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3 years ago

14

An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1000 K and a heat rejection at 400 K. During t

he heat addition the volume triples. Find the two specific heat transfers (q) in the cycle and the overall cycle efficiency

1 answer:

vivado [14]3 years ago

6 0

Answer:

W / n = - 9133 J / mol, W / n = 3653 J / mol , e = 0.600

Explanation:

The Carnot cycle is described by

$e= 1 - Q_{c} / Q_{H} = 1 - T_{c} / T_{H}$

In this case they indicate that the final volume is

V = 3V₀

In the part of the heat absorption cycle from the source is an isothermal expansion

W = n RT ln (V₀ / V)

W / n = 8.314 1000 ln (1/3)

W / n = - 9133 J / mol

During the part of the isothermal compression in contact with the cold focus, as in a machine the relation of volumes is maintained in this part is compressed three times

W / n = 8.314 400 (3)

W / n = 3653 J / mol

The efficiency of the cycle is

e = 1- 400/1000

e = 0.600

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An airplane cruises at 900 km/h relative to the air. It is flying from Denver, Colorado, due west to Reno, Nevada, a distance of

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26.5 minutes

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4 0

3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

a

The New angular speed is $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

8 0

3 years ago