# Chapter 10 : Miscellaneous Problems¶

## Example 10.1 Page No : 316¶

In [1]:
import math

#initialisation of variables
w = 62.4 		#lb/ft**3
x = 8.  		#ft
A = 16. 		#ft**2
X = 2.5 		#ft
X1 = 0.66 		#ft
x1 = 3.834 		#ft
x2 = 2.182 		#ft

#CALCULATIONS
P = w*x*A
y = A/3
P1 = w*x*A*0.5*X1
R = math.sqrt(P1**2+P**2)
m = P1/P
X2 = x1-x2
C = ((2./3)*A)-m*X
Y = m*X2+ C
print P1
#RESULTS
print  'Water pressure on vertical face  = %.f lbs'%(round(P,-3))
print  ' pressure which acts at the base  = %.2f ft'%(y)
print  ' Resultant  = %.f lbs'%(R)
print  ' x coordinate of the resultant  = %.3f ft'%(X2)
print  ' y coordinate of the resultant  = %.3f ft'%(Y)

2635.776
Water pressure on vertical face  = 8000 lbs
pressure which acts at the base  = 5.33 ft
Resultant  = 8411 lbs
x coordinate of the resultant  = 1.652 ft
y coordinate of the resultant  = 10.387 ft


## Example 10.2 Page No : 319¶

In [3]:
import math

#initialisation of variables
s = 13.6
h = 12. 		#in
u = 0.04
k = 1.
d = 6.   		#in
g = 32.2 		#ft/sec**2
w = 62.4 		#lbs/ft**3

#CALCULATIONS
h1 = h*(s-1)/12
hf = u*h1
hn = h1-hf
Q = k*math.pi/4*(d/12)**2*8.02*math.sqrt(hn)/(math.sqrt(16-k))
Q = Q*60*w/10        # fro, cusecs to GPM

#RESULTS
print  'discharge through flow = %.f ft G.P.M'%(Q)

# rounding off error

discharge through flow = 529 ft G.P.M


## Example 10.3 Page No : 321¶

In [3]:
import math

#initialisation of variables
za = 16. 		#ft
h1 = 2. 		#ft
h2 = 3. 		#ft
g = 32.2 		#ft/sec**2

#CALCULATIONS
vc = math.sqrt(2*g*(za-h1-h2))
vb = vc*(h1/(2*h1))**2
r = -h1-h2-(vb**2/(2*g))
r1 = r+34

#RESULTS
print  'pressure head at B = %.1f ft lb'%(r1)

pressure head at B = 28.3 ft lb


## Example 10.4 Page No : 322¶

In [6]:
import math

#initialisation of variables
g = 32.2 		#ft/sec**2
Cd = 0.62
a = 90. 		#degrees
H1 = 14. 		#in
H2 = 8. 		#in

#CALCULATIONS
Q = Q1-Q2

#RESULTS
print  'Discharge through notch = %.2f cuses'%(Q)

Discharge through notch = 2.13 cuses


## Example 10.5 Page No : 324¶

In [7]:
import math

#initialisation of variables
g = 32.2 		#ft/sec**2
Cd = 0.62
d = 5./4 		#in
h = 9. 		#ft

#CALCULATIONS
T = (2./3)*math.pi*(h)**(3./2)/(Cd*(math.pi/4)*math.sqrt(2*g)*(d/12)**2)

#RESULTS
print  'time required to lower water level = %.f secs'%(T)

# rounding off error

time required to lower water level = 1334 secs


## Example 10.6 Page No : 325¶

In [10]:
import math

#initialisation of variables
a = 60. 		#degrees
d = 4.  		#in
Cd = 0.62
h = 5. 	    	#ft
w = 30. 		#ft
g = 32.2 		#ft/sec**2

#CALCULATIONS
H2 = H1-h

#RESULTS
print  'time required to lower water level = %.f secs'%(T)


time required to lower water level = 1214 secs


## Example 10.7 Page No : 326¶

In [12]:
import math

#initialisation of variables
p1 = 40. 		#percent
p2 = 35. 		#percent
dh = 200. 		#ft
f = 0.1
g = 32.2 		#ft/sec**2
l = 2000. 		#ft
d = 1.   		#ft

#CALCULATIONS
hf1 = p1*dh/100
hf2 = p2*dh/100
hf3 = (100-p1-p2)*dh/100
hft = hf1+hf2+hf3
v1 = math.sqrt(2*g*hf1/(4*f*l))
Q = v1*math.pi*d**2/4
d2 = (Q*7*math.sqrt(3/(5*g)))**(2./3)
v3 = Q*4*(4./3)**2/math.pi
l3 = hf2*2*g*(3./4)/(4*f*v3**2)

#RESULTS
print  'proportion of the quantity folwing in the bypass to the whole pass = %d ft'%(l3)

# rounding off error

proportion of the quantity folwing in the bypass to the whole pass = 415 ft


## Example 10.8 Page No : 328¶

In [8]:
import math

#initialisation of variables
d = 1. 	    	#ft
l = 2000. 		#ft
f = 0.038
g = 32.2 		#/ft/sec**2
Q = 6.   		#cuses
l1 = 1500. 		#ft
r = 2.

#CALCULATIONS
v = 4*Q/(d**2*math.pi)
hf = 4*f*l*v**2/(2*g)
v1 = math.sqrt(hf*2*g/(4*f*l1+4*f*(l-l1)*r**2))
v3 = r*v1
Q1 = math.pi*d**2*v3/4
Q2 = math.pi*d**2*v1/4
r1 = Q2/Q1

#RESULTS
print  'proportion of the quantity folwing in the bypass to the whole pass = %.1f '%(r1)

proportion of the quantity folwing in the bypass to the whole pass = 0.5


## Example 10.9 Page No : 329¶

In [13]:
import math

#initialisation of variables
f = 0.01
d = 3. 		#in
l = 22. 		#ft
l1 = 20. 		#ft
w = 20. 		#ft
h = 5. 		#ft
h1 = 20. 		#ft
t = 4. 		#min
g = 32.2 		#ft/sec**2

#CALCULATIONS
h2 = h+h1
h3 = (h-(t*60*math.pi*math.sqrt(2*g/h)/(l1*w*2*64)))**2-4
dh = h2-h3
Q = dh*l1*w

#RESULTS
print  'Quantiy discharged = %.f cuses '%(round(Q,-2))

Quantiy discharged = 1800 cuses


## Example 10.10 Page No : 332¶

In [10]:
import math

#initialisation of variables
g = 32.2     		#ft/sec**2
sct = 1.6
sl = 0.8
K = 0.98
dh1 = 4.      		#ft
W = 62.4     		#lbs/ft**3
d1 = 8.       		#in
d2 = 6.       		#in

#CALCULATIONS
dp = dh1*((sct/sl)-1)
C = math.sqrt(2*g)*math.pi*(d1/24)**2 /math.sqrt((d1**2/d2**2)**2 -1)
Q = C*K*math.sqrt(dh1)

#RESULTS
print  'Discharge passing through the pipe = %.1f cuses '%(Q)
#The answer given in textbook is wrong. Please verify it.

Discharge passing through the pipe = 3.7 cuses