Question

A steam turbine plant developing 120 MW of electrical output is equipped with reheating and regenerative feed heating arrangement consisting of two feed heaters—one surface type on H.P. side and other direct contact type on L.P. side. The steam conditions before the steam stop valve are 100 bar and 530°C. A pressure drop of 5 bar takes place due to throttling in valves. 

Steam exhausts from the H.P. turbine at 25 bar. A small quantity of steam is bled off at 25 bar for H.P. surface heater for feed heating and the remaining is reheated in a reheater to 550°C and the steam enters at 22 bar in L.P. turbine for further expansion. Another small quantity of steam is bled off at pressure 6 bar for the L.P. heater and the rest of steam expands up to the back pressure of 0.05 bar. The drain from the H.P. heater is led to the L.P. heater and the combined feed from the L.P. heater is pumped to the high-pressure feed heater and finally to the boiler with the help of boiler feed pump. 

The component efficiencies are : Turbine efficiency 85%, pump efficiency 90%, generator efficiency 96%, boiler efficiency 90% and mechanical efficiency 95%. It may be assumed that the feed-water is heated up to the saturation temperature at the prevailing pressure in feed heater. Work out the following : 

(i) Sketch the feed heating system and show the process on T-s and h-s diagrams. 

(ii) Amounts of steam bled off. 

(iii) Overall thermal efficiency of turbo-alternator considering pump work. 

(iv) Specific steam consumption in kg/kWh.

Answer 1

(i) The schematic arrangement including feed heating system, and T-s and h-s diagrams of the process are shown in  respectively

(ii) Amounts of bled off. The enthalpies at various state points as read from h-s diagram/steam tables, in kJ/kg, are :

h₁ = h₂ = 3460 

h₃′ = 3050, and 

∴ h₃ = 3460 – 0.85(3460 – 3050) = 3111.5 

h₄ = 3585 h₅′ = 3140, and 

∴ h₅ = 3585 – 0.85(3585 – 3140) = 3207 

h₆′ = 2335, and 

∴ h₆ = 3207 – 0.85 (3207 – 2335) = 2466 

h₇ = 137.8 kJ/kg (h_f at 0.05 bar) 

h₈ = h₁₀ = 962 kJ/kg (h_f at 25 bar) and 

h₉ = 670.4 (h_f at 6 bar).

Enthalpy balance for surface heater :

m₁h₃ + h₉ = m₁h₈ + h₁₀, neglecting pump work

Enthalpy balance for contact heater :

m₂h₅ + (1 – m₁ – m₂)h₇ + m₁h₈ = h₉, neglecting pump work

m₂ × 3207 + (1 – 0.13566 – m₂) × 137.8 + 0.13566 × 962 = 670.4

m₂ = 0.1371 kg.

Pump Work. Take specific volume of water as 0.001 m³/kg.

(W_pump)_L.P. = (1 – m₁ – m₂)(6 – 0.05) × 0.001 × 10²

= (1 – 0.13566 – 0.1371) × 5.95 × 0.1 = 0.4327 kJ/kg.

(W_pump)_H.P. = 1 × (100 – 6) × 0.001 × 10² = 9.4 kJ/kg

Total pump work (actual) = \(\cfrac{0.4327+9.4}{0.9}\) = 10.925 kJ/kg

Turbine output (indicated) 

= (h₂ – h₃) + (1 – m1)(h₄ – h₅) + (1 – m₁ – m₂)(h₅ – h₆) 

= (3460 – 3111.5) + (1 – 0.13566)(3585 – 3207) + (1 – 0.13566 – 0.1371)(3207 – 2466)

= 1214.105 kJ/kg 

Net electrical output = (Indicated work – Pump work) × η_mech. × η_gen. 

= (1214.105 – 10.925) × 0.9 × 0.96 = 1039.55 kJ/kg 

[Note. All the above calculations are for 1 kg of main (boiler) flow.]

Main steam flow rate 

Amounts of bled off are : 

(a) Surface (high pressure) heater, 

= 0.13566 kg/kg of boiler flow 

or = 0.13566 × 4.155 × 105 

= 5.6367 × 104 kg/h. 

(b) Direct contact (low pressure) heater 

= 0.1371 kg/kg of boiler flow 

or = 0.1371 × 4.155 × 10⁵

= 5.697 × 10⁴ kg/h.

(b) Direct contact (low pressure) heater 

= 0.1371 kg/kg of boiler flow 

or = 0.1371 × 4.155 × 10⁵

= 5.697 × 104 kg/h.

(iii) Overall thermal efficiency, η_overall :

Heat input in boiler

Heat input in reheater

(iv) Specific steam consumption :

Specific steam consumption = \(\cfrac{4.155\times10^5}{120\times10^3}\) = 3.4625 kg/kWh