Question 3.16: Estimate the dryness fraction of steam in a cylinder at 0.7 ...
Estimate the dryness fraction of steam in a cylinder at 0.7 of the stroke from the following data :
r.p.m. = 100 ; cut-off = 0.5 stroke ; steam condensed/min = 44.95 kg/min ; clearance = 8% ; swept volume = 0.1062 m³ ; pressure of steam at 0.7 stroke = 4.13 bar ; pressure of steam at 0.8 of return stroke on compression curve = 1.31 bar ; volume of 1 kg of steam at 4.13 bar = 0.438 m³ ; volume of 1 kg of steam at 1.31 bar = 1.296 m³. (N.U.)
Mass of steam used per stroke or cylinder feed
= \frac{44.95}{100} = 0.4495 kg.
Clearance volume = 0.08 V_{s} = 0.08 × 0.1062 = 0.008496 m³
Volume at M on compression curve,
V_{M} = 0.008496 + (1 – 0.8) × 0.1062 = 0.02974 m³
∴ Mass of cushion steam = \frac{V_{M} }{v} = \frac{0.02974}{1.296} = 0.02295 kg
∴ Total mass of steam during expansion stroke
= 0.4495 + 0.02295 = 0.4725 kg
Volume of steam at L = 0.008494 + 0.7 × 0.1062 = 0.08283 m³
∴ Dryness fraction at L
= \frac{Volume of steam in the cylinder at L}{Volume of steam at L , if dry } = \frac{0.08283}{0.4725 × 0.438} = 0.4.
