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Air contained within a piston–cylinder assembly undergoes three processes in series: Process 1–2: Compression during which the pressure–volume relationship is pV = constant from p1 = 10 lbf/in.2, V1 = 4 ft3 to p2 = 50 lbf/in.2 Process 2–3: Constant volume from state 2 to state 3 where p = 10 lbf/in.2 Process 3–1: Constant pressure expansion to the initial state. Sketch the processes in series on p–V coordinates. Evaluate (a) the volume at state 2, in ft3, and (b) the work for each process, in Btu. 2.26 WP A 0.15-m-d

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DayyanKhan

Answer:

The solution to this question is attached below:

Ver imagen DayyanKhan
Ver imagen DayyanKhan
Ver imagen DayyanKhan

The volume at state 2 is 0.80 ft^3 and the work done for each of the process are -12.572kJ, 0 kJ, 6.2493kJ

Data;

  • p1 = 10 lb f/in^2 = 68.947 kPa
  • v1 = 4 ft^3 = 0.1133 m^3
  • p2 = 50 lb f/in^2 = 344.737 kPa
  • p3 = 10 lb f/in^2 =68.947 kPa

The volume at state 2

[tex]p_1v_1=p_2v_2[/tex]

Let's substitute the values and solve for v2

[tex]p_1v_1 = p_2v_2\\69.947*0.1133 = 344.737*v_2\\v_2 = \frac{68.947*0.1133}{344.737}\\ v_2 = 0.02266m^3\\v_2 = 0.80 ft^3[/tex]

The work for each process

for process 1-2

[tex]w_1_-_2 = p_1v_1 * \ln[\frac{p_1}{p_2]}\\ w_1_-_2= 68.947*0.1133*\ln[\frac{68.9947}{344.737]}= -12.572kJ[/tex]

The negative sign indicates compression process.

for process 2-3

The volume of this process is constant

[tex]W_2_-_3 = 0[/tex]

for process 3-4

[tex]W_3_-_4 = p(v_1 - v_2)\\W_3_-_4= 68.947(0.1133-0.02266)\\W_3_-_4 = 6.2493kJ\\[/tex]

Learn more on work done in a piston-cylinder here;

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