To solve the problem, we apply the Bernoulli equation and the continuity equation. Here are the steps:

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Key Given Data:

• Density of oil: $\rho_{\text{oil}} = 915 \, \text{kg/m}^3$
• Volume flow rate at $P$: $Q = 1.20 \times 10^{-2} \, \text{m}^3/\text{s}$
• Radius at $P$: $r_P = 0.10 \, \text{m}$
• Radius at $Q$: $r_Q = 1.20 \, \text{m}$
• Height difference ($h$) between $Q$ and $P$: $h = 1.25 \, \text{m}$
• Atmospheric pressure at $P$: $P_P = 0 \, \text{gauge pressure}$

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Step 1: Continuity Equation

The continuity equation relates the flow rate and areas: $A_P v_P = A_Q v_Q$ where:

• $A_P = \pi r_P^2$: Cross-sectional area at $P$
• $A_Q = \pi r_Q^2$: Cross-sectional area at $Q$
• $v_P, v_Q$: Velocities at $P$ and $Q$, respectively.

The flow rate $Q$ is given by: $Q = A_P v_P = \pi r_P^2 v_P$ Thus: $v_P = \frac{Q}{A_P} = \frac{1.20 \times 10^{-2}}{\pi (0.10)^2} \approx 3.82 \, \text{m/s}$

Now, using the continuity equation to find $v_Q$: $A_P v_P = A_Q v_Q \implies v_Q = \frac{A_P}{A_Q} v_P = \frac{\pi (0.10)^2}{\pi (1.20)^2} v_P$ $v_Q = \frac{(0.10)^2}{(1.20)^2} v_P = \frac{0.01}{1.44} \times 3.82 \approx 0.0265 \, \text{m/s}$

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Step 2: Bernoulli's Equation

The Bernoulli equation is: $P_Q + \frac{1}{2} \rho v_Q^2 + \rho g z_Q = P_P + \frac{1}{2} \rho v_P^2 + \rho g z_P$ where:

• $P_Q$: Pressure at $Q$ (to find),
• $P_P = 0 \, \text{(gauge pressure)}$,
• $v_P = 3.82 \, \text{m/s}$,
• $v_Q = 0.0265 \, \text{m/s}$,
• $z_Q - z_P = h = 1.25 \, \text{m}$,
• $\rho = 915 \, \text{kg/m}^3$.

Rewriting: $P_Q = P_P + \frac{1}{2} \rho v_P^2 - \frac{1}{2} \rho v_Q^2 + \rho g h$ Substitute known values: $P_Q = 0 + \frac{1}{2} (915) (3.82)^2 - \frac{1}{2} (915) (0.0265)^2 + (915)(9.81)(1.25)$

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Step 3: Calculate Terms

1. Kinetic energy at $P$: $\frac{1}{2} \rho v_P^2 = \frac{1}{2} (915) (3.82)^2 \approx 6672.9 \, \text{Pa}$

2. Kinetic energy at $Q$: $\frac{1}{2} \rho v_Q^2 = \frac{1}{2} (915) (0.0265)^2 \approx 0.32 \, \text{Pa}$

3. Potential energy difference: $\rho g h = (915)(9.81)(1.25) \approx 11241.1 \, \text{Pa}$

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Step 4: Final Calculation

$P_Q = 0 + 6672.9 - 0.32 + 11241.1 \approx 17913.7 \, \text{Pa}$

Thus, the pressure at $Q$ is approximately: $\boxed{17914 \, \text{Pa}}$

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Let me know if you want more details or have further questions.

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Related Questions:

1. How would the result change if the oil had a different density?
2. What would happen if the height difference $h$ were doubled?
3. How does the flow velocity at $P$ compare with that at $Q$?
4. What assumptions are we making in using Bernoulli's equation?
5. How is the atmospheric pressure factored into the calculation?

Tip:

Always ensure that units are consistent when applying Bernoulli’s equation, especially for density, height, and velocities