Answer:
The valve should be closed at 1081 kPa ⇒ 11 bar
Explanation:
[tex]Q_{C_V} + m_ih_i = m_2u_2 - m_1u_1 + W_{C_V}[/tex]
[tex]Q_{C_V} = W_{C_V} = 0[/tex]
[tex]m_1 = \frac{V}{v_1} = \frac{2}{0.49927} = 4.006 kg[/tex]
[tex]m_i = m_2 - m_1 = 15 - 4.006 = 10.994 kg[/tex]
[tex]u_1 = 1057.5, h_i = 1509.9[/tex]
[tex]u_2 = m_ih_i + m_1u_1[/tex]
[tex]m_2 = \frac{[10.994*1509.9] + [4.006*1057.5]}{15} = 1389.1 kJ/kg[/tex]
[tex]v_2 = \frac{V}{m_2} = \frac{2}{15} = 0.1333 m^3/kg[/tex]
Therefore, v₂, u₂ fix state 2.
By trial and error, P₂ = 1081 kPa & T₂ = 50.4° C
1081 kPa ⇒ 11 bar