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siniylev [52]
3 years ago
15

low-velocity steam (with negligible kinetic energy) enters an adiabatic nozzle at 300 C and 3 MPa. the steam leaves the nozzle a

t 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.
Physics
1 answer:
Zanzabum3 years ago
7 0

Answer:

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

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 + \frac{v^2_1}{2}= h_2 + \frac{v^2_2}{2}

h_1-h_2 = \frac{v^2_2}{2}

h_2 = h_1 - \frac{v^2_2}{2}

h_2 = 2992.35 - \frac{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 = \frac{A_2v_2}{V_2}

Making A the subject of the formula; we have:

A_2 = \frac{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 = \frac{mv_2}{V_2}

A_2 = \frac{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

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