Question

A rigid, well-insulated tank contains wet steam (20:80 wt% liquid:vapor) at 200^◦C. The wet steam is withdrawn from the tank until the fluid remaining in the tank is at 150^◦C. The process is carried out slowly such that internal gradients are negligible inside the tank. Determine at the end of this process:

(a) The pressure in the tank.

(b) The mass fraction of vapor and liquid in the tank.

(c) The fraction of total water (liquid and vapor) present initially that was withdrawn.

Answer 1

Let the wt of the content in the tank be 1 kg

initially, at 200⁰C, for saturated vapour,

Pressure in the tank is 15.54 bar

If wt of the tank is 100kg then 0.80 kg is vapor 0.20kg is liquid

from the steam tables the specific volume for the two states are:

vf = 1.156 x 10-3 m3/kg,

vg = 0.127 m3/kg

Therefore the volume of each phase must equal the mass of the phase " multiplied " by its specific volume:

Vf = 0.20 x 1.156 x 10-3 = 0.0002312 m3

Vg = 0.80 x 0.127 = 0.1016 m3

total Volume = 10.16 + 0.02312 = 0.10183 m3

At 150⁰C,

Pressure in the tank is 4.758 bar

If volume of the tank is 0.10183 m3 then let x wt % is liquid [1-x] wt % is vapor

from the steam tables the specific volume for the two states are:

vf = 1.091 x 10-3 m3/kg,

vg = 0.393 m3/kg

let x wt% of liq is available

Therefore the volume of the phase be:

Vf = x*1.091 x 10-3 m3

Vg = [1-x]*0.393 m3

total volume = 0.10183 m3 =  x*1.091 x 10-3 +  [1-x]*0.393

x = 0.74

0.74 wt% of liquid is available and 0.26 wt% of vapor is available at 150⁰C