Luigi is attempting to fix some Mushroom Kingdom plumbing. At one point in the plumbing, a horizontal pipe is carrying an incompressible frictionless fluid of density 1.31*10^ 3 kg/m^ 3 . The pipe widens from d_{1} = 2.59 cm at left to d_{2} = 6.61 cm at right , as shown below.If the fluid enters the narrower left end of the pipe at a speed of 2.33 m/s , by how much must the pressure drop or rise (in Pa) at the wider right end of the pipe? Be sure to include the appropriate sign!Explain how you solved the problem involving a fluid passing through a pipe. Be sure to state what your known and unknown quantities are, what concepts were applied , and what equations were used! ### Fluid Dynamics Example: Mushroom Kingdom Plumbing Problem#### Problem StatementLuigi is attempting to fix some Mushroom Kingdom plumbing. At one point in the plumbing, a horizontal pipe is carrying an incompressible, frictionless fluid of density \(1.31 \times 10^3 \text{ kg/m}^3\). The pipe widens from \(d_1 = 2.59 \, \text{cm}\) at the left to \(d_2 = 6.61 \, \text{cm}\) at the right, as shown below.![Diagram of plumbing issue]#### Explanation of DiagramThe diagram shows a pipe with two distinct widths. The narrower left end has a diameter \(d_1\) of \(2.59 \, \text{cm}\) and the wider right end has a diameter \(d_2\) of \(6.61 \, \text{cm}\).#### QuestionIf the fluid enters the narrower left end of the pipe at a speed of \(2.33 \, \text{m/s}\), by how much must the pressure drop or rise (in Pa) at the wider right end of the pipe? Be sure to include the appropriate sign!#### Solution StepsTo solve for the pressure difference, you will likely need to apply the principle of conservation of mass and Bernoulli's equation.1. **Conservation of Mass: Continuity Equation** \[ A_1 v_1 = A_2 v_2 \] Where \( A_1 = \pi (d_1/2)^2 \) and \( A_2 = \pi (d_2/2)^2 \).2. **Bernoulli’s Equation:** \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \] Where: - \( P_1 \) and \( P_2 \) are the pressures at the two ends. - \( v_1 \) is the speed at the narrow end (2.33 m/s). - \(v_2 \) is required for the calculation. - \( \rho \) is the fluid density (1.31 x 10^3 kg/m\(^3\)).By rearr

Name: Luigi is attempting to fix some Mushroom Kingdom plumbing. At one poin
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Description: Luigi is attempting to fix some Mushroom Kingdom plumbing. At one point in the plumbing, a horizontal pipe is carrying an incompressible frictionless fluid of density 1.31*10^ 3 kg/m^ 3 . The pipe widens from d_{1} = 2.59 cm at left to d_{2} = 6.61 c

Given Data: The density of the fluid is: ρ=1.31x103 kg/m3 The diameter of the narrower left end is:…

Answered: Luigi is attempting to fix some…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Luigi is attempting to fix some Mushroom Kingdom plumbing. At one point in the plumbing, a horizontal pipe is carrying an incompressible frictionless fluid of density 1.31*10^ 3 kg/m^ 3 . The pipe widens from d_{1} = 2.59 cm at left to d_{2} = 6.61 cm at right , as shown below. If the fluid enters the narrower left end of the pipe at a speed of 2.33 m/s , by how much must the pressure drop or rise (in Pa) at the wider right end of the pipe? Be sure to include the appropriate sign! Explain how you solved the problem involving a fluid passing through a pipe. Be sure to state what your known and unknown quantities are, what concepts were applied , and what equations were used!

### Fluid Dynamics Example: Mushroom Kingdom Plumbing Problem

#### Problem Statement
Luigi is attempting to fix some Mushroom Kingdom plumbing. At one point in the plumbing, a horizontal pipe is carrying an incompressible, frictionless fluid of density \(1.31 \times 10^3 \text{ kg/m}^3\). The pipe widens from \(d_1 = 2.59 \, \text{cm}\) at the left to \(d_2 = 6.61 \, \text{cm}\) at the right, as shown below.

![Diagram of plumbing issue]

#### Explanation of Diagram
The diagram shows a pipe with two distinct widths. The narrower left end has a diameter \(d_1\) of \(2.59 \, \text{cm}\) and the wider right end has a diameter \(d_2\) of \(6.61 \, \text{cm}\).

#### Question
If the fluid enters the narrower left end of the pipe at a speed of \(2.33 \, \text{m/s}\), by how much must the pressure drop or rise (in Pa) at the wider right end of the pipe? Be sure to include the appropriate sign!

#### Solution Steps
To solve for the pressure difference, you will likely need to apply the principle of conservation of mass and Bernoulli's equation.

1. **Conservation of Mass: Continuity Equation**
\[
A_1 v_1 = A_2 v_2
\]
Where \( A_1 = \pi (d_1/2)^2 \) and \( A_2 = \pi (d_2/2)^2 \).

2. **Bernoulli’s Equation:**
\[
P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2
\]
Where:
- \( P_1 \) and \( P_2 \) are the pressures at the two ends.
- \( v_1 \) is the speed at the narrow end (2.33 m/s).
- \(v_2 \) is required for the calculation.
- \( \rho \) is the fluid density (1.31 x 10^3 kg/m\(^3\)).

By rearr

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