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laiz [17]
3 years ago
12

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?
Physics
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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Answer:

a = 3 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} =v_{o} +a*t

where:

Vf = final velocity = 60 [m/s]

Vo = initial velocity = 20 [m/s]

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Now replacing these values we have:

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Answer:

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Explanation:

Density is equal to the ratio between the mass of an object and its volume:

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In our problem, we know:

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So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:

V=\frac{m}{d}=\frac{44 g}{2.70 g/mL}=16.3 mL

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