Question #103880
high pressure steam enters a turbine with a velocity of 38.0 meters/sec and leaves the exhaust trunk with a velocity of 305 meters/sec. calculate the kinetic energy of the steam in KJ/kg of the steam in KJ/kg and N-m/kg at a. at the entrance to the turbine and b. the exhaust trunk
1
Expert's answer
2020-03-02T09:12:47-0500

As per the given question,

Velocity of the steam at the entry point (v1)(v_1) = 38 m/sec

Velocity of the steam at the exhaust trunk (v2)(v_2) =305 m/sec

We know that density of the steam is =0.6 kg/m3kg/m^3

a)

So, kinetic energy of the steam at the entrance to the turbine=mv22\dfrac{m v^2}{2}

if m=1 kg and energy per kg will be =mv2m=v22=\dfrac{mv^2}{m}=\dfrac{v^2}{2}

=3822=722J/kg\dfrac{38^2}{2}=722J/kg

or 0.722KJ/kg0.722 K J/kg

b)

Kinetic energy at the exhaust trunk,

=mv222=\dfrac{m v_2^2}{2}

if m=1 kg and energy per kg will be =mv22m=v222=\dfrac{mv_2^2}{m}=\dfrac{v_2^2}{2}

=30522==\dfrac{305^2}{2}= 46512.5 J/kg

or 46.5125 KJ/kg



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