# Chapter 13 : Miscellaneous Problems¶

## Example 13.1 Page No : 282¶

In [2]:
import math

#initialisation of variables
W= 5000 	#lb
vr= 6
e= 0.95
ep = 0.75
d= 9 	#in
D= 45 	#ft
t= 2 	#min
v= 4.5 	#ft/sec

#CALCULATIONS
L= W*vr/(e*ep)
Pr= L/(math.pi*d**2/4)
s= D/vr
V= s*math.pi*ep**2/(4*t*60)
T= D/v
V1= s*math.pi*ep**2/4
V2= V*T
V3= V1-V2

#RESULTS
print  'Pressure on ram = %.f ln/in**2 '%(Pr)
print ' Pump duty = %.4f cusec'%(V)
print  ' Minimum capacity if accumulator = %.3f ft**3 '%(round(V3))

# rounding off error

Pressure on ram = 662 ln/in**2
Pump duty = 0.0258 cusec
Minimum capacity if accumulator = 3.000 ft**3


## Example 13.2 Page No : 283¶

In [5]:
import math

#initialisation of variables
P1= 1100. 	#lb/in**2
P2= 85. 	#lb/in**2
f= 0.01
g= 32.2 	#ft/sec**2
l= 1600. 	#ft
r= 1./8
W= 2500. 	#lb
d= 6.    	#in

#CALCULATIONS
L= W*d
P= L*2.31/(math.pi*(d/2)**2)
s1= P1*2540/1100
s2= P2*196/85
vp= math.sqrt((s1-s2-P)/(4*f*l/(2*g*r)))
V= vp/16
Vl= V*d
Vp= math.sqrt((s1/3)/(4*f*l/(2*g*r)))
vl= Vp*d/16
Hr= s1-(s1/3)-s2
Lr= Hr*math.pi*(d/2)**2/(2.31*d)

#RESULTS
print "In case 1 velocity of load = %.2f ft/sec"%(Vl)
print " In case 2 velocity of load = %.2f ft/sec"%(vl)
print  ' Load to be lifted = %d lb '%(Lr)

# note : roundin off error

In case 1 velocity of load = 4.45 ft/sec
In case 2 velocity of load = 3.87 ft/sec
Load to be lifted = 3054 lb


## Example 13.3 Page No : 284¶

In [1]:
import math

#initialisation of variables
bhp= 1500 	#h.p
e= 0.86
h1= 300 	#ft
h2= 15 	#ft
w= 62.4 	#lb/ft**3
t= 30 	#days
t1= 10 	#hr
t2= 3 	#months
f= 0.005
l= 1000 	#ft

#CALCULATIONS
WHP= bhp/e
Ha= h1-h2
W= WHP*550
Q= W/(Ha*w)
Qt= Q*36009*t1*t*t2
Qp= Qt/(3600*t*45)
d= (f*l*(Q/2)**2/(t1*h2))**(1./5)

#RESULTS
print "Minimum size of basin required = %.1e cusecs"%Qt
print "Pump Discharge : %.f cusecs"%(Qp/10)
print  'Diameter = %.2f ft '%(d)

Minimum size of basin required = 1.7e+09 cusecs
Pump Discharge : 36 cusecs
Diameter = 1.89 ft


## Example 13.4 Page No : 285¶

In [7]:
import math

#initialisation of variables
l= 140 	#ft
P= 70 	#percent
V= 3*10**8 	#ft**3
w= 62.4 	#lb/ft**3
SBD= 4.9*10**8 	#ft**3
Q= 162 	#cuses
s= 12.2*10**6 	#ft**3/day

#CALCULATIONS
O= Q*w*l*(P/1000.)/550.

#RESULTS
print  'Size of reservoir= %.2e ft**3'%(SBD)
print  ' output = %.f h.p '%(O)
print  ' output = %d h.p '%(Q)

# rounding off error

Size of reservoir= 4.90e+08 ft**3
output = 180 h.p
output = 162 h.p


## Example 13.5 Page No : 287¶

In [6]:
import math

#initialisation of variables
Q= 140 	#cuses
w= 62.4 	#lb/ft**3
l= 140 	#ft
P= 70 	#percent
k= 1.6
v= 3*10**8

#CALCULATIONS
rv= k*v
HP= Q*l*w*(P/1000.)/550.

#RESULTS
print  'Required size of reservoir = %.1e ft**3 '%(rv)
print  ' horsepower = %.f h.p '%(HP)

Required size of reservoir = 4.8e+08 ft**3
horsepower = 156 h.p


## Example 13.6 Page No : 288¶

In [8]:
import math

#initialisation of variables
P= 10. 	#lb/in**2
r1= 0.5 	#ft
r= 0.25 	#ft
f= 42.3 	#ft/sec
b= 1./40
Tt= 1400. 	#lb

#CALCULATIONS
Q= 2*math.pi*r*b*f
p1= 34+P
Fu= p1*math.pi*(r-(r/4))*144/2.3
Fr= Fu-Tt

#RESULTS
print  'Quantity = %.2f cusecs '%(Q)
print  ' Resultant force on the plate = %.f lb '%(round(Fr,-1))

Quantity = 1.66 cusecs
Resultant force on the plate = 220 lb


## Example 13.7 Page No : 289¶

In [8]:
import math

#initialisation of variables
r= 0.5 	#ft
N= 300
w= 62.4 	#lb/ft**3
g= 32.2 	#ft/sec**2

#CALCULATIONS
A= N*2*math.pi/60
Ft= math.pi*A**2*r**4*w/(4*g)

#RESULTS
print  'total force = %.1f lb '%(Ft)

total force = 93.9 lb


## Example 13.8 Page No : 292¶

In [10]:
import math

#initialisation of variables
d= 4. 	#in
h= 12. 	#in
h1= 9. 	#in
g= 32. 	#ft/sec**2

#CALCULATIONS
H= 2*(1-(h1/h))
A= math.sqrt((H*2*g/((d/24)**2)))
A1= math.sqrt((H*2*g*2/((d/24)**2)))

#RESULTS
print  'speed when the axial is zero = %.f radn/sec '%(A)
print  ' speed when the axial is zero = %.f radn/sec '%(A1)

speed when the axial is zero = 34 radn/sec
speed when the axial is zero = 48 radn/sec


## Example 13.10 Page No : 295¶

In [9]:
import math

#initialisation of variables
P= 14.7 	#lb/in**2
T= 15. 	#C
v= 350. 	#ft/sec
R= 0.714

#CALCULATIONS
P1= P*144
r= 3091*(273+T)
d1= P1/r
r1= r+(v**2/7)
P2= (r1*d1/(P1**R))**(1/(1-R))/144
dP= P2-P
T2= r1/3091
dT= T2-(273+T)

#RESULTS
print  'rise in pressure = %.f lb/in**2 '%(dP)
print  ' rise in temperature = %.1f C '%(dT)

# rounding off error

rise in pressure = 1 lb/in**2
rise in temperature = 5.7 C


## Example 13.11 Page No : 297¶

In [13]:
import math

#initialisation of variables
T= 27. 	#C
P = 33. 	#lb/in**2
p1= 14.7 	#lb/in**2
w= 250. 	#lb
g= 32.2 	#ft/sec**2
Cd= 0.99
r= 1.4

#CALCULATIONS
w1= P*144/(96*(273+T))
d= p1*144/(96*(273+T))
W= d*w/60
d= math.sqrt(W*4/(Cd*math.pi*math.sqrt(2*g*P*144*(r/(r-1))*w1*(0.528**(2/1.4)-0.528**(2.4/1.4)))))*12

#RESULTS
print  'Diameter = %.3f in '%(d)

Diameter = 0.722 in


## Example 13.12 Page No : 299¶

In [2]:
import math

#initialisation of variables
sp= 13.6
hm= 800. 	#mm
d= 3. 	#in
r= 1.4
R= 1385. 	#ft-lb/lb/C
w= 62.4 	#lb/ft**3
T= 15. 	#C
hm1= 765. 	#mm
r1= 9.
g= 32.2 	#ft/sec**2

#CALCULATIONS
p1= hm*sp*w/304.8
r2= (273+T)*R
w1= p1/r2
k= hm/hm1
v1= math.sqrt((2*g*r*r2*(1-k**0.286))/((1-r)*(r1**2*k**1.43-1)))
W= v1*w1*3600*(math.pi/64)

#RESULTS
print  'Weight flowing = %.f lb/hr '%(W)

#The answer is a bit different due to rounding off error in textbook

Weight flowing = 115 lb/hr


## Example 13.13 Page No : 301¶

In [12]:
import math

#initialisation of variables
p= 160. 	#lb/in**2
d= 1./3 	#ft
T= 15.   	#C
R= 96.
V= 120. 	#ft**3
f= 0.004
a= 60*math.pi
l= 10560. 	#ft
g= 32.2 	#ft/sec**2

#CALCULATIONS
p1= p*144
w1= p*144/(R*(273+T))
v1= V*36/a
p2= math.sqrt(p1**2-((2*4*f*p1*w1*v1**2*l)/(2*g*d)))/144
v2= p*v1/p2

#RESULTS
print  ' pressure = %.1f lb/in**2 '%(p2)
print  ' velocity = %.1f ft/sec '%(v2)

#The answer is a bit different due to rounding off error in textbook

 pressure = 134.0 lb/in**2
velocity = 27.4 ft/sec