A rigid insulated tank is initially divided into two sections by a partition. One side contains...

A rigid insulated tank is initially divided into two sections by a partition. One side contains 1.0 kg of saturated-liquid water initially at 6.0 MPa, and the other side is evacuated. The partition is broken, and the fluid expands into the entire container. The total volume is such that the final equilibrium pressure is 3.0 MPa. Determine a) the initial volume of the saturated liquid in liters, b) the total volume of the entire tank in liters,

Expert Solution

We have to use saturated steam tables. then we can find online or used a thermodynamics reference book. If we use this to the specific volumes of both the liquid and the vapor. Specific volume says how much space (volume) a phase of matter will occupy per mass.
The table gives value:

A Specific Volume of Steam (v_g): 0.0318556 m3/kg at 6.0 Mpa

Now, quality is (mass vapor/mass total). So...

1 kg of saturated liquid, This means that 1 kg are vapor and 0 kg are liquid.

Now, multiply the the mass of vapor by its specific volume and the mass of liquid by its specific volume and add to get total volume.

1kg x 0.0318556 (m³/kg) = 0.0318556 m³

a) Total Volume = 0.0318556 m³ at 6.0 MPa

simillarly,

b) half to the above volume becouse the same container was divided in to two parts.

therefore

Total Volume = 0.06447 m³ at 3.0 MPa

queen_honey_blossom answered 2 years ago

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