# Variables
#Given
W = -2.25*745.7; #work done on system in J/s
Q = -3400.*(10.**3)/3600; #heat transferred to the surrounding in J/s
# Calculations
#To find the change in internal energy
#Using equation 2.4 (Page no. 26)
U = Q-W; #change in internal energy in J/s
# Results
print 'Internal energy of system increases by %f J/s'%U
#Given
T = 298.; #temperature in K
P = 101.; #pressure in kPa
n_iron = 2.; #moles of iron reacted
Q = -831.08; #heat liberated in kJ
R = 8.314; #ideal gas constant
# Calculations and Results
#To find heat liberated work done and change in internal energy
print 'Heat liberated during the reaction is %f kJ'%Q
n_oxygen = 1.5; #moles of oxygen reacted
#Using ideal gas equation P(Vf-Vi)=nRT and W=P(Vf-Vi)
W = -1.5*R*T; #work done by system in J
#Using equation 2.4 (Page no. 26)
U = (Q*10**3)-W; #change in internal energy in J
print 'Work done by gas is %f J'%W
print 'Change in internal energy is %6.3e J'%U
# Variables
#Given
u = 20.; #speed of car in m/s
z = 30.; #height vertically above the bottom of hill in m
m = 1400.; #mass of car in kg
g = 9.81; #acceleration due to gravity
# Calculations
#To find the heat energy dissipated by brakes
#Using equation 2.3 (Page no. 26)
KE = -0.5*m*(u**2); #change in kinetic energy in J
PE = -m*g*z; #change in potential energy in J
Q = -(KE+PE); #heat dissipated by brakes in J
# Results
print 'Heat dissipated by brakes is %3.2e J'%Q
# Variables
#Given:
#For step 1
W1 = -50.; #work received in J
Q1 = -25.; #heat gven out in J
# Calculations and Results
U1 = Q1-W1; #internal energy change in J
print 'Change in internal energy for constant pressure process is %i J'%U1
#For step 2
W2 = 0.; #work done for constant volume process is zero
Q2 = 75.; #heat received in J
U2 = Q2; #internal energy change in J
print 'Change in internal energy for constant volume process is %i J'%U2
#For step 3
Q3 = 0.; #no heat exchange in adiabatic process
#Since the process is cyclic
#U3+U2+U1 = 0;
U3 = -(U1+U2);
W3 = -U3; #work done in J
print 'Work done during adiabatic process is %i J'%W3
# Variables
#Given:
n_water = 10.**3; #moles of water
T = 373.; #tempearture(K)
P = 101.3; #pressure(kPa)
sv_liquid = 0.00104; #specific volume of liquid(m**3/kmol)
sv_vapour = 1.675; #specific volume of vapour(m**3/kmol)
Q = 1.03*10**3; #heat added in kJ
#To find change in internal energy and enthalpy
W = P*n_water*(sv_vapour-sv_liquid)*10**-3; #expansion work done in kJ
U = Q-W; #change in internal energy in kJ
#For constant pressure process
H = Q; #enthalpy change in kJ
# Results
print 'Change in internal energy is %f kJ'%U
print 'Change in enthalpy is %3.2e J'%H
# Variables
#Given:
T = 233.; #temperature in K
VP = 1.005*10**3; #vapour pressure of CO2 in kPa
sv_liquid = 0.9*10**-3; #specific volume of liquid CO2 in m**3/kg
sv_vapour = 38.2*10**-3; #specicific volume of CO2 vapour in m**3/kg
L = 320.5; #latent heat of vaporisation of CO2 in kJ/kg
#Assuming at these conditions CO2 is saturated liquid so
H1 = 0; #enthalpy in liquid state
# Calculations
#To find internal energy of saturated liquid and internal energy and enthalpy of saturated vapour
#For saturated liquid
U1 = H1-(VP*sv_liquid); # internal energy in liquid state in kJ/kg
#For saturated vapour
Hv = H1+L; #enthalpy of saturated vapour in kJ/kg
Uv = Hv-(VP*sv_vapour); #internal energy in vapour state in kJ/kg
# Results
print 'Internal Energy of saturated liquid is %f kJ/kg'%U1
print 'Enthalpy of vapour state is %f kJ/kg'%Hv
print 'Internal Energy of vapour state is %f kJ/kg'%Uv
# Variables
#Given:
I = 0.5; #current in Amperes
V = 12.; #voltage in volts
t = 5*60.; #time in sec
m = 0.798; #mass of water vaporised in g
M = 18.; #molecular mass of water in g
R = 8.314*10**-3 #ideal gas constant
T = 373 #temperature
# Calculations
#To calculate molar internal energy change and molar enthalpy change
Q = (I*V*t/1000.); #electric energy supplied in kJ
#Referring equation 2.10 (Page no. 29)
H = (Q*M)/m; #molar enthalpy change in kJ/mole
#BY ideal gas equation PV=RT
#Referring equation 2.9 for constant pressure process (Page no. 29)
U = H-(R*T); #molar internal energy change in kJ/mole
# Results
print 'Molar Enthalpy change during the process is %i kJ/mole'%H
print 'Molar Interanl Energy change during the process is %f kJ/mole'%U
# Variables
#Given:
m = 1650.; #mass of steam used in kg/hr
H1 = 3200.; #enthalpy at 1368 kPa and 645 K in kJ/kg
H2 = 2690.; #enthalpy at 137 kPa and 645 K in kJ/kg
#To determine the theoretical horsepower developed
#Using equation 2.13 (Page no.32)
Q = 0; #since the process is adiabatic
z = 0; #assuming that inlet and discharge of turbine are at same level
u = 0; #feed and discharge velocities being equal
# Calculations
Ws = -(H2-H1);
Wj = Ws*10**3*m/3600.; #work done by turbine in J
W = Wj/745.7; #work done by turbine in hp
# Results
print 'Work done by turbine is %f hp'%W
# Variables
#Given:
m = 25.*10**3; #mass flow rate of water in kg/h
P = 2.; #power supplied by motor in hp
q = 42000.; #heat given in kJ/min
z = 20.; #elevation in m
T = 368.; #temperature in K
To = 273.; #standard temperature in K
Cp = 4.2; #specific heat of water in kJ/kg K
g = 9.81 #acceleration due to gravity(m/s^2)
# Calculations
#To find temperature of water delivered to second storage tank
W = (P*745.7*10**-3*3600)/m; #work done per kg of water pumped in kJ/kg
Q = q*60./m; #heat given out per kg of fluid
PE = g*z*10**-3; #change in potential energy in kJ/kg
#Using equation 2.13 (Page no. 32)
H = -Q+W-PE;
#H = H2-H1
H1 = Cp*(T-To);
H2 = H1+H;
#Let T1 be the temperature at second storage tank
T1 = To+(H2/Cp);
# Results
print 'Temperature of water at second storage tank is %i K'%T1
# Variables
#Given:
D1 = 25.; #internal diameter of pipe in mm
u1 = 10.; #upstream velocity in m/s
D2 = 50.; #downstream diameter of pipe in mm
# Calculations and Results
#(a)
#Let A1 nad A2 be upstream and downstream crosssectional areas of pipe
u2 = ((D1/D2)**2)*u1; #downstream velocity in m/s
H = 0.5*(u1**2-u2**2); #change in enthalpy in J/kg
print 'Change in enthalpy is %f J/kg'%H
#(b)
#For maximum enthalpy change
u2 = 0;
Hmax = 0.5*u1**2; #(J/kg)
print 'Maximum enthalpy chnage for a sudden enlargement in pipe is %f J/kg'%Hmax
# Variables
#At inlet:
T1 = 293.; #Temperature(K)
P1 = 300+136.8; #Pressure(kPa)
#At exit:
T2 = 453.; #Temperature(K)
P2 = 136.8; #Pressure(kPa)
Cp = 29.4; #specific heat capacity at constant pressure in kJ/kmol
m = 1000.; #mass of hydrogen in kg
M = 2.02; #molecular mass of hydrogen
# Calculations
#To determine heat transfer rates
#Neglecting the kinetic nd potential energy changes
#Assuming the process to be occuring through a number of steps
#Step 1 be isothermal and step 2 be isobaric
H1 = 0; #change in enthalpy for step 1
H2 = (m/M)*Cp*(T2-T1)/1000; #change in enthalpy for step 2 in kJ
H = H2+H1;
Q = H; #heat transferred in coils in kJ
# Results
print 'Heat transferred in coils is %f kJ'%Q
#Given:
m = 10.; #mass of air in kg
P1 = 100.; #initial pressure(kPa)
T1 = 300.; #initial temperature(K)
T2 = 600.; #final temperature(K)
R = 8.314; #ideal gas constant(kJ/kmol K)
Cp = 29.099; #specific heat capacity at constant pressure (kJ/kmol K)
Cv = 20.785; #specific heat capacity at constsant volume (kJ/kmol K)
M = 29.; #molecular weight of air
# Calculations and Results
#To determine change in internal energy enthalpy heat supplied and work done
n = m/M; #number of moles of gas(kmol)
V1 = (n*R*T1)/P1; #initial volume of air (m**3)
#(a)
#Constant volume process
V2 = V1 #final volume
#Change in internal energy U = n*intg(CvdT)...so
U = n*Cv*(T2-T1); #change in internal energy(kJ)
Q = U; #heat supplied(kJ)
W = 0; #work done
H = U+(n*R*(T2-T1)); #change in enthalpy(kJ)
print 'For constant volume process'
print 'Change in internal energy is %i kJ'%U
print 'Heat supplied is %i kJ'%Q
print 'Work done is %i kJ'%W
print 'Change in enthalpy is %i kJ'%H
# (b)
# Constant pressure process
# Change in enthalpy H = n*intg(CpdT)...so
H = n*Cp*(T2-T1); #change in enthalpy(kJ)
Q = H; #heat supplied(kJ)
U = H-(n*R*(T2-T1)); #change in internal energy(kJ)
W = Q-U; #work done(kJ)
print 'For constant pressure process'
print 'Change in internal energy is %i kJ'%U
print 'Heat supplied is %i kJ'%Q
print 'Work done is %i kJ'%W
print 'Change in enthalpy is %i kJ'%H
# Variables
#Given:
R = 8.314; #ideal gas constant(kJ/kmol K)
Cv = 20.8; #specific heat capacity at constant volume(kJ/kmol K)
Cp = 29.1; #specific heat capacity at constant pressure(kJ/kmol K)
P1 = 10.; #initial pressure(bar)
T1 = 280.; #initial temperature in K
P2 = 1.; #final pressure(bar)
T2 = 340.; #final temperature(K)
# Calculations
#To determine the change in internal energy and change in enthalpy
#Solution
n = 1 #basis: 1 kmol of ideal gas
V1 = (n*R*T1)/(P1*100); #initial volume in m**3
V2 = (n*R*T2)/(P2*100); #final volume in m**3
Po = P2;
Vo = V1;
To = (Po*100*Vo)/(n*R);
U1 = Cv*(To-T1);
H1 = U1+(V1*100*(P2-P1));
W1 = 0;
Q1 = U1;
H2 = Cp*(T2-To);
U2 = H2-100*(V2-V1);
Q2 = H2;
W2 = Q2-U2;
#For actual process
U = U1+U2; #change in internal energy(kJ)
H = H1+H2; #change in enthalpy(kJ)
# Results
print 'Change in internal energy is %f kJ'%U
print 'Change in enthalpy is %f kJ'%H