Question

Three kilograms (3.0 kg) of saturated steam at 162oC is held in a closed piston-cylinder apparatus with a weighted piston that exerts a constant force on the steam. The piston-cylinder apparatus loses energy to the surroundings such that at a certain point 50% of the steam has condensed to a saturated liquid inside the cylinder (NOTE: the vapor and liquid are still saturated at the new condition).

a. For the process described above, determine the heat exchange with the surroundings. Assume the transition is isothermal. Identify the direction of heat transfer.

b. At the point in time where 50% of the steam has condensed to a saturated liquid inside this piston-cylinder apparatus (as above), part of the weight on the piston is removed such that the absolute pressure inside the cylinder is reduced to one-half of its former value. Assume that the saturated steam and saturated liquid in the cylinder either condense or vaporize appropriately to reach a new equilibrium condition. Assume that no heat is lost to the surroundings during this transition. Determine the mass of the saturated liquid water and the mass of saturated steam present in the cylinder at this new pressure.

Answer #1

Mass of steam = 3 kg

Temperarure of saturated steam = 162^{0}C

Mass of steam condensed = 1.5 kg

From steam table, we have

Pressure = 6.52 bar

Heat ,Q lost to the surroundings is given by

Heat has flown from piston cylinder to the outside environment

b)

Let the initial pressure be p=6.52 bar

Hence final presssure will be p/2 = 3.26 bar

From steam table for satureated steam

Temperature = 133.5^{0}C

As we know that , pressure of the system is directly proportional to mass of steam in the system

Hence mass of steam = 0.75 kg

And mass of saturated water = 1.5 + 0.75 = 2.25 kg

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