Question

A piston-cylinder device with 15 lbm of steam is heated from a temperature and pressure of 500 °F and 350 psia to a new temperature of 600°F. The weight of the piston is supported by the pressure of the steam and moves as the temperature increases. What is the work done by the steam in this process?

Answer 1

Answer:

The work done is 1050 Btu

Solution:

As per the question:

Mass of steam, M = 15 lbm

Temperature, T = $500^{\circ}F$

Temperature, T' = $600^{\circ}F$

Pressure, P = 350 psia

Since, the process results in the change in volume at constant pressure.

Now, work done by the steam is given by;

W = $\int_{V}^{V'} P dU$

And

U = mV

So,

W = $m\int_{V}^{V'} P dV$

W = $mP(V' - V)$           (1)

Now, using the super heated vapor table:

At 350 psia and $500^{\circ}F$,  V = $1.5 ft^{3}/lb$

At 350 psia and $600^{\circ}F$,  V' = $1.7 ft^{3}/lb$

Now, using these values in eqn(1):

W = $15\times 350(1.7 - 1.5) = 1050 Btu$