Answer:
Explanation:
Given that:
At state 1:
Pressure P₁ = 20 bar
Volume V₁ = 0.03 [tex]\mathbf{m^{3}}[/tex]
From the tables at saturated vapour;
Temperature T₁ = 212.4⁰ C ; [tex]v_1 = vg_1[/tex] = 0.0996 [tex]\mathbf{m^{3}}[/tex] / kg
The mass inside the cylinder is m = 0.3 kg, which is constant.
The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg
At state 2:
Temperature T₂ = 200⁰ C
Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 [tex]\mathbf{m^{3}}[/tex] / kg
From temperature T₂ = 200⁰ C
[tex]v_f_2 = 0.0016 \ m^3/kg[/tex]
[tex]vg_2 = 0.127 \ m^3/kg[/tex]
Since [tex]vf_2 < v_2<vg_2[/tex] , the saturated pressure at state 2 i.e. P₂ = 15.5 bar
Mixture quality [tex]x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}[/tex]
[tex]x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}[/tex]
[tex]x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}[/tex]
[tex]\mathsf{x_2 =0.78}[/tex]
At temperature T₂, the specific internal energy [tex]u_f_2 = 850.6 \ kJ/kg[/tex] , also [tex]ug_2 = 2594.3 \ kJ/kg[/tex]
Thus,
[tex]u_2 = uf_2 + x_2 (ug_2 -uf_2)[/tex]
[tex]u_2 =850.6 +0.78 (2594.3 -850.6)[/tex]
[tex]u_2 =850.6 +1360.086[/tex]
[tex]u_2 =2210.686 \ kJ/kg[/tex]
At state 3:
Temperature [tex]T_3=T_2 = 200 ^0 C ,[/tex]
[tex]V_3 = 2V_1 = 0.06 \ m^3[/tex]
Specific volume [tex]v_3 = 0.2 \ m^3/kg[/tex]
Thus; [tex]vg_3 =vg_2 = 0.127 \ m^3/kg[/tex] ,
SInce [tex]v_3 > vg_3[/tex], therefore, the phase is in a superheated vapour state.
From the tables of superheated vapour tables; at [tex]v_3 = 0.2 \ m^3/kg[/tex] and T₃ = 200⁰ C
The pressure = 10 bar and v =0.206 [tex]\ m^3/kg[/tex]
The specific internal energy [tex]u_3[/tex] at the pressure of 10 bar = 2622.3 kJ/kg
The changes in the specific internal energy is:
[tex]u_2-u_1[/tex]
= (2210.686 - 2599.2) kJ/kg
= -388.514 kJ/kg
≅ - 389 kJ/kg
[tex]u_3-u_2[/tex]
= (2622.3 - 2210.686) kJ/kg
= 411.614 kJ/kg
≅ 410 kJ/kg
We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.
