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A horizontal turbine takes in steam with an enthalpy of h = 2.80 MJ/kg at 45 m/s. A steam-water mixture exits the turbine with an enthalpy of h = 1.55 MJ/kg at 20 m/s. If the heat loss to the surroundings from the turbine is 300 J/s, determine the power the fluid supplies to the turbine. The mass flow rate is 0.85 kg/s.

User Ryan King
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Answer:

The power that fluid supplies to the turbine is 1752.825 kilowatts.

Step-by-step explanation:

A turbine is a device that works usually at steady state. Given that heat losses exists and changes in kinetic energy are not negligible, the following expression allows us to determine the power supplied by the fluid to the turbine by the First Law of Thermodynamics:


-\dot Q_(loss) - \dot W_(out) + \dot m \cdot \left[(h_(in)-h_(out)) + (1)/(2)\cdot (v_(in)^(2)-v_(out)^(2)) \right] = 0

Output power is cleared:


\dot W_(out) = -\dot Q_(loss) + \dot m \cdot \left[(h_(in)-h_(out))+(1)/(2)\cdot (v_(in)^(2)-v_(out)^(2)) \right]

If
\dot Q_(loss) = 0.3\,kW,
\dot m = 0.85\,(kg)/(s),
h_(in) = 2800\,(kJ)/(kg),
h_(out) = 1550\,(kJ)/(kg),
v_(in) = 45\,(m)/(s) and
v_(out) = 20\,(m)/(s), then:


\dot W_(out) = -0.3\,kW + \left(0.85\,(kg)/(s) \right)\cdot \left\{\left(2800\,(kJ)/(kg)-1550\,(kJ)/(kg) \right)+(1)/(2)\cdot \left[\left(45\,(m)/(s) \right)^(2)-\left(20\,(m)/(s) \right)^(2)\right] \right\}


\dot W_(out) = 1752.825\,kW

The power that fluid supplies to the turbine is 1752.825 kilowatts.

User RealMarkusSchmidt
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