Question

The net power output of an ideal regenerative-reheat steam cycle is 80MW. Steam enters the h.p. turbine at 80 bar, 500°C and expands till it becomes saturated vapour. Some of the steam then goes to an open feedwater heater and the balance is reheated to 400°C, after which it expands in the I.p. turbine to 0.07 bar. Compute (a) the reheat pressure, (b) the steam flow rate to the h.p. turbine, and (c) the cycle efficiency. Neglect pump work.

Answer 1

From S.T of 80 bar 500°C

h₁ = 3398.3 kJ/kg

s₁ = 6.724 kJ/kg – K

s₂ = 6.725 at 6.6 bar so

Reheat pr. 6.6 bar

∴ h₂ = 2759.5 kJ/kg

h₃ = 3270.3 + 0.6(3268.7 – 3270.3) = 3269.3 kJ/kg 

s₃ = 7.708 + 0.6 (7.635 – 7.708) 

= 7.6643 kJ/kg – K

At 0.07 bar 

h_f = 163.4, h_fg = 2409.1

h_f = 0.559, s_fg = 7.71

∴ If quality is x then 7.6642 = 0.559 + x × 7.717

⇒ x = 0.9207

∴ h₄ = 163.4 + 0.9207 × 2409.1 = 2381.5 kJ/kg

h₅ = 163.4 kJ/kg ≈ h6, 

h₇ = 686.8 kJ/kg ≈h8

∴ From Heat balance of heater m × h₂ + (1 – m) h₆ = h₇

∴ m = 0.2016 kg/kg of steam at H.P

∴ (1 – m) = 0.7984

W_T = h₁ – h₂ + (1 – m) (h₃ – h₄) = 1347.6kJ/kg

W_P neglected

Q = (h₁ – h₈) + (1 – m) (h₃ – h₂) = 3118.5 kJ/kg at H.P

(a) Reheat pr. 6.6 bar

(b) Steam flow rate at H.P = \(\frac{80 \times 10^3}{1347.6}\) kg/s = 59.36 kg/s

(c) Cycle efficiency (η) = \(\frac WQ = \frac{1347.6}{3118.5}\) × 100% = 43.21%