Question

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.

Answer 1

245°C

Step-by-step explanation:

Properties of steam at 90°C

`s_f= 1.19(KJ)/(Kg-K) , s_g=7.47(KJ)/(Kg-K)`

`v_f=0.001035(m^3)/(Kg) ,v_g=2.39(m^3)/(Kg)`

to find dryness fraction

`v=v_f+x(v_g-v_f)(m^3)/(Kg)`

v is the specific volume,
`v=(1.8)/(32.933)(m^3)/(kg)`

0.0546=0.001035+x(2.39-0.001035)

x=0.0224

When all water will covert in vapor steam then

volume
`vapor steam=0.0546(m^3)/(kg)`

`v_g=0.0546(m^3)/(kg)`

Now from steal table we can say that temperature corresponding to vapor volume
`v_g=0.0546(m^3)/(kg)` is 245°C.