Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s and the exit conditions ar

Question

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s and the exit conditions are 30 kPa, 92 percent quality, and 50 m/s. The mass flow rate of the steam is 12.6 kg/s. The properties of the steam are v1 = 0.086442 m^3/kg, h1 = 3446 kJ/kg, and h2 = 2437.7 kJ/kg. Determine:

a. the change in kinetic energy.
b. the power output.
c. the turbine inlet area.

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Thu Nguyệt 2 weeks 2021-08-30T01:59:00+00:00 1 Answers 0 views 0

Answers ( )

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    2021-08-30T02:00:52+00:00

    Answer:

    a) \Delta \dot K = 24.570\,kW, b) \dot W_{out} = 12729.15\,kW, c) A_{in} = 0.0136\,m^{2}

    Explanation:

    A turbine is a device which works usually in steady state and assumption of being adiabatic means no heat interactions between steam through turbine and surroudings and produce mechanical work from fluid energy. Changes in gravitational energy can be neglected. This system can be modelled after the First Law of Thermodynamics:

    -\dot W_{out} + \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})   = 0

    a) Change in kinetic energy

    \Delta \dot K = \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2} - v_{out}^{2})

    \Delta \dot K = \frac{1}{2} \cdot \left(12.6\,\frac{kg}{s} \right) \cdot \left[\left(80\,\frac{m}{s} \right)^{2}-\left(50\,\frac{m}{s} \right)^{2}\right]

    \Delta \dot K = 24570\,W

    \Delta \dot K = 24.570\,kW

    b) Power output

    \dot W_{out} = \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})

    \dot W_{out} = \left(12.6\,\frac{kg}{s}\right)\cdot \left(3446\,\frac{kJ}{kg} - 2437.7\,\frac{kJ}{kg} \right) + 24.570\,kW

    \dot W_{out} = 12729.15\,kW

    c) Turbine inlet area

    Turbine inlet area can be found by using the following expressions:

    \dot V_{in} = \dot m \cdot \nu_{in}

    \dot V_{in} = \left(12.6\,\frac{kg}{s}\right) \cdot \left(0.086442\,\frac{m^{3}}{kg} \right)

    \dot V_{in} = 1.089\,\frac{m^{3}}{s}

    A_{in} = \frac{\dot V_{in}}{v_{in}}

    A_{in} = \frac{1.089\,\frac{m^{3}}{s} }{80\,\frac{m}{s} }

    A_{in} = 0.0136\,m^{2}

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