kg of steam at a pressure of 17 bar and dryness 0.95 is heated at a constant pressure until it is completely dry. Determine: (i) Increase in volume (ii) Quantity of heat added.

3.0kviews

written 8.4 years ago by

teamques10 ★ 68k

• modified 5.2 years ago

thermodynamics

ADD COMMENT EDIT

1 Answer

316views

written 8.4 years ago by

teamques10 ★ 68k

Given: m = 1 kg, P = 17 bar, x = 0.95

Solution:

Now from steam tables,

$h_1 = h_f + x xh_{fg} = 871.8 + 0.95 x 1921.5 = 2697.32 kJ/kg$

$h_2 = 2793.4 kJ/kg$

Quantity of Heat added = $m × (h_2 - h_1) = 96.08 kJ$

Now from steam tables

$V_1 =xx V_g = 0.11664 × 0.95 m^3/kg$

$V_2 = V_g = 0.11664 m^3/kg$

Therefore, Increase in Volume = $m × (V_2 - V_1) = 5.832 × 10^{-3} m^3$

ADD COMMENT EDIT

Please log in to add an answer.