Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.

(a) the reversible power output of turbine is 5810 kw

(b) The second-law efficiency of he turbine = 86.05%

Explanation:

In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.

h1 = 3658 kJ/kg s1=7.167 kJ/kgK

In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state

h2 = 2682kl/kg S2= 7.694 kJ/kg

Assuming that the energy balance equation given  

Wout=m [h1-h2+(v1²-v2²) /2]

Let

W =5 MW

V1= 80 m/s  V2= 140 m/s

h1 = 3658kJ/kg  h2 = 2682 kJ/kg

∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)

m = 5.158kg/s

Consider the energy balance equation given  

Wrev,out =Wout-mT0(s1-s2)

Substitute Wout =5 MW m = 5.158kg/s 7

s1=  7.167 kJ/kg-K            s2= 7.694kJ/kg-K and 25°C .

Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)

= 5810 kW

(a) Therefore, the reversible power output of turbine is 5810 kw.

The given values of quantities were substituted and the reversible power output are calculated.

(b) Calculating the second law efficiency of the turbine:  

η=Wout/W rev,out

Let Wout =  5 MW and Wrev,out = 5810 kW  

η=(5 MW x 1000 kW)/(1 MW *5810)  

η= 86.05%

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

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