Answer to Question #263452 in Mechanical Engineering for Rekteat

Question

Water at 2.5 Mpa and 200°C is heated at constant temperature up to a

quality of 80%. Find (a) the quantity of heat received by the water, (b) the

change in internal energy, and (c) the work of nonflow process. (HINT:

table 4 needed)

Answer 1

Given T₁=200°C, P₁=2.5MPa

From the steam tables, at P₁=2.5MPa and T₁=200°C

u₁=849.9 kJ/kg

h₁=852.8kJ/kg

At superheated temp,

T₂=200°C, u_fg=1743.7

x=0.8, h_fg=1739.8

u₂=u₁+xu_fg

u₂=849.9+0.8*1743.7

=2242.42kJ/kg

h₂=h₁+xh_fg

h₂=852.8+0.8*1739.8

=2404.1kJ/kg

(a) Q=h₂-h₁

=2404.1-852.8

=1551.3kJ/kg

(b) ∆u=u₂-u₁

=2242.42-849.9

=1396.5kJ/kg

(c) Q=∆u+W

From which W=Q-∆u

=1551.3-1386.5

=154.8kJ/kg