A cylinder with diameter D=3cm has two pistons mounted, the upper one, mp1 = 25

Question

A cylinder with diameter D=3cm has two pistons mounted, the

upper one, mp1 = 25 kg, initially resting on the stops. The lower

piston, mp2 = 5 kg, has air below it, with a spring in vacuum

connecting the two pistons. The spring force is zero when the lower

piston stands at the bottom, and when the lower piston hits the stops

the volume is 150 L. The air, initially at 20o

C, 250 kPa, and with a

volume V = 50 L, receives 200 kJ of heat from an external source. Find the Final state of the process and what if the amount of heat received by the system is 120 kJ.

Explanation / Answer

Initial pressure of 250 kPa results in some compression of spring.

Area A = pi/4 *d^2 = 3.14 / 4 * 0.03^2 = 0.0007065 m^2

Force = pressure*area

Force = 250*10^3 *0.0007065

Force = 176.625 N

Initial position of lowr piston = Voume / area

= (50*10^-3) / 0.0007065

= 70.77 m

Spring stiffness k = Force / initial compression

k = 176.625 / 70.77

k = 2.496 N/m

As heat is supplied, pressure increases and is baanced by spring reaction. This will occur till the spring reaction

= Force due to piston + atm pressure

= 5*9.81 + 25*9.81 + 10^5 *0.0007065

= 364.95 N

This wil result when spring compression s = 364.95 / 2.496 = 146.2 m

So, volume will be 0.0007065*146.2 = 0.1033 m^3 = 103.3 L

Pressure = 364.95 / 0.0007065 = 516.56 kPa

From here on, it'll follow constant pressure process until piston hits the stops.

Work done in first phase = 1/2*(p1 + p2)*(v2 - v1)

= 1/2 * (250 + 516.56)*(103.3 - 50)

= 20428 J = 20.4 kJ

Work done in second phase = p2*(v3 - v2)

= 516.56*(150 - 103.3)

= 24123 J = 24.1 kJ

Total work done = 20.4 + 24.1 = 44.5 kJ

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