Question

Consider a steam turbine, with inflow at 500oC and 7.9 MPa. The machine has a total-to-static efficiency ofηts=0.91, and the pressure at the outflow is 16kPa. Power extracted by the turbine is 38 MW. Assuming heat transfer and kinetic energy in the machine is negligible, find the mass flow rate and static enthalpy at the outflow.

Answer 1

Answer: $\dot m_{in} = 23.942 \frac{kg}{s}$, $\dot H_{out} = 39632.62 kW$

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to $\dot V = 1 \frac{m^{3}}{s}$. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

Principle of Mass Conservation

$\dot m_{in} - \dot m_{out} = 0$

First Law of Thermodynamics

$- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0$

This 2 x 2 System can be reduced into one equation as follows:

$-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0$

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:

Inflow (Superheated Steam)

$\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}$

The mass flow rate can be calculated by using this expression:

$\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}$

$\dot m_{in} = 23.942 \frac{kg}{s}$

Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:

$h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}$

$h_{out} = 1655.36 \frac{kJ}{kg}$

The enthalpy rate at outflow is:

$\dot H_{out} = \dot m \cdot h_{out}$

$\dot H_{out} = 39632.62 kW$