A steam turbine generates energy at the rate of 318.8 kJ/kg of steam. The steam

Question

A steam turbine generates energy at the rate of 318.8 kJ/kg of steam. The steam at the inlet of the turbine is at 8 MPa, 480oC, and at a velocity of 196.2 m/s. The steam exits the turbine at 2 MPa, 240oC, and a velocity of 70.8 m/s. Heat transfer to the surroundings occurs where the outer surface temperature is 339.2oC. Determine the rate of entropy production (kJ/kg-K) within the turbine. Assume steady state conditions prevail.

We uses these charts for this class

http://mece.utpa.edu/~tarawneh/Thermodynamics/Thermo Booklet.pdf

Explanation / Answer

From steam properties table, we get corresponding to 8 MPa, 480 deg C, enthalpy h1 = 3348.4 kJ/kg and entropy s1 = 6.6586 kJ/kg-K

and corresponding to 2 MPa,, 240 deg C, we get h2 = 2876.5 and s2 = 6.3409 kJ/kg-K

Steady Flow Energy Equation states: W = (h1 + v1^2 /2) - (h2 + v2^2 /2) - q where q is heat transfer from turbine

So, 318.8*10^3 = (3348.4*10^3 + 196.2^2 /2) - (2876.5*10^3 + 70.8^2 /2) - q

q = 169.84 kJ/kg

Entropy production = s1 - s2 - q/T

Entropy production = 6.6586 - 6.3409 - 169.84 / (339.2 + 273.15)

Entropy production = 0.040341 kJ/kg-K

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