Steam to drive an old-fashioned steam locomotive is supplied at a constant gauge pressure of 1.75x10° N/m (about 250 psi) to a piston with a 0.200-m radius. (a) By calculating PAV, find the work done by the steam when the piston moves 0.800 m. Note that this is the net work output, since gauge pressure is used. (b) Now find the amount of work by calculating the force exerted times the distance traveled. Is the answer the same as in part (a)?

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09-12-2022
Physics

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Steam to drive an old-fashioned steam locomotive is supplied at a constant gauge pressure of 1.75x10° N/m (about 250 psi) to a piston with a 0.200-m radius. (a) By calculating PAV, find the work done by the steam when the piston moves 0.800 m. Note that this is the net work output, since gauge pressure is used. (b) Now find the amount of work by calculating the force exerted times the distance traveled. Is the answer the same as in part (a)?

Answer :

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qwcherry qwcherry · Answer 1 · 2022-12-18T08:00:12-05:00

The work done by the steam when the piston moves 0.800 m is 1.75×10⁵ J and the amount of work by calculating the force exerted times the distance traveled is 1.85×10⁵ J. From this, the work done by the second part is more than the initial one. So they are not the same.

The work produced by the compression or expansion of fluid is known as pressure-volume work. Gas molecules gain kinetic energy when energy is added, which causes the gas to expand and affect its surroundings. The work is calculated using the formula W=pΔV where p is pressure, W is work done and ΔV is the change in the volume.

Given the gauge pressure (Pg) of the steam locomotive is 1.75×10⁶ N/m², the radium of piston r is 0.200 m, and the distance moved by the piston l is 0.800 m.

a) First, calculate the ΔV.

ΔV = area×distance

ΔV = πr²×l

ΔV = π×(0.20)²×0.80

ΔV = 0.1 m³

Then, the work done will be,

W = 1.75×10⁶× 0.1

W = 1.75×10⁵ J

The required answer is 1.75×10⁵ J.

b) The absolute pressure is given by the sum of gauge pressure and the atmospheric pressure, i.e., P=P₀+Pg. Then,

P = 1.01×10⁵ + 1.75×10⁶

P = 18.5×10⁵ Pa.

The force is calculated as,

F = P×A

F = 18.5×10⁵×π×(0.20)²

F = 232477 N

The work done by the piston is calculated as,

W = Force×distance

W = 232477×0.80

W = 1.85×10⁵ J

The required answer is 1.85×10⁵ J.

The answer for the first and second parts are not the same.

To know more about pressure-volume work:

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