A rigid tank with a volume of 1.8 m3 contains 32.9333 kg of saturated liquid–vapor mixture of water at 90°C. Now the water is slowly heated. Determine the temperature at which the liquid in the tank is completely vaporized. Use data from the steam tables.

Respuesta :

Answer:

245°C

Explanation:

Properties of steam at 90°C

[tex]s_f= 1.19\frac{KJ}{Kg-K} , s_g=7.47\frac{KJ}{Kg-K}[/tex]

[tex]v_f=0.001035\frac{m^3}{Kg} ,v_g=2.39\frac{m^3}{Kg}[/tex]

to find dryness fraction

[tex]v=v_f+x(v_g-v_f)\frac{m^3}{Kg}[/tex]  

v is the specific volume,[tex]v=\dfrac{1.8}{32.933}\dfrac{m^3}{kg}[/tex]

0.0546=0.001035+x(2.39-0.001035)

x=0.0224

When all water will covert in vapor steam then

volume[tex]vapor steam=0.0546\dfrac{m^3}{kg}[/tex]

[tex]v_g=0.0546\dfrac{m^3}{kg}[/tex]

Now from steal table we can say that temperature corresponding to vapor volume [tex]v_g=0.0546\dfrac{m^3}{kg}[/tex] is 245°C.