Students have made a device capable to operate in a mode that is close to the ideal Brayton cycle. T — Physical Chemistry — Thermodynamics Chemistry Question
Brayton cycle
Students have made a device capable to operate in a mode that is close to the ideal Brayton cycle. This thermodynamic cycle has once been proposed for development of internal combustion engines. The device consists of a cylinder with 1 mole of helium fitted with a computer-controlled movable piston. A Peltier element which can heat or cool the gas is mounted in the cylinder wall. The device can operate in the following modes:
1) reversible adiabatic expansion or compression, 2) reversible isobaric cooling or heating.
Through a number of cooling and compression steps, helium is going from the initial state with the pressure of 1 bar and the temperature of 298 K into the final state with the pressure of 8 bar and the temperature of 298 K. (The total number of cooling and compression stages can be from two up to infinity).
What is the minimum work that should be done on the gas for this? Compare this value to the work during a reversible isothermal compression.
Model Answer
As can be seen from the figure, there are many possible ways to go from point A (1 bar, 298 K) to point B (8 bar, 298 K) using only adiabatic and isobaric segments. The work W is equal to the area under the path. It is clear that W is minimal if we complete the process in two stages: isobaric cooling and then adiabatic compression.
We will derive a general formula to calculate the work of transformation from (p1, T1) to (p2, T2) in two stages. If for the reversible adiabatic process pV^(5/3) = const, then T/p^(2/5) = const. After the isobaric stage, the pressure is still p1 and the temperature is T2(p1/p2)^(2/5). The work at the adiabatic stage is W = ∆U = 3/2 R (T2 - T2(p1/p2)^(2/5)), and at the first stage W = p1(V - V1) = R(T2(p1/p2)^(2/5) - T1). In total, W = 5/2 R (T2(p1/p2)^(2/5) - T1) + 3/2 R(T2 - T2(p1/p2)^(2/5)). If T1 = T2, then W = 5/2 RT1 ((p1/p2)^(2/5) - 1). Thus, W = 5/2 × 8.314 × 298 × ((1/8)^(2/5) - 1) = -3500 J.
In a reversible isothermal compression, W = RT ln(p2/p1) = -5150 J.
What is the maximum work that can be done on the gas in this process?
Model Answer
The maximum work is done when the first stage is adiabatic and the second one is isobaric. We can use the same formula for the reverse process and obtain the work with the opposite sign.
W = 5/2 RT (1 - (p1/p2)^(2/5)) = 5/2 × 8.314 × 298 × (1 - (1/8)^(2/5)) = 8040 J.
Let the process be accomplished in three steps. At each step helium is first cooled and then compressed. At the end of each step the pressure increases twice and the temperature returns to the value of 298 K. What is the total heat removed from the gas by a Peltier element?
Model Answer
According to the first law of thermodynamics, Q = W + ∆U = W. The total work done on the gas during three steps is: W = 3 × 5/2 RT (1 - (1/2)^(2/5)) = 4500 J.
Once the gas is compressed, it is returned to the initial state (1 bar and 298 K) in two stages (heating and expansion).
What is the range of possible values of the formal efficiency η for the resulting cycle? η is the ratio of the useful work done by the gas to the amount of heat given to the gas during the heating stage.
Model Answer
The maximum efficiency is achieved when the area of the cycle is the largest, i.e. when we complete the cycle in four steps: cooling, compression, heating, expansion.
Then η = (8040 – 3500) / 8040 = 0.565. All the efficiencies from 0 to 0.565 are possible, if we go in more steps.
In one of the experiments, the gas has been compressed from 1 bar and 298 K to 8 bar and 298 K in several steps (like in question 3). At the end of each step the pressure is increased by x times and the temperature returns to 298 K. Then helium has been returned to the initial state in two stages – heating and expansion.
Theoretical value of η for this cycle is 0.379. How many steps were used?
Model Answer
The work W done on gas during cooling and compression stages can be found from equation η = (8040 – W) / 8040 = 0.379; W = 4993 J.
If the number of steps is n, then x^n = 8. Since the work at each step is the same, the total work is: W = n × 5/2 RT (1 – (1/8)^(2/5n)). After some calculations with different integer n, we find that n = 13.
In fact, Peltier elements also consume electric energy during the cooling stage. Assume that they consume as much energy as is removed from the gas.
What is the maximum possible efficiency of the considered cycle, taking into account energy consumption during cooling?
Hint: In reversible adiabatic process for helium pV^(5/3) = const. Isochoric molar heat capacity of helium is 3/2R.
Model Answer
η = (8040 – 3500) / (8040 + 3500) = 0.393.