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Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C with a velocity of 80 m/s. Calculate the velocity of the water at the tube inlet and the inlet volume flow rate. The specific volumes of water at the inlet and exit are 0.001017 m3/kg and 0.05217 m3/kg.

Respuesta :

Answer:

v_1 - 1.55 m/s

flow rate is  = 0.0206 m^3/s

Explanation:

Given data:

inside diameter is 130 mm

[tex]p_1 = 7MPa[/tex]

[tex]T_1 =65 degree C[/tex]

[tex]p_2 = 6 MPa[/tex]

[tex]T_2 = 450 degree C[/tex]

velocity v_2 = 80 m/s

specific volume [tex]\alpha_1 = 0.001017 m^3/kg[/tex]

[tex]\alpha_2 = 0.05217 m^3/kg[/tex]

initial velocity can be calculated by equating mass flow rate at inlet and outlet point

[tex]v_1  = \frac{\alpha_1}{\alpha_2} v_2[/tex]

      [tex] =\frac{0.001017}{0.05217} \times 80 = 1.55 m/s[/tex]

inital flow rate is calculated as

[tex]\dot V_1 =A_1 v_1[/tex]

            [tex] = \frac{D^2}{4} \pi v_1[/tex]

[tex]= \frac{0.13^2}{4} \pi 1.55 = 0.0206 m^3/s[/tex]

annyksl

The water velocity will be equal to [tex]1.55m/s[/tex], while the flow will be equal to [tex]0.0206 m^3/s.[/tex]

How to get to this result?

  • To calculate the speed of water we use this equation:

[tex]v_1=(\frac{a_1}{a_2})*v_2[/tex]

  • From the values presented in the question, the equation will be calculated as follows:

[tex]v_1=(\frac{ 0.001017}{0.05217 })*80\\v_1=1.55 m/s[/tex]

  • To calculate the flow we will need the equation:

[tex]V_1=(\frac{D^2}{4})*\pi *v_1[/tex]

  • When substituting the values, the result of the equation will be:

[tex]V1=(\frac{0.13^2}{4})*\pi *1.55\\V_1= 0.0206m^3/s[/tex]

More information about calculating the flow in the link:

https://brainly.com/question/4902109

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