| Basic Thermodynamics ~ J. Pohl © |
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Vacuum Launched Rocket
To place scientific packages into Earth orbit is expensive. A great part of the cost is the fuel consumed with lift-off. The schematic shows a system proposed as the first, lift-off stage.
Setup: The tall circular launch tube is a shaft that extends downward into a large mine with numerous accesses. The rocket, positioned in the tube snugly, is supported at the bottom of the shaft. A strong, deployable hatch covers the top of the shaft and beneath the rocket, sealing the shaft below, is an explosively destructible shield. Prior to a launch, vacuum machines slowly and economically extract air from the shaft until a pressure of 3 kPa is attained.
Fire and Lift-Off: To initiate launch the seal at the bottom of the shaft is destroyed. Atmospheric air rushes into the shaft, encounters the bottom of the rocket and shoves it upward. Air ahead of the rocket is compressed. The top cover is designed to "spring-open" once the pressure of air within the shaft above the rising rocket attains 101.3 kPa. Launch is initiated upon destruction of the bottom seal.
Will atmospheric pressure move the rocket upward?
♦ With the rocket as system, we apply Newton's Second Law of Motion, in the vertical direction. Once the bottom is removed, atmospheric pressure will act across the bottom area and 3 kPa will act over its top. If the acceleration at that instant is positive or upward, the rocket will move upward.
We have numbers for most of the terms:
Thus upon "fire" (at t = 0+) the rocket will have an upward acceleration of 45 m/s².
At what elevationposition of the rocket will the top cover deploy?
♦ We take the air in the shaft above the rocket as system. We set the elevation at the bottom of the rocket to be zero (Z = 0m). As the rocket moves upward the air trapped above it will be compressed. The sketch (right) shows the system in its initial state (1). It is likely that the initial temperature of the air equals that of its surroundings and that the sum of heats of the air would be zero.
Since compression of the air trapped above (as the rocket moves upward) will be quick. It is reasonable to assume that compression to be adiabatic (i.e. Σ Q = 0).
Although the air will experience friction during its event, we have no way (at this level of investigation) to account for that effect. To proceed toward a solution we are obliged to assume the event to be entirely frictionless for the air and that no contact friction occurs between the rocket and the shaft wall.
When the rocket attains a new position the pressure of the air above it will have attained 101.3kPa. That second state, (2) is calculated, below.
Entering numbers, we obtain:
Subject to our assumptions, when the top of the rocket is 60 meters below the exit plane of the shaft, the top cover must deploy.
Calculate the rocket upward velocity at the instant the upper cover deploys.
♦ For this calculation we must choose as system, either the rocket or the rocked and the air above it in the shaft. With the rocket as system, a calculation of work at its nose or upper boundary in contact with the "air above" must be done.
Whenever possible, we avoid calculation of work. So we select the rocket and air together as our system. The sketch (right) shows states (1) and (2) of the rocket and air. The increment form of the energy equation for these two masses is:
Before firing, the sum of heats is zero (ΣQ = 0) because everything is in thermal equilibrium. The event is fast so we assume the sum of heats for the event to be very small, (ΣQ1-2 = 0) Next we expand the terms of the energy equation and strike out the small terms,
as " /."
As the rocket moves up the shaft, there is one work effect - the pressure of atmospheric air acting over the bottom of the rocket as it moves upward. The event ends when the top of the rocket is 60 meters below the exit. We set up the work integral carefully and write the energy changes explicitly.
The temperature difference of the air can be determined. We assumed its process adiabatic and frictionless. In a previous example (Air Pistol) the following expression for the second temperature which uses the property, γ = 7/5 (for an ideal diatomic gas) was derived. . We combine these facts to calculate the second temperature of the gas.
In addition, the mass of the trapped air is needed.
The numbers are entered into the equation.
Applying the numbers carefully:
Watch the units!
That seems a rather substantial speed.. Check the numbers, please.
