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vazorg [7]
3 years ago
14

A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

. Consider a reciprocating compressor running at 150 rpm . During a compression stroke, 2.00 mol of air is compressed. The initial temperature of the air is 390 K, the engine of the compressor is supplying 7.9 kW of power to compress the air, and heat is being removed at the rate of 1.1 kW.Calculate the temperature change per compression stroke.
Physics
1 answer:
Stella [2.4K]3 years ago
6 0

Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

Angular frequency = 150 rpm

Stroke = 2.00 mol

Initial temperature = 390 K

Supplied power = -7.9 kW

Rate of heat = -1.1 kW

We need to calculate the time for compressor

Using formula of compression

\terxt{time for compression}=\text{time for half revolution}

\terxt{time for compression}=\dfrac{1}{2}\times T

\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}

Put the value into the formula

\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60

\terxt{time for compression}=0.2\ sec

We need to calculate the rate of internal energy

Using first law of thermodynamics

U=Q-W

\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}

Put the value into the formula

\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)

\dfrac{\Delta U}{\Delta t}=6.8\ kW

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}

\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}

Put the value into the formula

\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}

\Delta\theta=32.48^{\circ}

Hence, The temperature change per compression stroke is 32.48°.

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Answer:

2f

Explanation:

The formula for the object - image relationship of thin lens is given as;

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Where;

s is object distance from lens

s' is the image distance from the lens

f is the focal length of the lens

Total distance of the object and image from the lens is given as;

d = s + s'

We earlier said that; 1/s + 1/s' = 1/f

Making s' the subject, we have;

s' = sf/(s - f)

Since d = s + s'

Thus;

d = s + (sf/(s - f))

Expanding this, we have;

d = s²/(s - f)

The derivative of this with respect to d gives;

d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²

Equating to zero, we have;

(2s/(s - f)) - s²/(s - f)² = 0

(2s/(s - f)) = s²/(s - f)²

Thus;

2s = s²/(s - f)

s² = 2s(s - f)

s² = 2s² - 2sf

2s² - s² = 2sf

s² = 2sf

s = 2f

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A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
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(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

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Solving for t, we find

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Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

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