Question

Steam at 70 bar and 450°C is supplied to a steam turbine. After expanding to 25 bar in high pressure stages, it is reheated to 420°C at the constant pressure. Next ; it is expanded in intermediate pressure stages to an appropriate minimum pressure such that part of the steam bled at this pressure heats the feed water to a temperature of 180°C. The remaining steam expands from this pressure to a condenser pressure of 0.07 bar in the low pressure stage. The isentropic efficiency of H.P. stage is 78.5%, while that of the intermediate and L.P. stages is 83% each. From the above data, determine : 

(i) The minimum pressure at which bleeding is necessary. 

(ii) The quantity of steam bled per kg of flow at the turbine inlet. 

(iii) The cycle efficiency. Neglect pump work.

Answer 1

The schematic arrangement of the plant is shown in  (a) and the processes are represented on T-s and h-s diagrams as shown in  (b) and (c) respectively.

(i) The minimum pressure at which bleeding is necessary : 

It would be assumed that the feed water heater is an open heater. Feed water is heated to 180°C. So p_sat at 180°C ~− 10 bar is the pressure at which the heater operates. 

Thus, the pressure at which bleeding is necessary is 10 bar.  

From the h-s chart (Mollier chart), we have :

From steam tables, we have :

h_f6 = 163.4 kJ/kg ; h_f8 = 762.6 kJ/kg 

h₁ – h₂′ = 0.785(h₁ – h₂) = 0.785(3285 – 2980) = 239.4 kJ/kg 

∴ h₂′ = h₁ – 239.4 = 3285 – 239.4 = 3045.6 kJ/kg

(a) Schematic arrangement of the plant

(b) diagram

(c) diagram (Pump work not shown)

(ii) The quantity of steam bled per kg of flow at the turbine inlet, m : 

Considering energy balance for the feed water heater, we have : 

m × h₄′ + (1 – m) h_f7 = 1 × h_f8 

m × 3072.5 + (1 – m) × 163.4 = 1 × 762.6

3072.5 m + 163.4 – 163.4 m = 762.6

= 0.206 kg of steam flow at turbine inlet.

(iii) Cycle efficiency, η_cycle :