To solve the problem, we need to calculate the volume flow rate (Q) and the mass flow rate (ṁ).

Step 1: Interpret the Given Information

1. Flow rate for bottles: $Q = \frac{\text{Total volume per hour}}{\text{time}} = \frac{75000 \times 500 \text{ mL}}{1 \text{ hour}}$. Since $1 \text{ mL} = 10^{-6} \text{ m}^3$: $Q = 75000 \times 500 \times 10^{-6} \text{ m}^3/\text{hour}.$ Convert to m³/s: $Q = \frac{75000 \times 500 \times 10^{-6}}{3600} \text{ m}^3/\text{s}.$ Simplifying: $Q = 10.42 \times 10^{-3} \text{ m}^3/\text{s} = 0.01042 \text{ m}^3/\text{s}.$

2. Volume Flow Rate: $Q = 0.01042 \, \text{m}^3/\text{s}$.

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Step 2: Mass Flow Rate

The mass flow rate $\dot{m}$ is given by: $\dot{m} = \rho_{\text{oil}} \cdot Q.$ Given $\rho_{\text{oil}} = 912 \, \text{kg/m}^3$ and $Q = 0.01042 \, \text{m}^3/\text{s}$: $\dot{m} = 912 \cdot 0.01042 = 9.50 \, \text{kg/s}.$

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Step 3: Verify Velocity at Points XX and YY (Bernoulli's Principle)

The velocity at each point can be calculated using the continuity equation: $A_X v_X = A_Y v_Y,$ where $A_X = \pi (d_X/2)^2$ and $A_Y = \pi (d_Y/2)^2$.

1. Cross-sectional areas:

   • For $d_X = 5.80 \, \text{cm} = 0.058 \, \text{m}$: $A_X = \pi (0.058/2)^2 = \pi (0.029)^2 = 2.64 \times 10^{-3} \, \text{m}^2.$
   • For $d_Y = 1.50 \, \text{cm} = 0.015 \, \text{m}$: $A_Y = \pi (0.015/2)^2 = \pi (0.0075)^2 = 1.77 \times 10^{-4} \, \text{m}^2.$

2. Velocities: Using $Q = A \cdot v$, we solve for $v_X$ and $v_Y$:

   • At $X$: $v_X = \frac{Q}{A_X} = \frac{0.01042}{2.64 \times 10^{-3}} = 3.95 \, \text{m/s}.$
   • At $Y$: $v_Y = \frac{Q}{A_Y} = \frac{0.01042}{1.77 \times 10^{-4}} = 58.9 \, \text{m/s}.$

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Final Answer

1. Volume Flow Rate $Q$: $0.01042 \, \text{m}^3/\text{s}$.
2. Mass Flow Rate $\dot{m}$: $9.50 \, \text{kg/s}$.

Let me know if you'd like further clarifications or detailed steps!

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Related Questions:

1. How does the velocity difference between $X$ and $Y$ impact pressure drop in the pipe?
2. Can Bernoulli’s equation be applied to calculate the pressure difference?
3. What is the kinetic energy difference between $X$ and $Y$?
4. How would the mass flow rate change if the density of oil increased?
5. What happens to the flow rates if $d_X$ is reduced by 50%?

Tip:

Always convert flow rates to consistent units (e.g., $\text{m}^3/\text{s}$) before proceeding with calculations!