import math
# Variables
R = 287.; #J/kg K
V1 = 40.; #m**3
V2 = 40.; #m**3
p1 = 1.*10**5; #Pa
p2 = 0.4*10**5; #Pa
T1 = 298.; #K
T2 = 278.; #K
# Calculations
m1 = p1*V1/R/T1;
m2 = p2*V2/R/T2;
#Let mass of air removed be m
m = m1-m2;
# Results
print ("Mass of air removed = %.3f")% (m),("kg")
V = m*R*T1/p1;
print ("Volume of gas removed = %.3f")% (V), ("m**3")
Mass of air removed = 26.716 kg Volume of gas removed = 22.849 m**3
# Variables
V = 0.04; #m**3
p = 120.*10**5; #Pa
T = 293.; #K
R0 = 8314.;
# Calculations and Results
print ("(i) kg of nitrogen the flask can hold")
M = 28; #molecular weight of Nitrogen
R = R0/M;
m = p*V/R/T;
print ("kg of nitrogen = %.3f")% (m), ("kg")
print ("(ii) Temperature at which fusible plug should melt")
p = 150.*10**5; #Pa
T = p*V/R/m; #K
t = T-273; #0C
print ("Temperature = %.3f")% (t),("°C")
(i) kg of nitrogen the flask can hold kg of nitrogen = 5.517 kg (ii) Temperature at which fusible plug should melt Temperature = 93.250 °C
import math
# Variables
p1 = 1.*10**5; #Pa
T1 = 293.; #K
d = 6.; #m; diameter of the spherical balloon
p2 = 0.94*p1;
T2 = T1;
cv = 10400.; #J/kg K
R = 8314/2.;
r = 3.; #m
# Calculations and Results
print ("(i) Mass of original gas escaped")
mass_escaped = (p1-p2)/p1*100;
print ("%mass_escaped = "), (mass_escaped), ("%")
print ("(ii)Amount of heat to be removed ")
T2 = 0.94*T1;
m = p1*4/3*math.pi*r**3/R/T1;
Q = m*cv*(T1-T2)/10**6;
print ("Q = %.3f")% (Q),("MJ")
(i) Mass of original gas escaped %mass_escaped = 6.0 % (ii)Amount of heat to be removed Q = 1.698 MJ
from numpy import *
import math
# Variables
m = 28.; #kg
V1 = 3.; #m**3
T1 = 363.; #K
R0 = 8314.;
M = 28.; #Molecular mass of N2
R = R0/m;
V2 = V1;
T2 = 293.; #K
# Calculations and Results
print ("(i) Pressure (p1) and specific volume (v1) of the gas")
p1 = m*R*T1/V1/10**5; #bar
print ("Pressure = %.3f")% (p1), ("bar")
v1 = V1/m;
print ("specific volume = %.3f")% (v1), ("m**3/kg")
#cp-cv = R/1000;
#cp-1.4cv = 0;
#solving the above two eqns
A = [[1,-1],[1,-1.4]];
B = [R/1000,0];
X = linalg.inv(A)*B;
cp = X[0,0]
print ("cp = %.3f")% (cp), ("kJ/kg K")
cv = X[1][0];
print ("cv = %.3f")% (cv),("kJ/kg K")
print ("(iii) Final pressure of the gas after cooling to 20°C")
p2 = p1*T2/T1;
print ("p2 = %.3f")% (p2), ("bar")
du = cv*(T2-T1);
print ("Increase in specific internal energy = %.3f")% (du), ("kJ/kg")
dh = cp*(T2-T1);
print ("Increase in specific Enthalpy = %.3f")%(dh), ("kJ/kg")
v2 = v1;
ds = cv*math.log(T2/T1) + R*math.log(v2/v1);
print ("Increase in specific entropy = %.3f")%(ds),("kJ/kg K")
W = 0; #constant volume process
Q = m*du+W;
print ("Heat transfer = %.3f")%(Q), ("kJ")
(i) Pressure (p1) and specific volume (v1) of the gas Pressure = 10.060 bar specific volume = 0.107 m**3/kg cp = 1.039 kJ/kg K cv = 0.742 kJ/kg K (iii) Final pressure of the gas after cooling to 20°C p2 = 8.120 bar Increase in specific internal energy = -51.963 kJ/kg Increase in specific Enthalpy = -72.748 kJ/kg Increase in specific entropy = -0.159 kJ/kg K Heat transfer = -1454.950 kJ
import math
print ("Part (a)")
# Variables
R = 0.287; #kJ/kg K
y = 1.4;
m1 = 1.; #kg
p1 = 8.*10**5; #Pa
T1 = 373.; #K
p2 = 1.8*10**5; #Pa
cv = 0.717; #kJ/kg K
n = 1.2;
# Calculations and Results
#pv**1.2 = consmath.tant
print ("(i) The final specific volume, temperature and increase in entropy")
v1 = R*10**3*T1/p1;
v2 = v1*(p1/p2)**(1./n);
print ("v2 = %.3f")%(v2), ("m**3/kg")
T2 = p2*v2/R/10**3; #K
t2 = T2-273; #0C
print ("Final temperature = %.3f")% (t2), ("0C")
ds = cv*math.log(T2/T1) + R*math.log(v2/v1);
print ("ds = %.3f")%(ds), ("kJ/kg K")
W = R*(T1-T2)/(n-1);
print ("Work done = %.3f")% (W), ("kJ/kg")
Q = cv*(T2-T1) + W;
print ("Heat transfer = %.3f")%(Q),("kJ/kg")
print ("Part (b)")
print ("(i) Though the process is assumed now to be irreversible and adiabatic, the end states are given to be the same as in (a). Therefore, all the properties at the end of the process are the same as in (a).")
print ("(ii) Adiabatic process")
Q = 0;
print ("Heat transfer = %.3f")%(Q), ("kJ/kg")
W = -cv*(T2-T1);
print ("Work done = %.3f")%(W),("kJ/kg")
Part (a) (i) The final specific volume, temperature and increase in entropy v2 = 0.464 m**3/kg Final temperature = 17.897 0C ds = 0.179 kJ/kg K Work done = 117.818 kJ/kg Heat transfer = 58.950 kJ/kg Part (b) (i) Though the process is assumed now to be irreversible and adiabatic, the end states are given to be the same as in (a). Therefore, all the properties at the end of the process are the same as in (a). (ii) Adiabatic process Heat transfer = 0.000 kJ/kg Work done = 58.868 kJ/kg
# Variables
d = 2.5; #m; diameter
V1 = 4./3*math.pi*(d/2)**3; #volume of each sphere
T1 = 298.; #K
T2 = 298.; #K
m1 = 16.; #kg
m2 = 8.; #kg
V = 2.*V1; #total volume
m = m1+m2;
R = 287.; #kJ/kg K
# Calculations
p = m*R*T1/V/10**5; #bar
# Results
print ("pressure in the spheres when the system attains equilibrium = %.3f")%(p),("bar")
pressure in the spheres when the system attains equilibrium = 1.254 bar
# Variables
m = 6.5/60; #kg/s
import math
cv = 0.837; #kJ/kg K
p1 = 10*10**5; #Pa
p2 = 1.05*10**5; #Pa
T1 = 453; #K
R0 = 8.314;
M = 44.; #Molecular mass of CO2
# Calculations and Results
R = R0/M;
cp = cv+R;
y = cp/cv;
T2 = T1*(p2/p1)**((y-1)/y);
print T2
t2 = T2-273;
print ("Final temperature = %.3f")%(t2),("0C")
v2 = R*10**3*T2/p2; #m**3/kg
print ("final specific volume = %.3f")%(v2), ("m**3/kg")
ds = 0; #Reversible and adiabatic process
print ("Increase in entropy = "), (ds)
Q = 0; #Adiabatic process
print ("Heat transfer rate from turbine = "), (Q)
W = m*cp*(T1-T2);
print ("Power delivered by the turbine = %.3f")% (W), ("kW")
# Note : answers wont match with the book because of the rounding error.
299.106840283 Final temperature = 26.107 0C final specific volume = 0.538 m**3/kg Increase in entropy = 0 Heat transfer rate from turbine = 0 Power delivered by the turbine = 17.104 kW
import math
# Variables
p1 = 8.*10**5; #Pa
V1 = 0.035; #m**3
T1 = 553.; #K
p2 = 8.*10**5; #Pa
V2 = 0.1; #m**3
n = 1.4;
R = 287.; #J/kg K
T3 = 553.; #K
cv = 0.71; #kJ/kg K
# Calculations and Results
m = p1*V1/R/T1;
T2 = p2*V2/m/R;
p3 = p2/((T2/T3)**(n/(n-1)));
V3 = m*R*T3/p3;
print ("(i) The heat received in the cycle")
#constant pressure process 1-2
W_12 = p1*(V2-V1)/10**3; #kJ
Q_12 = m*cv*(T2-T1) + W_12; #kJ
#polytropic process 2-3
W_23 = m*R/10**3*(T2-T3)/(n-1);
Q_23 = m*cv*(T3-T2) + W_23;
Q_received = Q_12 + Q_23;
print ("Total heat received in the cycle = "),(Q_received), ("kJ")
print ("(ii) The heat rejected in the cycle")
#Isothermal process 3-1
W_31 = p3*V3*math.log(V1/V3)/10**3; #kJ
Q_31 = m*cv*(T3-T1) + W_31;
print ("Heat rejected in the cycle = %.3f")% (-Q_31), ("kJ")
n = (Q_received - (-Q_31))/Q_received*100;
print ("Efficiency of the cycle = %.3f")% (n), ("%")
(i) The heat received in the cycle Total heat received in the cycle = 182.0 kJ (ii) The heat rejected in the cycle Heat rejected in the cycle = 102.883 kJ Efficiency of the cycle = 43.471 %
# Variables
v = 44.; #m**3/kg-mol
T = 373.; #K
# Calculations and Results
print ("(i) Using Van der Waals’ equation")
a = 362850.; #N*m**4/(kg-mol)**2
b = 0.0423; #M**3/kg-mol
R0 = 8314.; #J/kg K
p = ((R0*T/(v-b)) - a/v**2);
print ("Pressure umath.sing Van der Waals equation = %.3f")%(p), ("N/m**2")
print ("(ii) Using perfect gas equation")
p = R0*T/v;
print ("Pressure using perfect gas equation = %.3f")% (p), ("N/m**2")
(i) Using Van der Waals’ equation Pressure umath.sing Van der Waals equation = 70360.445 N/m**2 (ii) Using perfect gas equation Pressure using perfect gas equation = 70480.045 N/m**2
# Variables
V = 3.; #m**3
m = 10.; #kg
T = 300.; #K
# Calculations and Results
R0 = 8314.;
M = 44.;
R = R0/M;
p = m*R*T/V;
print ("Pressure Using perfect gas equation = %.3f")% (p),("N/m**2")
a = 362850; #Nm**4/(kg-mol)**2
b = 0.0423; #m**3/(kg-mol)
v = 13.2; #m**3/kg-mol
p = R0*T/(v-b) - a/v**2;
print ("Pressure Using Van der Waals’ equation = %.3f")%(p), ("N/m**2")
A0 = 507.2836;
a = 0.07132;
B0 = 0.10476;
b = 0.07235;
C = 66*10**4;
A = A0*(1-a/v);
B = B0*(1-b/v);
e = C/v/T**3;
p = R0*T*(1-e)/v**2*(v+B) - A/v**2;
print ("Pressure Using Beattie Bridgeman equation = %.3f")%(p), ("N/m**2")
Pressure Using perfect gas equation = 188954.545 N/m**2 Pressure Using Van der Waals’ equation = 187479.533 N/m**2 Pressure Using Beattie Bridgeman equation = 190090.365 N/m**2
import math
from scipy.integrate import quad
# Variables
a = 139250; #Nm**4/(kg-mol)**2
b = 0.0314; #m**3/kg-mol
R0 = 8314; #Nm/kg-mol K
v1 = 0.2*32; #m**3/kg-mol
v2 = 0.08*32; #m**3/kg-mol
T = 333; #K
print ("(i) Work done during the process")
# Calculations
def f21( v):
return R0*T/(v-b) - a/v**2
W = quad(f21, v1, v2)[0]
# Results
print ("W = %.3f")% (W),("Nm/kg-mol")
print ("(ii) The final pressure")
p2 = R0*T/(v2-b) - a/v2**2;
print ("p2 = %.3f")%(p2), ("N/m**2")
(i) Work done during the process W = -2524722.415 Nm/kg-mol (ii) The final pressure p2 = 1073651.290 N/m**2
# Variables
pr = 20;
Z = 1.25;
Tr = 8.0;
Tc = 282.4; #K
# Calculations
T = Tc*Tr;
# Results
print ("Temperature = %.3f")%(T),("K")
Temperature = 2259.200 K
# Variables
p = 260.*10**5; #Pa
T = 288.; #K
pc = 33.94*10**5; #Pa
Tc = 126.2; #K
R = 8314./28;
# Calculations
pr = p/pc;
Tr = T/Tc;
Z = 1.08;
rho = p/Z/R/T;
# Results
print ("Density of N2 = %.3f")% (rho), ("kg/m**3")
Density of N2 = 281.517 kg/m**3
# Variables
p = 200.*10**5; #Pa
pc = 73.86*10**5; #Pa
Tc = 304.2; #K
pr = p/pc;
Z = 1;
Tr = 2.48;
# Calculations
T = Tr*Tc;
# Results
print ("Temperature = "), (T), ("K")
Temperature = 754.416 K
import math
# Variables
d = 12.; #m; diameter of spherical balloon
V = 4./3*math.pi*(d/2)**3;
T = 303.; #K
p = 1.21*10**5; #Pa
pc = 12.97*10**5; #Pa
Tc = 33.3; #K
R = 8314./2;
# Calculations
pr = p/pc;
Tr = T/Tc;
Z = 1;
m = p*V/Z/R/T;
# Results
print ("Mass of H2 in the balloon = %.3f")% (m), ("kg")
Mass of H2 in the balloon = 86.917 kg
# Calculations
Z_cp = 3./2-9./8;
# Results
print ("Z_cp = "),(Z_cp)
Z_cp = 0.375