Brayton Cycle Efficiency & Back Work Ratio

Hello! Let's dive into our next lesson on thermodynamics.

In our last session, we established the framework for the ideal Brayton cycle. We identified its four distinct processes, visualized them on P-v and T-s diagrams, and wrote down the fundamental energy equations for each component using specific enthalpy ($h$).

Today, we will build directly on that foundation. Our learning goal is to calculate the thermal efficiency and back work ratio for an ideal Brayton cycle. These two metrics are crucial for evaluating the performance of a gas turbine engine, making this a vital step in understanding the powerplants central to aerospace engineering.

1. Thermal Efficiency of the Ideal Brayton Cycle

The thermal efficiency (${\eta }_{th}$) of any heat engine is a measure of how much of the heat energy we put in gets converted into useful net work. The definition is:

${\eta }_{th}=\frac{W_{net}}{Q_{in}}$

From our last lesson, we know that for a cycle, the net work is the net heat transfer, so $W_{net}=Q_{in}-Q_{out}$. This allows us to write the efficiency as:

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

Using the enthalpy expressions for the Brayton cycle components, we have $q_{in}=h_3-h_2$ and $q_{out}=h_4-h_1$. To make our analysis more direct, we'll employ the cold-air-standard assumptions:

1. The working fluid is air, which behaves as an ideal gas.
2. The specific heats ($c_p$ and $c_v$) are constant and evaluated at room temperature.

With these assumptions, $\Delta h=c_p\Delta T$, so our efficiency equation becomes:

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

This is correct, but not very practical, as it requires knowing all four temperatures. Our goal is to express efficiency in terms of a primary design parameter of the engine: the pressure ratio, $r_p$, defined as $r_p=P_2/P_1$.

To do this, we analyze the two isentropic processes:

• Process 1-2 (Isentropic Compression): For an ideal gas, we have the relation $\frac{T_2}{T_1}={\left(\frac{P_2}{P_1}\right)}^{(k-1)/k}=r_p^{(k-1)/k}$.
• Process 3-4 (Isentropic Expansion): Similarly, $\frac{T_3}{T_4}={\left(\frac{P_3}{P_4}\right)}^{(k-1)/k}$. Since P3 = P2 and P4 = P1, we have $\frac{T_3}{T_4}={\left(\frac{P_2}{P_1}\right)}^{(k-1)/k}=r_p^{(k-1)/k}$.

From these two results, we can see that $\frac{T_2}{T_1}=\frac{T_3}{T_4}$. A little algebraic rearrangement gives us $\frac{T_4}{T_1}=\frac{T_3}{T_2}$.

Now, let's substitute this back into our temperature-based efficiency formula. We can factor $T_1$ from the numerator and $T_2$ from the denominator:

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

Since $\frac{T_4}{T_1}=\frac{T_3}{T_2}$, the terms in the parentheses are identical and cancel out, leaving:

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

Finally, we substitute the isentropic relation for $T_1/T_2$: $\frac{T_1}{T_2}={\left(\frac{P_1}{P_2}\right)}^{(k-1)/k}=\frac{1}{r_p^{(k-1)/k}}$.

This gives us the final, powerful result for the thermal efficiency of an ideal Brayton cycle:

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

Here, $k$ is the specific heat ratio ($k=c_p/c_v$), which is approximately 1.4 for air. This equation reveals a crucial insight: for an ideal Brayton cycle, the thermal efficiency depends only on the pressure ratio and the properties of the working fluid.

The following resource confirms this derivation and shows the key diagrams we discussed in the last lesson.

2. Back Work Ratio

While a high thermal efficiency is desirable, it doesn't tell the whole story. A significant portion of the work produced by the turbine is not available as net output; it must be used to drive the compressor. The back work ratio ($r_{bw}$) quantifies this.

The formula for the back work ratio is:
$r_{bw}=\frac{W_{comp,in}}{W_{turb,out}}$

Using the cold-air-standard assumptions, we can express this in terms of temperatures:
$r_{bw}=\frac{c_p(T_2-T_1)}{c_p(T_3-T_4)}=\frac{T_2-T_1}{T_3-T_4}$

For gas turbines, this ratio is typically very high, often between 40% and 80%. This is because compressing a gas requires a large amount of energy compared to pumping a liquid (as in a steam power plant, where $r_{bw}$ is only 1-2%). This high back work ratio is a defining characteristic of gas turbine engines.

3. Worked Example: Analyzing an Ideal Brayton Cycle

Let's apply these formulas to a concrete problem. This will be very similar to the analysis you'll perform for actual aerospace systems.

Problem:
Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa and 300 K. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. Assuming $k=1.4$, calculate:

1. The thermal efficiency of the cycle.
2. The back work ratio.

Solution:

1. Calculate Thermal Efficiency (${\eta }_{th}$)

This is a direct application of the formula we derived.

• Given: $r_p=10$, $k=1.4$.
• The exponent is $(k-1)/k=(1.4-1)/1.4\approx 0.2857$.
• ${\eta }_{th}=1-\frac{1}{10^{0.2857}}=1-\frac{1}{1.9307}\approx 1-0.5179$

${\eta }_{th}\approx 0.482\text{or}48.2\%$

2. Calculate Back Work Ratio ($r_{bw}$)

To find the back work ratio, we need temperatures T1, T2, T3, and T4.

• Given: $T_1=300$ K and $T_3=1400$ K.
• Find $T_2$: Using the isentropic relation for the compressor:
  $T_2=T_1\cdot r_p^{(k-1)/k}=300\cdot (10{)}^{0.2857}\approx 300\cdot 1.9307$
  $T_2\approx 579.2\text{ K}$
• Find $T_4$: Using the isentropic relation for the turbine:
  $T_4=T_3/r_p^{(k-1)/k}=1400/(10{)}^{0.2857}\approx 1400/1.9307$
  $T_4\approx 725.1\text{ K}$
• Now, calculate the back work ratio:
  $r_{bw}=\frac{T_2-T_1}{T_3-T_4}=\frac{579.2-300}{1400-725.1}=\frac{279.2}{674.9}$
  $r_{bw}\approx 0.414\text{or}41.4\%$

This means that over 41% of the work generated by the turbine is used just to power the compressor.

To see this problem solved from start to finish, you can consult the following resource.

4. The Drive for Higher Pressure Ratios

Our efficiency formula makes it clear: increasing the pressure ratio is the most direct way to increase the efficiency of an ideal Brayton cycle. This theoretical principle has been a primary driver of gas turbine development for decades.

As the video shows, engine manufacturers are in a constant race to develop materials and designs that can withstand higher pressures and temperatures, all in the pursuit of greater efficiency. Modern turbofan engines can have overall pressure ratios exceeding 50:1, a huge leap from the engines of the mid-20th century.

Conclusion

Today, we've translated the theoretical framework of the Brayton cycle into practical performance metrics. You are now equipped to perform a first-order analysis of an ideal gas turbine engine.

Key Takeaways:

• The thermal efficiency of an ideal Brayton cycle depends only on the pressure ratio ($r_p$) and the specific heat ratio ($k$): ${\eta }_{th}=1-1/r_p^{(k-1)/k}$.
• Higher pressure ratios lead to higher ideal thermal efficiency.
• The back work ratio ($r_{bw}$) is the fraction of turbine work needed to drive the compressor, and it is characteristically high for gas turbines.

Preview of the Next Lesson:
Our analysis so far has been for an ideal cycle. In reality, compressors and turbines are not perfectly efficient. In our next lesson, we will learn how to apply isentropic efficiencies to analyze the performance of real turbines and compressors in a cycle. This will allow us to create a more realistic model of engine performance.