4. Oil at a rate of 13ft/s is flowing steadily from tank A to tank B through 40 pipe (The ID of 3-in schedule 40 pipe is 3.068 in, the absolute roughness is 0.0018 in) The oil has a density of 62.3 lbm/ft³ and a viscosity of 50 cP. The levels of the free surfaces are the same in both tanks. Tank B is vented to the atmosphere. Neglecting the expansion and contraction losses, What is the gauge pressure in Tank A required to produce this flow rate? Tank A L=1000 ft Vent Tank B
4. Oil at a rate of 13ft/s is flowing steadily from tank A to tank B through 40 pipe (The ID of 3-in schedule 40 pipe is 3.068 in, the absolute roughness is 0.0018 in) The oil has a density of 62.3 lbm/ft³ and a viscosity of 50 cP. The levels of the free surfaces are the same in both tanks. Tank B is vented to the atmosphere. Neglecting the expansion and contraction losses, What is the gauge pressure in Tank A required to produce this flow rate? Tank A L=1000 ft Vent Tank B
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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i have calculate friction and reynolds but dont know how to calc the guage pressure needed
Transcribed Image Text:**Transcription and Explanation for Educational Website**
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**Problem Statement:**
4. Oil at a rate of 13 ft/s is flowing steadily from tank A to tank B through 1000 ft of 3-in schedule 40 pipe (The ID of 3-in schedule 40 pipe is 3.068 in, the absolute roughness is 0.0018 in). The oil has a density of 62.3 lbm/ft³ and a viscosity of 50 cP. The levels of the free surfaces are the same in both tanks. Tank B is vented to the atmosphere. Neglecting the expansion and contraction losses, what is the gauge pressure in Tank A required to produce this flow rate?
**Diagram Explanation:**
- **Tanks and Flow:**
- There are two tanks, labeled as Tank A and Tank B.
- Oil flows from Tank A to Tank B through a horizontal pipeline.
- **Pipeline Details:**
- The pipeline is 1000 ft in length.
- It is a 3-inch schedule 40 pipe, with an inner diameter (ID) of 3.068 inches and an absolute roughness of 0.0018 inches.
- **Flow Characteristics:**
- The flow rate of oil is 13 ft/s.
- Tank B is vented to the atmosphere, indicating atmospheric pressure in Tank B.
- **Physical Properties:**
- The oil has a density of 62.3 lbm/ft³.
- The oil viscosity is given as 50 centipoise (cP).
The task is to determine the gauge pressure in Tank A needed to achieve the specified flow rate, considering that Tank B is at atmospheric pressure and ignoring expansion and contraction losses.
![**Transcription for Educational Website**
This image contains two equations used in fluid mechanics for calculating the friction factor and Reynolds number in a pipe system.
1. **Friction Factor Equation:**
\[
f = 0.00135 \left[ 1 + \left( 20,000 \cdot \left( \frac{\epsilon}{D} + \frac{10^6}{R} \right) \right)^{\frac{1}{3}} \right]
\]
Breaking it down:
- \( f = 0.00135 \left[ 1 + \left( 20,000 \cdot \left( \frac{0.0018 \, \text{in}}{3.068 \, \text{in}} \right) + \frac{10^6}{6162.63} \right) \right] \)
- The result of the calculation is \( f = 0.000888 \).
2. **Reynolds Number Equation:**
\[
R_e = \frac{\rho DV}{\mu}
\]
Components in this equation:
- The variables are replaced by given values:
\[
R_e = \left( 3.068 \, \text{in} \times \frac{1 \, \text{ft}}{12 \, \text{in}} \right) (13 \, \text{ft/s}) \left( 62.3 \, \frac{\text{lbm}}{\text{ft}^3} \right)
\]
- The denominator is:
\[
50 \, \text{cP} \times 6.72 \times 10^{-4} \, \frac{\text{lbm}}{\text{ft.s}}
\]
- The calculated result is \( R_e = 6162.63 \).
These equations are essential for analyzing the flow characteristics and predicting the behavior of fluids in various engineering applications. The friction factor is used to determine the resistance offered by the pipe, while the Reynolds number helps to identify the flow regime, whether laminar or turbulent.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd47fc2c3-8f4b-4085-b234-bc909be774e3%2F329db1a7-9fea-4f27-96b5-ea0d62c5eab1%2Fijk03kn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Transcription for Educational Website**
This image contains two equations used in fluid mechanics for calculating the friction factor and Reynolds number in a pipe system.
1. **Friction Factor Equation:**
\[
f = 0.00135 \left[ 1 + \left( 20,000 \cdot \left( \frac{\epsilon}{D} + \frac{10^6}{R} \right) \right)^{\frac{1}{3}} \right]
\]
Breaking it down:
- \( f = 0.00135 \left[ 1 + \left( 20,000 \cdot \left( \frac{0.0018 \, \text{in}}{3.068 \, \text{in}} \right) + \frac{10^6}{6162.63} \right) \right] \)
- The result of the calculation is \( f = 0.000888 \).
2. **Reynolds Number Equation:**
\[
R_e = \frac{\rho DV}{\mu}
\]
Components in this equation:
- The variables are replaced by given values:
\[
R_e = \left( 3.068 \, \text{in} \times \frac{1 \, \text{ft}}{12 \, \text{in}} \right) (13 \, \text{ft/s}) \left( 62.3 \, \frac{\text{lbm}}{\text{ft}^3} \right)
\]
- The denominator is:
\[
50 \, \text{cP} \times 6.72 \times 10^{-4} \, \frac{\text{lbm}}{\text{ft.s}}
\]
- The calculated result is \( R_e = 6162.63 \).
These equations are essential for analyzing the flow characteristics and predicting the behavior of fluids in various engineering applications. The friction factor is used to determine the resistance offered by the pipe, while the Reynolds number helps to identify the flow regime, whether laminar or turbulent.
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