Question

Steam enters a turbine operating at steady state with a mass flow of 10 kg/min, a specific enthalpy of 3100 kJ/kg, and a velocity of 30 m/s. At the exit, the specific enthalpy is 2300 kJ/kg and the velocity is 45 m/s. The elevation of the inlet is 3 m higher than at the exit. Heat transferred away from the turbine is 1.1 kJ/kg of stream flow. Let g = 9.81 m/s^2.
Determine the power developed by the turbine, in kW.

Answer 1

Answer:

$\dot W_{out} = 133.327\,kW$

Explanation:

The model for the turbine can be derived by means of the First Law of Thermodynamics:

$-\dot Q_{out}-\dot W_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right] =0$

The work produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} +\dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) + g\cdot (z_{in}-z_{out})\right]$

The mass flow and heat transfer rates are, respectively:

$\dot m = (10\frac{kg}{min})\cdot (\frac{1\,min}{60\,s} )$

$\dot m = 0.167\,\frac{kg}{s}$

$\dot Q_{out} = (0.167\,\frac{kg}{s} )\cdot (1.1\times 10^{3}\,\frac{J}{kg} )$

$\dot Q_{out} = 183.7\,W$

Finally:

$\dot W_{out} = -183.7\,W + (0.167\,\frac{kg}{s} )\cdot \left(8\times 10^{5}\,\frac{J}{kg} -562,5\,\frac{J}{kg} +29.43\,\frac{J}{kg} \right)$

$\dot W_{out} = 133.327\,kW$