Question

Fill out the following table for substance water:

	P [kPa]	T [°C]	v [m3/kg]	X
(a)	500	20	0.001002	......
(b)	500	.......	0.20	0.532
(c)	1400	200	0.14302	.......
(d)	8581	300	........	0.8

Answer 1

(a)  P > P_sat so it is compressed liquid

(b)  v_f < v < v_g so two phase L + V

(c) Only one of the two look-up is needed 

200°C P < P_sat = => superheated vapor 

1400 kPa T > T_sat = 195°C 

subtable for 1400 kPa gives the state properties 

(d)  since quality is given it is two-phase

v = v_f + x × v_fg = 0.001404 + 0.8 × 0.02027 = 0.01762 m³/kg

	P [kPa]	T [°C]	v [m3/kg]	X
(a)	500	20	0.001002	Undefined
(b)	500	151.86	0.20	0.532
(c)	1400	200	0.14302	Undefined
(d)	8581	300	0.01762	0.8