A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Respuesta :

Answer:

0.0072 m³/s

Explanation:

Using Bernoulli's law

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

flow rate is constant

A₁v₁ = A₂v₂

A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²

A₂ = πr₂² = π (0.0225)² = 0.00159 m²

v₁  = (A₂ / A₁)v₂

v₁  = (0.00159 m²/ 0.0028278  m²) v₂ = 0.562  v₂

substitute v₁  into the Bernoulli's equation

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

500 ( 1 - 0.3161 ) v₂²  = (31.0 - 24 ) × 10³ Pa

341.924 v₂² = 7000

v₂² = 20.472

v₂ = √ 20.472 = 4.525 m/s

volume follow rate = 0.00159 m² ×  4.525 m/s = 0.0072 m³/s

[tex]0.0072 \;\rm m^{3}/s[/tex]The volume flow rate at the exit of the pipe is [tex]0.0072 \;\rm m^{3}/s[/tex].

Given data:

The  initial diameter of horizontal pipe is, d = 6.0 cm.

The final diameter of pipe is, d' = 4.5 cm.

The gauge pressure at inner section is, P = 31.0 kPa.

The gauge pressure at outer section is, P' = 24.0 kPa.

Applying the Bernoulli's concept, which says the total pressure energy and kinetic energy throughout the flow remains constant .

So, for the horizontal pipe, the expression is,

[tex]P + \dfrac{1}{2} \rho v^{2} = P' + \dfrac{1}{2} \rho v'^{2}[/tex] ............................................(1)

Here, [tex]\rho[/tex] is the density of water throughout the flow, which remains constant.

Now, apply the continuity equation as,

[tex]A\times v = A' \times v'\\\\(\pi/4 \times d^{2}) \times v = (\pi/4 \times d'^{2}) \times v'\\\\ d^{2} \times v = d'^{2} \times v'\\\\v/v' = 0.045^{2}/0.006^{2}\\\\v = 0.5625\times v'[/tex]

Now substitute the value in equation (1) as,

[tex]P -P' = \dfrac{1}{2} \rho v'^{2} - \dfrac{1}{2} \rho v^{2}\\\\(31 -24) \times 10^{3} \;\rm Pa = \dfrac{1}{2} \times 1000 v'^{2} - \dfrac{1}{2} \times 1000 (0.5625 v')^{2}\\v' = 4.52 \;\rm m/s[/tex]

Then the flow rate is calculated as,

[tex]Q' = A' \times v'\\\\Q' = (\pi /4) d'^{2} \times v'\\Q' = (\pi /4) \times 0.06'^{2} \times 4.52\\\\Q' = 0.0072 \;\rm m^{3}/s[/tex]

Thus, the required value of volume flow rate is, [tex]0.0072 \;\rm m^{3}/s[/tex].

Learn more about the Bernoulli's theorem here:

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