Answer:
A) 127923.08KJ
B) 131.347°C
C) 51.083kg
Explanation:
A) This will be done in steps
STEP1: CALCULATE THE PROPERTIES OF THE STEAM IN THE INITIAL STATE:
The first stage is a superheat, because the steam is dry. Using superheat steam table. At
T₁ = 275 °C and p₁ = 15 bar
From steam table for superheated steam you get the following properties in that state:
specific volume v₁ = 0.150285m²/kg
specific enthalpy h₁ = 2978.67kJ/kg
Because the specific volume equals volume of the vessel divided by mass of water, we have;
v₁ = V/m
you can compute the mass of water vapor enclosed in the vessel:
m = V/v₁ = 10m³ / 0.150285m²/kg = 66.54 kg
STEP 2: CALCULATE THE PROPERTIES OF THE STEAM AT THE FINALE STATE:
In final state there is wet steam at
p₂ = 1.8 bar
Using the steam table for saturated steam, because the steam is wet, you get:
Saturated steam temperature T₂ = 131.347 °C
specific volume
liquid water vl₂ = 1.01071×10⁻³ m³/kg
water vapor vg₂ = 0.643419m³/kg
specific enthalpy
liquid water hl₂ = 552.141kJ/kg
water vapor hg₂ = 2721.93 kJ/kg
STEP 3: CALCULATE THE DRYNESS FRACTION:
Specific volume and enthalpy of the gas mixture are given by:
v₂ = vl₂ + (vg₂ - vl₂)∙x
h₂ = hl₂ + (hg₂ - hl₂)∙x
where x is the dryness (or vapor quality), which is the mass fraction of water vapor in the mixture:
x = m_vapor/m
During the cooling process both volume of the tank and mass of water enclosed does not change. Hence the specific volume of water is the same
v₁ = v₂
v₁ = vl₂ + (vg₂ - vl₂)∙x
So the dryness fraction in final state is:
x = (v₁ - vl₂) / (vg₂ - vl₂)
= (0.150285 m²/kg - 1.071×10⁻³ m³/kg) / (0.643419 m³/kg - 1.071×10⁻³ m³/kg)
= 0.232295
STEP 4: CALCULATE THE SPECIFIC ENTHALPY OF THE WET STEAM:
h₂ = hl₂ + (hg₂ - hl₂)∙x
= 552.141kJ/kg + (2721.93kJ/kg - 552.141kJ/kg) × 0.2322945
= 1056.171kJ/kg
STEP 5: CALCULATE THE HEAT REMOVED FROM THE TANK.
The heat removed from the tank equals mass water times change in specific enthalpy:
Q = m∙∆h = m∙(h₁ - h₂) =
= 66.54kg × (2978.67 kJ/kg - 1056.171kJ/kg)
= 127923.08kJ
Therefore the heat removed from the tank is 127923.08kJ
B) From the saturated steam table, the temperature of the wet steam is;
T₂ = 131.347°C
C) From the calculation above, the fraction of water vapor is x. So the fraction of liquid water is (1 - x).
Therefore the mass of liquid water which has condensed form the super hated steam in initial state will be;
ml = (1 - x)∙m
= (1 - 0.232295) × 66.54kg
= 51.083 kg
