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Work Done in Isothermal Reversible Process

Steam enters ...

Question

Steam enters a turbine at 30 bar, $300^{o} C$ (u=2750 ,kJ/kg, h=2993,kJ/kg) and exits the turbine as saturated liquid at 15 kPa (u = 225 kJ/kg, h = 226 kJ/kg). Heat loss to the surrounding is 50 kJ/kg of steam flowing through the turbine. Neglecting changes in kinetic energy and potential energy, the work output of the turbine (in kJ/kg of steam) is
2717

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Solution

The correct option is A 2717

Given data:
At inlet, $u_{1} = 2750 k J / k g; h_{1} = 2993 k J / k g$
At exit, $u_{2} = 225 k J / k g; h_{2} = 226 k J / k g$
Specific work done,
$w_{1 - 2} = (h_{1} - h_{2}) + q_{1 - 2}$
$= (2993 - 226) - 50$
$= 2717 k J / k g$

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