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Chapter 5: Problem 11

A rocket sled weighs 3 tons (6000 lbf) including 1 ton of fuel and rests on a level section of ground. At time \(t=0\), the solid fuel of the rocket is ignited and burns at the rate of \(150 \mathrm{lbm} / \mathrm{s}\). The exit velocity of the exhaust gas relative to the rocket is \(3500 \mathrm{ft} / \mathrm{s}\). Neglecting friction and air resistance, what is the velocity of the sled at the instant at which all the fuel is burned? Be sure to state all assumptions clearly.

Short Answer

Expert verified
Velocity of the sled when all fuel is burned is approximately 1389.85 ft/s.

Step by step solution

01

Define Given Variables and Known Values

The total weight of the rocket sled is 3 tons, equal to 6000 lbf. This includes 1 ton of fuel, which is 2000 lbf. The initial burn rate of the fuel is 150 lbm/s. The velocity of the exhaust gases, denoted as exit velocity, is 3500 ft/s.
02

Understand the Rocket Equation

The rocket equation, also known as the Tsiolkovsky rocket equation, is given by:\[ v_f = v_i + v_e \ln\left(\frac{m_i}{m_f}\right) \]where:- \(v_f\) is the final velocity,- \(v_i\) is the initial velocity (0 ft/s here, as the sled starts from rest),- \(v_e\) is the exhaust velocity of the gas (3500 ft/s here),- \(m_i\) is the initial total mass of the sled,- \(m_f\) is the final total mass of the sled.
03

Set Initial and Final Masses

The initial mass \(m_i\) is equal to the total weight before ignition divided by the gravitational acceleration to convert to mass, which is \(m_i = \frac{6000}{32.2} \approx 186.34\, \text{lbm}\). The final mass \(m_f\) includes the sled without fuel: \(m_f = \frac{4000}{32.2} \approx 124.22\, \text{lbm}\).
04

Calculate Final Velocity

Substitute the known values into the rocket equation:\[ v_f = 0 + 3500 \times \ln\left(\frac{186.34}{124.22}\right) \]Calculate the natural logarithm and multiply by the exhaust velocity:\[ \ln\left(\frac{186.34}{124.22}\right) \approx 0.3971 \]Thus, the final velocity is:\[ v_f = 3500 \times 0.3971 \approx 1389.85 \text{ ft/s} \]
05

State Assumptions

The solution assumes there is no friction or air resistance affecting the sled. Initial velocity is zero at the beginning (since the sled starts from rest). All the fuel is burned uniformly without any loss or inefficiency in the propulsion system.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tsiolkovsky Rocket Equation
The Tsiolkovsky Rocket Equation is a fundamental principle used to determine a rocket's velocity as it sheds mass by burning fuel. Here is the equation: \[v_f = v_i + v_e \ln\left(\frac{m_i}{m_f}\right)\] This formula allows us to calculate the final velocity \(v_f\) by considering the initial velocity \(v_i\), the exhaust velocity \(v_e\), and the ratio of the initial mass \(m_i\) to the final mass \(m_f\).
  • \(v_i\) is the starting velocity, assumed to be zero if the rocket starts from rest.
  • \(v_e\) is the speed at which exhaust leaves the rocket.
  • \(m_i\) and \(m_f\) represent the mass of the rocket before and after fuel burn.
This equation highlights the relationship between mass and velocity in rockets, showing how shedding mass (fuel) results in acceleration, crucial for spacecraft escape from gravitational forces.
Exhaust Velocity
Exhaust Velocity is a key factor in rocket propulsion, representing the speed at which gases leave the rocket. It significantly influences the rocket's overall velocity. In our exercise, the exhaust velocity \(v_e\) is 3500 ft/s.
  • Exhaust velocity propels a rocket forward by the principle of conservation of momentum.
  • The higher the exhaust velocity, the greater the thrust and thus the rocket's acceleration.

The exhaust velocity hinges on the rocket engine's design and fuel properties. It affects the efficiency of the rocket's thrust, making it crucial for overcoming Earth's gravity and achieving the necessary speed for space travel. Considering this, a high exhaust velocity is desirable for faster and more efficient propulsion.
Burn Rate
The Burn Rate indicates how quickly a rocket consumes its fuel, affecting both the duration of propulsion and the rate of acceleration. In the current exercise, the burn rate is given as 150 lbm/s.
  • Burn rate determines how rapidly the total mass of the rocket decreases.
  • The quicker the fuel burns, the faster the sled accelerates due to decreasing mass.

This concept is crucial in the context of Tsiolkovsky's equation because it determines the time it takes to reach a specific velocity. The burn rate, combined with the exhaust velocity, defines the 'power' of a rocket, thereby setting a pace for achieving particular velocity marks.
Initial and Final Mass
Initial and Final Mass are two critical components in the rocket equation, affecting how much velocity can be gained when fuel is burnt.
  • Initial mass \(m_i\) considers the rocket's total mass, including fuel, at launch. Here, it's around 186.34 lbm.
  • Final mass \(m_f\) reflects the mass after the fuel is burnt out, here about 124.22 lbm.

Why are these masses important? The rocket equation shows a logarithmic relationship between these masses and the velocity. As the fuel burns, leading to a lower mass, the rocket accelerates.
Mass changes dramatically affect launching and operating costs in modern rocketry. Lighter rockets can go further with less fuel. Therefore, managing these masses is crucial to optimizing a rocket's efficiency and velocity.

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Most popular questions from this chapter

A toy balloon of \(100 \mathrm{~g}\) mass is filled with air of density \(1.030 \mathrm{~kg} / \mathrm{m}^{3}\), by a small hose \(5 \mathrm{~mm}\) in diameter oriented in a vertical direction. The balloon is released, letting air escape through the hose. Neglecting friction, calculate the rate at which air escapes if the initial acceleration of the balloon is zero.

An isentropic gas moves through a horizontal pipe of constant cross-sectional area. If the inlet pressure is \(12 \mathrm{psi}\) with a velocity of \(70 \mathrm{ft} / \mathrm{s}\), and the exit pressure is \(1 \mathrm{psi}\), calculate the exit velocity of gas, given the ratio of specific heats \(k=1.65\)

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Water is moving through a nozzle at a volume rate of flow of \(3 \mathrm{~m}^{3} / \mathrm{s}\). A pinhole leak in the nozzle exists. If the velocity at a downstream area of \(8 \mathrm{~cm}^{2}\) is \(12 \mathrm{~m} / \mathrm{s}\), how much fluid is lost every 10 seconds?

A submarine moves steadily through water at a depth of \(350 \mathrm{~m}\) and a.speed of \(5 \mathrm{~m} / \mathrm{s}\). (a) Find the static pressure \(p_{\infty}\) far away from the submarine at the \(350 \mathrm{~m}\) depth \((\mathrm{kPa})\). (b) Find the pressure on the nose of the submarine \((\mathrm{kPa})\).
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