Chapter 12: THE MOMENTUM AND MECHANICAL ENERGY EQUATIONS

Example 12.01, page: 273

In [1]:
from __future__ import division
import math

#Initialization  of  Variable
V = 10 #ft/s
A = 0.06 #ft2
p = 1.94 #slug/ft3

#calculations:
Fax = -1*p*A*V**2
Fay = p*A*V**2

#Results
print  "resultant  force  in  x  direction  is", round(Fax,2),"(1-cos(theta)) lbf"
print  "resultant  force  in  y  direction  is", round(Fay,2),"sin(theta) lbf"

resultant  force  in  x  direction  is -11.64 (1-cos(theta)) lbf
resultant  force  in  y  direction  is 11.64 sin(theta) lbf

Example 12.02, page: 275

In [2]:
from __future__ import division
import math

# Initialization  of  Variable
p = 1.94 #slug/ft3
P1 = 30 #psi
V = 50 #ft/s
P2 = 24 #psi
A = 0.1 #ft2
theta  = 180 #deg

#calcualtions:
#force in x dirn
Fax = 0
#mass flow rate
mdot = p*A*V
#force in y dirn
Fay = -2*mdot*V - (P1 + P2)*144*A

#Results
print  "resultant  force  in  x  direction is", Fax,"lbf"
print  "resultant  force  in  y  direction is", round(Fay,0),"lbf"

resultant  force  in  x  direction is 0 lbf
resultant  force  in  y  direction is -1748.0 lbf

Example 12.03, page: 277

In [3]:
from __future__ import division
import math

# Initialization  of  Variable
D1 = 16 #mm
Q = 0.6 #lt/s
Wn = 1 #N
Ww = 0.03 #N
D2 = 5 #mm
P1 = 464 #kPa
P2 = 0 #kPa
p = 999 #kg/m3

#calculations:
#mass flow rate
mdot = p*Q/1000
#areas
A1 = math.pi*1E-6*(D1**2)/4
A2 = math.pi*1E-6*(D2**2)/4
#volumetric flow rate
V1 = Q*1E-3/A1
V2 = Q*1E-3/A2
Fa = mdot*(V1 - V2) + Wn + Ww + P1*1000*A1 - P2*A2

#Results
print  "anchoring force is",round(Fa,1),"N"

anchoring force is 77.8 N

Example 12.05, page: 284

In [4]:
from __future__ import division
import math

# Initialization  of  Variable
l = 6 #m
D = 0.1 #m
P1_r = 3 #m
P2_r = 0.5 #m
z1 = 0 #m
z2 = 2 #m

#calculations:
hL = P1_r - P2_r + z1 - z2

#Results
print  "head  loss  in  terms  of  height  of  water is", hL, "m"

head  loss  in  terms  of  height  of  water is 0.5 m

Example 12.06, page: 284

In [5]:
from __future__ import division
import math

# Initialization  of  Variable
z1 = 100 #m
V2 = 6 #m/s
g = 9.81 #m/s2
r = 9.8E3 #N/m3
D = 1 #m

#calculations:
#Volumetric flow rate
Q = math.pi*D**2*V2/4
hL = 0 #no head loss
#turbine head
ht = z1 - (V2**2)/(2*g) - hL
#power output
Wtdot = r*Q*ht/1000

#Results
print  "max power  output is", round(Wtdot,0),"kW" 

max power  output is 4533.0 kW

Example 12.07, page: 285

In [6]:
from __future__ import division
import math

# Initialization  of  Variable
Wpdot = 10 #hp
r = 62.4 #lbf/ft3
Q = 2 #ft3/s
z2 = 30 #ft
z1 = 0

#calculations:
#pump head
hp = Wpdot*550/(r*Q)
#head Loss
hL = z1 - z2 + hp
#on a power basis, the head loss
hLp = r*Q*hL/550

#Results
print  "head  loss is", round(hL,1),"ft or", round(hLp,2),"hp"

head  loss is 14.1 ft or 3.19 hp

Example 12.08, page: 286

In [7]:
from __future__ import division
import math

# Initialization  of  Variable
A2 = 0.001 #m2
P1 = 1 #Mpa
P0 = 1 #Mpa
T1 = 360 #K
T0 = 360 #K
k = 1.4
Pb1 = 500 #kPa
Pb2 = 784 #kPa
V1  = 0
R = 8314/28.97 #J/kg-K

#calculations:
Pcr = 0.528*P0*1000
if Pb1 < Pcr:
    M1 = 1

#exit Temp
T2 = T0/(1 + (k - 1)*M1**2/2)
#exit Velocity
V2 = M1*(k*R*T2)**0.5
#mass flow rate
mdot1 = Pcr*1000*A2*V2/(R*T2)
#Mach at 2
M2 = (2/(k - 1)*((P0*1E6/(Pb2*1E3))**((k-1)/k) - 1))**0.5
#exit Temp
T2 = T0/(1 + (k - 1)*M2**2/2)
#exit Velocity
V2 = M2*(k*R*T2)**0.5
#mass flow rate
mdot2 = Pb2*1000*A2*V2/(R*T2)

#Results
print  "the  mass flow rate and exit  mach  no at back pressure of 500 kPa are",round(mdot1,2),"kg/s and", M1
print  "the  mass flow rate and exit  mach  no at back pressure of 784 kPa are",round(mdot2,2),"kg/s and", round(M2,1)

the  mass flow rate and exit  mach  no at back pressure of 500 kPa are 2.13 kg/s and 1
the  mass flow rate and exit  mach  no at back pressure of 784 kPa are 1.79 kg/s and 0.6

Example 12.09, page: 286

In [8]:
from __future__ import division
import math

# Initialization  of  Variable
A2 = 2.4 #in2
P1 = 100 #lbf/in2
P0 = 100 #lbf/in2
T1 = 500 #degR
T0 = 500 #degR
At = 1 #in2
k = 1.4
V1  = 0
R = 1548/28.97 #ft.lbf/lb-degR

#calculations:
#a)
#mach at throat
Mta = 0.7
#From Table 12.2
#At/A* = At_As
At_Asa = 1.09437
A2_Asa = A2*At_Asa/At
#from Table 12.2
M2a = 0.24
T2_T0a = 0.988
P2_P0a = 0.959
T2a = T0*T2_T0a
P2a = P0*P2_P0a
#velocity at exit
V2a = M2a*(k*R*T2a*32.2)**0.5
#mass flow rate
mdota = P2a*A2*V2a/(R*T2a)

#b)
#from Table 12.2
M2b = 0.26
T2_T0b = 0.986
P2_P0b = 0.953
T2b = T0*T2_T0b
P2b = P0*P2_P0b
#velocity at exit
V2b = M2b*(k*R*T2b*32.2)**0.5
#mass flow rate
mdotb = P2b*A2*V2b/(R*T2b)

#c)
#from Table 12.2
M2c = 2.4
P2_P0c = 0.0684
P2c = P0*P2_P0c
#mass flow rate (nozzle is choked same as b)
mdotc = mdotb

#d)
#from Table 12.2
Mxd = 2.4
Pxd = 6.84 #lbf/in2
Myd = 0.52
M2d = Myd
Pyd = 6.5533*Pxd
P2d = Pyd
#mass flow rate (nozzle is choked same as b)
mdotd = mdotb

#e)
Ax = 2 #in2
Axs = At
#from table 12.2
Ax_Axs = 2 #Ax/Ax*
Mxe = 2.2
#Poy/Pox
Poy_Pox = 0.62812
#A2/Ay*
A2_Ays = (A2/Axs)*Poy_Pox
M2e = 0.43
#P2/Poy
P2_Poy = 0.88
Pox = P0
P2e = P2_Poy*Poy_Pox*Pox
#mass flow rate (nozzle is choked same as b)
mdote = mdotb

#Results
print  "a)the  mass flow rate is",round(mdota,2),"lb/s or", round(mdota/32.1740,4),"slug/s, exit pressure is",round(P2a,1),"lbf/in2 and exit mach number is",M2a
print  "b)the  mass flow rate is",round(mdotb,2),"lb/s or", round(mdotb/32.1740,4),"slug/s, exit pressure is",round(P2b,1),"lbf/in2 and exit mach number is",M2b
print  "c)the  mass flow rate is",round(mdotc,2),"lb/s or", round(mdotc/32.1740,4),"slug/s, exit pressure is",round(P2c,2),"lbf/in2 and exit mach number is",M2c
print  "d)the  mass flow rate is",round(mdotd,2),"lb/s or", round(mdotd/32.1740,4),"slug/s, exit pressure is",round(P2d,2),"lbf/in2 and exit mach number is",M2d
print  "e)the  mass flow rate is",round(mdote,2),"lb/s or", round(mdote/32.1740,4),"slug/s, exit pressure is",round(P2e,1),"lbf/in2 and exit mach number is",M2e

a)the  mass flow rate is 2.28 lb/s or 0.071 slug/s, exit pressure is 95.9 lbf/in2 and exit mach number is 0.24
b)the  mass flow rate is 2.46 lb/s or 0.0765 slug/s, exit pressure is 95.3 lbf/in2 and exit mach number is 0.26
c)the  mass flow rate is 2.46 lb/s or 0.0765 slug/s, exit pressure is 6.84 lbf/in2 and exit mach number is 2.4
d)the  mass flow rate is 2.46 lb/s or 0.0765 slug/s, exit pressure is 44.82 lbf/in2 and exit mach number is 0.52
e)the  mass flow rate is 2.46 lb/s or 0.0765 slug/s, exit pressure is 55.3 lbf/in2 and exit mach number is 0.43