Caster oil, which has a density of 0.96 × 103 kg/m3 at room temperature, is forced through a pipe of circular cross section by a pump that maintains a gauge pressure of 950 Pa. The pipe has a diameter of 2.6 cm and a length of 65 cm. The castor oil emerging from the free end of the pipe at atmospheric pressure is collected. After 90 s, a total of 1.23 kg has been collected. What is the coefficient of viscosity of the castor oil at this temperature?

To determine the coefficient of viscosity of castor oil under the given conditions, we can apply the principles of fluid dynamics, specifically using the Hagen-Poiseuille equation. This equation relates the flow rate of a viscous fluid through a pipe to the pressure difference, the viscosity of the fluid, and the dimensions of the pipe. Let's break down the problem step by step.

Understanding the Parameters

First, we need to gather all the relevant information:

• Density of castor oil (ρ): 0.96 × 10³ kg/m³
• Gauge pressure (ΔP): 950 Pa
• Diameter of the pipe (D): 2.6 cm = 0.026 m
• Length of the pipe (L): 65 cm = 0.65 m
• Mass of castor oil collected (m): 1.23 kg
• Time (t): 90 s

Calculating the Flow Rate

Next, we need to find the flow rate (Q) of the castor oil. The flow rate can be calculated using the mass collected and the time:

Q = m / t = 1.23 kg / 90 s = 0.01367 kg/s

Converting Flow Rate to Volume Flow Rate

Since the density of castor oil is known, we can convert the mass flow rate to volume flow rate (Q_v):

Q_v = Q / ρ = 0.01367 kg/s / (0.96 × 10³ kg/m³) = 1.42 × 10^-5 m³/s

Applying the Hagen-Poiseuille Equation

The Hagen-Poiseuille equation for laminar flow through a circular pipe is given by:

ΔP = (8μLQ_v) / (πR⁴)

Where:

• μ = dynamic viscosity (Pa·s)
• R = radius of the pipe = D/2 = 0.026 m / 2 = 0.013 m

Rearranging for Viscosity

We can rearrange the equation to solve for the viscosity (μ):

μ = (ΔP * πR⁴) / (8LQ_v)

Substituting the Values

Now, substituting the known values into the equation:

μ = (950 Pa * π * (0.013 m)⁴) / (8 * 0.65 m * 1.42 × 10^-5 m³/s)

Calculating Each Component

Calculating R⁴:

R⁴ = (0.013 m)⁴ = 2.56 × 10^-8 m⁴

Now substituting this back into the viscosity equation:

μ = (950 * π * 2.56 × 10^-8) / (8 * 0.65 * 1.42 × 10^-5)

Calculating the numerator:

Numerator = 950 * π * 2.56 × 10^-8 ≈ 7.63 × 10^-5 Pa·m⁴

Calculating the denominator:

Denominator = 8 * 0.65 * 1.42 × 10^-5 ≈ 7.37 × 10^-5 m³/s

Final Calculation

Now we can find μ:

μ = (7.63 × 10^-5) / (7.37 × 10^-5) ≈ 10.35 Pa·s

Result

The coefficient of viscosity of the castor oil at this temperature is approximately 10.35 Pa·s.