Chapter 05:First law applied to Flow Processes

Ex5.1:pg-97

In [1]:
# Part(a)
import math
V1 = 0.95 # Inlet volume flow rate in m**3/kg

P1 = 100 # Pressure at inlet in kPa

v1 = 7 # velocity of flow at inlet in m/s

V2 = 0.19 # Exit volume flow rate in m**3/kg

P2 = 700 # Pressure at exit in kPa 

v2 = 5 # velocity of flow at exit in m/s

w = 0.5 # mass flow rate in kg/s

u21 = 90  # change in internal energy in kJ/kg

Q = -58  # Heat transfer in kW

W =  - w*( u21 + (P2*V2-P1*V1) + ((v2**2-v1**2)/2) ) + Q  # W = dW/dt  

print "\n Example 5.1"

print "\n The rate of work input is ",abs(W) ," kW"

#The answers given in textbook is wrong

# Part (b)

A = (v2/v1)*(V1/V2)  # A = A1/A2

d_ratio = math.sqrt(A)  # d = d1/d2

print "\n The ratio of the inlet pipe diameter and outet pipe diameter is ",d_ratio ," "

#The answers vary due to round off error
 Example 5.1

 The rate of work input is  116.0  kW

 The ratio of the inlet pipe diameter and outet pipe diameter is  0.0  

Ex5.2:pg-98

In [2]:
import math
V1 = 0.37 # volume flow rate at inlet in m**3/kg

P1 = 600# Inlet pressure in kPa

v1 = 16 # Inlet velocity of flow in m/s

V2 = 0.62 # volume flow rate at exit in m**3/kg 

P2 = 100# Exit pressure in kPa

v2 = 270 # Exit velocity of flow in m/s

Z1 = 32 # Height of inlet port from datum in m

Z2 = 0 #Height of exit port from datum in m

g = 9.81  # Acceleration due to gravity

Q = -9  # Heat transfer in kJ/kg

W = 135  # Work transfer in kJ/kg

U12 = (P2*V2-P1*V1) + ((v2**2-v1**2)/2000) + (Z2-Z1)*g*1e-3 + W - Q  # Change in internal energy in kJ



print "\n Example 5.2"

print "\n The internal energy decreases by ",round(U12) ," kJ"

#The answers vary due to round off error
 Example 5.2

 The internal energy decreases by  20.0  kJ

Ex5.3:pg-99

In [3]:
import math

P1 = 4 # Boiler pressure in MPa

t1 = 400 # Exit temperature at boiler in degree Celsius

h1 = 3213 # Enthalpy at boiler exit in kJ/kg

V1 = 0.073 # specific volume at boiler exit in m**3/kg

P2 = 3.5 # Pressure at turbine end in MPa

t2 = 392 # Turbine exit temperature in degree Celsius

h2 = 3202 # Enthalpy at turbine exit in kJ/kg

V2 = 0.084 # specific volume at turbine exit in m**3/kg

Q = -8.5 # Heat loss from pipeline in kJ/kg

v1 = math.sqrt((2*(h1-h2+Q)*1e3)/(1.15**2-1)) # velocity of flow in m/s

A1 = (math.pi/4)*0.2**2 # Area of pipe in m**2

w = (A1*v1)/V1 # steam flow rate in Kg/s



print "\n Example 5.3"

print "\n The steam flow rate is ",w ," Kg/s"

#The answers vary due to round off error
 Example 5.3

 The steam flow rate is  53.5854836932  Kg/s

Ex5.4:pg-100

In [4]:
import math
h1 = 313.93 # Enthalpy of water at heater inlet in kJ/kg

h2 = 2676 # Enthalpy of hot water at temperature 100.2 degree Celsius

h3 = 419 #Enthalpy of water at heater inlet in kJ/kg

w1 = 4.2 # mass flow rate in kg/s



print "\n Example 5.4"

w2 = w1*(h3-h1)/(h2-h3)# Steam rate  

print "\n The amount of heat that should be supplied is ",w2*3600 ," Kg/h"



#The answers vary due to round off error
 Example 5.4

 The amount of heat that should be supplied is  703.880549402  Kg/h

Ex5.5:pg-100

In [5]:
import math
t1 = 15 # Heat exchanger inlet temperature in degree Celsius

t2 = 800 # Heat exchanger exit temperature in degree Celsius

t3 = 650 # Turbine exit temperature in degree Celsius

t4 = 500 # Nozzle exit temperature in degree Celsius

v1 = 30 # Velocity of steam at heat exchanger inlet in m/s

v2 = 30# Velocity of steam at turbine inlet in m/s

v3 = 60 # Velocity of steam at nozzle inlet in m/s

w = 2 # mass flow rate in kg/s

cp = 1005 # Specific heat capacity of air in kJ/kgK



print "\n Example 5.5"

Q1_2 = w*cp*(t2-t1) # rate of heat transfer

print "\n The rate of heat transfer to the air in the heat exchanger is ",Q1_2/1e3 ," kJ/s"



W_T = w*( ((v2**2-v3**2)/2) + cp*(t2-t3)) # power output from the turbine

print "\n The power output from the turbine assuming no heat loss is ",W_T/1000 ," kW"

v4 = math.sqrt( (v3**2) + (2*cp*(t3-t4)) ) # velocity at the exit of the nozzle

print "\n The velocity at the exit of the nozzle is ",v4 ," m/s"

#The answers vary due to round off error
 Example 5.5

 The rate of heat transfer to the air in the heat exchanger is  1577.85  kJ/s

 The power output from the turbine assuming no heat loss is  298  kW

 The velocity at the exit of the nozzle is  552.358579186  m/s

Ex5.6:pg-102

In [16]:
import math

ha = 260  # Enthalpy of air in kJ/kg

hg = 912  # Enthalpy of gas in kJ/kg

Va = 270  # Velocity of air in m/s

wf = 0.0190  # mass of fuel in Kg

wa = 1 # mass of air in Kg

Ef = 44500  # Chemical energy of fuel in kJ/kg

Q = 21  # Heat loss from the engine in kJ/kg



print "\n Example 5.6"

Eg = 0.05*wf*Ef/(1+wf)  # As 5% of chemical energy is not released in reaction

wg = wa+wf # mass of flue gas

Vg = math.sqrt(2000*(((ha+(Va**2*0.001)/2+(wf*Ef)-Q)/(1+wf))-hg-Eg)) 



print "\n Velocity of exhaust gas is ",Vg ," m/s"

#Answer given in textbook is wrong
 Example 5.6

 Velocity of exhaust gas is  541.409855832  m/s

Ex5.8:pg-103

In [6]:
import math
# Given that

V = 0.12 # Volume of tank in m**3

p = 1 # Pressure in MPa

T = 150 # Temperature in degree centigrade

P = 0.1 # Power to peddle wheel in kW

print "\n Example 5.8"

u0 = 0.718*273 # Internal energy at 0 degree Celsius

# Function for internal energy of gas

def f1(t):
    u = u0+(0.718*t)
    pv = 0.287*(273+t)
    return (u,pv)
       
U,PV=f1(T)
       
       
hp = U+PV # At 150 degree centigrade
m_a = P/hp
       
print "\n The rate at which air flows out of the tank is ",round(m_a*3600,2) ," kg/h"

#The answers vary due to round off error
 Example 5.8

 The rate at which air flows out of the tank is  0.85  kg/h