Steam enters a turbine operating at steady state with a specific enthalpy of 1407.6 Btu/lb and expands to the turbine exit where the specific enthalpy is 1236.4 Btu/lb. The mass flow rate is 5 lb/s. During this process, heat transfer to the surroundings occurs at a rate of 40 Btu/s. Neglecting kinetic and potential energy effects, the power developed by the turbine is

Answer:

The power developed by the turbine is 816 BTU per second.

Explanation:

Thermodynamically speaking, a turbine produces work at the expense of fluid energy. By First Law of Thermodynamics, the energy balance of a turbine working at steady state is:

[tex]-\dot Q -\dot W +\dot m\cdot (h_{in}-h_{out}) = 0[/tex] (1)

Where:

[tex]\dot Q[/tex] - Heat transfer rate, measured in BTU per second.

[tex]\dot W[/tex] - Power developed by the turbine, measured in BTU per second.

[tex]\dot m[/tex] - Mass flow rate, measured in pounds per second.

[tex]h_{in}[/tex], [tex]h_{out}[/tex] - Specific enthalpies at inlet and outlet, measured in BTU per pound.

If we know that [tex]\dot Q = 40\,\frac{BTU}{s}[/tex], [tex]\dot m = 5\,\frac{lbm}{s}[/tex], [tex]h_{in} = 1407.6\,\frac{BTU}{lbm}[/tex] and [tex]h_{out} = 1236.4\,\frac{BTU}{lbm}[/tex], then the power developed by the turbine is:

[tex]\dot W = -\dot Q + \dot m \cdot (h_{in}-h_{out})[/tex]

[tex]\dot W = 816\,\frac{BTU}{s}[/tex]

The power developed by the turbine is 816 BTU per second.

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