Aturbine operating adiabatically is fed with steam at 400 °c and 8.0 mpa at the rate of 1000 kg/h. process steam saturated at 0.5 mpa is withdrawn from an intermediate location in the turbine at a rate of 300 kg/h, and the remaining steam leaves the turbine saturated at 0.1 mpa. the kinetic energies and differences in potential energies of all streams are negligible. what is the power output of the turbine?

Explanation:

We select the enthalpy of steam at state 1 at and 8.0 MPa from the steam tables as follows.

                      = 3138 kJ/kg

Also, we select the enthalpy of steam at state 2 at 0.5 MPa from the steam tables as follows.

                      = 2748.6 kJ/kg

At state 3 also, from the steam tables at state 3 at 0.1 MPa.

                      = 2675.1 kJ/kg

Hence, calculate the mass flow rate at state 3 as follows.

                              = 1000 kg/h - 300 kg/h

                              = 700 kg/h

Now, we will calculate the power output of the turbine as follows.

                             = 1000 kg/h (3138 kJ/kg - 2748.6 kJ/kg) + 700 kg/h (2784.6 kJ/kg - 2675.1 kJ/kg)

                            = 440850 kJ/h

It is known that 1 kJ/h = 0.000278 kW.

Therefore,            

                          = 122.56 kW

Thus, we can conclude that the power output of the turbine is 122.56 kW.

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation:

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