Respuesta :
Answer:
a) The change in Kinetic energy, KE = -1.95 kJ
b) Power output, W = 10221.72 kW
c) Turbine inlet area, [tex]A_1 = 0.0044 m^2[/tex]
Explanation:
a) Change in Kinetic Energy
For an adiabatic steady state flow of steam:
[tex]KE = \frac{V_2^2 - V_1^2}{2} \\[/tex].........(1)
Where Inlet velocity, V₁ = 80 m/s
Outlet velocity, V₂ = 50 m/s
Substitute these values into equation (1)
[tex]KE = \frac{50^2 - 80^2}{2} \\[/tex]
KE = -1950 m²/s²
To convert this to kJ/kg, divide by 1000
KE = -1950/1000
KE = -1.95 kJ/kg
b) The power output, w
The equation below is used to represent a steady state flow.
[tex]q - w = h_2 - h_1 + KE + g(z_2 - z_1)[/tex]
For an adiabatic process, the rate of heat transfer, q = 0
z₂ = z₁
The equation thus reduces to :
w = h₁ - h₂ - KE...........(2)
Where Power output, [tex]W = \dot{m}w[/tex]..........(3)
Mass flow rate, [tex]\dot{m} = 12 kg/s[/tex]
To get the specific enthalpy at the inlet, h₁
At P₁ = 10 MPa, T₁ = 450°C,
h₁ = 3242.4 kJ/kg,
Specific volume, v₁ = 0.029782 m³/kg
At P₂ = 10 kPa, [tex]h_f = 191.81 kJ/kg, h_{fg} = 2392.1 kJ/kg[/tex], x₂ = 0.92
specific enthalpy at the outlet, h₂ = [tex]h_1 + x_2 h_{fg}[/tex]
h₂ = 3242.4 + 0.92(2392.1)
h₂ = 2392.54 kJ/kg
Substitute these values into equation (2)
w = 3242.4 - 2392.54 - (-1.95)
w = 851.81 kJ/kg
To get the power output, put the value of w into equation (3)
W = 12 * 851.81
W = 10221.72 kW
c) The turbine inlet area
[tex]A_1V_1 = \dot{m}v_1\\\\A_1 * 80 = 12 * 0.029782\\\\80A_1 = 0.357\\\\A_1 = 0.357/80\\\\A_1 = 0.0044 m^2[/tex]
