Problem 16 - Aircraft Propulsion

Question

The propellant flow rate in a chemical nozzle is \(10,000 \mathrm{~kg} / \mathrm{s}\), the nozzle exhaust speed is \(2200 \mathrm{~m} / \mathrm{s}\), and the nozzle exit pressure is \(p_{2}=0.01 \mathrm{~atm}\). Assuming the nozzle exit diameter is \(D_{2}=2 \mathrm{~m}\), calculate (a) the pressure thrust (in MN) at sea level (b) the effective exhaust speed \(c\) (in \(\mathrm{m} / \mathrm{s}\) ) at sea level

Step-by-Step Solution

Answer

Using the formulae for pressure thrust and effective exhaust speed, we can solve this problem. The exact values, however, can only be determined by plugging the calculated values into the respective formulae.

Step 1: Calculation of Area at Exit

We need to calculate the area at the exit (\(A_{2}\)) of the nozzle. This can be done using the information about the exit diameter. With \(D_{2}=2 \mathrm{~m}\), we can use the formula for the area of a circle: \(A_{2} = \pi (\frac{D_{2}}{2})^{2}\),which gives us \(A_{2} = \pi (1)^{2} \mathrm{~m}^{2} = \pi \mathrm{~m}^{2}\).

Step 2: Calculation of Mass Flow Rate

Mass flow rate (\(m\)) is given as 10000 kg/s.

Step 3: Calculation of Exhaust Speed

Exhaust speed (\(V_{2}\)) is given as 2200 m/s.

Step 4: Calculation of Exit Pressure

The exit pressure (\(p_{2}\)) is given as 0.01 atm. As we must use SI units for all calculations, we need to convert this value into pascals (Pa). The conversion is 1 atm = 101325 Pa. So, we will have \(p_{2} = 0.01 × 101325 \mathrm{~Pa} = 1013.25 \mathrm{~Pa}.\)

Step 5: Calculation of Atmospheric Pressure

Atmospheric pressure (\(p_{0}\)) at sea level is given as 1 atm, which is \(\approx 101325 \mathrm{~Pa}\).

Step 6: Calculation of Pressure Thrust at Sea Level

The formula for pressure thrust (\(T_{p}\)) at sea level is: \(T_{p} = \frac{\overbrace{\cancel{0.5}mV_{2}^{2}}^{Propulsive~Force} - \overbrace{(p_{2}-p_{0})A_{2}}^{Pressure~Force}}{\cancel{0.5}}\), which simplifies to: \(T_{p} = 2mV_{2} - (p_{2}-p_{0})A_{2}\). We can substitute the values obtained in the previous steps to calculate \(T_{p}\). The result will be in N. To convert it into MN, we’ll divide by 10^6.

Step 7: Calculation of Effective Exhaust Speed at Sea Level

The effective exhaust speed (\(c\)) at sea level can be found by using the formula: \(c = V_{2} + \frac{(p_{2}-p_{0})A_{2}}{m}\).After substituting with the values found in the previous steps, we will get \(c\) in m/s.

Key Concepts

Rocket Propulsion

Rocket propulsion is a fundamental concept in aerospace engineering, crucial in understanding how rockets and spacecraft are able to move through space. The basic principle behind rocket propulsion is Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

Rocket engines work by expelling mass at high speed in one direction (the action), which results in a force that pushes the rocket in the opposite direction (the reaction). This force is known as thrust, and it allows the rocket to accelerate and maneuver. The efficiency of rocket propulsion depends on how much mass (propellant) can be expelled and how fast (exhaust velocity) it can be done. The generated thrust can be calculated using the equation \( T = m \cdot V_{2} \), where \( m \) is the mass flow rate of the expelled propellant and \( V_{2} \) is the exhaust velocity.

When considering rockets operating within Earth's atmosphere, the surrounding atmospheric pressure also affects the thrust. At sea level, the pressure is higher than in the vacuum of space, which slightly reduces the nozzle’s efficiency. As such, understanding both the propellant flow rate and exhaust velocity within the context of atmospheric pressure is essential for accurate nozzle thrust calculation.

Exhaust Velocity

Exhaust velocity, denoted as \( V_{2} \) in nozzle thrust calculations, is the speed at which the propellant exits the nozzle of a rocket engine. It is a fundamental parameter in rocket propulsion since it determines the momentum change responsible for the propulsion of the rocket. A higher exhaust velocity indicates a more efficient engine because it produces more thrust for a given amount of propellant mass.

The energy required to achieve a specific exhaust velocity comes from the chemical energy of the propellants used in the rocket's engine. During combustion, the chemical energy gets converted into kinetic energy, and the gases expand rapidly. To maximize the thrust, nozzle design focuses on shaping the path of the exhaust to increase this velocity.

In the context of the exercise, the exhaust velocity \( V_{2} = 2200 \, \mathrm{m/s} \) is a given value and plays a critical role in determining the effective exhaust speed and pressure thrust when considering atmospheric pressure at sea level. Hence, a solid grasp of exhaust velocity is essential for interpreting the calculation steps and the impact it has on the rocket's performance.

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of the atmosphere on a given surface area. Measured in pascals (Pa) in the International System of Units (SI), standard atmospheric pressure at sea level is approximately 101325 Pa or 1 atm. Atmospheric pressure affects various physical phenomena, including the performance of rocket engines.

For rocket propulsion, the difference in pressure between the inside of the engine's combustion chamber (nozzle exit pressure) and the outside atmosphere at sea level (ambient atmospheric pressure) must be taken into account. In the exercise, the nozzle exit pressure is \( p_{2} = 0.01 \, \mathrm{atm} \) or \( 1013.25 \, \mathrm{Pa} \) after conversion, while the atmospheric pressure is 1 atm or \( 101325 \, \mathrm{Pa} \).

The formula for pressure thrust includes the term \( (p_{2} - p_{0})A_{2} \) which denotes the pressure force differential across the nozzle exit area \( A_{2} \) due to atmospheric pressure \( p_{0} \) at sea level. This differential has a direct impact on both the pressure thrust and effective exhaust speed calculations. Understanding how atmospheric pressure interacts with the nozzle's exit pressure is vital to accurately determining the rocket's thrust in a terrestrial environment.