A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the top, the pipe has a diameter of 30 mm and a pressure gauge indicates a pressure of 860 kPa. At the bottom the diameter is 85 mm and a pressure gauge reading is 1 MPa. Assume the losses are negligible and determine the flov rate. Does the flow direction matter?

Respuesta :

Answer:

[tex]Q=7.3\times 10^{-3} m^3/s[/tex]

Explanation:

Given that

At top[tex]d_2=30 mm,P_2=860 KPa ,P_1=1000 KPa,d_1=85 mm[/tex]

[tex]\rho =900\dfrac{Kg}{m^3}[/tex]

We know that

[tex]\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2[/tex]

[tex]A_1V_1=A_2V_2[/tex]

[tex]\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2[/tex]

[tex]\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2[/tex]

[tex]V_2=8.02V_1[/tex]

[tex]Z_2=12 sin60^{\circ}[/tex]

[tex]\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}[/tex]

So [tex]V_1=1.30[/tex]m/s

We know that flow rate Q=AV

[tex]Q=A_1V_1[/tex]

By putting the values

[tex]A_1=\dfrac{\pi}{4}d^2[/tex]

[tex]Q=7.3\times 10^{-3} m^3/s[/tex]

To find the flow rate we do not need the direction of flow,because we are just doing balancing of energy at inlet and at the exits of pipe.

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