21.3.3 The Brayton Cycle

The heat engines of Examples 21.1 and 21.2 have been educational but not realistic. As an example of a more realistic heat engine we’ll look at the thermodynamic cycle known as the Brayton cycle. It is a reasonable model of a gas turbine engine. Gas turbines are used for electric power generation and as the basis for jet engines in aircraft and rockets. The Otto cycle, which describes the gasoline internal combustion engine, and the Diesel cycle, which, not surprisingly, describes the diesel engine, will be the subject of homework problems.

A photograph of a jet engine is shown. — A jet engine uses a modified Brayton cycle.

FIGURE 21.14a is a schematic look at a gas turbine engine, and FIGURE 21.14b is the corresponding pV diagram. To begin the Brayton cycle, air at an initial pressure $p_{1}$ is rapidly compressed in a compressor. This is an adiabatic process, with $Q = 0$ , because there is no time for heat to be exchanged with the surroundings. Recall that an adiabatic compression raises the temperature of a gas by doing work on it, not by heating it, so the air leaving the compressor is very hot.

The figure labeled ("a") illustrates a gas turbine engine following a Brayton cycle.

The figure labeled (b) illustrates a pressure versus volume graph. The graph represents a cyclic process.

Full Alternative Text

The hot gas flows into a combustion chamber. Fuel is continuously admitted to the combustion chamber where it mixes with the hot gas and is ignited, transferring heat to the gas at constant pressure and raising the gas temperature yet further. The high-pressure gas then expands, spinning a turbine that does some form of useful work. This adiabatic expansion, with $Q = 0$ , drops the temperature and pressure of the gas. The pressure at the end of the expansion through the turbine is back to $p_{1}$ , but the gas is still quite hot. The gas completes the cycle by flowing through a device called a heat exchanger that transfers heat energy to a cooling fluid. Large power plants are often sited on rivers or oceans in order to use the water for the cooling fluid in the heat exchanger.

This thermodynamic cycle, called a Brayton cycle, has two adiabatic processes—the compression and the expansion through the turbine—plus a constant-pressure heating and a constant-pressure cooling. There’s no heat transfer during the adiabatic processes. The hot-reservoir temperature must be $T_{H} \geq T_{3}$ for heat to be transferred into the gas during process $2 \to 3$ . Similarly, the heat exchanger will remove heat from the gas only if $T_{C} \leq T_{1}$ .

The thermal efficiency of any heat engine is

η = \frac{W_{out}}{Q_{H}} = 1 - \frac{Q_{C}}{Q_{H}}

Heat is transferred into the gas only during process $2 \to 3$ . This is an isobaric process, so $Q_{H} = n C_{P} Δ T = n C_{P} (T_{3} - T_{2})$ . Similarly, heat is transferred out only during the isobaric process $4 \to 1$ .

We have to be careful with signs. $Q_{41}$ is negative because the temperature decreases, but $Q_{C}$ was defined as the amount of heat exchanged with the cold reservoir, a positive quantity. Thus

Q_{C} = | Q_{41} | = | n C_{P} (T_{1} - T_{4}) | = n C_{P} (T_{4} - T_{1})

(21.11)

With these expressions for $Q_{H}$ and $Q_{C}$ , the thermal efficiency is

η_{Brayton} = 1 - \frac{T_{4} - T_{1}}{T_{3} - T_{2}}

(21.12)

This expression isn’t useful unless we compute all four temperatures. Fortunately, we can cast Equation 21.12 into a more useful form.

You learned in Chapter 19 that $p V^{γ} = constant$ during an adiabatic process, where $γ = C_{P} / C_{V}$ is the specific heat ratio. If we use $V = n R T / p$ from the ideal-gas law, $V^{γ} = {(n R)}^{γ} T^{γ} p^{-}^{γ} . {(n R)}^{γ}$ is a constant, so we can write $p V^{γ} = constant$ as

p^{(1 - γ)} T^{γ} = constant

(21.13)

Equation 21.13 is a pressure-temperature relationship for an adiabatic process. Because ${(T^{γ})}^{1 / γ} = T$ , we can simplify Equation 21.13 by raising both sides to the power $1 / γ$ . Doing so gives

p^{(1 - γ) / γ} T = constant

(21.14)

during an adiabatic process.

Process $1 \to 2$ is an adiabatic process; hence

p_{1}^{(1 - γ) / γ} T_{1} = p_{2}^{(1 - γ) / γ} T_{2}

(21.15)

Isolating $T_{1}$ gives

T_{1} = \frac{p_{2}^{(1 - γ) / γ}}{p_{1}^{(1 - γ) / γ}} T_{2} = {(\frac{p_{2}}{p_{1}})}^{(1 - γ) / γ} T_{2} = {(\frac{p_{max}}{p_{min}})}^{(1 - γ) / γ} T_{2}

(21.16)

If we define the pressure ratio $r_{p}$ as $r_{p} = p_{max} / p_{min}$ , then $T_{1}$ and $T_{2}$ are related by

T_{1} = r_{p}^{(1 - γ) / γ} T_{2}

(21.17)

The algebra of getting to Equation 21.17 was a bit tricky, but the final result is fairly simple.

Process $3 \to 4$ is also an adiabatic process. The same reasoning leads to

T_{4} = r_{p}^{(1 - γ) / γ} T_{3}

(21.18)

If we substitute these expressions for $T_{1}$ and $T_{4}$ into Equation 21.12, the efficiency is

\begin{array}{l} η_{B} = 1 - \frac{T_{4} - T_{1}}{T_{3} - T_{2}} = 1 - \frac{r_{p}^{(1 - γ) / γ} T_{3} - r_{p}^{(1 - γ) / γ} T_{2}}{T_{3} - T_{2}} = 1 - \frac{r_{p}^{(1 - γ) / γ} (T_{3} - T_{2})}{T_{3} - T_{2}} \\ \begin{matrix} = & 1 - \end{matrix} r_{p}^{(1 - γ) / γ} \end{array}

Remarkably, all the temperatures cancel and we’re left with an expression that depends only on the pressure ratio. Noting that $(1 - γ)$ is negative, we can make one final change and write

η_{B} = 1 - \frac{1}{r_{p}^{(γ - 1) / γ}}

(21.19)

FIGURE 21.15 is a graph of the efficiency of the Brayton cycle as a function of the pressure ratio, assuming $γ = 1.40$ for a diatomic gas such as air.

The figure illustrates an eta subscript Brayton versus r subscript p graph.

FIGURE 21.15 Full Alternative Text

21.3.3 The Brayton Cycle

FIGURE 21.14 A gas turbine engine follows a Brayton cycle.

FIGURE 21.15 The efficiency of a Brayton cycle as a function of the pressure ratio $r_{p}$ .