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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.

Respuesta :

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

[tex]\terxt{time for compression}=\text{time for half revolution}[/tex]

[tex]\terxt{time for compression}=\dfrac{1}{2}\times T[/tex]

[tex]\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}[/tex]

Put the value into the formula

[tex]\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60[/tex]

[tex]\terxt{time for compression}=0.2\ sec[/tex]

We need to calculate the rate of internal energy

Using first law of thermodynamics

[tex]U=Q-W[/tex]

[tex]\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}[/tex]

Put the value into the formula

[tex]\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)[/tex]

[tex]\dfrac{\Delta U}{\Delta t}=6.8\ kW[/tex]

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

[tex]\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}[/tex]

[tex]\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}[/tex]

Put the value into the formula

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

[tex]\Delta\theta=32.48^{\circ}[/tex]

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

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