Chapter 4: The Conservation Law for Momentum¶

ILLUSTRATIVE EXAMPLE 4.1, Page number: 39¶

In :
#Variable declaration:
Vx_in = 420                     #Entry Velocity in X direction (m/s)
Vx_out = 0                      #Exit Velocity in X direction (m/s)
Vy_in = 0                       #Entry Velocity in Y direction (m/s)
Vy_out = 420                    #Exit Velocity in Y direction (m/s)
m =  0.15                       #Rate of water entrained by the steam (kg/s)
lb = 1.0/4.46                   #Pound force in a newton force

#Calculations:
Mx_out = m*Vx_out               #Rate of change of momentum at entry in x-direction (kg.m)
Mx_in = m*Vx_in                 #Rate of change of momentum at exit in x-direction  (kg.m)
My_out = m*Vy_out               #Rate of change of momentum at entry in y-direction (kg.m)
My_in = m*Vy_in                 #Rate of change of momentum at exit in y-direction  (kg.m)
Fxgc = (Mx_out - Mx_in)*lb      #Force in X direction (lbf)
Fygc = (My_out - My_in)*lb      #Force in X direction (lbf)

#Results:
if Fxgc < 1:
print "The x-direction supporting force acting on the 90° elbow is :",round(-Fxgc,1)," lbf acting toward the left. "
else:
print "The x-direction supporting force acting on the 90° elbow is :",round(Fxgc,1)," lbf acting toward the right. "
if Fygc < 1:
print "The y-direction supporting force acting on the 90° elbow is :",round(-Fygc,1)," lbf acting downwards. "
else:
print "The y-direction supporting force acting on the 90° elbow is :",round(Fygc,1)," lbf acting upwards. "
The x-direction supporting force acting on the 90° elbow is : 14.1  lbf acting toward the left.
The y-direction supporting force acting on the 90° elbow is : 14.1  lbf acting upwards.

ILLUSTRATIVE EXAMPLE 4.2 Page number: 40¶

In :
from math import sqrt,degrees,atan2

#Variable declaration:
Fx = -63                                #Force component in X direction (N)
Fy = 63                                 #Force component in Y direction (N)
lbf = 0.22481                           #Pound-forrce in unit newton (lbf)

#Calculations:
Fr = sqrt(Fx**2 + Fy**2)*lbf            #The resultant supporting force (lbf)
u = degrees(atan2(Fy,Fx))               #Angle between the positive x axis and the direction of the force (degrees)

#Result:
if (0<u<90):
print "The supporting force is :",round(Fr,1)," lbf acting at",u,"° i.e in the “northeast” direction."
elif (90<u<180):
print "The supporting force is :",round(Fr,1)," lbf acting at",u,"° i.e in the “northwest” direction."
elif (180<u<270):
print "The supporting force is :",round(Fr,1)," lbf acting at",u,"° i.e in the “southwest” direction."
elif (270<u<360):
print "The supporting force is :",round(Fr,1)," lbf acting at",u,"° i.e in the “southeast” direction."
The supporting force is : 20.0  lbf acting at 135.0 ° i.e in the “northwest” direction.

ILLUSTRATIVE EXAMPLE 4.3, Page number: 42¶

In :
#Variable declaration:
R1_in = 10000                   #Rate of fuel fed into the boiler (lb/h)
R2_1n = 20000                   #Rate of air fed into the boiler (lb/h)
R3_in = 2000                    #Rate of methane fed into the boiler (lb/h)

#Calculations:
m_in = R1_in + R2_1n + R3_in    #Rate of mass in (lb/h)
m_out = m_in                    #Rate of mass out (lb/h)

#Result:
print "The rate of the product gases exit from the incinerator is :",round(m_in)," lb/h"
The rate of the product gases exit from the incinerator is : 32000.0  lb/h

ILLUSTRATIVE EXAMPLE 4.4, Page number: 42¶

In :
#Variable declaration:
E1 = 65                             #Efficiency of spray tower (%)
E2 = 98                             #Efficiency of packed column (%)
m_in = 76                           #Mass flow rate of HCl entering the system (lb/h)

#Calculations:
m1_out = (1 - E1/100.0)*m_in        #Mass flow rate of HCl leaving the spray tower (lb/h)
m2_out = (1 - E2/100.0)*m1_out      #Mass flow rate of HCl entering the packed column (lb/h)
E = (m_in - m2_out)/m_in            #Overall fractional efficiency (%)

#Result:
print "The mass flow rate of HCl leaving the spray tower is :",round(m1_out,2)," lb/h HCL"
print "The mass flow rate of HCl entering the packed column is :",round(m2_out,3)," lb/h HCL"
print "The overall fractional efficiency is :",round(E*100,1)," %"
The mass flow rate of HCl leaving the spray tower is : 26.6  lb/h HCL
The mass flow rate of HCl entering the packed column is : 0.532  lb/h HCL
The overall fractional efficiency is : 99.3  %

ILLUSTRATIVE EXAMPLE 4.5, Page number: 43¶

In :
#Variable declaration:
m1 = 1000                   #Flowrate data 1 (lb/min)
m2 = 1000                   #Flowrate data 2 (lb/min)
m4 = 200                    #Flowrate data 4 (lb/min)

#Calculations:
m5 = m1 + m2 - m4           #Flowrate data 5 (lb/min)
m6 = m2                     #Flowrate data 6 (lb/min)
m = m5 - m6                 #Flowrate of water lost in operation (lb/min)

#Result:
print "The amount of water lost by evaporation in the operation is", round(m)," lb/min"
The amount of water lost by evaporation in the operation is 800.0  lb/min

ILLUSTRATIVE EXAMPLE 4.6, Page number: 44¶

In :
#Variable declaration:
q1 = 1000.0                   #Volumetric flowrate from tank 1 (gal/day)
q2 = 1000.0                   #Volumetric flowrate from tank 2 (gal/day)
q3 = 2000.0                   #Volumetric flowrate from tank 3 (gal/day)
q4 = 200.0                    #Volumetric flowrate from tank 4 (gal/day)
q5 = 1800.0                   #Volumetric flowrate from tank 5 (gal/day)
q6 = 1000.0                   #Volumetric flowrate from tank 6 (gal/day)
C1 = 4.0                      #Phosphate concentration in tank 1 (ppm)
C2 = 0.0                      #Phosphate concentration in tank 2 (ppm)
C3 = 2.0                      #Phosphate concentration in tank 3 (ppm)
C4 = 20.0                     #Phosphate concentration in tank 4 (ppm)
C5 = 0.0                      #Phosphate concentration in tank 5 (ppm)
C6 = 0.0                      #Phosphate concentration in tank 6 (ppm)
Cf = 120000.0                 #conversion factor for water (gal/10**6lb)

#Calculations:
C1q1 = C1*q1/Cf             #Data 1 (lb/day)
C2q2 = C2*q2/Cf             #Data 2 (lb/day)
C3q3 = C3*q3/Cf             #Data 3 (lb/day)
C4q4 = C4*q4/Cf             #Data 4 (lb/day)
C5q5 = C5*q5/Cf             #Data 5 (lb/day)
C6q6 = C6*q6/Cf             #Data 6 (lb/day)

#Results:
if (((C1q1 + C2q2) == C3q3) and  C3q3 == (C4q4 + C5q5) and C5q5 == C6q6 and C2q2 == C6q6):
print "The data appear to be consistent ."
else:
print "The data appear to be inconsistent ."
The data appear to be consistent .

ILLUSTRATIVE EXAMPLE 4.7, Page number: 48¶

In :
#Variable declaration:
Dz = 3000                       #Height (ft)
V0 = 500000                     #Flowrate of water (gal/min)
n = 30                          #Turbine efficiency (%)
m = 0.3048                      #Meters in a feet
m3 = 0.00378                    #Meters-cube in a gallon
g = 9.8                         #Gravitational acceleration (m/s^2)
gc = 1                          #Conversion factor
MW = 10**(-6)                   #Megawatt in newton-meter-per-second

#Calculations:
V1 = (V0*m3)*1000.0/60.0        #The mass flow rate of the water in kilograms/second (kg/s)
DPE = V1*g*Dz*m/gc*MW           #The loss in potential energy (MW)
AP = n/100.0*DPE                #The actual power output (MW)

#Result:
print "The power generated by the lake located is :",round(AP,1)," MW"
The power generated by the lake located is : 84.7  MW

ILLUSTRATIVE EXAMPLE 4.8, Page number: 50¶

In :
#Variable declaration:
n = 111.4                   #Flowrate of air stream (lbmol/min)
H1 = 1170                   #Average heat capacity at 200°F (Btu/lbmol)
H2 = 4010                   #Average heat capacity at 600°F (Btu/lbmol)

#Calculation:
Q = n*(H2 - H1)             #The heat transfer rate (Btu/min)

#Result:
print "The heat transfer rate required is:",round(Q/10**5,2)," x 10**5 Btu/min"
The heat transfer rate required is: 3.16  x 10**5 Btu/min

ILLUSTRATIVE EXAMPLE 4.9, Page number: 50¶

In :
#Variable declaration:
n = 600                     #The mass flow rate of fluid (lbmol/min)
Cp_AV = 0.271               #Heat capacity (Btu/lbmol . °F)
T1 = 200                    #Initial temperature(°F)
T2 = 600                    #Final temperature(°F)

#Calcultaion:
Q = n*Cp_AV*(T2 - T1)       #The required heat rate (Btu/min)

#Result:
print "The required heat rate is :",round(Q,-2)," Btu/min"
The required heat rate is : 65000.0  Btu/min

ILLUSTRATIVE EXAMPLE 4.10, Page number: 51¶

In :
#Variable declaration:
T_c1 = 20                               #Initial cold fluid temperature (°C)
T_h1 = 82                               #Initial hot fluid temperature (°C)
T_h2 = 94                               #Final hot fluid temperature (°C)

#Calculation:
T_c2 = (T_h2 - T_h1 + T_c1)             #Final cold fluid temperature (°C)

#Result:
print "The heat transfer rate is:",round(T_c2)," °C"
print "There is a printing mistake in book regarding unit of the final result."
The heat transfer rate is: 32.0  °C
There is a printing mistake in book regarding unit of the final result.

ILLUSTRATIVE EXAMPLE 4.11, Page number: 51¶

In :
#Variable declaration:
Q = -5.5*10**6                  #The heat transferred out from the gas (W)
Cp = 1090.0                     #The average heat capacity of the gas (J/(kg . °C))
m = 9.0                         #The gas mass flow rate (kg/s)
T1 = 650                        #The gas inlet temperature (°C)

#Calculation:
T2 = Q/(m*Cp)+T1                #The gas outlet temperature (°C)

#Result:
print "The gas outlet temperature is :",round(T2)," °C"
The gas outlet temperature is : 89.0  °C

ILLUSTRATIVE EXAMPLE 4.12, Page number: 52¶

In :
#Variable declaration:
n = 3500.0                      #Inlet flowrate of water (gal/min)
Cp_W = 75.4                     #Heat capacity of water (J/(gmol . °C)
p = 62.4                        #Density of water (lb/ft^3)
M = 24*60.0                     #Minutes in a day (min/day)
G = 7.48                        #Gallons in a feet cube (gal/ft^3)
gm = 454.0                      #Grams in a pound (g/lb)
J = 1054.0                      #Joules in a Btu (J/Btu)
g = 18.0                        #Grams in a gmol (g/gmol)
F = 1.8                         #Degree fahrenheit in a degree celcius (°F)
Ti = 38.0                       #Initial temperature (°F)
Tf = 36.2                       #Final temperature (°F)

#Calculations:
T= Ti-Tf                        #Temperature loss (°F)
m = n*p*M/G                     #Mass flow rate of water (lb/day)
Cp = Cp_W*gm/J/g/F              #Heat capacity in cosistent units (Btu/(lb.°F))
Q = m*Cp*T                      #Rate of heat flow from water (Btu/day)

#Result:
print "The rate of Btu removed from the water per day is :",round(Q/10**8,2)," x 10**8 Btu/day ."
print "There is a calculation mistake in the book regarding the final result."
The rate of Btu removed from the water per day is : 0.76  x 10**8 Btu/day .
There is a calculation mistake in the book regarding the final result.