# 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

# 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)

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);


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);


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

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

p = 457 psf


## Example 2.10 Page No : 52¶

In [15]:
import math

# 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)

# 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