Steam enters a turbine at 4 mpa, 600°c and 115 m/s. it exits at 10 kpa, with a quality of 92% and the same velocity as the inlet. the power output from the turbine is 12,500 kw. determine: a)- the mass flow rate through the turbine, in kg/s. b)-the outlet area, in m2

9.75 kg/s

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

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