Chapter no 4: How Fluids flow

Example 4.1, Page No 64

#initialisation of variables
a1=12.56       #Area(in^2)
s1=24          #stroke(in^3)
a2=3.14        #Area(in^2) displacement of ram
s2=24          #stroke in^3
Ve=0.785       #Extension volume in^3


#CALCULATIONS
Ve=a1*s1       #extension volume
Vs=a2*s2       
Vr=Ve-Vs       #Retraction volume
Vt=Ve+Vr       #Total volume

#RESULTS
print('Displacement of the cylinder would be = %.2f in^3' %Vt)

Displacement of the cylinder would be = 527.52 in^3

Example 4.4, Page No 75

#initialisation of variables
A=0.785          #area
Q=100            #gal/min flow rate
D=1              #in diameter
v=0.05           #viscocity Newts

#CALCULATIONS
V=Q/(A*3.12)    #velocity
Nr=(12*V*D)/v   #reynolds number

#RESULTS
print('The velocity would be = %.2f lbf-ft' %V)
print('The Reynolds number would be = %.2f Hence the flow is turbulent' %Nr)

The velocity would be = 40.83 lbf-ft
The Reynolds number would be = 9799.12 Hence the flow is turbulent

Example 4.5, Page No 76

#initialisation of variables
Nr=4000         #reynolds number
Nr2=2000        #reynolds number
m=0.27          #P viscocity
D=2.54          #cm diameter
p=0.85          #pressure gm/cm^3
m2=27           #cP viscocity
D2=1            #in diameter
#CALCULATIONS
V=(Nr*m)/(D*p)    #velocity

# for lower critical velocity
V2=(Nr2*m)/(D*p)
v=m2/p
vn=0.001552*v
vh=(Nr*vn)/(12*D2)
vl=(Nr2*vn)/(12*D2)

#RESULTS
print('The velocity would be = %.2f cm/sec' %V)
print('The lower velocity would be = %.2f cm/sec' %V2)
print('Using reynolds formula the velocity would be = %.2f ft/sec' %vh)
print('Using reynolds formula the lower velocity would be = %.2f ft/sec' %vl)

The velocity would be = 500.23 cm/sec
The lower velocity would be = 250.12 cm/sec
using reynolds formula the velocity would be = 16.43 newts
using reynolds formula the lower velocity would be = 8.22 cm/sec