Five kilograms of H2O are contained in a closed rigid tank at an initial pressure of 20 bar and a quality of 50%. Heat transfer occurs until the tank contains only saturated vapor. Determine the volume of the tank, in m^3 , and the final pressure, in bar. Also, show the process on both a P-v and T-v diagram.

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Answer:

The volume of tank is 0.25191 [tex]m^3[/tex] and the final pressure is 38.476 bar.

Explanation:

mass=m=5 kg

Pressure 1=P_1=20 bar

quality=x=50% =0.5

From the steam tables  for Pressure of 20 bar and the quality as 0.5

v_f=0.001177 m^3/kg

v_g=0.099587 m^3/kg

Now the specific volume is given as

[tex]v=v_f+x(v_g-v_f)\\v=0.001177+0.5(0.099587-0.001177)\\v=0.0504 m^3/kg[/tex]

So the Total volume is given as

[tex]v=V/m\\V=vm\\V=0.05040\times 5\\V=0.25191 m^3[/tex]

As the tank is rigid so its volume will remain same and the volume is 0.25191 m^3.

For the final pressure as the tank is rigid

v_1=v_2=0.0504 m^3/kg and saturated vapor table gives

P_sat=38.476 bar.

So the volume of tank is 0.25191 [tex]m^3[/tex] and the final pressure is 38.476 bar.

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