Question

Superheated steam at 15 MPa and 600 C enters the turbine of a steam power plant and is expanded to an exit pressure of 15 kPa. Determine the work output if the turbine has an isentropic efficiency of 87.6%.

Superheated steam at 15 MPa and 600 C enters the turbine of a steam
power plant and is expanded to an exit pressure of 15 kPa. Determine
the work output if the turbine has an isentropic efficiency of 87.6%.

Answer 1

The work output of the turbine can be calculated using the isentropic efficiency and the enthalpy change of the steam. The isentropic efficiency is 87.6%. The work output is 1032.34 kJ/kg.

Step-by-step explanation:

To determine the work output of the turbine, we need to use the isentropic efficiency of the turbine.

The isentropic efficiency is given as 87.6%, which means that the actual work output of the turbine is 87.6% of the ideal or isentropic work output.

To calculate the ideal work output, we can use the enthalpy change between the initial and final states of the steam. We can find the enthalpy values from the steam tables.

First, we need to determine the enthalpy of the steam at the initial state. Using the steam tables, we can find that the enthalpy at 15 MPa and 600°C is 3407.5 kJ/kg.

Next, we need to determine the enthalpy of the steam at the final state. Using the steam tables, we can find that the enthalpy at 15 kPa is 2498.7 kJ/kg.

Now, we can calculate the ideal work output using the formula:

Work = (Enthalpy Initial - Enthalpy Final) / Isentropic Efficiency

Work = (3407.5 - 2498.7) / 0.876

= 1032.34 kJ/kg