Chapter 6:Differential analysis of fluid flow

Example 6.4 Page no.296

In [1]:
p1=30                       #kPa
d=1000                    #kg/(m**3)
r1=1                        #m
r2=0.5                     #m
#applying energy equation between points (1) and (2) and using the equation V**2=16*(r**2)
V1=(16*(r1**2))**0.5          #m/sec
V2=(16*(r2**2))**0.5         #m/sec
p2=((p1*1000)+(d*((V1**2)-(V2**2)))/2)/1000#kPa

#result
print "The pressure at point (2) =",round(p2,3),"kpa"
The pressure at point (2) = 36.0 kpa

Example 6.5 Page no.301

In [14]:
ang2=math.pi/6                      #radians
#vp=-2*math.log(r)

#calculation
#vr=d(vp)/d'r
#vr=(-2)/r
#vang=(1/r)*(d(vp)/d(ang))
import math
from scipy import integrate
def f1(dtheta):
    R=1
    return((-2/R)*R)
q=integrate.quad(f1,0.0,(3.14/6))

#result
print "Volume rate of flow (per unit length) into the opening = ",round(q[0],2),"ft**2/s"
Volume rate of flow (per unit length) into the opening =  -1.05 ft**2/s

Example 6.7 Page no.310

In [1]:
%matplotlib inline
Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].
For more information, type 'help(pylab)'.
In [1]:
h=200                                 #ft
U=40                                  #mi/hr
d=0.00238                          #slugs/ft**3

#calculation
import math
#V**2= (U**2)*(1 + (2*b*math.cos(ang)/r) + ((b**2)/(r**2)))
#at point 2, ang=math.pi/2
#r=b*(math.pi-ang)/math.sin(ang)=(math.pi*b/2)
V=U*(1+(4/(math.pi**2)))**0.5      #mi/hr
y2=h/2                                           #ft
#bernoulli equation
#p1-p2= d*((V2**2)-(V1**2)) + (sw*(y2-y1))
V1=U*(5280/3600)
V2=V*(5280/3600)
pdiff=((d*((V2**2)-(V1**2))/2) + (d*32.2*(y2)))/144#psi

#result
print "elevation of point above the plane",round(y2,2),"ft"
print "The magnitude of velocity at (2) for a 40 mi/hr approaching wind =",round(V,1),"mi/hr"
print "The pressure difference between points (1) and (2)=",round(pdiff,2),"psi"

#Plot
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111)

u=[0,20,40,60,80,100]
p=[0.05,0.055,0.0647,0.08,0.10,0.12]
xlabel("U  (mph)") 
ylabel("p1-p2  (psi)") 
plt.xlim((0,100))
plt.ylim((0,0.14))
ax.plot([40], [0.0647], 'o')
ax.annotate('(40mph,0.0647psi)', xy=(40,0.06))

a=plot(u,p)
show(a)
elevation of point above the plane 100.0 ft
The magnitude of velocity at (2) for a 40 mi/hr approaching wind = 47.4 mi/hr
The pressure difference between points (1) and (2)= 0.06 psi

Example 6.10 Page no.331

In [2]:
d=1.18*1000                                          #kg/m**3
vis=0.0045                                             #Ns/m**2, viscosity
Q=12.0                                                     #ml/sec
dia1=4.0                                                   #mm
l=1.0                                                         #m
dia2=2.0                                                   #mm

#calculation
import math
V=Q/(1000000*math.pi*((dia1/1000)**2)/4)       #mean velocity, m/sec
Re=(d*V*dia1/1000)/vis

#result
print "  The Reynolds number is ",round(Re,0),"is well below critical value of 2100 so flow is laminar."
pdiff=(8*vis*(l)*(12*10**-6)/(math.pi*(dia1/2000)**4))*10**-3        #kPa
print "a)The pressure drop along a 1 m length of the tube which is far from the tube entrance ",round(pdiff,),"kpa"

#for flow in the annulus
V1=Q/(1000000*math.pi*(((dia1/1000)**2)-((dia2/1000)**2))/4)     #mean velocity, m/sec
Re1=d*((dia1-dia2)/1000)*V1/vis

#Result
print "b) The Reynolds number is ",round(Re1)," is well below critical value of 2100 so flow is laminar."
r1=dia1/2000
r2=dia2/2000
pdiff1=((8*vis*(l)*(12*10**-6)/(math.pi))*((r1**4)-(r2**4)-((((r1**2)-(r2**2))**2)/(math.log(r1/r2))))**(-1))*10**-3  #kPa

#result
print "The pressure drop along a 1 m length of the symmetric annulus =",round(pdiff1,1),"kpa"

#plot
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111)

T=[0,0.001,0.005,0.01,0.2,0.35,0.5]
K=[1.1,1.17,1.22,1.28,2.2,3.8,7.94]
xlabel("Ri/Ro") 
ylabel("Pannulus/Ptube") 
plt.xlim((0,0.5))
plt.ylim((1,8))
ax.plot([0.5], [7.94], 'o')
ax.annotate('(0.5,7.94)', xy=(0.5,7.9))
a=plot(T,K)
show(a)

T1=[0,0.001,0.005,0.01]
K1=[1.1,1.14,1.20,1.28]
xlabel("Ri/Ro") 
ylabel("Pannulus/Ptube") 
plt.xlim((0,0.01))
plt.ylim((1,1.3))
a1=plot(T1,K1)
show(a1)
  The Reynolds number is  1002.0 is well below critical value of 2100 so flow is laminar.
a)The pressure drop along a 1 m length of the tube which is far from the tube entrance  9.0 kpa
b) The Reynolds number is  668.0  is well below critical value of 2100 so flow is laminar.
The pressure drop along a 1 m length of the symmetric annulus = 68.2 kpa