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

v₂ = 35.57 m/s

Step-by-step 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)+(v_(1)^(2))/(2) \right )` =
`\left ( h_(2)+(v_(2)^(2))/(2) \right )`

`v_(2)^(2)= 2\left [ h_(1)-h_(2)+(v_(1)^(2))/(2) \right ]`

`v_(2)^(2)= 2\left [ 2964.8-2782.14+(30^(2))/(2) \right ]`

`v_(2)^(2)=`2 X 632.66

v₂ = 35.57 m/s

Therefore, outlet velocity, v₂ = 35.57 m/s