Please check over these answers (for white boxes only) for both parts; if incorrect, please step-by-step corrections. ### Understanding Pressure Drop in PipesPressure drop is influenced by the flow rate and the roughness of the pipe. Here, we explore the characteristics of pressure drop for a pipe with a 7.0-inch diameter.---#### Mean Velocity**What is the mean velocity for the new 7.0 in. diameter pipe?**- **$ V_0 = 8.93 \, \text{ft/s} $**---#### Reynolds Number**What is the Reynolds number for the new 7.0 in diameter pipe?**- **$ \text{Re} = 43300 $**---#### Friction Factor**What is the friction factor for the new 7.0 in diameter pipe?**- **$ f_0 = 0.0134 $**---#### Pressure Drop**What is the pressure drop per mile for the new 7.0 in diameter pipe?**- **$ \Delta p_0 = 0.0288 \, \text{psi/mile} $**---This data provides valuable insights into the fluid dynamics within the pipe, influenced by its diameter and other factors. Understanding these parameters helps in optimizing the design and operation of piping systems for efficient fluid transport. **Title: Reducing Pressure Drop in Pipes Using a Plastic Liner****Introduction:**In fluid mechanics, the pressure drop in a pipe is influenced by factors such as flow rate and surface roughness. Here, we'll explore a case study involving an old, rusty 7.8-inch diameter pipe and the effect of inserting a smooth plastic liner with a 7.0-inch inner diameter.**Problem Statement:**Water flows at a rate of 2.4 cubic feet per second (ft³/s) in the original 7.8-inch diameter pipe with a relative roughness of 0.010. By inserting a plastic liner, can the pressure drop per mile be reduced?**Diagram Explanation:**- **Old Pipe:** 7.8-inch diameter.- **New Pipe with Liner:** 7.0-inch inner diameter, smoother surface.**Calculations:**1. **Mean Velocity (Vo):** For the current 7.8-inch diameter pipe, the mean velocity is calculated as: \[ V_o = 7.23 \text{ ft/s} \]2. **Reynolds Number (Re):** Indicates whether the flow is laminar or turbulent. For the current 7.8-inch diameter pipe: \[ \text{Re} = 388528 \]3. **Friction Factor (fo):** A dimensionless number indicating the pipe’s resistance to flow: \[ f_o = 0.038 \]4. **Pressure Drop per Mile (Δpo):** Calculated for the 7.8-inch diameter pipe: \[ \Delta p_o = 0.0477 \text{ psi/mile} \]**Conclusion:**By applying this analysis and determining the pressure drop reduction potential with the smoother liner, one can evaluate if the lined pipe meets the requirement to carry 2.4 ft³/s with a decreased pressure drop as compared to the old pipe. Further calculations would be needed for the lined pipe to complete the comparative analysis.

Answered: Water flows at a rate of 2.4 ft³/s in…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Please check over these answers (for white boxes only) for both parts; if incorrect, please step-by-step corrections.

### Understanding Pressure Drop in Pipes

Pressure drop is influenced by the flow rate and the roughness of the pipe. Here, we explore the characteristics of pressure drop for a pipe with a 7.0-inch diameter.

---

#### Mean Velocity

**What is the mean velocity for the new 7.0 in. diameter pipe?**

- **$ V_0 = 8.93 \, \text{ft/s} $**

---

#### Reynolds Number

**What is the Reynolds number for the new 7.0 in diameter pipe?**

- **$ \text{Re} = 43300 $**

---

#### Friction Factor

**What is the friction factor for the new 7.0 in diameter pipe?**

- **$ f_0 = 0.0134 $**

---

#### Pressure Drop

**What is the pressure drop per mile for the new 7.0 in diameter pipe?**

- **$ \Delta p_0 = 0.0288 \, \text{psi/mile} $**

---

This data provides valuable insights into the fluid dynamics within the pipe, influenced by its diameter and other factors. Understanding these parameters helps in optimizing the design and operation of piping systems for efficient fluid transport.

**Title: Reducing Pressure Drop in Pipes Using a Plastic Liner**

**Introduction:**

In fluid mechanics, the pressure drop in a pipe is influenced by factors such as flow rate and surface roughness. Here, we'll explore a case study involving an old, rusty 7.8-inch diameter pipe and the effect of inserting a smooth plastic liner with a 7.0-inch inner diameter.

**Problem Statement:**

Water flows at a rate of 2.4 cubic feet per second (ft³/s) in the original 7.8-inch diameter pipe with a relative roughness of 0.010. By inserting a plastic liner, can the pressure drop per mile be reduced?

**Diagram Explanation:**

- **Old Pipe:** 7.8-inch diameter.
- **New Pipe with Liner:** 7.0-inch inner diameter, smoother surface.

**Calculations:**

1. **Mean Velocity (Vo):**

For the current 7.8-inch diameter pipe, the mean velocity is calculated as:
\[
V_o = 7.23 \text{ ft/s}
\]

2. **Reynolds Number (Re):**

Indicates whether the flow is laminar or turbulent. For the current 7.8-inch diameter pipe:
\[
\text{Re} = 388528
\]

3. **Friction Factor (fo):**

A dimensionless number indicating the pipe’s resistance to flow:
\[
f_o = 0.038
\]

4. **Pressure Drop per Mile (Δpo):**

Calculated for the 7.8-inch diameter pipe:
\[
\Delta p_o = 0.0477 \text{ psi/mile}
\]

**Conclusion:**

By applying this analysis and determining the pressure drop reduction potential with the smoother liner, one can evaluate if the lined pipe meets the requirement to carry 2.4 ft³/s with a decreased pressure drop as compared to the old pipe. Further calculations would be needed for the lined pipe to complete the comparative analysis.

Expert Solution

Step 1: Determine the given data

Given;

$diameter space of space pipe space left parenthesis straight D right parenthesis equals 7.8 space in equals 7.8 divided by 12 equals 0.65 space ft discharge left parenthesis straight Q right parenthesis equals 2.4 space ft cubed divided by straight s density space of space water left parenthesis straight rho right parenthesis equals 1000 kg divided by straight m cubed equals 62.4 space ft divided by straight s squared K i n e m a t i c space viscosity space of space water left parenthesis nu right parenthesis equals 1.21 cross times 10 to the power of negative 5 end exponent space f t squared divided by s diameter space of space t h e space n e w space pipe space left parenthesis straight D right parenthesis equals 7 space in equals 0.5833 space ft$

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