Answer:1747.53KJ/kg
Explanation:
Given data
Initial Temprature[tex]\left ( T_i\right )=200^{\circ}[/tex]
Quality[tex]\left ( \chi\right )=0.25[/tex]
Final pressure[tex]\left ( P_2\right )=8MPa[/tex]
Given tank is rigid therefore specific Volume remains same
now calculating properties at [tex]200^{\circ}[/tex]
[tex]\nu _f=0.001156m^3/kg[/tex]
[tex]\nu {fg}=0.126434m^3/kg[/tex]
[tex]h_f=852.26KJ/kg[/tex]
[tex]h_{fg}=1939.8KJ/kg[/tex]
Now Specific volume of steam at [tex]200^{\circ}[/tex]
[tex]\nu _1=\nu _f+\chi \times \nu {fg}[/tex]
[tex]\nu _1=0.001156+0.25\times 0.126434[/tex]
[tex]\nu _1=0.03276m^3/kg[/tex]
[tex]h_1=h_f+\chi \times h_{fg}[/tex]
[tex]h_1=852.26+0.25\times 1939.8[/tex]
[tex]h_1=1337.21 KJ/kg[/tex]
Now [tex]\nu_1=\nu_2[/tex]
[tex]0.03276=\nu_2[/tex]
At 8 MPa and [tex]\nu =0.03276m^3/kg[/tex]
[tex]\nu _g=0.02356[/tex]
Since [tex]\nu _2\ is\ greater\ than \nu _g[/tex]
therefore final state of steam is superheated at [tex]T=381^{\circ}[/tex]
at this state [tex]h_2=3084.74 KJ/kg[/tex](obtained from steam table)
Therefore heat added
[tex]Q=h_2-h_1=3084.74-1337.21=1747.53 KJ/kg[/tex]