Question

EXAMPLE 3-9 The Use of Steam Tables to Determine Properties Determine the missing properties and the phase descriptions in the following table for water P, kPa X u, kJ/kg T, C Phase description 200 0.6 1600 1000 2950 75 500 850 125 OBOSS 0.0

Answer 1

(1) Given data: P = 200 kPa; x = 0.6

The water exists as a saturated liquid-vapor mixture. Therefore, the given pressure is the saturation pressure i.e. . Using the steam table

P= Psat = 200kPa

Psat = 200kPa + T = Tsat = 120.21°C

Psat = 200kPa u= 504.49kJ/kg; ug = 2529.1kJ/kg

u= u + x(ug-u) = 504.49 +0.6 x (2529.1 - 504.49) = 1719.26kJ/kg

(2) Given data: T = 125⁰C; u = 1600 kJ/kg

Using the steam table,

T = 125°C → up=524.82kJ/kg; ug = 2534.3kJ/kg

zu <u<uq
, water exists as a saturated liquid-vapor mixture. Therefore, the pressure is the saturation pressure at the given temperature
T = Tsat = 125°C

Tsat = 125°C → P= Psat = 232.24kPa

u= us +(4, - u) = 1600 = 524.82 + (2534.3 – 524.82).=r=0.535

(3) Given data: P = 1000 kPa; u = 2950 kJ/kg

Using the steam table,

p= 1000kPa → Uq=2582.7kJ/kg

, water exists as a superheated vapor. From the steam table (for superheated vapor)

on <n.

P= 1000kPa, T = 390°C → u= 2941.4kJ/kg

P= 1000kPa, T = 400°C → u= 2957.9kJ/kg

Using linear interpolation, we can find the temperature as follows

T – 390 390 – 400 2950 - 2941.4 2941.4 - 2957.9 = T = 395.21°C

(4) Given data: T = 75⁰C, P = 500 kPa

Tsat = 75°C → Psat = 38.6kPa

Psat = 500kPa → Tsat = 151.831°C

T<Tsat
and
PP rsat
, water exists as a subcooled liquid. From the steam table (for subcooled water)

T = 75°C, P = 500kPa u=313.89kJ/kg

In case the properties of subcooled are not available, the properties of saturated liquid at the given temperature can be taken as an approximation.

T = 75°C, → u=u= 313.99kJ/kg

(5) Given data: P = 850 kPa; x = 0

As x = 0, water exists as a saturated liquid. Therefore, the temperature is the saturation temperature at

P= Psat = 850kPa

Psat = 840kPa Tsat = 172.44°C;u= 728.84kJ/kg

Psat = 860kPa Tsat = 173.428°C; u = 733.15kJ/kg

Once again using linear interpolation

T – 172.44 172.44 – 173.428 850 - 840 = T = 172.934°C 840 - 860

u – 728.84 728,84 - 733,15 -728.84 850 - 840 840 - 860 u=730.995kJ/kg