Question

Steam enters an adiabatic nozzle at l MPa, 260 C, 30 m/s and exits at 0.3 MPa and 160 'C. Calculate the velocity at the exit.

Answer 1

Answer:

v₂ = 35.57 m/s

Explanation:

Given :

Inlet steam pressure, P₁ = 1 MPa

Inlet steam temperature, T₁ = 260°C

Inlet velocity of steam, V₁ = 30 m/s

Outlet steam pressure, P₂ = 0.3 MPa

Outlet steam temperature, T₂ = 160°C

Now,

From steam table at pressure 1 Mpa and temperature 260°C, enthalpy, h₁ = 2964.8 kJ/kg

From steam table at pressure 0.3 Mpa and temperature 160°C, enthalpy, h₂ = 2782.14 kJ/kg

Therefore, for an open system from 1st law of thermodynamics, we get

Energy in = Energy out

E₁ = E₂

$\left ( h_{1}+\frac{v_{1}^{2}}{2} \right )$ = $\left ( h_{2}+\frac{v_{2}^{2}}{2} \right )$

$v_{2}^{2}= 2\left [ h_{1}-h_{2}+\frac{v_{1}^{2}}{2} \right ]$

$v_{2}^{2}= 2\left [ 2964.8-2782.14+\frac{30^{2}}{2} \right ]$

$v_{2}^{2}=$2 X 632.66

v₂ = 35.57 m/s

Therefore, outlet velocity, v₂ = 35.57 m/s