Old spec L5 worked example 2 - heating wet steam in a rigid tankTutorialOld spec ENGR2172:212 min

ENGR217 Lecture 5 steam tables worked example 2 (slides 12-16)

A rigid tank of volume 2.5 m32.5\ \mathrm{m^3} contains 5 kg of wet steam at 75 ∘C75\ ^\circ\mathrm{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=0.001026v_f = 0.001026, vg=4.1291 m3/kgv_g = 4.1291\ \mathrm{m^3/kg}, uf=313.99u_f = 313.99, ug=2475.3 kJ/kgu_g = 2475.3\ \mathrm{kJ/kg}. Pressure table: at 350 kPa, Ts=138.86 ∘CT_s = 138.86\ ^\circ\mathrm{C}, vg=0.52422 m3/kgv_g = 0.52422\ \mathrm{m^3/kg}, ug=2548.5 kJ/kgu_g = 2548.5\ \mathrm{kJ/kg}; at 375 kPa, Ts=141.30 ∘CT_s = 141.30\ ^\circ\mathrm{C}, vg=0.49133 m3/kgv_g = 0.49133\ \mathrm{m^3/kg}, ug=2550.9 kJ/kgu_g = 2550.9\ \mathrm{kJ/kg}.

Formulas you may need
  • First law: ΔU=Q−W\Delta U = Q - W, with W=0W = 0 for a rigid tank (on the formula sheet)
  • Wet steam: v=(1−x)vf+xvgv = (1 - x)v_f + xv_g, u=(1−x)uf+xugu = (1 - x)u_f + xu_g (on the formula sheet)
  • Linear interpolation: y−y1y2−y1=x−x1x2−x1\dfrac{y - y_1}{y_2 - y_1} = \dfrac{x - x_1}{x_2 - x_1} (learn this)
  1. (a)
    Find the initial dryness fraction and specific internal energy.
  2. (b)
    Find the final pressure and temperature when the steam is just dry saturated.
  3. (c)
    Find the heat required.