Question

low-velocity steam (with negligible kinetic energy) enters an adiabatic nozzle at 300 C and 3 MPa. the steam leaves the nozzle at 2 MPa with a velocity of 400 m/s. the mass flow rate is 0.4 kg/s. determine the quality or temperature (C) of the steam leaving the nozzle and the exit area of the nozzle in mm^2.

Answer 1

the quality of the steam = 1.05

the temperature at
`P_2 = 2 \ MPa` =
`380.733^0 \ C`

the exit area of the nozzle =
`146.33 \ mm^2`

Step-by-step explanation:

Given that:

the inlet temperature of steam
`T_1` =
`300^0 \ C`

Inlet pressure of steam
`P_1= 3 \ MPa`

Initial Velocity
`v_1` at the inlet = 0 m/s

Exit pressure of the steam
`P_2 = 2 \ MPa`

Exit velocity of steam
`v_2 \ = 400 \ m/s`

Mass flow rate m = 0.4 kg/s

Now, from steam tables at
`T_1` =
`300^0 \ C` and
`P_1= 3 \ MPa`

`h_1 = 2992. 35 \ kJ/kg\\s_1 = 6.53535 \ kJ/kg\\`

At
`P_2 = 2 \ MPa`

`sf_2 = 2.4474 kJ/kgK\\s_g = 6.3409 \ kJ/kg`

To determine the condition of the steam at exit ;

`s_1 = s_2`

therefore,

`6.53535 = s_(f2) +x_2sg-sf_2`

`6.53535 = 2.4474 +x_2(6.3409-2.4474)`

`x_2 = 1.05`

Thus , the quality of the steam = 1.05

However, By using the energy balance equation to determine the temperature of the steam; we have:

`h_1 + (v^2_1)/(2)= h_2 + (v^2_2)/(2)`

`h_1-h_2 = (v^2_2)/(2)`

`h_2 = h_1 - (v^2_2)/(2)`

`h_2 = 2992.35 - (250^2)/(2000)`

`h_2 = 2912.35 \ kJ/kg`

From steam tables ; at enthalpy
`h_2` and
`P_2 = 2 \ MPa`; the corresponding temperature
`T_2 = 380.733^0 \ C`

Thus, the temperature at
`P_2 = 2 \ MPa` =
`380.733^0 \ C`

Finally, to calculate the exit area of the nozzle in mm²

we use the mass flow rate relation:

`m = (A_2v_2)/(V_2)`

Making A the subject of the formula; we have:

`A_2 = (mv_2)/(V_2)`

From the superheated steam tables at pressure
`P_2 = 2 \ MPa`; specific volume
`v_2 = 0.14633 \ m^3/kg`

We have:

`A_2 = (mv_2)/(V_2)`

`A_2 = (0.4*0.14633)/(400)`

`A_2 = 1.4633*10^(-4) m^2`

`A_2 = 146.33 \ mm^2`

Thus, the exit area of the nozzle =
`146.33 \ mm^2`