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ludmilkaskok [199]

4 years ago

9

A gas contained within a piston-cylinder undergoes the follow change in states: Process 1: Constant volume from p1 = 1 bar V1 =

2.6 m3 to state 2 with p2 = 2.7 bar Process 2: Compression to V3 = 1.5 m3, which the pressure-volume relationship is pV = constant. Process 3: Constant pressure to state 4, where V4 = 0.5 m3. Sketch the processes on p-V graph and evaluate the work for each process in kJ.

1 answer:

laiz [17]4 years ago

7 0

Answer:

<u>Process 1:</u>W=0

<u>Process 2:</u>W= -386.13 KJ

<u>Process 3:</u>W= -468 KJ

Explanation:

Process 1: $P_1=1 bar,V_1=2.6m^3$

Process 2: $P_2=2.7bar,V_2=2.6m^3$

Process 3: $V_3=1.5 m^3$

$V_4=0.5 m^3$

<u>Process 1:</u>

Work (W)=0 ,because it is constant volume process.

<u>Process 2:</u>

It is constant temperature process so PV=C

$P_2V_2=P_3V_3$

$P_3=\dfrac{P_2V_2}{V_3}$

$P_3=\dfrac{2.7\times 2.6}{1.5}$

$P_3=4.68$ bar

So work in constant temperature process

W= $P_2V_2\ ln\dfrac{V_3}{V_2}$

W= $270\times 2.6\ ln\dfrac{1.5}{2.6}$ (1 bar=100KPa)

W= -386.13 KJ

Negative sign means it is compression process.

<u>Process 3:</u>

It is a constant pressure.

So work W= $P_3(V_4-V_3)$

W=468(0.5-1.5) KJ

W= -468 KJ

Negative sign means it is compression process.

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3 years ago

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Now we find the quality,

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3 years ago

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Which gives;

$560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ]$

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Answer:

hello your question lacks the required image attached to this answer is the image required

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3 years ago

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6 0

3 years ago