Answer:
a.Internal temperature change is 90.49°F
b.Internal energy change is -2.43Btu/lbm
Explanation:
The energy balance for this steady-flow system can be expressed as:
[tex]\dot E_i_n-\dot E_o_u_t=\bigtriangleup \dot E_s_y_s=0\\\dot E_i_n=\dot E_o_u_t[/tex]
Therefore,
[tex]\dot mh_1=\dot mh_2\\h_1=h_2[/tex]
#Since:[tex]\dot Q \approx =\dot W=\bigtriangleup ke\approx =\bigtriangleup pc\approx =0[/tex]
From tableA-1
[tex]p_2=120pisa\\x_1=0\ \ \ \ \ \ \ \ \ \ ->h_1=41.79Btu/lbm, u_1=41.49Btm/lbm,T_1=90.49\textdegree F[/tex]
[tex]p_2=20pisa\\\ \ \ \ \ \ \ \ \ \ \\h_2=h_1=41.79Btu/lbm, u_2=38.96Btm/lbm,T_2=-2.43\textdegree F\\\\\therefore \bigtriangleup T=T_2-T_1[/tex]
#Replacing the values in the equation:
[tex]\bigtriangleup T=-2.43-90.49 ->|\bigtriangleup T|=92.92\textdegree F[/tex]
We have that [tex]\bigtriangleup u=u_2-u_1\\\bigtriangleup u=38.96-41.49=\\\bigtriangleup u=-2.43Btu/lbm[/tex]
