7.41. A 10.0-m3 tank contains steam at 275°C and 15.0 bar. The tank and its contents are cooled until the pressure drops to 1.8 bar. Some of the steam condenses in the process. (a) How much heat was transferred from the tank? (b) What is the final temperature of the tank contents? (c) How much steam condensed (kg)?

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

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