In a three-stage steam turbine steam enters at 35 bar and 400°C and exhausts at 0.05 bar, 0.9 dry. If the work developed per stage is equal, determine :
(i) Condition of steam at entry to each stage.
(ii) The stage efficiencies.
(iii) The reheat factor. ( iv) Internal turbine efficiency.
Assume condition line to be straight.

Question

Accepted Answer

Initial condition of steam = 35 bar, 400°C
Exhaust condition of steam = 0.05 bar, 0.9 dry

(i) Condition of steam at entry to each stage :
Refer Fig. 55.

• Locate points [latex] L_{1} and L_{4}[/latex] corresponding to entry and exhaust conditions of steam.
• Since the condition line is straight (given), points [latex] L_{1} and L_{4}[/latex] are joined by a straight line.
Heat drop due to expansion from [latex] L_{1} to L_{4}[/latex] = 3222 – 2315 = 907 kJ/kg
• Since the work developed per stage is equal,
∴ Useful work/stage = [latex] \frac{907}{3} [/latex] = 302.3 kJ/kg i.e., ([latex] h_{L_1} - h_{N_1}[/latex]) = 302.3 kJ/kg

Produce horizontal to cut condition line at [latex]L_{2}. L_{1} N_{1} [/latex] produced cuts the pressure line through [latex]L_{2} at M_{1}. Thus L_{2}, N_{1} and M_{1} [/latex] are located.

Proceeding for other stages likewise, we get the following results :
Points Pressure Temp. or quality of steam
[latex]L_{2} [/latex] 6.2 bar 234°C
[latex]L_{3} [/latex] 0.73 bar 0.98 dry

[latex] h_{L_1 M_1} = h_{L_1 } - h_{M_1} [/latex] = 3222 – 2800 = 422 kJ/kg

[latex] h_{L_2 M_2} = h_{L_2 } - h_{M_2} [/latex] = 2920 – 2525 = 395 kJ/kg

[latex] h_{L_3 M_3} = h_{L_3 } - h_{M_3} [/latex] = 2615 – 2235 = 380 kJ/kg

(ii) Stage efficiencies :

Efficiency of stage 1, [latex] η_{1} = \frac{ h_{L_1 } - h_{N_1 } }{ h_{L_1 } - h_{M_1 }} = \frac{302.3}{422} [/latex] = 0.7163 or 71.63%.

Efficiency of stage 2, [latex] η_{2} = \frac{ h_{L_2 } - h_{N_2 } }{ h_{L_2 } - h_{M_2 }} = \frac{302.3}{395} [/latex] = 0.7653 or 76.53%.

Efficiency of stage 3, [latex] η_{3} = \frac{ h_{L_3 } - h_{N_3} }{ h_{L_3} - h_{M_3}} = \frac{302.3}{380} [/latex]
= 0.7955 = 79.55%.

(iii) Reheat factor :

Reheat factor = [latex] \frac{Cumulative drop}{Isentropic enthalpy drop} = \frac{h_{L_1 M_1} + h_{L_2 M_2} + h_{L_3 M_3} }{ h_{L_1} - h_{S} } [/latex]

= [latex] \frac{422 + 395 + 380 }{3222 - 2090} [/latex]= 1.057.

(iv) Internal turbine efficiency :

Internal turbine efficiency = [latex] \frac{h_{L_1} - h_{L_4} }{h_{L_1} - h_{S} } = \frac{3222 - 2315}{3222 - 2090} [/latex] = 0.801 or 80.1%.

Question 6.40: In a three-stage steam turbine steam enters at 35 bar and 40...

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