.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 vessel holds 2 kg water at 0.2 MPa. What are the quality, temperature, and internal energy?

Part a: The volume of vessel is 4.7680 and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

From the steam tables at 80 °C

Specific volume of saturated vapours=v_g=3.40527

Specific volume of saturated liquid=v_f=0.00102

Now the relation  of total specific volume for a specific dryness value is given as

Substituting the values give

Now the volume of vessel is given as

So the volume of vessel is 4.7680.

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680 and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

So the quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

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