/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 161 \( \mathrm{A}\) manned space cap... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 161

\( \mathrm{A}\) manned space capsule travels in level flight above the Earth's atmosphere at initial speed \(U_{0}=8.00 \mathrm{km} / \mathrm{s}\). The capsule is to be slowed by a retro-rocket to \(U=5.00 \mathrm{km} / \mathrm{s}\) in preparation for a reentry maneuver. The initial mass of the capsule is \(M_{0}=1600 \mathrm{kg}\). The rocket consumes fucl at \(m=8.0 \mathrm{kg} / \mathrm{s},\) and exhaust gases leave at \(V_{e}=3000 \mathrm{m} / \mathrm{s}\) relative to the capsule and at negligible pressure. Evaluate the duration of the retro-rocket firing needed to accomplish this. Plot the final speed as a function of firing duration for a time range \(\pm 10 \%\) of this firing time.

Short Answer

Expert verified
The duration of the retro-rocket firing needed to accomplish the slowing from initial to final speed must be calculated by understanding and applying the principle of conservation of linear momentum for the rocket and its exhaust gases, and integrating the equation between the initial and final states of speed and rocket mass.

Step by step solution

01

State the Principle of Conservation of Momentum

The rocket propels forward by the principle of conservation of linear momentum. As the rocket fuel is expelled backward with a certain momentum, the rocket gains momentum in the forward direction. The momentum the rocket gains is equal to the momentum of the exhaust gases in the opposite direction. Hence, the force due to the fuel being expelled can be written as \( F = -dm/dt * Ve \). Here, \( Ve \) is the speed of exhaust gases, and \( dm/dt \) is the rate of change of mass of the spacecraft.
02

Set Up the Equations for Rocket Motion

In the given problem, there is no external force acting on the rocket. Thus the force \( F \) gives rise to the change in velocity \( U \) of the rocket. The rate of change in velocity is given by \( F = M * dU/dt \). Therefore, \( M * dU/dt = -dm/dt * Ve \). This equation links the rate of change of mass of the rocket with the rate of change of its velocity.
03

Solve for Time

To solve for \( t \), the time of retro-rocket firing, we can integrate the above equation between the initial and final states of speed and rocket mass respectively. This gives us \( \int_{U0}^{U} M dU = - \int_{M0}^{Mf} Ve dm \). Solving this equation considering \( Mf = M0 - m*t \) (Because the mass of the rocket decreases due to fuel consumption), we find the time \( t \) required.

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Most popular questions from this chapter

A small rocket motor is used to power a "jet pack" device to lift a single astronaut above the Moon's surface. The rocket motor produces a uniform exhaust jet with constant specd, \(V_{e}=3000 \mathrm{m} / \mathrm{s},\) and the thrust is varied by changing the jet size. The total initial mass of the astronaut and the jet pack is \(M_{0}=200 \mathrm{kg}, 100 \mathrm{kg}\) of which is fuel and oxygen for the rocket motor. Find (a) the exhaust mass flow rate required to just lift off initially, (b) the mass flow rate just as the fuel and oxygen are used up, and (c) the maximum anticipated time of flight. Note that the Moon's gravity is about 17 percent of Earth's.

Air enters a duct, of diameter \(D=25.0 \mathrm{mm},\) through a well-rounded inlet with uniform speed, \(U_{1}=0.870 \mathrm{m} / \mathrm{s}\) At a downstream section where \(L=2.25 \mathrm{m},\) the fully developed velocity profile is \\[ \frac{u(r)}{U_{c}}=1-\left(\frac{r}{R}\right)^{2} \\] The pressure drop between these sections is \(p_{1}-p_{2}=1.92 \mathrm{N} /\) \(\mathrm{m}^{2}\), Find the total force of friction exerted by the tube on the air.

Viscous liquid from a circular tank, \(D=300 \mathrm{mm}\) in diameter, drains through a long circular tube of radius \(R=50 \mathrm{mm}\) The velocity profile at the tube discharge is \\[ u=u_{\max }\left[1-\left(\frac{r}{R}\right)^{2}\right] \\] Show that the average speed of flow in the drain tube is \(\bar{V}=\frac{1}{2} u_{\max } .\) Evaluate the rate of change of liquid level in the tank at the instant when \(u_{\max }=0.155 \mathrm{m} / \mathrm{s}\)

The narrow gap between two closely spaced circular plates initially is filled with incompressible liquid. At \(t=0\) the upper plate, initially \(h_{0}\) above the lower plate, begins to move downward toward the lower plate with constant speed, \(V_{0}\) causing the liquid to be squeezed from the narrow gap. Neglecting viscous effects and assuming uniform flow in the radial direction, develop an expression for the velocity field between the parallel plates, Hint: Apply conservation of mass to a control volume with the outer surface located at radius \(r\) Note that even though the speed of the upper plate is constant, the flow is unsteady. For \(V_{0}=0.01 \mathrm{m} / \mathrm{s}\) and \(h_{0}=2\) \(\mathrm{mm},\) find the velocity at the exit radius \(R=100 \mathrm{mm}\) at \(t=0\) and \(t=0.1\) s. Plot the exit velocity as a function of time, and explain the trend.

\(\mathrm{A}\) nozzle for a spray system is designed to produce a fiat radial sheet of water. The sheet leaves the nozzle at \(V_{2}=\) \(10 \mathrm{m} / \mathrm{s},\) covers \(180^{\circ}\) of arc, and has thickness \(t=1.5 \mathrm{mm}\). The nozzle discharge radius is \(R=50 \mathrm{mm}\). The water supply pipe is \(35 \mathrm{mm}\) in diameter and the inlet pressure is \(p_{1}=150 \mathrm{kPa}\) (abs). Evaluate the axial force exerted by the spray nozzle on the coupling.
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