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Chapter 14: Problem 103

At what flow rate might turbulence begin to develop in a water main with a 0.200 -m diameter? Assume a \(20^{\circ} \mathrm{C}\) temperature.

Short Answer

Expert verified
Turbulence might begin to develop in the water main at a flow rate of approximately 0.0316 \(m^3/s\).

Step by step solution

01

Find the properties of water at 20掳C

We need the density (蟻) and dynamic viscosity (渭) of water at 20掳C. Based on standard water properties, these values are: \(蟻 = 998 kg/m^3\) \(渭 = 1.002 脳 10^{鈭�3} Ns/m^2\)
02

Determine the critical Reynolds number

The critical Reynolds number value that generally describes the transition from laminar to turbulent flow is 2000.
03

Calculate flow velocity when turbulence begins based on Reynolds number

Rearrange the Reynolds number formula to solve for flow velocity (v). \(v = \frac{Re * 渭}{蟻 * D}\) Plug in the values for Re, 渭, 蟻, and D: \(v = \frac{2000 * (1.002 脳 10^{鈭�3} Ns/m^2)}{(998 kg/m^3)*(0.200 m)}\) Now, we can calculate the flow velocity: \(v 鈮� 1.005 m/s\)
04

Calculate the flow rate at which turbulence starts

Now we need to calculate the flow rate (Q) using the flow velocity (v) and the pipe's cross-sectional area (A). The formula for flow rate is: \(Q = A * v\) The pipe has a diameter of 0.200 m, so its radius (r) is: \(r = \frac{D}{2} = \frac{0.200 m}{2} = 0.100 m\) Now we calculate the cross-sectional area (A) of the pipe using the formula: \(A = 蟺 * (r^2)\) \(A = 蟺 * (0.100 m)^2 鈮� 0.0314 m^2\) Finally, we can find the flow rate (Q) when turbulence begins: \(Q = A * v 鈮� 0.0314 m^2 * 1.005 m/s 鈮� 0.0316 m^3/s\) So, turbulence might begin to develop in the water main at a flow rate of 0.0316 m鲁/s.

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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 dimensionless value used in fluid dynamics to predict the flow pattern of a fluid. It helps determine whether the flow will be laminar or turbulent. A flow is considered laminar when the Reynolds number is less than 2000, and turbulent when it exceeds 4000. This number is crucial for engineers and scientists when analyzing fluid behavior in different conditions.

The formula for calculating the Reynolds number (Re) is:
  • Re = \( \frac{\rho v D}{\mu} \)
Where:
  • \( \rho \) = density of the fluid (kg/m鲁)
  • \( v \) = velocity of the fluid (m/s)
  • \( D \) = characteristic length, typically the diameter for pipe flow (m)
  • \( \mu \) = dynamic viscosity of the fluid (Ns/m虏)
By rearranging this formula, you can find flow velocity or even the criteria for the transition point between laminar and turbulent flow modes.
Laminar flow
Laminar flow refers to a smooth, ordered movement of fluid where all the particles follow parallel paths. This type of flow is characterized by a low Reynolds number, typically less than 2000. In pipes, laminar flow means the fluid moves in straight lines with little mixing between layers.

Key aspects of laminar flow:
  • Fluid moves in parallel layers, known as streamlines.
  • Pressure loss is mainly due to viscous forces, making it predictable and easy to calculate.
  • Common in small diameter pipes or very slow flows.
The predictable nature of laminar flow makes it ideal for scenarios requiring precise control over fluid motion, such as in chemical reactions or microfluidics.
Turbulent flow
In contrast to laminar flow, turbulent flow is chaotic and involves eddies and vortices. It occurs when the Reynolds number exceeds 4000 and is characterized by a three-dimensional, swirling motion. This type of flow increases mixing within the fluid, enhancing heat and mass transfer but also introducing complexities in flow behavior.

Key characteristics of turbulent flow:
  • High energy dissipation due to fluid instability.
  • Increased resistance and pressure drop within pipelines.
  • Better mixing, useful in industrial applications involving heat exchangers.
Understanding the transition to turbulent flow is critical for designing efficient piping systems, as it can significantly affect pressure and energy use.
Water properties
The properties of water, such as density and viscosity, change with temperature and influence fluid flow behavior. At 20掳C, water's density is approximately 998 kg/m鲁, and its dynamic viscosity is about 1.002 x 10^{-3} Ns/m虏. These properties are essential when calculating Reynolds number, predicting whether a flow will be laminar or turbulent.

Water properties are not constant; they are affected by various parameters:
  • Density (\( \rho \)): Changes very little with temperature, thus maintaining consistent mass per unit volume.
  • Dynamic viscosity (\( \mu \)): Determines the internal friction and significantly affects the flow rate.
Knowing these properties is crucial for performing accurate fluid dynamic calculations, especially in systems involving heating or cooling applications.
Flow rate calculation
Flow rate calculation determines how much fluid moves through a system over time, typically measured in cubic meters per second (m鲁/s). It is vital for designing systems like water mains, as it informs decisions on pipe sizing and pump selection.

The formula for calculating flow rate (Q) is:
  • Q = A * v
Where:
  • \( Q \) = flow rate (m鲁/s)
  • \( A \) = cross-sectional area of the pipe (m虏)
  • \( v \) = velocity of the fluid (m/s)
By using the pipe's diameter to find its cross-sectional area, you can multiply by the flow velocity to determine the flow rate. Understanding this principle helps in ensuring pipelines operate efficiently and meet demand without unnecessary energy expenditure.

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Most popular questions from this chapter

Doppler ultrasound can be used to measure the speed of blood in the body. If there is a partial constriction of an artery, where would you expect blood speed to be greatest: at or after the constriction? What are the two distinct causes of higher resistance in the constriction?

Water supplied to a house by a water main has a pressure of \(3.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) early on a summer day when neighborhood use is low. This pressure produces a flow of \(20.0 \mathrm{L} / \mathrm{min}\) through a garden hose. Later in the day, pressure at the exit of the water main and entrance to the house drops. and a flow of only \(8.00 \mathrm{L} / \mathrm{min}\) is obtained through the same hose. (a) What pressure is now being supplied to the house, assuming resistance is constant? (b) By what factor did the flow rate in the water main increase in order to cause this decrease in delivered pressure? The pressure at the entrance of the water main is \(5.00 \times 10^{5} \mathrm{N} / \mathrm{m}^{2},\) and the original flow rate was \(200 \mathrm{L} /\) min. (c) How many more users are there, assuming each would consume \(20.0 \mathrm{L} / \mathrm{min}\) in the morning?

(a) Estimate the time it would take to fill a private swimming pool with a capacity of 80,000 L using a garden hose delivering 60 L/min. (b) How long would it take if you could divert a moderate size river, flowing at \(5000 \mathrm{m}^{3} / \mathrm{s}\) into the pool?

Based on Bernoulli's equation, what are three forms of energy in a fluid? (Note that these forms are conservative, unlike heat transfer and other dissipative forms not included in Bernoulli's equation.)

Mercury is commonly supplied in flasks containing \(34.5 \mathrm{kg}\) (about 76 lb.). What is the volume in liters of this much mercury?
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