Question 12.27: Steam at 70 bar and 450°C is supplied to a steam turbine. Af...
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.
The schematic arrangement of the plant is shown in Fig. 12.42 (a) and the processes are represented on T-s and h-s diagrams as shown in Figs. 12.42 (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 p_{\text {sat }} \text { at } 180^{\circ} C \simeq 101 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 :
h_1=3285 kJ/kg ; h_2=2980 kJ/kg ; h_3=3280 kJ/kg ; h_4=3030 kJ/kg
h_3-h_4^{\prime}=0.83\left(h_3-h_4\right)=0.83(3280-3030)=207.5 kJ/kg
∴ h_4^{\prime}=h_3-207.5=3280-207.5=3072.5 kJ/kg
h_5=2210 kJ/kg
h_4^{\prime}-h_5^{\prime}=0.83\left(h_4^{\prime}-h_5\right)=0.83(3072.5-2210) \simeq 715.9 kJ/kg
∴ h_5^{\prime}=h_4^{\prime}-715.9=3072.5-715.9=2356.6 kJ/kg
From steam tables, we have :
h_{f 6}=163.4 kJ/kg ; h_{f 8}=762.6 kJ/kg
h_1-h_2^{\prime}=0.785\left(h_1-h_2\right)=0.785(3285-2980)=239.4 kJ/kg
∴ h_2^{\prime}=h_1-239.4=3285-239.4=3045.6 kJ/kg
(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 \times h_4^{\prime}+(1-m) h_{f 7}=1 \times h_{f B}m \times 3072.5+(1-m) \times 163.4=1 \times 762.6 \left(\because \quad h_{f 7}=h_{f 6}\right)
3072.5 m + 163.4 – 163.4 m = 762.6
∴ m=\frac{(762.6-163.4)}{(3072.5-163.4)}
= 0.206 kg of steam flow at turbine inlet.
(iii) Cycle efficiency, \eta_{\text {cycle }} :
\eta_{\text {cycle }}=\frac{\text { Work done }}{\text { Heat supplied }}=\frac{1\left(h_1-h_2\right)+1\left(h_3-h_4\right)+(1-m)\left(h_4{ }^{\prime}-h_5{ }^{\prime}\right)}{\left(h_1-h_{f 8}\right)+\left(h_3-h_2{ }^{\prime}\right)}= \frac{(3285-3045.6)+207.5+(1-0.206)(715.9)}{(3285-762.6)+(3280-3045.6)}=\frac{1015.3}{2756.8}
= 0.3683 or 36.83%.
