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A liquid of density 1290 kg/m 3 1290 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.83 m/s 9.83 m/s and the pipe diameter d 1 d1 is 12.1 cm 12.1 cm . At Location 2, the pipe diameter d 2 d2 is 17.7 cm 17.7 cm . At Location 1, the pipe is 8.35 m higher than it is at location 2. Ignoring viscosity, calculate the difference between fluid pressure at location 2 and the fluid pressure at location 1.

Respuesta :

Answer:

[tex]\Delta P=1060184.8946\ Pa[/tex]

[tex]P_1=124651.2383\ Pa[/tex]

Explanation:

Given:

  • density of liquid, [tex]\rho=1290\ kg.m^{-3}[/tex]
  • speed of flow at location 1, [tex]v_1=9.83\ m.s^{-1}[/tex]
  • diameter of pipe at location 1, [tex]d_1=0.121\ m[/tex]
  • diameter of pipe at location 2, [tex]d_2=0.177\ m[/tex]
  • height of pipe at location 1, [tex]z_1=8.35\ m[/tex]

We know the Bernoulli's equation of in-compressible flow:

[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} + z=constant[/tex] ........................(1)

Cross sectional area of pipe at location 2:

[tex]A_2=\pi \frac{d_2^2}{4}[/tex]

[tex]A_2=\pi\times \frac{0.177^2}{4}[/tex]

[tex]A_2=0.0246\ m^2[/tex]

Cross sectional area of pipe at location 1:

[tex]A_1=\pi \frac{d_1^2}{4}[/tex]

[tex]A_1=\pi\times \frac{0.121^2}{4}[/tex]

[tex]A_1=0.0115\ m^2[/tex]

Using continuity equation:

[tex]A_1.v_1=A_2.v_2[/tex]

[tex]0.0115\times 9.83=0.0246\times v_2[/tex]

[tex]v_2=4.5953\ m.s^{-1}[/tex]

Now apply continuity eq. on both the locations:

[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} + z_1= \frac{P_2}{\rho.g} +\frac{v_2^2}{2g} + z_2[/tex]

[tex](P_2-P_1) = \rho.g [\frac{v_1^2}{2g} + z_1-\frac{v_2^2}{2g} ][/tex]

[tex]\Delta P=1290\times 9.8 [\frac{9.83^2}{19.6} + 8.35-\frac{4.5953^2}{19.6} ][/tex]

[tex]\Delta P=154266.016\ Pa[/tex]...................................Ans (a)

Now the mass flow rate at location 1:

[tex]\dot{m_1}=\rho\times \dot{V}[/tex]

[tex]\dot{m_1}=1290\times (0.0115\times 9.83)[/tex]

[tex]\dot{m_1}=145.828\ kg.s^{-1}[/tex]

Now pressure at location 1:

[tex]P_1=\frac{\dot{m_1}\times v_1}{A_1}[/tex]

[tex]P_1=\frac{145.828\times 9.83}{0.0115}[/tex]

[tex]P_1=124651.2383\ Pa[/tex] ...................................Ans (b)

okpalawalter8

The difference between fluid pressure at location 2 and fluid pressure at location 1 is mathematically given as

dP = 114 kPa

What is the difference between fluid pressure at location 2 and fluid pressure at location 1.?

Question Parameter(s):

Generally, the Bernoulli's equation   is mathematically given as

P + ρ*g*y + v² =pipe  constant

Where

A1*v1 = A2*v2

π*(0.105/2)²*9.91 = π*(0.167/2)²*v2

v2 = 3.9 m/s

Therefore

P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²

dP = 1290*9.8*9.01 + 9.91² - 3.9²

dP = 114 kPa

In conclusion, difference between fluid pressure is

dP = 114 kPa

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