Question

A steam power plant of 110 MW capacity is equipped with regenerative as well as reheat arrangement. The steam is supplied at 80 bar and 55°C of superheat. The steam is extracted at 7 bar for feed heating and remaining steam is reheated to 350°C, and then expanded to 0.4 bar in the L.P. stage. Assume indirect type of feed heaters. Determine : 

(i) The ratio of steam bled to steam generated, 

(ii) The boiler generating capacity in tonnes of steam/hour, and 

(iii) Thermal efficiency of the cycle. Assume no losses and ideal processes of expansion

Answer 1

The schematic arrangement of the plant is shown in  (a) and the processes are represented on h-s chart in  (b). 

Given : Capacity of plant = 110 MW ; 

t₁ = 350°C t_s at 80 bar ~− 295°C + 55°C = 350°C) 

p₂ = p₃ = 7 bar ; t₃ = 350°C ; p₄ = 0.4 bar

• Locate point 1 corresponding to the condition p₁ = 80 bar and t₁ = 350°C, on the h-s chart. 
• Locate point 2 by drawing vertical line through point 1 till it cuts the 7 bar pressure line. 
• Locate point 3 as the cross point of 7 bar and 350°C temperature line. 
• Locate point 4 by drawing vertical line through the point 3 till it cuts the 0.4 bar pressure line.

From h-s chart, we find :

h₁ = 2985 kJ/kg ; h₂ = 2520 kJ/kg ; 

h₃ = 3170 kJ/kg ; h₄ = 2555 kJ/kg. 

Also, from steam tables, we have : 

h_f2 (at 7 bar) = 697.1 kJ/kg ; h_f4 (at 0.4 bar) = 317.7 kJ/kg.

(a) Schematic arrangement of the plant

(b) diagram

(i) The ratio of steam bled to steam generated : 

Consider energy/heat balance of feed heater : 

Heat lost by m kg of steam = Heat gained by (1 – m) kg of condensed steam 

m(h₂ – h_f2) = (1 – m) (h_f2 – h_f4) 

m(2520 – 697.1) = (1 – m) (697.1 – 317.7) 

1822.9 m = (1 – m) × 379.4 

∴ m = 0.172 kg 

Amount of steam bled per kg of steam supplied to the turbine = 0.172 kg

∴ \(\cfrac{Steam \,generated}{Steam \,bled}\) = \(\cfrac1{0.172}\) = 5.814.

(ii) The boiler generating capacity : 

If m_s is the mass of steam supplied to the power plant per second, then the work developed is given by : 

m_s(h₁ – h₂) + m_s (1 – m) (h₃ – h₄) = 110 × 10³ 

or, m_s (2985 – 2520) + m_s (1 – 0.172) (3170 – 2555) = 110 × 10³ 

or, m_s (465 + 509.22) = 110 × 10³ 

∴ m_s = 112.91 kg/s or 406.48 tonnes/hour

(iii) Thermal efficiency of the cycle, η_thermal :