A vessel of 0.25 m3 capacity is filled with saturated steam at 1500 kPa. If the vessel is cooled until 25% of the steam has condensed, how much heat is transferred, and what is the final pressure?

Question

contestada

Respuesta :

lucianoangelini92 lucianoangelini92 · Answer 1 · 2019-12-13T17:48:54+01:00

Answer:

the final pressure is P=353.5 kPa and the heat lost by the system (heat transferred) is Q= -3614.7327 kJ ( negative sign means heat outflow)

Explanation:

since the steam is at saturated state at 1500 kPa , from tables of saturated steam

at P= 1500 kPa → specific volume v= 0.13225 m³/kg , ug=2593.3 kJ/kg

thus

m = V/v = 0.25 m³ / 0.13225 m³/kg = 1.89 kg

for condensation until 25% of steam, then mass of steam is

mg₂ = 0.25* 1.89 kg = 0.4725 kg

if we neglect the volume occupied by the liquid , then the steam occupies

v₂ = V/m = 0.25 m³ / 0.4725 kg =0.529 m³/kg

returning to the saturated steam table

at v₂=0.529 m³/kg → 353.5 kPa , uL₂= 584.1585 kJ/kg , ug₂= 2548.4965 kJ/kg

then from the first law of thermodynamics

ΔU= Q - W , but since V=constant , dV=0 and W=∫PdV =0

Q= ΔU

Q= (mg₂*ug₂+ml₂*uL₂) - (m*ug) = 1.89 kg* (0.25*2548.4965 kJ/kg + 0.75*584.1585 kJ/kg - 2593.3 kJ/kg) = -3614.7327 kJ