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Chapter 5: Problem 56

Gasoline at \(20^{\circ} \mathrm{C}\) is to be delivered in a 25 -m-long tube where the difference in pressure between the entrance and exit of the tube is \(5 \mathrm{kPa}\). If laminar flow conditions are to be maintained in the tube, (a) what is the maximum tube diameter than can be used and (b) what is the volume flow rate through the tube when using this maximum diameter?

Short Answer

Expert verified
The maximum tube diameter for laminar flow is critical and is calculated first, applying it yields the maximum volume flow rate.

Step by step solution

01

Identify the Reynolds number for laminar flow

For a flow to be considered laminar, the Reynolds number must be less than 2000. The Reynolds number, \(Re\), is calculated by the formula:\[Re = \frac{\rho V D}{\mu}\]where:- \(\rho\) is the density of the fluid,- \(V\) is the velocity of the fluid,- \(D\) is the diameter of the tube,- \(\mu\) is the dynamic viscosity of the fluid. We will use this condition to find the maximum diameter (Step 4).
02

Use the Hagen-Poiseuille equation

To establish a relation between diameter, flow rate, and pressure drop, we'll use the Hagen-Poiseuille equation for laminar flow:\[Q = \frac{\pi (\Delta P) D^4}{128 \mu L}\]where:- \(Q\) is the volume flow rate,- \(\Delta P\) is the pressure drop (5000 Pa),- \(D\) is the diameter,- \(\mu\) is the dynamic viscosity,- \(L\) is the length of the tube (25 m).
03

Gather physical property data for gasoline

At \(20^{\circ} \mathrm{C}\), the properties for gasoline are:- Density \(\rho = 680 \text{ kg/m}^3\)- Dynamic viscosity \(\mu = 5.4 \times 10^{-4} \text{ kg/ms}\).These values are essential to calculate both the Reynolds number and volume flow rate.
04

Calculate the maximum diameter using Reynolds number

To maintain laminar flow, \(Re < 2000\). Rearrange the Reynolds number formula to solve for \(D\):\[2000 = \frac{\rho V_{max} D}{\mu}\]Since \(V_{max} = \frac{Q_{max}}{A}\) and \(A = \frac{\pi D^2}{4}\), use the Hagen-Poiseuille equation:\[V_{max} = \frac{\Delta P D^2}{32 \mu L}\]Substitute this into the \(Re\) equation and solve for \(D\).Insert the numeric values:\[D^3 = \frac{2000 \cdot 32 \cdot 5.4 \times 10^{-4} \cdot 25}{680 \cdot 5000 / \pi}\]Solve for \(D\) numerically.
05

Calculate the volume flow rate with the maximum diameter

Once we have the maximum diameter \(D_{max}\), substitute it back into the Hagen-Poiseuille equation to find the maximum volume flow rate:\[Q_{max} = \frac{\pi (5000) (D_{max})^4}{128 (5.4 \times 10^{-4}) (25)}\]Calculate \(Q_{max}\) using \(D_{max}\) from Step 4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reynolds Number
The Reynolds number is a critical dimensionless value used to determine the flow regime in a fluid system. In essence, it helps us differentiate between laminar and turbulent flow. In the realm of this exercise, the primary focus is on ensuring the flow remains laminar. Laminar flow is characterized by smooth and orderly motion of the fluid, where all particles of the fluid move in layers or parallel paths. To maintain this type of flow within a tube, the Reynolds number
  • should be less than 2000.
Reynolds number is calculated using the formula: \[ Re = \frac{\rho V D}{\mu} \]where:
  • \(\rho\) is the fluid density,
  • \(V\) is the fluid velocity,
  • \(D\) is the tube diameter,
  • \(\mu\) represents dynamic viscosity.
Understanding the Reynolds number is essential as it dictates the conditions under which the tube's diameter should be calculated to ensure a smooth, laminar flow of gasoline.
Hagen-Poiseuille Equation
The Hagen-Poiseuille equation is a fundamental principle in fluid dynamics, specifically used for modeling laminar flow through a cylindrical pipe. It's particularly useful because it provides a relationship among several factors such as the tube diameter, fluid's viscosity, length of the tube, and the pressure drop. The equation is given by:\[Q = \frac{\pi (\Delta P) D^4}{128 \mu L}\]where:
  • \(Q\) is the volume flow rate,
  • \(\Delta P\) signifies the pressure drop along the tube,
  • \(D\) is the diameter of the tube,
  • \(\mu\) is the dynamic viscosity,
  • \(L\) represents the length of the tube.
This equation allows us to compute the maximum diameter that retains laminar flow, given the pressure drop and other parameters. To solve the exercise, we use this equation along with the derived values of the tube's properties and dimensional constraints to determine the permissible diameter and the resulting flow rate.
Dynamic Viscosity
Dynamic viscosity is a fundamental property of fluids that measures its resistance to flow when subjected to an external force or pressure difference. Think of it as a fluid's internal "thickness" or resistance to movement. In the provided exercise, it's crucial because it affects how easily gasoline flows through the tube. Dynamic viscosity \(\mu\) is a factor in both Reynolds number and the Hagen-Poiseuille equation:- For Reynolds number: It denotes the fluid's drag on the internal flow layers.- In Hagen-Poiseuille equation: It acts to reduce the flow rate for higher viscosity, impacting the overall volume throughput.Gasoline at \(20^{\circ} \mathrm{C}\) has a dynamic viscosity of \(5.4 \times 10^{-4} \text{ kg/ms}\). This low viscosity means gasoline can flow relatively easier compared to more viscous fluids like oil. Understanding dynamic viscosity is essential for choosing the correct tube diameter to maintain the desired laminar flow condition.

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Most popular questions from this chapter

The nozzle shown in Figure 5.50 is designed such that the velocity, \(\mathbf{v}\), at any point near the centerline of the nozzle under steady- state conditions can be estimated by $$ \mathbf{v}=\frac{4}{0.8-x} \mathbf{i} \mathrm{m} / \mathrm{s} $$ where \(x\) is the distance in meters from the center of the nozzle entrance, measured along the centerline of the nozzle. The centerline velocities at the entrance and exit of the nozzle are \(4 \mathrm{~m} / \mathrm{s}\) and \(20 \mathrm{~m} / \mathrm{s}\), respectively, and the length of the nozzle is \(0.6 \mathrm{~m}\). What is the acceleration of a fluid element at the center of the midsection of the nozzle, where \(x=0.2 \mathrm{~m} ?\)

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A two-dimensional velocity field in the \(x z\) plane is described by the velocity components \(u=2\left(x^{2}-z^{2}\right)-6 x \mathrm{~m} / \mathrm{s}\) and \(w=4 x z-6 z \mathrm{~m} / \mathrm{s}\), where \(x\) and \(z\) are the Cartesian coordinates in meters. The gravity force acts in the negative \(z\) -direction, and the fluid has a density of \(998 \mathrm{~kg} / \mathrm{m}^{3}\). Determine the pressure gradients in the \(x\) and \(z\) -directions at the point \(x=2 \mathrm{~m}, z=1 \mathrm{~m}\).

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