Question

Two hundred litres of wet steam at 250°C has an entropy of 25 kJ/K. Determine the steam quality and mass of steam.

Answer 1

Answer:93 %

Step-by-step explanation:

Given

Volume(V)=200 L

entropy=25 KJ/K

At
`T_(sat)=250 ^(\circ)`

Properties are

`s_f=2790 J/kg-K`

`s_g=6070 J/kg-K`

`\rho _g=20 kg/m^3`

`\rho _f=799 kg/m^3`

Entropy at \eta quality is given by

`(S)/(m)=s_f+\chi s_(fg)`

`(25)/(m)=2790+\chi \left ( 6070-2790\right )`

`(25)/(m)=2790+\chi \left ( 3280\right )----1`

Also for mass

`\left ( (m)/(V)\right )^(-1)=\rho _f^(-1)+\chi \left ( \rho _g^(-1)-\rho _f^(-1)\right )`

`\left ( (m)/(0.2)\right )^(-1)=799^(-1)+\chi \left ( 20^(-1)-799^(-1)\right )`

`\left ( (0.2)/(m)\right )=1.25* 10^(-3)+\chi \left ( 0.04875\right )---2`

From 1 and 2

we get

m=4.3 kg

`\chi =0.93` or 93%