Respuesta :
Answer:
The amount of shaft work the turbine cam do per second is 3660.29 kJ
Explanation:
The given parameters are;
Pressure at entry p₁ = 300 kPa
The mass flow rate, [tex]\dot {m}[/tex] = 5.0 kg/sec
The initial temperature, T₁ = 175°C
Therefore;
The enthalpy at 300 kPa and 175°C, h₁ = 2,804 kJ/kg
At the turbine exit, we have;
The pressure at exit, p₂ = 100 kPa
The quality of the steam at exit, in percentage, x₂ = 60%
Therefore, for the enthalpy, h₂ for saturated steam at 100 kPa and quality 60%, we have;
h₂ = 417.436 + 0.6 × 2257.51 = 1771.942 kJ/kg
Heat loss from the turbine, [tex]h_l[/tex]= 1,500 kW
By energy conservation principle we have;
dE/dt = [tex]\dot Q[/tex] - [tex]\dot W[/tex] + ∑[tex]m_i \cdot (h_i[/tex]+ [tex]ke_i[/tex] +[tex]pe_i[/tex]) - ∑[tex]m_e \cdot (h_e[/tex] + [tex]ke_e[/tex] +[tex]pe_e[/tex] )
0 = -[tex]h_l[/tex] - [tex]\dot W[/tex] + [tex]m_i \cdot h_i[/tex] - [tex]m_e \cdot h_e[/tex]
[tex]\dot W[/tex] = [tex]\dot {m}[/tex] × (h₁ - h₂) - [tex]h_l[/tex] = 5.0×(2,804 - 1771.942) - 1500 = 3660.29 kJ/s
The rate of work of the shaft = 3660.29 kJ/s
The amount of shaft work the turbine cam do per second = 3660.29 kJ.
