Steam enters a steam turbine at a pressure of 15 bar and 350°C with a velocity of 60 m/s. The steam leaves the turbine at 1.2 bar and with a velocity of 180 m/s. Assuming the process to be reversible adiabatic, determine the work done per kg of steam flow through the turbine.

Neglect the change in potential energy.

Question

Steam enters a steam turbine at a pressure of 15 bar and 350°C with a velocity of 60 m/s. The steam leaves the turbine at 1.2 bar and with a velocity of 180 m/s. Assuming the process to be reversible adiabatic, determine the work done per kg of steam flow through the turbine.

Neglect the change in potential energy.

Accepted Answer

Initial pressure of steam, [latex]p_1=15[/latex] bar

Initial temperature of steam, [latex]t_1=t_{s u p}=350^{\circ} C[/latex]

Initial velocity of steam, [latex]C_1=60[/latex] m/s

Final pressure, [latex]p_2=1.2[/latex] bar

Final velocity, [latex]C_2=180[/latex] m/s

Process of expansion : Reversible adiabatic

As the process is reversible adiabatic, it will be represented by a vertical line on T-s diagram by 1-2 as it is also constant entropy process.

The condition at point ‘2’ can be calculated by equating the entropy at point ‘1’ and point ‘2’,

i.e., [latex]s_1=s_2[/latex] .....per kg of steam

[latex]7.102=s_{f_2}+x_2\left(s_{g_2}-s_{f_2} ight)[/latex]

= [latex]1.3609+x_2(7.2984-1.3609)[/latex]

∴ [latex]x_2=\frac{7.102-1.3609}{7.2984-1.3609}=0.967[/latex]

[latex]h_2=h_{f_2}+x_2 h_{f_2}=439.4+0.967 imes 2244.1=2609.44[/latex] kJ/kg

[latex]h_1[/latex] (at 15 bar and 350°C) = 3147.5 kJ/kg

Applying the first law energy equation for steady flow process,

[latex]h_1+\frac{C_1^2}{2}=h_2+\frac{C_2^2}{2}+W[/latex]

i.e., [latex]W=\left(h_1-h_2 ight)+\left(\frac{C_1^2-C_2^2}{2} ight)[/latex]

= [latex]3147.5-2609.44+\left(\frac{60^2-180^2}{2 imes 10^3} ight)[/latex]

= 3147.5 – 2609.44 – 14.4 = 523.66 kJ/kg.

Hence work done per kg of steam = 523.66 kJ/kg.

Question 4.59: Steam enters a steam turbine at a pressure of 15 bar and 350...

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