Question

The cylindrical container shown in the figure has a radius of \(1.00 \mathrm{~m}\) and contains motor oil with a viscosity of \(0.300 \mathrm{~Pa} \mathrm{~s}\) and a density of \(670 . \mathrm{kg} \mathrm{m}^{-3}\). Oil flows out of the \(20.0 \mathrm{~cm}\) long, \(0.200 \mathrm{~cm}\) diameter tube at the bottom of the container. How much oil flows out of the tube in a period of \(10.0 \mathrm{~s}\) if the container is originally filled to a height of \(0.500 \mathrm{~m} ?\)

Short answer

The amount of oil that flows out of the tube in a period of 10.0 seconds is approximately \(7.377\times 10^{-6}\) cubic meters.

Step-by-step solution

List out given values and symbols for variables

We are given the following values: - Radius of the cylindrical container \(R = 1.00 m\) - Height of the oil \(h = 0.500 m\) - Viscosity of motor oil \(\eta= 0.300 Pa \cdot s\) - Density of motor oil \(\rho = 670 kg \cdot m^{-3}\) - Length of tube \(l = 20.0 cm = 0.200 m\) - Diameter of tube \(d = 0.200 cm = 0.00200m\) - Time period \( t = 10.0 s\) Now, let's assign symbols for the unknowns that we need to find: - Velocity of motor oil in tube \(v\) - Volumetric flow rate \(Q\)

Apply Hagen-Poiseuille Equation

To find the oil flow velocity through the tube, we will use Hagen-Poiseuille equation: $$v = \frac{R^2(P_1-P_2)}{8\eta l}$$ Where \(R\) is the radius of the tube, \(\eta\) is the viscosity of motor oil, \(l\) is the length of the tube and \((P_1-P_2)\) is the pressure difference. To find the pressure difference, we need to use Bernoulli's Equation.

Apply Bernoulli's Equation and determine pressure difference

We can consider the top of the oil inside the container and the oil at the bottom of the tube as points 1 and 2. Applying Bernoulli's equation: $$P_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2$$ Since the tube is narrow and we've assumed the top of the oil in the container as point 1, the velocities of the oil are very small at points 1 and 2 relative to the height \(h\). Therefore, we can ignore the velocity terms: $$P_1+\rho gh_1 = P_2 + \rho gh_2$$ Now we can calculate the pressure difference \((P_1-P_2)\): $$(P_1-P_2) = \rho g (h_1 - h_2)$$ As tube is positioned at the bottom of the container, \(h_2=0\), so we get: $$(P_1-P_2) = \rho g h_1$$

Calculate the Velocity of Oil Flow in the Tube

Now, we have all the required values to calculate the flow velocity \(v\) using the Hagen-Poiseuille equation: $$v = \frac{R^2(P_1-P_2)}{8\eta l}$$ Substitute the values and \((P_1-P_2) = \rho g h_1\): $$v= \frac{\left(\frac{d}{2}\right)^2 (\rho g h_1)}{8\eta l}$$ Calculate the flow velocity \(v\): $$v= \frac{\left(\frac{0.00200m}{2}\right)^2(670 kg \cdot m^{-3}\cdot 9.81 m \cdot s^{-2} \cdot 0.500 m)}{8 \cdot 0.300 Pa \cdot s \cdot 0.200 m}$$ $$v\approx 0.236 m \cdot s^{-1}$$

Calculate Volumetric Flow Rate

Now that we have the flow velocity, we can calculate the volumetric flow rate \(Q\). The volumetric flow rate is given by: $$Q = v \cdot A$$ where \(A\) is the cross-sectional area of the tube. Since the tube is cylindrical, the cross-sectional area is a circle: $$A = \pi \cdot(\frac{d}{2})^2$$ Now, calculate the volumetric flow rate \(Q\): $$Q = 0.236 m \cdot s^{-1} \cdot \pi\left(\frac{0.00200 m}{2}\right)^2$$ $$Q\approx 7.377\times 10^{-7} m^3 \cdot s^{-1}$$

Calculate the Amount of Oil Flowed Out in 10 seconds

Finally, to find out how much oil flows out of the tube in 10.0 seconds, multiply the volumetric flow rate by the given time: $$V = Q \cdot t$$ $$V = 7.377\times 10^{-7} m^3 \cdot s^{-1} \cdot 10.0 s$$ $$V \approx 7.377\times 10^{-6} m^3$$ So, the amount of oil that flows out of the tube in a period of 10.0 seconds is \(7.377\times 10^{-6}\) cubic meters.