Problem 10 - Aircraft Propulsion

Question

A liquid propellant rocket uses a hydrocarbon fuel and oxygen as propellant. The heat of reaction for the combustion is \(Q_{\mathrm{R}}=18.7 \mathrm{MJ} / \mathrm{kg}\). The specific impulse is \(335 \mathrm{~s}\) and the flight speed is \(2500 \mathrm{~m} / \mathrm{s}\). Neglecting the propellant kinetic power at the injector plate, calculate (a) effective exhaust speed \(c\) in \(\mathrm{m} / \mathrm{s}\) (b) propulsive efficiency \(\eta_{\mathrm{p}}\) (c) overall efficiency \(\eta_{\circ}\)

Step-by-Step Solution

Answer

The effective exhaust speed is calculated by multiplying the specific impulse by gravity and the propulsive efficiency is determined using the formula where exhaust speed and flight speed are taken into account. The overall efficiency is calculated using the formula which incorporates the flight speed and the heat of reaction per unit mass.

Step 1: Calculation of Effective Exhaust Speed

The effective exhaust speed, \(c\), can be calculated by multiplying the specific impulse, \(I_{sp}\), by the acceleration due to gravity, \(g\). Known values: - \(I_{sp} = 335 \, s\)- \(g = 9.81 \, m/s^2\)\(c = g \cdot I_{sp}\)

Step 2: Calculation of Propulsive Efficiency

The propulsive efficiency, \(\eta_p\), can be calculated using the formula: \(\eta_p = \frac{2}{1 + \frac{c}{v}}\) where \(c\) is the effective exhaust speed and \(v\) is the flight speed. Given that the flight speed is \(v = 2500 \, m/s\), we need to substitute these values into the formula.

Step 3: Calculation of Overall Efficiency

The overall efficiency, \(\eta_0\), can be calculated using formula: \(\eta_0 = 0.5 \cdot \frac{v^2}{Q_R}\) where \(v\) is the flight speed and \(Q_R\) is the heat of reaction per unit mass. Here, the heat of reaction is given as \(Q_R = 18.7 \, MJ/kg = 18700 \, kJ/kg = 18700 \times 10^3 \, J/kg\). Substitute these values into the formula.

Key Concepts

Effective Exhaust Speed

In rocket science, the effective exhaust speed (\(c\)) is a pivotal concept. It helps us understand how well the rocket engine converts fuel into thrust. Essentially, this speed is the velocity at which the exhaust gases leave the rocket engine. The formula we use to determine the effective exhaust speed is:
\[c = g \, I_{sp}\]
where \(g\) is the acceleration due to gravity, approximately \(9.81 \, m/s^2\) on Earth's surface, and \(I_{sp}\) is the specific impulse measured in seconds.

• The specific impulse represents how efficiently the rocket uses its propellant.
• By multiplying \(I_{sp}\) by \(g\), we obtain \(c\), the effective exhaust speed.

In our example, with a specific impulse of \(335 \, s\), the effective exhaust speed computes to:
\[c = 9.81 \, m/s^2 \times 335 \, s = 3284.35 \, m/s\]
This value reflects the speed at which exhaust gases exit the rocket, impacting the thrust and overall performance.

Propulsive Efficiency

Propulsive efficiency, denoted as \(\eta_p\), indicates how efficiently a rocket converts the energy from propulsion into useful work, like moving forward. The formula to calculate this efficiency is:
\[\eta_p = \frac{2}{1 + \frac{c}{v}}\]
Here, \(c\) is the effective exhaust speed and \(v\) is the rocket's flight speed. This relationship tells us:

• If \(c\) equals \(v\), the efficiency is 100% since all energy is effectively used for propulsion.
• Differences between \(c\) and \(v\) lead to losses and thus lower efficiency.

With our given values, substituting \(c = 3284.35 \, m/s\) and \(v = 2500 \, m/s\), we calculate:
\[\eta_p = \frac{2}{1 + \frac{3284.35}{2500}} = \frac{2}{1 + 1.31374} = 0.6054\]
This means approximately 60.54% of the thrust is effectively used to move the rocket. This metric is crucial for designing efficient rockets!

Overall Efficiency

Overall efficiency, symbolized as \(\eta_0\), is about how well a rocket system converts stored chemical energy (from fuel) into mechanical energy for flight. The equation to find this efficiency is:
\[\eta_0 = 0.5 \cdot \frac{v^2}{Q_R}\]
where \(v\) represents flight speed, and \(Q_R\) is the heat of reaction, or energy, released per kilogram of fuel. \(Q_R\) is provided in megajoules, converted to joules for calculations:
\[Q_R = 18.7 \, MJ/kg = 18.7 \times 10^6 \, J/kg\]
Plugging in the numbers:

• \(v = 2500 \, m/s\)
• \(Q_R = 18.7 \times 10^6 \, J/kg\)

Calculate:
\[\eta_0 = 0.5 \cdot \frac{(2500)^2}{18.7 \times 10^6} = 0.5 \cdot \frac{6250000}{18.7 \times 10^6} \approx 0.167\]
Thus, the overall efficiency is about 16.7%, demonstrating how much chemical energy is converted to kinetic energy. Optimizing this efficiency is key for rocket propulsion advancements.