A rigid tank of volume 2.5 m3 contains 5 kg of wet steam at 75 ∘C. How much heat is
required to completely convert this into dry saturated steam? What would be the final pressure and temperature?
Saturated water data: at 75 C, vf=0.001026, vg=4.1291 m3/kg, uf=313.99,
ug=2475.3 kJ/kg. Pressure table: at 350 kPa, Ts=138.86 ∘C, vg=0.52422 m3/kg,
ug=2548.5 kJ/kg; at 375 kPa, Ts=141.30 ∘C, vg=0.49133 m3/kg,
ug=2550.9 kJ/kg.
Formulas you may need
- First law: ΔU=Q−W, with W=0 for a rigid tank (on the formula sheet)
- Wet steam: v=(1−x)vf+xvg, u=(1−x)uf+xug (on the formula sheet)
- Linear interpolation: y2−y1y−y1=x2−x1x−x1 (learn this)
- (a)
Find the initial dryness fraction and specific internal energy.
- (b)
Find the final pressure and temperature when the steam is just dry saturated.
- (c)