Answer:
a) the mass flow rate of the steam is [tex]\mathbf{m_1 =6.92 \ kg/s}[/tex]
b) the exit velocity of the steam is [tex]\mathbf{V_2 = 562.7 \ m/s}[/tex]
c) the exit area of the nozzle is [tex]A_2[/tex] = 0.0015435 m²
Explanation:
Given that:
A steam with 5 MPa and 400° C enters a nozzle steadily
So;
Inlet:
[tex]P_1 =[/tex] 5 MPa
[tex]T_1[/tex] = 400° C
Velocity V = 80 m/s
Exit:
[tex]P_2 =[/tex] 2 MPa
[tex]T_2[/tex] = 300° C
From the properties of steam tables at [tex]P_1 =[/tex] 5 MPa and [tex]T_1[/tex] = 400° C we obtain the following properties for enthalpy h and the speed v
[tex]h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg[/tex]
From the properties of steam tables at [tex]P_2 =[/tex] 2 MPa and [tex]T_1[/tex] = 300° C we obtain the following properties for enthalpy h and the speed v
[tex]h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg[/tex]
Inlet Area of the nozzle = 50 cm²
Heat lost Q = 120 kJ/s
We are to determine the following:
a) the mass flow rate of the steam.
From the system in a steady flow state;
[tex]m_1=m_2=m_3[/tex]
Thus
[tex]m_1 =\dfrac{V_1 \times A_1}{v_1}[/tex]
[tex]m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}[/tex]
[tex]m_1 =\dfrac{0.4 }{0.057838 }[/tex]
[tex]\mathbf{m_1 =6.92 \ kg/s}[/tex]
b) the exit velocity of the steam.
Using Energy Balance equation:
[tex]\Delta E _{system} = E_{in}-E_{out}[/tex]
In a steady flow process;
[tex]\Delta E _{system} = 0[/tex]
[tex]E_{in} = E_{out}[/tex]
[tex]m(h_1 + \dfrac{V_1^2}{2})[/tex] [tex]= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})[/tex]
[tex]- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})[/tex]
[tex]- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]
[tex]- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]
[tex]- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]
[tex]V_2^2 = 316631.29 \ m/s[/tex]
[tex]V_2 = \sqrt{316631.29 \ m/s[/tex]
[tex]\mathbf{V_2 = 562.7 \ m/s}[/tex]
c) the exit area of the nozzle.
The exit of the nozzle can be determined by using the expression:
[tex]m = \dfrac{V_2A_2}{v_2}[/tex]
making [tex]A_2[/tex] the subject of the formula ; we have:
[tex]A_2 = \dfrac{ m \times v_2}{V_2}[/tex]
[tex]A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}[/tex]
[tex]A_2[/tex] = 0.0015435 m²
