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

`\dot W_(out) = 133.327\,kW`

Step-by-step 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))+(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))+(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(kg)/(min))\cdot ((1\,min)/(60\,s) )`

`\dot m = 0.167\,(kg)/(s)`

`\dot Q_(out) = (0.167\,(kg)/(s) )\cdot (1.1* 10^(3)\,(J)/(kg) )`

`\dot Q_(out) = 183.7\,W`

Finally:

`\dot W_(out) = -183.7\,W + (0.167\,(kg)/(s) )\cdot \left(8* 10^(5)\,(J)/(kg) -562,5\,(J)/(kg) +29.43\,(J)/(kg) \right)`

`\dot W_(out) = 133.327\,kW`