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AysviL [449]
3 years ago
14

A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment

arm is fired orthogonal to the axis of interest for 1 minute, how far has the vehicle rotated when the firing is complete?Also, what details of a real system does this question neglect?
Engineering
1 answer:
Kazeer [188]3 years ago
4 0

To solve this problem we will apply the concepts related to translational torque, angular torque and the kinematic equations of angular movement with which we will find the angular displacement of the system.

Translational torque can be defined as,

\tau = Fd

Here,

F = Force

d = Distance which the force is applied

\tau = (1N)(1m)

\tau = 1N\cdot m

At the same time the angular torque is defined as the product between the moment of inertia and the angular acceleration, so using the previous value of the found torque, and with the moment of inertia given by the statement, we would have that the angular acceleration is

\tau = I\alpha

\alpha = \frac{\tau}{I}

\alpha = \frac{1N\cdot m}{100kg\cdot m^2}

\alpha = 0.01rad/s^2

Now the angular displacement is

\theta = \omega_0 t + \frac{1}{2}\alpha t^2

Here

\omega_0= Initial angular velocity

t = time

\alpha =Angular acceleration

\theta= Angular displacement

Time is given as 1 minute, in seconds will be

t = 1m = 60s

There is not initial angular velocity, then

\theta= \frac{1}{2}\alpha t^2

Replacing,

\theta= \frac{1}{2}(0.01)(60)^2

\theta = 18rad

The question neglects the effect of gravitational force.

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<h3>What are receptacles?</h3>

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1 year ago
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2 years ago
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simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl
Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

<u>a) Air temperature at turbine exit </u>

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17  for  Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) (\frac{783.05-778.18}{800.13-778.18} ) = 764.45K

<u>b) The net work output </u>

first we determine the actual work input to compressor

Wc = h2 - h1  ( calculated values )

     = 626.57 - 295.17 =  331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4  ( calculated values )

     = 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

         = 541.88 - 331.4  = 210.48 kJ/kg

<u>c) determine thermal efficiency </u>

лth = Wnet / qin  ------ ( 1 )

where ; qin = h3 - h2

<em>equation 1 becomes </em>

лth = Wnet / ( h3 - h2 )

      = 210.48 / ( 1324.93 - 626.57 )

      = 0.3014  =  30.14%

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