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3 years ago

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A pressure difference of 6.00 x 104 Pa is required to maintain a volume flow rate of 0.400 m3 /s for a viscous fluid flowing thr

ough a section of cylindrical pipe that has a radius 0.330 m. What pressure difference is required to maintain the same volume flow rate if the radius of the pipe is decreased to 0.110 m?

1 answer:

Olin [163]3 years ago

8 0

Answer:

(P₁ - P₂)f = 4.86*10⁶ Pa

Explanation:

To determine how variables affect the flow rate of an incompressible fluid undergoing laminate flow in a cylindrical tube, we use Poiseuille's equation.

Q = Πr⁴ / 8η * [(p₁ - p₂) / L]

Q = 0.40 m³/s

P₁ - P₂ = 6*10⁴ pa

r₁ = 0.33m

r₂ = 0.11m

According to Poiseuille's law, the pressure difference is inversely proportional to the radius of the pipe raised to power of 4.

P₁ - P₂ = 1 / r⁴

(P₁ - P₂)i / (P₁ - P₂)f = R⁴f / Ri⁴

Saving for (P₁ - P₂)f

(P₁ - P₂)f = (Rf⁴ / Ri⁴) * (P₁ - P₂)i

(P₁ - P₂)f = [(0.33)⁴ / (0.11)⁴] = 6*10⁴

(P₁ - P₂)f = 4.86*10⁶pa.

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Read 2 more answers

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The formula that describes this thermal expansion process is given by:

$\Delta V = \beta V_0 \Delta T$

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$\beta =$ coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

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$\Delta V_l$ = Change in the volume of liquid

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Then replacing with the previous equation we have:

$\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T$

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

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$\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)$

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