A 1.8-m3 rigid tank contains water steam at 220oC. One-third of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure of the steam, (b) the quality of the saturated mixture, and (c) the density of the mixture.

a) 2319.6 kPa

b) 0.027

c) 287.86 kg/m^3

Explanation:

The pressure is determined from table  in the appendix for the given temperature:  

 P_220=2319.6 kPa

to calculate the quality we need to determine the masses of the vapor and the liquid and for that we need the respective specific volumes which can also be found in table.  

m_liq  = V_liq/α_liq

           = 504.2 kg

m_vap = V_vap/α_vap

            = 13.94 kg

        q = m_vap/m_liq

          = 0.027

Finally, the density is simply calculated as follows:  

        p = m_tot/V_tot

           = 518.14 kg/1.8 m^3

           = 287.86 kg/m^3

can you make it more obvious. i can't see it properly

answer:

explanation:

sbbdbdbdbdbdndjd