Answer to Question #145944 in Molecular Physics | Thermodynamics for Sejal

Question

3. A turbine has a power output of 7500 kJ/sec when steam with a mass flow rate of 11.11
kg/sec enters the turbine at 773 K, 8MPa and leaves as saturated vapor at 40 kPa. If the
surrounding of the turbine is at 25 °C, find the rate of entropy generation at steady state.

Answer 1

The steam mass flow rate m = 11.11 kg/sec

P₁ = 8 MPa

T₁ = 773 K

P₂ = 40 kPa

$W_{out} = 7500 \;kJ/sec$

$T_{sur} = 25 + 273 = 298 \;K$

h₁ = 3399.5 kJ/kg

s₁ = 6.7266 kJ/kgK

And at 40 kPa

h₂ = h_g = 2636.1 kJ/kg

s₂ = s_g = 7.6691 kJ/kgK

$m_{in} - m_{out} = ∆m = 0$

$m_1 = m_2 = m$

$E_{in} - E_{out} = ∆E = 0$

$E_{in} = E_{out}$

$mh_1 = Q_{out} + W_{out} + mh_2$

$Q_{out} = m(h_1 - h_2) - W_{out}$

$Q_{out} = 11.11(3399.5 - 2636.1) - 7500 = 981.4 \;kJ/s$

$S_{in} - S_{out} + S_{gen} = ∆S_{sys} = 0$

$ms_1 - ms_2 - \frac{Q_{out}}{T_{sur}} + S_{gen} = 0$

$S_{gen} = m(s_2 - s_1) + \frac{Q_{out}}{T_{sur}}$

$S_{gen} = 11.11(7.6691 - 6.7266) + \frac{981.4}{298} = 10.47 + 3.29 = 13.76 \;kW/K$