Answer:
v₂ = 35.57 m/s
Explanation:
Given :
Inlet steam pressure, P₁ = 1 MPa
Inlet steam temperature, T₁ = 260°C
Inlet velocity of steam, V₁ = 30 m/s
Outlet steam pressure, P₂ = 0.3 MPa
Outlet steam temperature, T₂ = 160°C
Now,
From steam table at pressure 1 Mpa and temperature 260°C, enthalpy, h₁ = 2964.8 kJ/kg
From steam table at pressure 0.3 Mpa and temperature 160°C, enthalpy, h₂ = 2782.14 kJ/kg
Therefore, for an open system from 1st law of thermodynamics, we get
Energy in = Energy out
E₁ = E₂
=
2 X 632.66
v₂ = 35.57 m/s
Therefore, outlet velocity, v₂ = 35.57 m/s
