Question

1. steam to drive an old - fashioned steam locomotive is supplied at a constant gauge pressure of ( 1.75\times10^{6} n/m^{2}) (about 250 psi) to a piston with a 0.200 - m radius. (a) by calculating ( pdelta v), 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)?

Explanation:

Step1: Calculate the area of the piston

The area of a circle is \(A = \pi r^{2}\). Given \(r = 0.200\space m\), then \(A=\pi\times(0.200)^{2}= 0.04\pi\space m^{2}\approx0.126\space m^{2}\)

Step2: Calculate \(\Delta V\) for part (a)

The volume change \(\Delta V=A\times d\), where \(d = 0.800\space m\). So \(\Delta V=0.126\times0.800 = 0.101\space m^{3}\)

Step3: Calculate work done in part (a) using \(W = P\Delta V\)

Given \(P = 1.75\times 10^{6}\space N/m^{2}\), then \(W=(1.75\times 10^{6})\times0.101 = 1.77\times 10^{5}\space J\)

Step4: Calculate force in part (b)

Force \(F = P\times A\). Substituting \(P = 1.75\times 10^{6}\space N/m^{2}\) and \(A = 0.126\space m^{2}\), we get \(F=(1.75\times 10^{6})\times0.126=2.21\times 10^{5}\space N\)

Step5: Calculate work done in part (b) using \(W = F\times d\)

Given \(d = 0.800\space m\), then \(W=(2.21\times 10^{5})\times0.800=1.77\times 10^{5}\space J\)

Answer:

a) The work done by the steam is \(1.77\times 10^{5}\space J\)
b) The work done is \(1.77\times 10^{5}\space J\), and the answer is the same as in part (a)