Question

Steam enters an adiabatic turbine operating at steady state with an enthalpy of 3251.0 kJ/kg and leaves as a saturated mixture at 15 kPa with quality (dryness fraction) 0.9. The enthalpies of the saturated liquid and vapour at 15 kPa are h_­f = 225.94 kJ/kg and h_g = 2598.3 kJ/kg respectively. The mass flow rate of steam is 10 kg/s. Kinetic and potential energy changes are negligible. The power output of the turbine in MW is:
1. 6.5
2. 8.9
3. 9.1
4. 27.0

Answer 1

Correct Answer - Option 2 : 8.9

Concept:

Enthalpy of the saturated mixture is given by:

h = h_f + x (h_g – h­_f)

where h_f = enthalpy of saturated liquid, h_g = enthalpy of saturated vapour, x = dryness fraction

Calculation:

Given:

Mass flow rate of steam = 10 kg/s

Enthalpy of steam at turbine entrance (h₁) = 3251 kJ/kg

At the exit, the pressure is 15 kPa, x = 0.9

At P₂ = 15 kPa, h_f = 225.94 kJ/kg, h_g = 2598.3 kJ/kg

Let the enthalpy of the saturated mixture is h_2.

h₂ = h_f + x (h_g – h­_f)

h₂ = 225.94 + 0.9(2598.3 – 225.94)

h₂ = 2361.064 kJ/kg

Power output of turbine can be calculated by:

P = ṁ(h₁ – h₂)

P = 10 × (3251 – 2361.064)

P =  8899.36 kW = 8.89 MW ≃ 8.9 MW