Question

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occurs at an average outer surface temperature of 315 K at the rate of 30 kJ per kg of air flowing. Kinetic and potential energy effects are negligible. Assuming the air is modeled as an ideal gas with variations in specific heat, determine a) the rate power is developed, in kJ per kg of air flowing.b) the rate of entropy production within the turbine, in kJ/K per kg of air flowing.

Answer 1

Answer:

a) $w_{out} = 281.55\,\frac{kJ}{kg}$, b) $s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}$

Explanation:

a) The process within the turbine is modelled after the First Law of Thermodynamics:

$-q_{out} - w_{out} + h_{in}-h_{out} = 0$

$w_{out} = h_{in} - h_{out}-q_{out}$

$w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}$

$w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}$

$w_{out} = 281.55\,\frac{kJ}{kg}$

b) The entropy production is determined after the Second Law of Thermodynamics:

$-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0$

$s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}$

$s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)$

$s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)$

$s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}$