 495 psf = 3.44 psi

Example 2.5 Page No : 42

In [6]:
import math 
from scipy.integrate import quad 
		
# variables        
l1 = 4.;		# feet
b1 = 6.;		# feet
b2 = 6.;		# feet
l2 = 2.55;		# feet
t = 1.; 		# feet

# calculations 
F1 = 0.5*l1*b1*62.4*(0.5*l1 + t) ;		# lb
F2 = 0.25*math.pi*b2**2 *62.4*(l2 + t);		# lb
a1 = l1*b2**3 /(36*0.5*b2*0.5*l1*b1);		# feet
a2 = 70/((0.5*l2 + t)*28.3);		# feet
l_p = (F1*(0.5*l1 + a1)+F2*(l2+a2))/(F1+F2) +1;		#feet
x_p1 = (0.5*l1-a1) - a1*2/b2;		# feet

def f2(y): 
	 return (62.4/2)*(36-y**2)*(y+1)

M =  quad(f2,0,6)[0]

x_p2 = M/F2;		# feet
x_p = (x_p2*F2 - F1*x_p1)/(F1+F2);		# feet

# results 
print 'Total force on composite area is %d lb'%(F1+F2); 
print ' Vertical location of resultant force is %.2f ft below the water surface'%(l_p);
print ' Horizontal location of resultant force is %.3f ft right of the water surface'%(x_p);

#incorrect answer given in textbook
Total force on composite area is 8509 lb
 Vertical location of resultant force is 4.38 ft below the water surface
 Horizontal location of resultant force is 1.423 ft right of the water surface

Example 2.6 Page No : 45

In [3]:
import math
import numpy
		
# variables
l = 8.; 		#feet
b = 10.;	    	# feet

# calculations 
F_h = 0.5*l*b*62.4*(b+2.5);		# lb
x = 83.2/(40*(b+2.5));		# feet
F_v = (b+5)*62.4*40-(l*62.4*(25 - 0.25*math.pi*25));		# lb
F = math.sqrt(F_h**2 + F_v**2);		# lb
e = (2680*3.91 + 37440*(0.25*b))/F_v ;		# feet
theta = 180*numpy.arctan(F_v/F_h) /math.pi;		# degrees
x_p = 0.25*b-x;		# feet

# results 
print 'Magnitude of resultant force is %d lb'%(F);
print 'Theta = %d degrees'%(theta);
print 'Location is %.3f feet above and  %.2f feet to the right of B'%(x_p,e);

#there are errors in the answer given in textbook
Magnitude of resultant force is 46709 lb
Theta = 48 degrees
Location is 2.334 feet above and  2.99 feet to the right of B

Example 2.7 Page No : 48

In [5]:
# variables
A = 4000.;		# sq.ft
d1 = 10.;		# feet
d2 = 2.;		# inches
rho = 64.;		# lb/cuft

# calculations 
W = A*(d2/12)*rho;		# lb

# results 
print 'Weight of cargo = %d lb'%(round(W,-2));

		
Weight of cargo = 42700 lb

Example 2.8 Page No : 49

In [7]:
import math 

# variables
gam = 53.0;		# lb/cuft
D = 17.;		# inches
d = 12.;		# inches

# calculations 
V = (math.pi/6)*(D/d)**3;
V1 = 0.584;		#cuft
V2 = 0.711;		#cuft
W = V*gam;
F_B = V1*62.4;
F_ACA = (V2)*62.4;
F = W+F_ACA-F_B;

# results 
print 'The force exerted between sphere and orfice plate = %.1f lb'%(F);

#incorrect answer for W in textbook. Hence the answer differs
The force exerted between sphere and orfice plate = 86.8 lb

Example 2.9 Page No : 51

In [13]:
from scipy.integrate import quad 

# variables
v = 15;		# ft/sec**2
d = 5;		# ft

# calculations 
def f3(z): 
	 return -62.4*(v+32.2)/32.2

p =  quad(f3,0,-5)[0]

# results 
print 'p = %d psf'%(p);
p = 457 psf

Example 2.10 Page No : 52

In [15]:
import math 
from scipy.integrate import quad 

# variables
m = -0.229;		#slope
a_z = 1.96;		# ft/sec**2
a_x = 4*a_z;		# ft/sec**2
a = math.sqrt(a_x**2 + a_z**2);		# ft/sec**2

def f1(z): 
	 return -(32.2 + a_z)*(62.4/32.2)

p =  quad(f1,0,-2.75)[0]

# results 
print 'p = %.1f psf'%(p);

#there is an error in the answer given in textbook
p = 182.0 psf

Example 2.11 Page No : 54

In [9]:
import math 
		
# variables
l1 = 2.;		# feet
l2 = 3.;		# feet
rpm = 100;

# calculations 
p_A = (l1+l2)-(2./3)*(2*math.pi*rpm/60)**2 /(2*32.2);
p_B = (l1+l2)+(1./3)*(2*math.pi*rpm/60)**2 /(2*32.2);

# results 
print 'Pressure heads at point A and point B ae %.2f ft and %.2ft ft respectively'%(p_A,p_B);
Pressure heads at point A and point B ae 3.86 ft and 5.57t ft respectively