Question

A reversible polytropic process, begins with steam at p₁ = 10 bar, t₁ = 200°C, and ends with p₂ = 1 bar. The exponent n has the value 1.15. Find the final specific volume, the final temperature, and the heat transferred per kg of fluid.

Answer 1

p₁ = 10 bar = 1000 kPa

p₂ = 1 bar = 100 kPa

t₁ = 200º C = 473 K

From Steam Table

V₁ = 0.206 m³/s

h₁ = 2827.9 kJ/kg

As at 1 bar v_g = 1.694 m³/kg

∴ then steam is wet

∴ 1.5256 = 0.001043 + x (1.694 – 0.001043)

∴ x = 0.9

Final temperature = 99.6º C

= u₁ - u₂

= (h₁ - h₂) - (p₁v₁ - p₂v₂)

= (2827.9 – 2450.8) – (1000 × 0.206 – 100 × 1.5256)

= 323.7 kJ/kg

[h₂ = h_f2 + x h_fg2]

= 417.5 + 0.9 × 2257.9

= 2450.8

Work done (W) = \(\frac{p_1v_1 - p_2v_2}{n -1}\)

= 356.27 kJ/kg

∴ From first law of thermo dynamics Q₂ = (u₂ -  u₁) + W_1–2

= (–323.7 + 356.27)

= 32.567 kJ/kg