Question

Steam initially at 1.5 MPa, 300°C expands reversibly and adiabatically in a steam
turbine to 40°C. Determine the ideal work output of the turbine per kg of steam.​

Answer 1

Answer:

then convert stream into 15 degree Celsius and reversibly hundred degree Celsius which determine the ideal work put off per kg

Answer 2

Answer:

The answer to this question is 696 KJ/Kg.

Explanation:

The ideal work output of a steam turbine can be determined using the first law of thermodynamics and the specific enthalpy and specific entropy values of the steam at the initial and final conditions.

The first law of thermodynamics states that the change in internal energy (dU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W). For a reversible and adiabatic process, the heat added to the system is zero (Q = 0) and the change in internal energy is equal to the work done by the system (dU = W).

The specific enthalpy (h) and specific entropy (s) of steam can be found using thermodynamic tables. At 1.5 MPa and 300°C, the specific enthalpy of steam is 3485 kJ/kg and the specific entropy is 6.8 kJ/kg-K. At 40°C and 1.5 MPa, the specific enthalpy is 2789 kJ/kg and the specific entropy is 7.2 kJ/kg-K.

Therefore, the ideal work output of the turbine per kg of steam can be determined using the following equation:

$W=H1-H2=3485-2789=696KJ/KG$

So the ideal work output of the turbine per kg of steam is 696 kJ/kg.

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