Answer:
the quality of the steam = 1.05
the temperature at [tex]P_2 = 2 \ MPa[/tex] = [tex]380.733^0 \ C[/tex]
the exit area of the nozzle = [tex]146.33 \ mm^2[/tex]
Explanation:
Given that:
the inlet temperature of steam [tex]T_1[/tex] = [tex]300^0 \ C[/tex]
Inlet pressure of steam [tex]P_1= 3 \ MPa[/tex]
Initial Velocity [tex]v_1[/tex] at the inlet = 0 m/s
Exit pressure of the steam [tex]P_2 = 2 \ MPa[/tex]
Exit velocity of steam [tex]v_2 \ = 400 \ m/s[/tex]
Mass flow rate m = 0.4 kg/s
Now, from steam tables at [tex]T_1[/tex] = [tex]300^0 \ C[/tex] and [tex]P_1= 3 \ MPa[/tex]
[tex]h_1 = 2992. 35 \ kJ/kg\\s_1 = 6.53535 \ kJ/kg\\[/tex]
At [tex]P_2 = 2 \ MPa[/tex]
[tex]sf_2 = 2.4474 kJ/kgK\\s_g = 6.3409 \ kJ/kg[/tex]
To determine the condition of the steam at exit ;
[tex]s_1 = s_2[/tex]
therefore,
[tex]6.53535 = s_{f2} +x_2sg-sf_2[/tex]
[tex]6.53535 = 2.4474 +x_2(6.3409-2.4474)[/tex]
[tex]x_2 = 1.05[/tex]
Thus , the quality of the steam = 1.05
However, By using the energy balance equation to determine the temperature of the steam; we have:
[tex]h_1 + \frac{v^2_1}{2}= h_2 + \frac{v^2_2}{2}[/tex]
[tex]h_1-h_2 = \frac{v^2_2}{2}[/tex]
[tex]h_2 = h_1 - \frac{v^2_2}{2}[/tex]
[tex]h_2 = 2992.35 - \frac{250^2}{2000}[/tex]
[tex]h_2 = 2912.35 \ kJ/kg[/tex]
From steam tables ; at enthalpy [tex]h_2[/tex] and [tex]P_2 = 2 \ MPa[/tex]; the corresponding temperature [tex]T_2 = 380.733^0 \ C[/tex]
Thus, the temperature at [tex]P_2 = 2 \ MPa[/tex] = [tex]380.733^0 \ C[/tex]
Finally, to calculate the exit area of the nozzle in mm²
we use the mass flow rate relation:
[tex]m = \frac{A_2v_2}{V_2}[/tex]
Making A the subject of the formula; we have:
[tex]A_2 = \frac{mv_2}{V_2}[/tex]
From the superheated steam tables at pressure [tex]P_2 = 2 \ MPa[/tex]; specific volume [tex]v_2 = 0.14633 \ m^3/kg[/tex]
We have:
[tex]A_2 = \frac{mv_2}{V_2}[/tex]
[tex]A_2 = \frac{0.4*0.14633}{400}[/tex]
[tex]A_2 = 1.4633*10^{-4} m^2[/tex]
[tex]A_2 = 146.33 \ mm^2[/tex]
Thus, the exit area of the nozzle = [tex]146.33 \ mm^2[/tex]
