import math
#Initialization of variables
d = 1.2 #diameter - m
w = 1. #m
U = 60.*1000/3600 #speed - m/s
nu = 1.5e-5 #m**2/s
Cd = 0.4
rho = 1.22 #kg/m**3
#calculations
Rn = U*d/nu
A = d*w
Fd = Cd*0.5*rho*U**2 *A
M = 0.5*Fd
#results
print "Bending moment = %.2f h**2 N m"%(M)
import math
#Initialization of variables
d = 0.006 #diameter - m
U = 0.01 #m/s
gaml = 8000. #specific weight - N/m**3
gams = 7.9*10**3 *9.81
mu = 13.9
#calculations
mu = d**2 /18 *(gams - gaml)/U
RN = U*d*(gaml/9.81) /mu
#results
print "Viscosity of oil = %.1f Ns /m**2"%(mu)
print "Reynolds number of motion is = %.3f"%RN
# rounding off error. please check.
import math
#Initialization of variables
s = 2.7
gamw = 9810. #N/m**3
mu = 0.001 #Ns/m**2
d = 0.15*10**-3 #m
rho = 1000. #kg/m**3
#calculations
gams = s*gamw
U = d**2 *(gams-gamw)/(18*mu)
RN = U*d*rho/mu
Cd = (1+ 3./16 *RN)**0.5 *(24/RN)
U22 = 4./3 *d*(gams-gamw) /(Cd*rho)
U2 = math.sqrt(U22)
#results
print "Settling velocity of sand in case 1 = %.2f m/s"%(U)
print " Settling velocity of sand in case 2 = %.4f m/s"%(U2)
#The answer is a bit different due to rounding off error.
import math
#Initialization of variables
A = 2. #area - m**2
U = 100*1000./3600 #speed-m/s
Cd = 0.32
rho = 1.24
#calculations
Fd = Cd*0.5*rho*U**2 *A
P = Fd*U
#results
print "Power required = %.1f kW"%(P/1000)
import math
#Initialization of variables
ratio = 0.15
#calculations
VU = (1/(1-ratio))**(1./3)
percent = (VU-1)*100
#results
print "percent increase in speed = %.1f %%"%(percent)
import math
from sympy import Symbol,solve
#Initialization of variables
U = 50.*1000/3600 #speed - m/s
cd1 = 0.34
cd2 = 1.33
#calculations
print ("On solving for both convex and concave surfaces,")
w = Symbol("w")
ans = solve(1.98*(13.98 - 0.25*w) - (13.88 + 0.25*w))
w = ans[0]
N = w/(2*math.pi) *60
#results
print "rotational speed = %.1f rpm"%(N)
# note : value of w is slightly different because of sympy inbuilt method solve. but it is very accurate.