An isolated rigid tank has a partition dividing it in half

. An isolated rigid tank has a partition dividing it in half. One half contains water at 10 MPa, 300°C. The other half is a vacuum. The partition the ruptures, allowing the water to fill the entire tank. o Given only this information, what two properties do you know at the second state? Is this enough to fully define the state? o Finding other properties at the final state only from those two properties is difficult without a computer program. The final temperature you would find is 285.95°C. What is the change in entropy of this process?

Solution

a) If we assume that the process is adiabatic, the steam enthalpy will remain constant.

Now the from steam tables the enthalpy of water @ 10 MPa, 300 ^oC is 1343.1 kJ/kg and entropy is 3.2484 kJ/(kg-K)

Sp vol is 0.001398 m³/kg

Say the initial water volume is V m³

Hence water volume = V/2

Since initially the water is very close to saturation, it is expected that the part of water will flash to steam and both water and steam wwill exist inside the vessel.

Since water expands to twice the volume the specific volume of water steam mixture will double, i.e. 2x0.001398 =0.002796 m³/kg

And as mentioned, the enthalpy will remain same = 1343.1 kJ/kg

As per thermodynamic laws, two properties are sufficient to define a pure substance. hence the state of water-steam is adequately defined after expansion.

Conclusion: Properties after expansion : enthalpy 1343.1 kJ/kg, sp. vol 0.002796 m³/kg, the thermodynamic state after expansion is properly defined.

b) With final temperature of 285.95 ^oC there exists saturated steam water mixture with the following properties from steam tables:

h_l= 1268.0 kJ/kg, h_v = 2772.4 kJ/kg, s_l = 3.1231 kJ/(kg-K), s_v = 5.8138 kJ/(kg-K)

Let x be the mass fraction of steam

The enthalpy at two states is constant.

Therefore 1268.0(1-x) + 2772.4 x = 1343.1

Or x = (1343.1-1268.0)/(2772.4-1268.0)

= 0.0499

Therefore s₂ = 3.1231 (1-0.0499) + 5.8138 x 0.0499

= 3,2574 kJ/(kg-K)

Therefore change in entropy in process 1-2 , s₂ -s₁ = 3.2574 - 3.2484

= 0.009 kJ/(kg-K)

(Ref : ASME Steam Tables)