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Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turbine if the steam is exhausted as a saturated vapor?

Sagot :

Hamzaahmedsidn

Answer:

[tex]\eta_{turbine} = 0.603 = 60.3\%[/tex]

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = [tex]h_{g\ at\ 125KPa}[/tex] = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than [tex]s_g[/tex] and greater than [tex]s_f[/tex] at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

[tex]x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88[/tex]

Now, we will find [tex]h_{2s}[/tex](enthalpy at the outlet for the isentropic process):

[tex]h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg[/tex]

Now, the isentropic efficiency of the turbine can be given as follows:

[tex]\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%[/tex]

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