Question 12.21: (a) How does erosion of turbine blades occur ? State the met...
(a) How does erosion of turbine blades occur ? State the methods of preventing erosion of turbine blades.
(b) What do you mean by TTD of a feed water heater ? Draw temperature-path-line diagram of a closed feed water heater used in regenerative feed heating cycle.
(c) In a 15 MW steam power plant operating on ideal reheat cycle, steam enters the H.P. turbine at 150 bar and 600°C. The condenser is maintained at a pressure of 0.1 bar. If the moisture content at the exit of the L.P. turbine is 10.4%, determine :
(i) Reheat pressure ; (ii) Thermal efficiency ; (iii) Specific steam consumption ; and (iv) Rate of pump work in kW. Assume steam to be reheated to the initial temperature.
(a) The erosion of the moving blades is caused by the presence of water particles in (wet) steam in the L.P. stages. The water particles strike the leading surface of the blades. Such impact, if sufficiently heavy, produces severe local stresses in the blade material causing the surface metal to fail and flake off.
The erosion, if any, is more likely to occur in the region where the steam is wettest, i.e., in the last one or two stages of the turbine. Moreover, the water droplets are concentrated in the outer parts of the flow annuals where the velocity of impact is highest.
Erosion difficulties due to moisture in the steam may be avoided by reheating (see Fig. 12.31). The whole of steam is taken from the turbine at a suitable point 2, and a further supply of heat is given to it along 2-3 after which the steam is readmitted to the turbine and expanded along 3-4 to condenser pressure.
Erosion may also be reduced by using steam traps in between the stages to separate moisture from the steam.
(b) TTD means “Terminal temperature difference”. It is the difference between temperatures of bled steam/condensate and the feed water at the two ends of the feed water heater.
The required temperature-path-line diagram of a closed feed water heater is shown in Fig. 12.32.
(c) The cycle is shown on T-s and h-s diagrams in Figs. 12.33 and 12.34 respectively. The following values are read from the Mollier diagram :
h_1=3580 kJ/kg, h_2=3140 kJ/kg, h_3=3675 kJ/kg, and h_4=2335 kJ/kg
Moisture contents in exit from L.P. turbine = 10.4%
x_4=1-0.104=0.896(i) Reheat pressure : From the Mollier diagram, the reheat pressure is 40 bar.
(ii) Thermal efficiency, \eta_{\text {th }} :
Turbine work =\left(h_1-h_2\right)+\left(h_3-h_4\right)
= (3580 – 3140) + (3675 – 2335) = 1780 kJ/kg.
Assuming specific volume of water = 10^{-3} m³/kg, the pump work = 10^{-3}(150-0.1) = 0.15 kJ/kg, i.e., may be neglected in computing of \eta_{\text {th }}, h_5=h_4=191.8 kJ/kg, (h_f at 0.1 bar) from steam tables,
Q_{\text {input }}=\left(h_1-h_5\right)+\left(h_3-h_2\right)= (3580 – 191.8) + (3675 – 3140) = 3923.2 kJ/kg
\% \eta_{\text {th }}=\frac{1780}{3923.2} \times 100=45.37 \%.
(iii) Specific steam consumption :
Steam consumption
=\frac{15 \times 10^3}{1780}=8.427 kg/s
Specific steam consumption =\frac{8.427 \times 3600}{15 \times 10^3}= 2 . 0 2 2 5 kg/kWh.
(iv) Rate of pump work :
Rate of pump work = 8.427 × 0.15 = 1.26 kW.
