Question regarding steam in thermodynamic process, anyone know how to do?
10At 20 bar absolute, the total enthalpy of 2 kg of wet steam is found to be 4500 kJ.
Determine, using steam tables, the
(a) temperature of the steam in oC,
(b) dryness fraction of the steam,
(c) total internal energy of the steam in kJ.
2)Steam with a dryness fraction of 0.8 and a pressure of 2 bar absolute is contained in a
closed rigid vessel of volume 1 m3. Heat is then added until the pressure reaches 4
bar absolute. Determine, using steam tables, the
(a) specific volume of the steam (in m3/kg),
(b) final temperature of the steam (in oC),
(c) mass of the steam (in kg).
(Note: Closed rigid vessel implies constant volume and constant mass)
- schmisoLv 77 years agoFavorite Answer
For wet stream, i.e. a mixture of boiling water and water vapor, temperature and pressure are mutually dependent. All you need to is o search the saturated steam table for a boiling pressure of 20 bar abs. From online resource  I get:
T = 212.38 °C
The wet steam has a specific enthalpy of:
h = 4500 kJ / 2 kg = 2250 kJ∙kg⁻¹
The specific enthalpy of wet steam with dryness fraction x is given by:
h = (1 - x)∙h_f + x∙h_g = h_f + x∙(h_g - h_f)
h_f and h_g are the enthalpies of saturated liquid water and saturated water vapor at this temperature/pressure.
From the same source is get for saturated water at 20 bar:
h_lf = 908.50 kJ∙kg⁻¹
h_g = 2798.3 kJ∙kg⁻¹
∆hv = (h_g - h_f) = 1889.8 kJ∙kg⁻¹
When solve the relation above for x you get:
x = (h - h_f)/(h_g - h_f)
= (2250kJ∙kg⁻¹ - 980.5kJ∙kg⁻¹)/(1889.8 kJ∙kg⁻¹)
Internal energy and enthalpy are related as:
H = U + P∙V
U = H - P∙V = H - P∙m∙v
The formula for the specific volume (and all other specific properties) of wet steam is similar to that for its specific enthalpy:
v = (1 - x)∙v_f + x∙v_g = v_f + x∙(v_g - v_f)
From saturated steam table at 20 bar:
v_f = 1.1768×10⁻³ m³∙kg⁻¹
v_g = 99.585×10⁻³ m³∙kg⁻¹
So the steam has specific volume of:
v = 1.1768×10⁻³ m³∙kg⁻¹ + 0.672 ∙(99.585×10⁻³ m³∙kg⁻¹ - 1.1768×10⁻³ m³∙kg⁻¹)
= 67.28×10⁻³ m³∙kg⁻¹
U = H - P∙m∙v
= 4500 kJ - 20×10² kPa ∙ 2kg ∙ 67.28×10⁻³ m³∙kg⁻¹
= 4231 kJ
(note that kPa∙m³ = kJ)
Neither volume of the vessel nor the mass of water enclosed changes, that means the specific volume of water in the vessel is constant. Initially the vessel contained wet steam with x=0.8 and 2 bar.
From saturated steam table you get for wet steam ar 2 bar:
v_f = 1.0605×10⁻³ m³∙kg⁻¹
v_g = 885.68×10⁻³ m³∙kg⁻¹
v = (1 - x)∙v_f + x∙v_g
= (1 - 0.8)∙1.0605×10⁻³ m³∙kg⁻¹ + 0.8∙885.68×10⁻³ m³∙kg⁻¹
= 708.77×10⁻³ m³∙kg⁻¹
= 0.70877 m³∙kg⁻¹
After heating up the steam has the same specific volume.
Saturated steam at 4 bar has specific water vapor volume of
v_g = 0.46238 m³∙kg⁻¹
Since the specific volume of wet steam cannot exceed the specific volume of dry saturated steam, the steam must be in superheated state after the heating. When you search superheated steam table  for
P = 2bar abs
v = 0.70876 m³∙kg⁻¹
you find a temperature of
T = 345.61 °C
v = V/m
m = V/v = 1 m³ / 0.70876 m³∙kg⁻¹ = 1.411 kgSource(s):  http://www.efunda.com/materials/water/steamtable_s...  http://www.spiraxsarco.com/resources/steam-tables/...