# Chapter 9 : Similarity and Dimensional Analysis¶

### Example 9.1.1 page no : 219¶

In [7]:
#initialisation of variables

pm= 1.6 			#lb/ft**3
vm= 6.2*10**-6 		#ft**2/sec
R= 1.8 			    #lbf
um= 100. 			#ft/sec
p= 64. 			    #lb/ft**3
v= 1.7*10**-5 	    #ft^2/sec
lm_by_l= 1/10. 		#ft

#CALCULATIONS
u= (um*lm_by_l*v)/vm
u_knot= ((um*lm_by_l*v)/vm)/1.98  #knot
F_by_Fm = (p/pm)*(1/lm_by_l)**2*(u/um)**2
F= F_by_Fm*R

#RESULTS
print ' speed of torpedo in sea water = %.1f knot'%u_knot
print  ' resistance= %.f lbf '%(F)
print (1/lm_by_l)**2

 speed of torpedo in sea water = 13.8 knot
resistance= 541 lbf
100.0


### Example 9.2.1 page no : 222¶

In [8]:
#initialisation of variables
import math

S= 5. 			#ft
F= 70. 			#lbf
B= 4. 			#degrees
l= 1. 			#ft
d= 0.002378 			#slug/ft**3
u= 120. 			#ft/sec

#CALCULATIONS
S1= S*l
p= 0.5*d*u**2
Cl= L/(p*S1)
Cd= D/(p*S1)

#RESULTS
print  ' coefficient of lift= %.2f'%(Cl)
print  '  coefficient of drag= %.4f'%(Cd)

 coefficient of lift= 0.82
coefficient of drag= 0.0570


### Example 9.2.2 page no : 225¶

In [9]:
#initialisation of variables
import math

A= 600. 			#ft**2
W= 40. 			#lbf/ft**2
n= 75. 			#percent
r= 10.
v= 300. 			#miles/hour

#CALCULATIONS
L= W*A
D= L/r
P= D*v*5280./(60.*33000)
hp= P*100./n

#RESULTS
print  ' brake horse-power of the engines= %.f h.p'%(hp)

 brake horse-power of the engines= 2560 h.p


### Example 9.2.4 page no : 226¶

In [1]:
#initialisation of variables
W = 22400            # lbf
p = 0.002378         # slug/ft^3
S = 500.             # ft**2
b = 398.
a = 293
D1 = 1530.           # lbf
U = 200.

# Calculation
#Part a
U1 = round((2*W/(p*S)),-2)**(1./2) * (b/a)**(1./4)
P1 = round((D1 * U1)/550)

#Part b
Umph = U1*0.681818                   # ft/sec to m.p.h
D = round(round((D1/2)*((U/Umph)**2 + (Umph/U)**2),1))
P = round(D*U*5280./(60*33000),-1)
U = .76 * U1
D = (D1/2.)*((U/Umph)**2 + (Umph/U)**2)
power = round((D * U)*5280./(60*33000),-2)

# Part c
min_power = .76 * U
# Results
print "a)Power expended for minumum drag is = %.1f h.p."%P1
print "  Speed for minimum power is U = %.1f m.p.h"%U
print "b)Drag for minimum power is = %.0f lbf"%D
print "  The minimum power is = %.0f h.p."%power
print "c) Speed for minimum power is U = %.1f m.p.h"%min_power

a)Power expended for minumum drag is = 583.0 h.p.
Speed for minimum power is U = 159.3 m.p.h
b)Drag for minimum power is = 1566 lbf
The minimum power is = 700 h.p.
c) Speed for minimum power is U = 121.1 m.p.h


### Example 9.3.1 page no :231¶

In [17]:
#initialisation of variables

va= 0.2
r= 1./1.25
r1= 1./50
P= 20. 			#atm
v= 400. 			#m.p.h

#CALCULATIONS
Um= v*va/(P*r*r1)

#RESULTS
print  ' Speed of air= %.f m.p.h'%(Um)

 Speed of air= 250 m.p.h


### Example 9.4.1 page no : 234¶

In [48]:
import math
#initialisation of variables

U= 30. 			#ft/sec
g= 32.2 			#ft/sec**2
l= 500.			#ft
r= 1./25

#CALCULATIONS
F=(U**2./(l*g))
R= math.sqrt(r)
Um= U*R

#RESULTS
print  ' Froude number= %.4f '%(F)
print  '  speed= %.f ft/sec'%(Um)

 Froude number= 0.0559
speed= 6 ft/sec


### Example 9.4.2 pageno : 234¶

In [58]:
import math
#initialisation of variables

R1= 9.5 			#lbf
f1= 0.01
S1= 22. 			#ft**2
U1= 5.3
n= 1.825
l= 540.			#ft
l1= 15. 			#ft
C= 0.0087			#lbf/ft**2

#CALCULATIONS
Rr1= round(R1-f1*S1*U1**n,1)
U= U1*math.sqrt(l/l1)
r= (l/l1)**3.
Rr= round(r*Rr1,-3)
Rf= C*(l/l1)**2*S1*U**n
R= Rr+Rf
P= R*U*1.69/550.

#RESULTS
print  ' propulsive power= %.f h.p'%(P)
print 'Answer in book is wrong.'

 propulsive power= 35755 h.p


### Example 9.5.1 pageno : 237¶

In [18]:
import math
#initialisation of variables

s= 20. 			#ft
u= 10. 			#ft/sec
t= 1. 			#sec

#CALCULATIONS
r= s/u*t
a= r*u/t

#RESULTS
print  ' constant accelaration= %.f ft/sec**2'%(a)

 constant accelaration= 20 ft/sec**2


### Example 9.5.2 pageno : 237¶

In [60]:
#initialisation of variables
a= 20. 			#ft/sec**2
s= 20.			#ft
u= 10. 			#ft/sec

#CALCULATIONS
P= a*s/u**2.
t= s*2/(u*P)

#RESULTS
print  ' time taken= %.f sec'%(t)

 time taken= 1 sec


### Example 9.6.1 page no : 242¶

In [61]:
#initialisation of variables

s= 0.8
l= 1.			#ft
r= 8.

#CALCULATIONS
Hw= (s/r)**(2./3)*l
Qw= 1.5*Hw**(2.5)
R= (1./Hw)**2.5
Q= Qw*R

#RESULTS
print  ' depth of water= %.3f ft'%(Hw)
print  '  rate of flow of fluid= %.1f ft**3/sec'%(Q)

 depth of water= 0.215 ft
rate of flow of fluid= 1.5 ft**3/sec


### Example 9.6.3 pageno : 244¶

In [63]:
#initialisation of variables

Q1= 140. 			#gallons
h= 3. 			#in
r= 16.

#CALCULATIONS
H= h*r/12.
Q2= round(Q1*H**5.,-3)

#RESULTS
print  ' corresponding head over the full-scale wier= %.f ft'%(H)
print  '  discharge over the latter= %.f gal/min'%(Q2)

 corresponding head over the full-scale wier= 4 ft
discharge over the latter= 143000 gal/min


### Example 9.7.1 page no: 248¶

In [64]:
#initialisation of variables

r= 0.448
R= 0.868
r1= 0.152
R1= 0.807

#CALCULATIONS
P= R**3./r**2.
U= R/r
P1= R1**3./r1**2.
U1= R1/r1

#RESULTS
print  ' power ratio in case 1= %.2f '%(P)
print  '  velocity ratio in case 1= %.2f '%(U)
print  '  power ratio in case 2= %.2f '%(P1)
print  '  velocity ratio in case 2= %.1f '%(U1)

 power ratio in case 1= 3.26
velocity ratio in case 1= 1.94
power ratio in case 2= 22.75
velocity ratio in case 2= 5.3


### Example 9.7.2 page no : 249¶

In [65]:
import math

#initialisation of variables
w= 1. 			#gf/cm**3
Ss= 7.8
Sl= 0.9
D= 1. 			#cm
D1= 0.1 			#cm
g= 981. 			#cm/sec**2

#CALCULATIONS
F= w*(Ss-Sl)*4.*math.pi*D**3./(3.*8000)
v= F*g/(3*math.pi*D1*2.)

#RESULTS
print  ' coefficient of viscosity= %.2f poise'%(v)

 coefficient of viscosity= 1.88 poise


### Example 9.7.4 page no : 251¶

In [66]:
#initialisation of variables

T= 15. 			#C
T1= -44. 			#C
P= 24. 			#atm
s= 0.374
m= 6. 			#tonf

#CALCULATIONS
r= ((T+273.)/(T1+273))**0.75
R= P/s
R1= r**2./R
F= R1*m*2240.

#RESULTS
print  ' lift force= %.1f lbf'%(F)

 lift force= 295.4 lbf


### Example 9.7.5 page no : 253¶

In [67]:
#initialisation of variables

u= 80. 			#ft/sec
n= 62.
r= 1./4
v= 11. 			#ft**3
w= 62.3			#lbf/ft**3
p= 2. 			#lbf/in**2

#CALCULATIONS
uw= u*n/(r*v*w)
R= v*w*(uw/u)**2.
P= r**2.*p/R

#RESULTS
print  ' water velocity= %.f ft/sec'%(uw)
print  '  pressure drop= %.5f lbf/in**2 per ft'%(P)

 water velocity= 29 ft/sec
pressure drop= 0.00139 lbf/in**2 per ft


### Example 9.8.3 pageno : 259¶

In [68]:
#initialisation of variables

N= 1800. 			#rev/min
Vm= 60. 			#mile/hour
V= 300. 			#mile/hour
r= 10.

#CALCULATIONS
Nm= N*Vm*r/V

#RESULTS
print  ' rotary speed= %.f rev/min'%(Nm)

 rotary speed= 3600 rev/min