Question

A jet airplane is traveling \(184 \mathrm{~m} / \mathrm{s}\). In each second the engine takes in \(68.2 \mathrm{~m}^{3}\) of air having a mass of \(70.2 \mathrm{~kg}\). The air is used to hurn \(2.92 \mathrm{~kg}\) of fuel each second. The energy is nsed to compress the products of combustion and to eject them at the rear of the engine at \(497 \mathrm{~m} / \mathrm{s}\) relative to the plane. Find \((a)\) the thrust of the jet engine and \((b)\) the delivered power (horsepower).

Short answer

The thrust of the engine is calculated to be \(F\) newtons and the power delivered is \(P\) horsepower.

Step-by-step solution

Calculate the Momentum

Calculate the momentum of the air taken in by the engine and the momentum of the fuel combusted each second. The formula for momentum is given by mass times velocity. The total momentum intake per second can be calculated as follows: \(\Delta p_{\text{intake}} = m_{\text{air}} * v_{\text{air}} + m_{\text{fuel}} * v_{\text{fuel}}\), where \(m_{\text{air}}\) and \(m_{\text{fuel}}\) are the masses of the air and fuel respectively, and \(v_{\text{air}}\) and \(v_{\text{fuel}}\) are their respective velocities.

Calculate the Momentum Output

Calculate the momentum output from the combustion products being ejected from the engine. The momentum output per second can be calculated as follows: \(\Delta p_{\text{output}} = (m_{\text{air}} + m_{\text{fuel}}) * v_{\text{ejected}}\), where \(v_{\text{ejected}}\) is the velocity of the ejected combustion products relative to the plane.

Calculate the Thrust

Thrust is the change in momentum per second. Therefore, the thrust of the jet engine can be calculated using the formula: \(F = \Delta p_{\text{output}} - \Delta p_{\text{intake}}\). Substitute the values obtained from steps 1 and 2 into this formula to find the thrust.

Calculate the Power

Power is the work done per unit time. Work done by the thrust is equal to the force (thrust) times the distance travelled per unit time (velocity of the plane). Therefore, the power delivered by the engine can be calculated using the formula: \(P = F * v_{\text{plane}}\). Substitute the values of thrust obtained from step 3 and the given plane velocity into this formula to find the power.

Convert to Horsepower

The power obtained in step 4 will be in watts. Convert it to horsepower (hp) by dividing by 746 (since 1 hp = 746 watts).

Key concepts

Momentum

Momentum is a fundamental concept in physics and is crucial in understanding how jet engines work. It is defined as the product of an object's mass and its velocity. For the jet plane problem, we calculate momentum for the air being taken in and the fuel combusted:

• The air, with a mass of 70.2 kg, is accelerated to the plane's speed of 184 m/s.
• Fuel, while smaller in mass (2.92 kg), also contributes to the overall momentum.

Both air and fuel combine to form the total intake momentum each second. By using the formula \[\Delta p_{\text{intake}} = m_{\text{air}} \times v_{\text{air}} + m_{\text{fuel}} \times v_{\text{fuel}}\]it's clear how mass and velocity influence momentum. Understanding this helps us analyze the forces at play when the engine operates.

Thrust

Thrust is essentially the force that propels the airplane forward. It is directly related to how much the momentum changes due to the jet engine. To calculate thrust:

• First, find the momentum of the ejected combustion products, which is \'output momentum\'.
• Calculate how much momentum has increased by ejecting these products at a higher velocity than the speed of the plane.

The thrust is then the difference between the output and input momenta, given by:\[F = \Delta p_{\text{output}} - \Delta p_{\text{intake}}\]Understanding thrust helps us appreciate how efficiently the engine turns fuel and air into forward motion for the airplane.

Horsepower

Horsepower is a unit of power that is often used to quantify the power output of an engine. Once thrust is calculated, determining the power involves figuring out how much work the engine does over time.Applying the formula for power:\[P = F \times v_{\text{plane}}\]means we're multiplying thrust by the plane's velocity to see how much energy is converted into motion each second. Given the result in watts, to find horsepower (a more familiar unit in automotive and aviation circles), you convert by:\[\text{Horsepower} = \frac{P}{746}\]This conversion helps to relate jet engine output to everyday terms, making it easier for non-specialists to understand engine power.

Combustion

Combustion in a jet engine is a chemical process where fuel is burned in the air. It releases energy, which is then used to accelerate air through the engine and produce thrust. This involves:

• Intake of air to mix with fuel.
• Burning this mixture to produce high-speed exhaust gases.
• The resultant gas ejection at higher speed increases output momentum.

The combustion of 2.92 kg of fuel each second is a key factor in increasing the momentum output in the jet engine. This transformed energy pushes the airplane forward. Understanding combustion is vital since it directly affects thrust and power output, dictating how efficiently and powerfully the airplane can fly.

Jet propulsion

Jet propulsion is the method through which an airplane moves forward by expelling gas at high speed. It is rooted in Newton's Third Law of Motion: for every action, there is an equal and opposite reaction. The steps involved include:

• Air intake at high volumes and speed.
• Combustion of fuel with this air.
• Exhausting combustion products quickly at speeds higher than intake.

In the given exercise, these processes concatenate to elevate ejection velocities to 497 m/s. This helps change the overall momentum, creating the push or thrust that moves the jet forward. Grasping jet propulsion is key to both understanding modern aerodynamics and the practical application of physical laws in aircraft design.