Answer to Question #103880 in Molecular Physics | Thermodynamics for nicole

Question

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

Answer 1

As per the given question,

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

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

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

a)

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

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

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

or $0.722 K J/kg$

b)

Kinetic energy at the exhaust trunk,

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

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

​

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

or 46.5125 KJ/kg