Answer:
9.75 kg/s
Step-by-step explanation:
hf = enthalpy of saturated liquid
hfg = enthalpy of saturated liquid-vapor
Turbine inlet conditions are:
V1 = 115 m/s, P1= 4 MPa, T1 = 600°C
Turbine outlet conditions are:
V1=V2= 115 m/s, x=0.92, P2= 10 kPa, Work(out) = 5MW
In order to determine the mass flow rate, the change in enthalpy(Δh), kinetic energy(Δke), potential energy(Δpe) and work done per unit mass(w) must be calculated.
From steam tables, the steam at the inlet of the turbine is at a superheated vapor state and its enthalpy at the inlet is:
P1= 4 MPa, T1 = 600°C ∴ h1 = 3674.9 kJ/kg
At the outlet of the steam turbine we have a saturated vapor-liquid mixture at pressure 10 kPa. The enthalpy at this state is:
h2 = hf + x2×hfg
From the steam tables hf=191.81 kJ/kg and hfg=2392.1 kJ/kg
∴ h2 = 191.81 + 0.92×2392.1 = 2392.542 kJ/kg
∴ Δh = h2-h1 = 2392.542 - 3674.9 = -1282.38 kJ/kg
The change in kinetic energy is the difference in velocity divided by 2:
Δke = (V2^2-V1^2)/2 = 0
The change in potential energy is defined by the gravitational force multiplied by the difference in the distance between two points. In this instance there is no difference in distance between the inlet and outlet:
Δpe = g×(z2-z1) = 0
From the energy balance for a steady-flow system:
Energy in - Energy out = Rate of change in kinetic, potential energies = 0
Ein = Eout
m(h1 + V1^2/2 +gz1) = Wout + m(h2 +V2^2/2 + gz1)
simplifies per mass: h1 = w + h2
w = h1-h2 = -Δh = 1282.38 kj/kg
m = Wout/w = 12500/(-1282.38) = 9.75kg/s