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Lyrx [107]
4 years ago
15

NASA’s Ames research center in Mountain View, California has a centrifuge built to simulate high-g environments. It spins horizo

ntally in uniform circular motion, with the payload (possibly a human!) in a box at the end of a 29-foot arm. Its maximum rotation rate is 50 revolutions per minute (rpm). (a) What is the centripetal acceleration experienced by an object in the payload box when it is spinning at its maximum rate? (b) How many gs is this acceleration, and could a human survive it (take the maximum survivable g-force to be 8g). (c) What would the rotation rate be to produce this maximum g-force of 8g?
Engineering
1 answer:
AVprozaik [17]4 years ago
6 0

Answer:

a) a_{c}=795.06 ft/s, b) approximately 25g's, c) 28.46 rpm

Explanation:

In order to solve this problem, we must first do a drawing of the situation to better visualize it (look at attached picture).

a)

Now, the very first thing we must do is convert the frequency to angular speed, we can do this like this:

\omega = \frac{50rev}{s}*\frac{2\pi rad}{1 rev}*\frac{1min}{60s}

so we get that:

\omega=5.236 rad/s

once we have the angular speed, we can use it to find the centripetal acceleration by using the following formula:

a_{c}=\omega^{2}r

which yields:

a_{c}=(5.236rad/s)^{2}(29ft)=795.06 ft/s^{2}

b)

we can use it to figure out how many g's this represents, we know that:

1g=32.2ft/s^{2}

so, we can do the conversion

795.06ft/s^{2}*\frac{1g}{32.2ft/s^{2}}=24.69g's

so we have approximately 25 g's on the machine.

c)

In order to produce a maximum acceleration of 8g's we can do the following.

First, turn the 8g's to ft/s^{2}

8g*\frac{32.2ft/s^{2}}{1g}=257.6ft/s^{2}

next, we use the same formula, but this time we solve for the angular speed:

a_{c}=\omega^{2}r

which solves to:

\omega=\sqrt{\frac{a_{c}}{r}}

and substitute the provided data:

\omega=\sqrt{\frac{257.6ft/s^{2}}{29ft}}

which yields:

\omega=2.98rad/s

we can use this to find our frequency, like this:

f=2.98rad/s*\frac{1rev}{2\pi rad}*\frac{60s}{1min}=28.46rev/min

so the frequency the machine should have to reach an acceleration of 8g's is:

f=28.46rpm

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Knowing that v = –8 m/s when t = 0 and v = 8 m/s when t = 2 s, determine the constant k. (Round the final answer to the nearest
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dv=kt^2dt

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3 years ago
1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty per
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Answer:

temperature of first extraction 330.8°C

temperature of second extraction 140.8°C

power output=3168Kw

Explanation:

Hello!

To solve this problem we must use the following steps.

1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa

2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave

m1=m2+m3

As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s

solving

5=1+m3

m3=4kg/s

3.

we find the enthalpies and temeperatures in each of the states, using thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties

4.we find the enthalpy and entropy of state 1 using pressure and temperature

h1=Enthalpy(Water;T=T1;P=P1)

h1=3457KJ/kg

s1=Entropy(Water;T=T1;P=P1)

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4.

remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1

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5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1

h3=Enthalpy(Water;s=s1;P=P3)

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T3=Temperature(Water;P=P3;s=s1)

T3=140.8°C

6.

Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.

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W=m1(h1-h2)+m3(h2-h3)

W=5(3457-3116)+4(3116-2750)=3168Kw

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