Question

Determine the phase or phases in a system consisting of H2O at the following conditions and sketch p-v and t-v diagram showing the location of each state

(a)p=10 bar, T=179.9 degree C.
(b)p=10 bar, T=150 degree C.
(c)T=100 degree C, p=0.5 bar
(d)T=20 degree C, p=50 bar
(e)p=1 bar, T=-6 degree

Answer 1

Step-by-step explanation:

We will use the water property tables at respective states. Using Tables A-3, A-2, and A-5.

Part a

P = 10 bar and T = 179.9

From tables we find that the T saturated is also 179.9; Hence, the H20 exist in a two phase liquid-vapor mixture. T = Tsat@P

For p-v and T-v see attachment 1

Part b

P = 10 bar and T = 150

From tables we find that the T saturated is 179.9; Hence, the H20 exist in the sub-cooled liquid region. T < Tsat@P

For p-v and T-v see attachment 2

Part c

P = 0.5 bar and T = 100 C

From Temperature tables we find that the P saturated is 1.014 bar; Hence, the H20 exist in the super-heated region. P < Psat@T. Table A-2

For p-v and T-v see attachment 3

Part d

P = 50 bar and T = 20 C

From Temperature tables we find that the P saturated is 0.02339 bar; Hence, the H20 exist in the sub-cooled liquid region. P > Psat@T. Table A-2

For p-v and T-v see attachment 4

Part e

P = 1 bar and T = -6 C

From Temperature tables we find that the P saturated is 0.003689 bar; Hence, the H20 P > Psat@T solid because T is below triple point Temperature. Table A-5

For p-v and T-v see attachment 5