Chapter 5:Fluid control volume analysis¶
Example 5.1 Page no.195¶
In [1]:
v2=20.0 #m/s, nozzle velocity
dia2= 40.0 #mm, nozzle diameter
#m1=m2
#d1*Q1=D2*Q2 where d1=d2 is density of seawater
#hence Q1=Q2
#calculation
import math
Q=v2*(math.pi*((dia2/1000)**2)/4) #m**3/sec
#result
print "Flowrate=",round(Q,3),"m**3/s"
Example 5.2 Page no.196¶
In [2]:
v2=1000 #ft/sec, velocity
p1=100 #psia pressure inlet
p2=18.4 #psia pressure outlet
T1=540 #degree R, Temprature inlet
T2=453 #degree R Temprature outlet
dia=4 #inches, inside dia of pipe
#m1=m2
#d1*A1*v1=d2*A2*v2
#A1=A2 and d=p/(R*T) since air at pressures and temperatures involved behaves as an ideal gas
#calculation
v1=p2*T1*v2/(p1*T2)
#result
print "Velocity at section 1 =",round(v1,1),"ft/s"
Example5.3Page no.197¶
In [1]:
m1=22 #slugs/hr
m3=0.5 #slugs/hr
#-m1+m2+m3=0
m2=m1-m3
#result
print "Mass flowrate of the dry air and water \n vapour leaving the dehumidifier=",m2,"slugs/hr"
Example 5.5 Page no.198¶
In [1]:
%matplotlib inline
In [44]:
Q=9.0 #gal/min, Q=m/d ,flow rate
l=5.0 #ft, length
b=2.0 #ft breadth
H=1.5 #ft, height
#continuity equation to water: integral of m= d*((h*b*l)+(H-h)*A) where A is cross-sectional area of faucet
#m=d*(b*l-A)*dh/dt, where dh/dt= hrate
#m=d*Q
#since A<<l*b, it can be neglected and dh/dt=hrate and m/d=Q
hrate=Q/(b*l)
hrate_=hrate*1.604 #Inch/min
#result
print "Time rate of change of depth of water in tub =",round(hrate_,2),"inch/min"
#Plot
Dj=[0,0,10,20,30]
h=[0,1.5,1.6,1.8,2.7]
a=plot(Dj,h)
xlabel("Dj in")
ylabel("dh/dt in/min")
xlim=(0,30)
ylim=(0,3)
plt.xticks(np.arange(min(Dj), max(Dj)+1, 10))
plt.yticks(np.arange(min(h), max(h)+1,0.5))
show(a)
Example 5.6 Page no.201¶
In [5]:
v=971 #km/hr, aeroplane speed
v2=1050 #km/hr velocity of exhaust gases
A1=0.80 #m**2 intake area of jet engine
d1=0.736 #Kg/m**3 density
A2=0.558 #m**2 area of engine
d2=0.515 #Kg/m**3, density
#w1=v=intake velocity
#mass flow rate of fuel intake = d2*A2*w2 - d1*A1*w1
w2=v2+v
m=(d2*A2*w2 - d1*A1*v)*1000 #in book ,calculation mistake
#Result
print "The mass flow rate of fuel intake = ",round(m,1),"kg/h"
Example 5.7 Page no.202¶
In [6]:
Q=1000 #ml/s, flow rate
A2=30 #mm**2 area
rotv=600 #rpm, revolutionary speed
#mass in = mass out
w2=(Q*0.001*1000000)/(2*A2*1000)
#result
print "Average speed of water leaving each nozzle \nwhen sprinkle head is stationary and when it rotates with a constant speed of 600rpm =",round(w2,1),"m/s"
Example 5.8 Page no.203¶
In [7]:
Ap=500 #mm**2
Q2=300 #cm**3/min
Qleak=0.1*Q2 #cm**3/min
#A1=Ap
#mass conservation in control volume
#-d*A1*V + m2 + d*Qleak =0 m2=d*Q2
#V=(Q2+Qleak)/Ap
V=(Q2+Qleak)*1000/Ap
print "The speed at which the plunger should be advanced=",round(V,2),"mm/min"
Example 5.9 Page no.204¶
In [8]:
Given
Q=9 #gal/min
l=5 #ft
b=2 #ft
H=1.5 #ft
#deforming control volume
#hrate=Q/(l*b-A)
#A<<l*b
hrate=Q*12/(l*b*7.48)
print "Time rate of change of depth of water in tub =",round(hrate,3),"inch/min"
Example 5.11 Page no.208¶
In [9]:
dia1=16.0 #mm
h=30.0 #mm
dia2=5.0 #mm
Q=0.6 #litre/sec
mass=0.1 #kg
p1=464.0 #kPa
d=999.0 #kg/m**3
m=d*Q*10**-3 #kg/s
#calculation
import math
A1=math.pi*((dia1/1000)**2)/4 #m**2
w1=Q/(A1*1000) #m/s
A2=math.pi*((dia2/1000)**2)/4 #m**2
w2=Q/(A2*1000) #m/s
Wnozzle=mass*9.81 #N
volwater=((1/12)*(math.pi)*(h)*((dia1**2)+(dia2**2)+(dia1*dia2)))/(1000**3) #m**3
Wwater=d*volwater*9.81 #N
F=m*(w1-w2)+Wnozzle+(p1*1000*A1)+Wwater #N
#result
print "The anchoring force=",round(F,3),"N"
Example 5.12 Page no.212¶
In [10]:
A=0.1 #ft**2
v=50 #ft/s
p1=30 #psia
p2=24 #psia
d=1.94 #slugs/ft**3
#v1=v2=v and A1=A2=A
m=d*v*A
Fay=-m*(v+v)-((p1-14.7)*A*144)-((p2-14.7)*A*144)
#result
print "The y component of anchoring force is ",round(Fay,2),"lb" "\n and the x component of anchoring force is",0,"lb"
Example 5.13 Page no.214¶
In [11]:
p1=100.0 #psia
p2=18.4 #psia
T1=540.0 #degree R
T2=453.0 #degree R
V2=1000.0 #ft/s
V1=219.0 #ft/s
dia=4.0 #in
#m=m1=m2
import math
A2=math.pi*((4.0/12.0)**2.0)/dia #ft**2
#equation of state d*R*T=p
d2=p2*144/(1716*T2)
m=A2*d2*V2 #slugs/s
Rx=A2*144*(p1-p2)-(m*(V2-V1)) #lb
#result
print "Frictional force exerted by pipe wall on air flow=",round(Rx,0),"lb"
Example 5.15 Page no.216¶
In [12]:
v1=200.0 #m/s
v2=500.0 #m/s
A1=1.0 #m**2
p1=78.5 #kPa(abs)
T1=268.0 #K
p2=101.0 #kPa(abs)
#F=-p1*A1 + p2*A2 + m*(v2-v1)
#m=d1*A1*v1
#d1=(p1)/(R*T1)
d1=(p1*1000)/(286.9*T1)
m=round(d1,2)*v1*A1
F=-((p1-p2)*A1*1000) + m*(v2-v1)
#Result
print "The thrust for which the stand is to be designed=",round(F,3),"N"
Example 5.17 Page no.219¶
In [13]:
v1=100.0 #ft/sec
v0=20.0 #ft/sec
ang=45 #degrees
A1=0.006 #ft**2
l=1 #ft
#Calculation
import math
#m1=m2=m continuity equation
#d=density of water= constant
#w=speed of water relative to the moving control volume=constant=w1=w2
#w1=v1-v0
w=v1-v0
d=1.94 #slugs/ft**3
#-Rx=(w1)(-m1)+(w2*math.cos(ang))(m2)
Rx=d*(w**2)*A1*(1-math.cos(ang*math.pi/180))
#wwater=(specific wt of water)*A1*l
wwater=62.4*A1*l
Rz=(d*(w**2)*(math.sin(ang*math.pi/180))*A1)+wwater
R=((Rx**2)+(Rz**2))**0.5
angle=(math.atan(Rz/Rx))*180/(math.pi)
#Result
print "The force exerted by stream of water on vane surface=",round(R,3),"lb"
print "The force points right and down from the x direction at an angle of=",round(angle,3),"degree"
Example 5.18 Page no.225¶
In [3]:
Q=1000.0 #ml/sec
A=30.0 #mm**2
r=200.0 #mm
n=500.0 # rev/min
#v2 is tangential v2=vang2
m=(Q/1000000)*999 #kg/sec
#m=2*d*(A)*v2=d*Q
v2=(Q)/(2*A) #m/sec
#Torque reuired to hold sprinkler stationary
Tshaft=(-(r/1000)*(v2)*m) #Nm
#u2=speed of nozzle=r*omega
#v21=v2-u2
import math
omega=n*(2*math.pi)/60 #rad/sec
v21=v2-(r*omega/1000)
#resisting torque when sprinker is rotating at a constant speed of n rev/min
Tshaft1=(-(r/1000)*(v21)*m) #Nm
#when no resistintg torque is applied
#Tshaft=0
omega1=v2/(r/1000)
n1=(omega1)*60/(2*math.pi) #rpm
#result
print "Resisting torque required to hold the sprinker stationary=",round(Tshaft1,3),"Nm"
print "Resisting torque when sprinker is rotating at a constant speed of 500 rev/min=",round(Tshaft,3),"Nm"
print "Speed of sprikler when no resisting torque is applied=",round(n1,0),"rpm"
#Plot
w=[0,800]
T=[-3.3,0]
xlabel("w (rpm)")
ylabel("T (Nm)")
plt.xlim((0,800))
plt.ylim((0,-4))
a=plot(w,T)
show(a)
Example 5.19 Page no. 228¶
In [1]:
h=1.0 #in
Q=230.0 #ft**3/min
ang=30.0 #degrees
dia1=10.0 #in
dia2=12.0 #in
n=1725.0 #rpm
#m=d*Q
m=(2.38*10**-3)*Q/60
#u2=rotor blade speed
import math
u2=(dia2/2)*(n*2*(math.pi)/(12*60))
#calculation
#result
#m=d*A2*Vr2 and A2=2*math.pi*r2*h and r2=dia2**2
#hence, m=d*2*math.pi*r2*h*Vr2
#Vr2=w2*math.sin(ang)
w2=m*12*12/((2.38*10**-3)*2*(math.pi)*(dia2/2)*(h)*(math.sin(math.pi/6))) #ft**sec
V2=u2-(w2*(math.cos(math.pi/6))) #ft**sec
Wshaft=m*u2*V2/(550) #hp
#Result
print "The power required to run the fan=",round(Wshaft,4),"hp"
Example 5.20 Page no.234¶
In [17]:
Q=300.0 #gal/min
d1=3.5 #in.
p1=18.0 #psi
d2=1.0 #in.
p2=60.0 #psi
diffu=3000#ft*lb/slug
#calculation
import math
#energy equation
#m(u2-u1+(p1/d)-(p2/d)+((v2**2)-(v1**2))/2 + g*(z2-z1))=q-Wshaft
m=Q*1.94/(7.48*60) #slugs/sec
v1=Q*12*12/(math.pi*(d1**2)*60*7.48/4)
v2=Q*12*12/(math.pi*(d2**2)*7.48*60/4)
Wshaft=m*(diffu + (p2*144/1.94) - (p1*144/1.94) + (((v2**2)-(v1**2))/2))/550.0 #hp
#result
print "The power required by the pump=",round(Wshaft,1),"hp"
print "The internal energy change accounts for =",round(m*(diffu/550),1),"hp"
print "The pressure rise accounts for =",round(m*(((p2*144/1.94) - (p1*144/1.94))/550),1),"hp"
print "The kinetic energy change accounts for =",round(m*(((v2**2)-(v1**2))/(550*2)),1),"hp"
Example 5.21 Page no.234¶
In [18]:
v1=30 #m/s
h1=3348 #kJ/kg
v2=60 #m/s
h2=2550 #kJ/kg
#energy equation
#wshaftin=Wshaftin/m= (h2-h1 + ((v2**2)-(v1**2))/2)
#wshaftout=-wshaftin
wshaftout=h1-h2 + (((v1**2)-(v2**2))/2000)
#Result
print "The work output involved per unit mass of steam through-flow=",round(wshaftout,3),"kj/kg"
Example 5.22 Page no. 235¶
In [19]:
z=500 #ft
#energy equation
#T2-T1 = (u2 - u1)/c = g*(z2 - z1)/c c=specific heat of water = 1 Btu/(lbm* degree R)
diffT = 32.2*z/(778*32.2) #degree R
#result
print "The temperature change associated with this flow=",round(diffT,3),"degree R"
Example 5.23 Page no.237¶
In [5]:
dia=120.0 #mm
p=1.0 #kPa
#using energy equation
#Q=A2*v2=A2*((p1-p2)/(d*(1+Kl)/2)) d =density, Kl= loss coefficient
Kl1=0.05
Kl2=0.5
#for rounded entrance cyliindrical vent
import math
Q1=(math.pi*((dia/1000)**2)/4)*(p*1000*2/(1.23*(1+Kl1)))**0.5
#for cylindrical vent
Q2=(math.pi*((dia/1000)**2)/4)*(p*1000*2/(1.23*(1+Kl2)))**0.5
#result
print "The volume fowrate associated with the rounded entrance \ncylindrical vent configuration =",round(Q1,3),"m**3/s"
print "The volume fowrate associated with the cylindrical vent configuration =",round(Q2,3),"m**3/s"
#Plot
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111)
k=[0,0.005,0.5]
Q=[0.450,0.445,0.372]
xlabel("k (mm)")
ylabel("Q (m**3/s)")
plt.xlim((0,0.5))
plt.ylim((0,0.5))
ax.plot([0.05], [0.441], 'o')
ax.annotate('(0.05,0.445 m**3/s)', xy=(0.05,0.44))
ax.plot([0.5], [0.372], 'o')
ax.annotate('(0.5,0.372 m**3/s)', xy=(0.5,0.37))
a=plot(k,Q)
show(a)
Example 5.24Page no.239¶
In [21]:
p=0.4 #kW
dia=0.6 #m
v2=12 #m/s
v1=0 #m/s
#calculation
#energy equation
import math
Wuseful=(v2**2)/2
#wshaftin= Wshaftin/m
wshaftin=(p*1000)/(1.23*math.pi*(0.6**2)*12/4)
eff=Wuseful/wshaftin
#result
print "The work to air which provides useful effect=",round(Wuseful,3),"Nm/kg"
print "Fluid mechanical efficiency of this fan=",round(eff)
Example 5.25 Page no.241¶
In [7]:
#given
p=10.0 #hp
z=30.0 #ft
hl=15.0 #ft
#calculation
#energy equation
#hs=Wshaftin/(sw*Q) = hl+z
Q=(p*550)/((hl+z)*62.4)
wloss=62.4*Q*hl/550
#result
print "Flowrate =",round(Q,3),"ft**3/s"
print "Power loss=",round(wloss,3),"hp"
#plot
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111)
h=[0,5,15,25]
Q=[3,2.5,1.96,1.6]
xlabel("h (ft)")
ylabel("Q (ft**3/s)")
plt.xlim((0,25))
plt.ylim((0,3.5))
ax.plot([15], [1.96], 'o')
ax.annotate('(15ft,1.96ft**3/s)', xy=(15,2))
a=plot(h,Q)
show(a)
Example 5.26 Page no.243¶
In [1]:
m=0.1 #kg/min
dia1=60.0 #mm
alpha1=2.0
dia2=30.0 #mm
alpha2=1.08
p=0.1 #kPa
power=0.14 #W
#calculation
import math
wshaftin=power*60/m #Nm/kg
vavg1=m*1000*1000/(60*1.23*math.pi*dia1*dia1/4)
vavg2=m*1000*1000/(60*1.23*math.pi*dia2*dia2/4)
loss1=wshaftin-(p*1000/1.23)+((vavg1**2)/2)-((vavg2**2)/2) #Nm/kg
loss2=wshaftin-(p*1000/1.23)+(alpha1*(vavg1**2)/2)-(alpha2*(vavg2**2)/2) #Nm/kg
#Result
print "Loss for uniform velocity profile=",round(loss1,3),"Nm/kg"
print "Loss for actual velocity profile=",round(loss2,2),"Nm/kg"
Example 5.29 Page no.250¶
In [2]:
p1=100.0 #psia
T1=520.0 #degree R
p2=14.7 #psia
#for incompressible flow
d=p1*144/(1716*T1) #where d=density, calculated by assuminng air to behave like an ideal gas
#Bernoulli equation
v2=(2*(p1-p2)*144/(round(d,4)))**0.5 #ft/sec
print "The velocity of expanded air considering incompressible flow =",round(v2,2),"ft/s"
#for compressible flow
k=1.4 #for air
d1=d
d2=d1*((p2/p1)**(1/k))#where d2=density of expanded air
#bernoulli equation
V2_=((2*k/(k-1))*((p1*144/d1)-(p2*144/d2)))**0.5 #ft/s
#result
print "The velocity of expanded air considering compressible flow =",round(V2_,3),"ft/s"