molten plastic flows out of a tube 8 cm long at a rate of 13 cm^3/min? a molten plastic flows out of a tube 8 cm long at a rate of 13 cm^3/min where the pressure differential between the two ends of the tube is 18 cm of mercury. find the viscosity of the plastic. the i.d of the tube is 1.30 mm. the density of mercury is 13.6 g/ cm^3​

Respuesta :

Answer:

97 cP

Explanation:

Use the Hagen-Poiseuille equation, also known as Poiseuille's law:

ΔP = 8μLQ / πR⁴

where ΔP is the pressure drop,

μ is the dynamic viscosity,

L is the length of the pipe,

Q is the volumetric flow rate,

and R is the radius of the pipe.

Convert all units to SI units:

ΔP = ρgh

ΔP = (13.6 g/cm³) (9.8 m/s²) (18 cm) × (1 kg / 1000 g) (100 cm/m)²

ΔP = 23,990 N/m²

L = 8 cm × (1 m / 100 cm)

L = 0.08 m

Q = 13 cm³/min × (1 m / 100 cm)³ (1 min / 60 s)

Q = 2.167×10⁻⁷ m³/s

R = (1.30 mm / 2) × (1 m / 1000 mm)

R = 6.5×10⁻⁴ m

Now plug in and solve for μ.

ΔP = 8μLQ / πR⁴

μ = πR⁴ΔP / 8LQ

μ = π (6.5×10⁻⁴ m)⁴ (23,990 N/m²) / 8 (0.08 m) (2.167×10⁻⁷ m³/s)

μ = 0.097 Pa·s

μ = 97 mPa·s = 97 cP