Some comments on the MTX 311 exam, 2014
Question 1A is a rather standard question; calculate the exergy of a process flow. However, the
student should notice that we are pretty close to the critical point, so ideal gas correlations don’t
work. Therefore, enthalpy and entropy values have to be corrected.
Question 1B tests 2 things. Fistly, the student should realize that the potential energy of the bottle is
part of the total exergy, and that the potential energy is given in J/kg instead of kJ/kg. Thus, the
student has to convert the units. Also, since steel is incompressible, there is no Rln(P/P0) term in the
entropy and the P0(V-V0) term is also zero. Since the bottle is “empty”, the residual amount of
oxygen in the bottle can be ignored. However, sharp students might realize that the vapour still has
a bit of exergy and calculate this. Not-so smart students might feel tempted to use the gas constant
of oxygen to calculate the change in entropy of the steel vessel.
Question 1C tests the student’s ability to work with gas mixtures. Firstly, the student should realize
that both oxygen and propane are ideal gases, so the ideal gas law can be used to calculate the
specific volume and thus the mass of the two process flows. Using these mass flow rates, the
student can now calculate the average Cp using ideal gas mixture theory.
Question 1D is probably the most difficult question in this exam. The student has to calculate the
exergy of a combustable material at elevated pressure. Firstly, the student has to solve the reaction
equation (which is easy for methane) . After this, the student should realize that at this pressure,
methane is still an ideal gas so enthalpy deviation corrections are NOT necessary. Next, the entropy
of formation of methane at non-standard conditions has to be calculated; for this, the “pressure
correction term” has to be converted from kJ/kgK to kJ to kmol K. After this the student should
realize that at reference conditions, water is a liquid, even though water vapour is formed in the
reaction. Finally, the enthalpy and entropy change of combustion can be calculated. This can be
combined into the exergy (in kJ/kmolK), which must then be converted back to kJ/kgK.
2A and 2B are probably the easiest questions in the exam; the student has to calculate the
irreversibility and the second law efficiency of a simple process and a simple unit (the boiler).
However, the student must be able to read the process flow diagram, find the relevant process
streams and get the appropriate values from some tables which contain far more info than the
student needs and convert heat fluxes into available work fluxes.
2C Similar to 2B. however, the student has to calculate the second law efficiency and the
irreversibility of some machine which is slightly different to what they have ever seen before. In this
case, that is a turbine with two inlets.
2D is quite difficult again; the student has to determine the second law efficiency of a complex
power cycle. In this case, energy is exchanged with an external party so the standard way to
calculate power cycle second law efficiency does not work. The student has define appropriate
system boundaries and has to come up with a method to define second law efficiency based on the
system boundaries that he/she choose. For this reason, this question has several correct answers.
2E is a question of average difficulty. The student has to break down the irreversibility of the turbine
into a part which is caused by friction and a part which is caused by heat losses. The heat loss related
irreversibility is relatively easy to calculate, the irreversibility due to friction is (almost) impossible to
, find directly (although the student has some equations which seems to give him the right answer but
cannot be used because of the shaft work). However, the student should realize that friction and
heat losses are the only processes causing irreversibility; by subtracting the heat loss term from the
total irreversibility, the student can get the irreversibility due to friction.
Memo Exam MTX311 2014
1A Find exergy of a process flow of ethylene, 40g/s, 60°C, 25 bar.
Strategy: check whether ethylene is an ideal gas
if not, find enthalpy and entropy deviation using the deviation charts
calculate exergy using control volume equation: Ψ=mψ, ψ=(h-h0)-T0(s-s0)
check if ethylene at actual conditions is an ideal gas:
T=333.15K P=25 bar
Tr=T/Tc=1.182 Pc=50.4 bar
Tc=282.4K Pr=P/Pc=0.496
check if ethylene at reference/ground state is an ideal gas:
T=298.15K P=1 bar
Tr=T/Tc=1.056 Pc=50.4 bar
Tc=282.4K Pr=P/Pc=0.019
Real gas when Tr<2 AND Pr>0.2 both hold (with quite some margin), so clearly the compressed
ehtylene is a real gas, whereas ethylene at ground state is an ideal gas.
Look up Δhd and Δsd for compressed ethylene in deviation tables: Δhd/Tc ≈3 kJ/kmol∙K and Δsd= 2
kJ/kmol∙K
Δhd/Tc =3∙282.4=847.2 kJ/kmol=847.2/28=30.26 kJ/kg
Δsd= 2/28=0.0714 kJ/kg∙K
(h-h0)= Δhig -ΔΔhd=Cp(T-T0)- ΔΔhd=1.548(60-25)-(30.26-0)=23.9kJ/kg
(s-s0)= Δsig -ΔΔsd=Cpln(T/T0)-R ln(P/P0)- ΔΔsd=1.548(333.15/298.15)-0.2964ln(25/1)-0.0714=
-0.8537kJ/kgK
ψ=(h-h0)-T0(s-s0)=23.9-298.15∙(-0.8537)=278.5 kJ/kg
Ψ=mψ=0.04*278.5=11.14 kJ
1B what is the total exergy of an empty oxygen bottle at the top of mount Everest?
Use control mass equation:
Ψ=mψ, ψ=(u-u0)-T0(s-s0)+P0(V-V0)+Epot+Ekin
Where u-u0=Cv(T-T0) and s-s0 = Cpln(T/T0) (there is no R ln(P/P0)because steel is not an ideal gas!)
Epot=gh and Ekin=½v²=0
For steel, Cv=Cp=0.9 kJ/kgK
u-u0=Cv(T-T0)=0.9(-25-25)=-45kJ/kg
Question 1A is a rather standard question; calculate the exergy of a process flow. However, the
student should notice that we are pretty close to the critical point, so ideal gas correlations don’t
work. Therefore, enthalpy and entropy values have to be corrected.
Question 1B tests 2 things. Fistly, the student should realize that the potential energy of the bottle is
part of the total exergy, and that the potential energy is given in J/kg instead of kJ/kg. Thus, the
student has to convert the units. Also, since steel is incompressible, there is no Rln(P/P0) term in the
entropy and the P0(V-V0) term is also zero. Since the bottle is “empty”, the residual amount of
oxygen in the bottle can be ignored. However, sharp students might realize that the vapour still has
a bit of exergy and calculate this. Not-so smart students might feel tempted to use the gas constant
of oxygen to calculate the change in entropy of the steel vessel.
Question 1C tests the student’s ability to work with gas mixtures. Firstly, the student should realize
that both oxygen and propane are ideal gases, so the ideal gas law can be used to calculate the
specific volume and thus the mass of the two process flows. Using these mass flow rates, the
student can now calculate the average Cp using ideal gas mixture theory.
Question 1D is probably the most difficult question in this exam. The student has to calculate the
exergy of a combustable material at elevated pressure. Firstly, the student has to solve the reaction
equation (which is easy for methane) . After this, the student should realize that at this pressure,
methane is still an ideal gas so enthalpy deviation corrections are NOT necessary. Next, the entropy
of formation of methane at non-standard conditions has to be calculated; for this, the “pressure
correction term” has to be converted from kJ/kgK to kJ to kmol K. After this the student should
realize that at reference conditions, water is a liquid, even though water vapour is formed in the
reaction. Finally, the enthalpy and entropy change of combustion can be calculated. This can be
combined into the exergy (in kJ/kmolK), which must then be converted back to kJ/kgK.
2A and 2B are probably the easiest questions in the exam; the student has to calculate the
irreversibility and the second law efficiency of a simple process and a simple unit (the boiler).
However, the student must be able to read the process flow diagram, find the relevant process
streams and get the appropriate values from some tables which contain far more info than the
student needs and convert heat fluxes into available work fluxes.
2C Similar to 2B. however, the student has to calculate the second law efficiency and the
irreversibility of some machine which is slightly different to what they have ever seen before. In this
case, that is a turbine with two inlets.
2D is quite difficult again; the student has to determine the second law efficiency of a complex
power cycle. In this case, energy is exchanged with an external party so the standard way to
calculate power cycle second law efficiency does not work. The student has define appropriate
system boundaries and has to come up with a method to define second law efficiency based on the
system boundaries that he/she choose. For this reason, this question has several correct answers.
2E is a question of average difficulty. The student has to break down the irreversibility of the turbine
into a part which is caused by friction and a part which is caused by heat losses. The heat loss related
irreversibility is relatively easy to calculate, the irreversibility due to friction is (almost) impossible to
, find directly (although the student has some equations which seems to give him the right answer but
cannot be used because of the shaft work). However, the student should realize that friction and
heat losses are the only processes causing irreversibility; by subtracting the heat loss term from the
total irreversibility, the student can get the irreversibility due to friction.
Memo Exam MTX311 2014
1A Find exergy of a process flow of ethylene, 40g/s, 60°C, 25 bar.
Strategy: check whether ethylene is an ideal gas
if not, find enthalpy and entropy deviation using the deviation charts
calculate exergy using control volume equation: Ψ=mψ, ψ=(h-h0)-T0(s-s0)
check if ethylene at actual conditions is an ideal gas:
T=333.15K P=25 bar
Tr=T/Tc=1.182 Pc=50.4 bar
Tc=282.4K Pr=P/Pc=0.496
check if ethylene at reference/ground state is an ideal gas:
T=298.15K P=1 bar
Tr=T/Tc=1.056 Pc=50.4 bar
Tc=282.4K Pr=P/Pc=0.019
Real gas when Tr<2 AND Pr>0.2 both hold (with quite some margin), so clearly the compressed
ehtylene is a real gas, whereas ethylene at ground state is an ideal gas.
Look up Δhd and Δsd for compressed ethylene in deviation tables: Δhd/Tc ≈3 kJ/kmol∙K and Δsd= 2
kJ/kmol∙K
Δhd/Tc =3∙282.4=847.2 kJ/kmol=847.2/28=30.26 kJ/kg
Δsd= 2/28=0.0714 kJ/kg∙K
(h-h0)= Δhig -ΔΔhd=Cp(T-T0)- ΔΔhd=1.548(60-25)-(30.26-0)=23.9kJ/kg
(s-s0)= Δsig -ΔΔsd=Cpln(T/T0)-R ln(P/P0)- ΔΔsd=1.548(333.15/298.15)-0.2964ln(25/1)-0.0714=
-0.8537kJ/kgK
ψ=(h-h0)-T0(s-s0)=23.9-298.15∙(-0.8537)=278.5 kJ/kg
Ψ=mψ=0.04*278.5=11.14 kJ
1B what is the total exergy of an empty oxygen bottle at the top of mount Everest?
Use control mass equation:
Ψ=mψ, ψ=(u-u0)-T0(s-s0)+P0(V-V0)+Epot+Ekin
Where u-u0=Cv(T-T0) and s-s0 = Cpln(T/T0) (there is no R ln(P/P0)because steel is not an ideal gas!)
Epot=gh and Ekin=½v²=0
For steel, Cv=Cp=0.9 kJ/kgK
u-u0=Cv(T-T0)=0.9(-25-25)=-45kJ/kg
