Brayton Cycle Efficiency and Back Work Ratio

How Good is the Brayton Cycle? Deriving Thermal Efficiency

Remember the fundamental goal of any heat engine: to take in heat (from burning fuel) and convert as much of it as possible into useful work. The measure of success for this conversion is thermal efficiency, which we've always defined as:

${\eta }_{th}=\frac{\text{What you get}}{\text{What you pay for}}=\frac{W_{net}}{Q_{in}}$

Also, from the First Law applied to a full cycle, we know the net work is the difference between the heat you add and the heat you reject: $W_{net}=Q_{in}-Q_{out}$. Substituting this gives the familiar form:

${\eta }_{th}=\frac{Q_{in}-Q_{out}}{Q_{in}}=1-\frac{Q_{out}}{Q_{in}}$

To make this formula useful for the Brayton cycle, we need to find expressions for $Q_{in}$ and $Q_{out}$.

Calculating Heat In and Heat Out

Let's analyze the heat transfers on a per-mass basis (using lowercase $q$):

1. Heat Addition ($q_{in}$): This happens in the combustor, which we model as a constant pressure process (2 → 3). For a steady-flow process at constant pressure, the heat added is equal to the change in enthalpy. Assuming our air is an ideal gas with constant specific heat, this becomes:
   $q_{in}=h_3-h_2=c_p(T_3-T_2)$
   We use $c_p$ (specific heat at constant pressure) because the process occurs at constant pressure.

2. Heat Rejection ($q_{out}$): This happens in the cooler, which is also a constant pressure process (4 → 1). The heat rejected is:
   $q_{out}=h_4-h_1=c_p(T_4-T_1)$

Now we can substitute these into our efficiency equation:

${\eta }_{th}=1-\frac{c_p(T_4-T_1)}{c_p(T_3-T_2)}$

The $c_p$ terms cancel out, which is convenient! This leaves us with an expression based only on the temperatures at the four corners of our cycle:

${\eta }_{th}=1-\frac{T_4-T_1}{T_3-T_2}$

While correct, this formula isn't very practical for an engineer. You don't directly set the four temperatures; you design the machine with a certain pressure ratio and choose materials that can withstand a certain maximum temperature. We need to express the efficiency in terms of those design parameters.

Rewriting Efficiency in Terms of Pressure Ratio

This is where our knowledge of isentropic processes becomes critical. Let's define the pressure ratio, $r_p$, as the ratio of the high pressure to the low pressure:

$r_p=\frac{P_2}{P_1}$

Since the heat addition (2-3) and rejection (4-1) are at constant pressure, $P_3=P_2$ and $P_4=P_1$, so $r_p=P_3/P_4$ as well.

Now, let's look at the two isentropic processes:

• Process 1 → 2 (Isentropic Compression): We know the relation $\frac{T_2}{T_1}={\left(\frac{P_2}{P_1}\right)}^{\frac{k-1}{k}}$. So, $\frac{T_2}{T_1}=(r_p{)}^{\frac{k-1}{k}}$.
• Process 3 → 4 (Isentropic Expansion): Similarly, $\frac{T_3}{T_4}={\left(\frac{P_3}{P_4}\right)}^{\frac{k-1}{k}}$. So, $\frac{T_3}{T_4}=(r_p{)}^{\frac{k-1}{k}}$.

Notice something amazing? We have $\frac{T_2}{T_1}=\frac{T_3}{T_4}$. Let's rearrange this to $\frac{T_4}{T_1}=\frac{T_3}{T_2}$. This is a hidden symmetry in the Brayton cycle!

Now, let's go back to our efficiency formula and do some clever algebra:
${\eta }_{th}=1-\frac{T_4-T_1}{T_3-T_2}$
Let's factor out $T_1$ from the numerator and $T_2$ from the denominator:
${\eta }_{th}=1-\frac{T_1(T_4/T_1-1)}{T_2(T_3/T_2-1)}$
Because of the symmetry we just found ($\frac{T_4}{T_1}=\frac{T_3}{T_2}$), the terms in the parentheses are identical! They cancel out.

This leaves us with a beautifully simple result:
${\eta }_{th}=1-\frac{T_1}{T_2}$
And since we know $\frac{T_2}{T_1}=(r_p{)}^{\frac{k-1}{k}}$, we can flip this to get $\frac{T_1}{T_2}=\frac{1}{(r_p{)}^{\frac{k-1}{k}}}$.

Substituting this in gives us the final, powerful formula for the thermal efficiency of an ideal Brayton cycle:

${\eta }_{th,Brayton}=1-\frac{1}{(r_p{)}^{\frac{k-1}{k}}}$

This is a massive insight! For an ideal Brayton cycle, the efficiency depends only on the pressure ratio ($r_p$) and the nature of the gas (its specific heat ratio, $k$, which is about 1.4 for air). The higher the pressure ratio—the more you squeeze the air in the compressor—the more efficient the engine is. This is why jet engine manufacturers are in a constant race to develop compressors that can achieve ever-higher pressure ratios.

To solidify this derivation, I recommend reading through the analysis provided by MIT's Thermodynamics course.

The "Company Tax": Understanding the Back Work Ratio

The Brayton cycle has a unique characteristic that we need to talk about. The turbine at the back of the engine produces a huge amount of work. But, not all of that work is available to power a generator or produce thrust. A very large portion of it has to be immediately "reinvested" to power the compressor at the front.

Imagine a company where the star salesperson (the turbine) brings in $100,000 in revenue. But to make those sales, they need a huge marketing budget and a support team (the compressor) that costs $50,000. The company's net profit is only $50,000. The $50,000 spent on marketing is like the work the turbine sends "back" to the compressor.

This leads us to the concept of the Back Work Ratio (BWR).

The Back Work Ratio, $r_{bw}$, is formally defined as the ratio of the compressor work to the turbine work:

$r_{bw}=\frac{W_C}{W_T}$

On a per-mass basis, this is:
$r_{bw}=\frac{w_C}{w_T}=\frac{h_2-h_1}{h_3-h_4}=\frac{c_p(T_2-T_1)}{c_p(T_3-T_4)}=\frac{T_2-T_1}{T_3-T_4}$

Why is this important?
For gas turbines, the BWR is very high, typically ranging from 40% to 80%. This means that for every 10 units of energy the turbine produces, 4 to 8 of those units are immediately consumed just to keep the cycle running by powering the compressor.

This is because compressing a gas is very energy-intensive. Think about it: you're trying to force trillions of energetic, fast-moving air molecules into a much smaller space. It takes a lot of effort. This is fundamentally different from a steam power cycle (like in a coal or nuclear plant), where you pump a liquid (water). Pumping a dense liquid requires very little work compared to compressing a low-density gas. As a result, steam power plants have back work ratios of only 1-2%.

This high back work ratio is a defining feature and a major engineering challenge of gas turbines.

Putting It All Together: Worked Problems

Let's solidify these concepts with some calculations.

Worked Problem 1: Basic Efficiency Calculation

An ideal Brayton cycle uses air ($k=1.4$) and has a pressure ratio of 12. Calculate its thermal efficiency.

• Goal: Find ${\eta }_{th}$.
• Formula: ${\eta }_{th}=1-\frac{1}{(r_p{)}^{\frac{k-1}{k}}}$
• Given: $r_p=12$, $k=1.4$
• Step 1: Calculate the exponent.
  $\frac{k-1}{k}=\frac{1.4-1}{1.4}=\frac{0.4}{1.4}\approx 0.2857$
• Step 2: Calculate the pressure ratio term.
  $(r_p{)}^{\frac{k-1}{k}}=(12{)}^{0.2857}\approx 2.106$
• Step 3: Calculate the efficiency.
  ${\eta }_{th}=1-\frac{1}{2.106}=1-0.4748\approx 0.525$
• Answer: The thermal efficiency of the cycle is approximately 52.5%.

Worked Problem 2: Full Cycle Analysis and Back Work Ratio

An ideal Brayton cycle with a pressure ratio of 10 operates with air entering the compressor at 290 K and entering the turbine at 1400 K. For air, use $k=1.4$ and $c_p=1.005\text{kJ/kg⋅K}$. Find:
a) The temperatures at the end of compression ($T_2$) and expansion ($T_4$).
b) The work of the compressor and the work of the turbine per kg of air.
c) The back work ratio.
d) The net work output per kg of air.

• Given: $r_p=10$, $T_1=290\text{K}$, $T_3=1400\text{K}$, $k=1.4$, $c_p=1.005\text{kJ/kg⋅K}$.

• a) Find $T_2$ and $T_4$:

  • Step 1: Find $T_2$.
    $\frac{T_2}{T_1}=(r_p{)}^{\frac{k-1}{k}}=(10{)}^{\frac{0.4}{1.4}}=(10{)}^{0.2857}\approx 1.931$
    $T_2=T_1\times 1.931=290\text{K}\times 1.931=559.9\text{K}$
  • Step 2: Find $T_4$.
    $\frac{T_3}{T_4}=(r_p{)}^{\frac{k-1}{k}}\approx 1.931$
    $T_4=\frac{T_3}{1.931}=\frac{1400\text{K}}{1.931}=725.0\text{K}$

• b) Find Compressor Work ($w_C$) and Turbine Work ($w_T$):

  • Step 3: Find $w_C$.
    $w_C=c_p(T_2-T_1)=1.005(559.9-290)=1.005\times 269.9=271.2\text{kJ/kg}$
  • Step 4: Find $w_T$.
    $w_T=c_p(T_3-T_4)=1.005(1400-725.0)=1.005\times 675.0=678.4\text{kJ/kg}$

• c) Find the Back Work Ratio ($r_{bw}$):

  • Step 5: Calculate BWR.
    $r_{bw}=\frac{w_C}{w_T}=\frac{271.2}{678.4}\approx 0.40$
    So, the back work ratio is 40%.

• d) Find the Net Work Output ($w_{net}$):

  • Step 6: Calculate Net Work.
    $w_{net}=w_T-w_C=678.4-271.2=407.2\text{kJ/kg}$

• Answer:
  a) $T_2\approx 560\text{K}$, $T_4\approx 725\text{K}$.
  b) Compressor work is 271.2 kJ/kg, Turbine work is 678.4 kJ/kg.
  c) Back work ratio is 40%.
  d) Net work output is 407.2 kJ/kg.

Conclusion

Fantastic job today! We've added a crucial layer of quantitative analysis to our understanding of the Brayton cycle. We didn't just learn that it works; we learned how to measure how well it works.

Key Takeaways:

• The thermal efficiency of an ideal Brayton cycle is given by ${\eta }_{th}=1-1/(r_p{)}^{\frac{k-1}{k}}$.
• This means efficiency is primarily driven by the pressure ratio ($r_p$). Higher pressure ratios lead to higher efficiency.
• The Back Work Ratio ($r_{bw}=W_C/W_T$) is the fraction of the turbine's work that must be used to power the compressor.
• Gas turbines have a characteristically high back work ratio (40-80%) because compressing a gas is an energy-intensive process.

Preview of the Next Lesson

We've now completed our individual investigations of the three major ideal gas power cycles: Otto, Diesel, and Brayton. You understand the mechanics, diagrams, and performance metrics for each. In our next lesson, we'll bring it all together. We will focus on solving comparative problems for different ideal gas power cycles, learning how to evaluate which cycle is better under different constraints and for different applications. This will sharpen your engineering judgment and consolidate everything we've learned in this module.