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Someone claims that the volume flow rate in a circular pipe with laminar flow can be determined by measuring the velocity at the centerline in the fully developed region, multiplying it by the cross-sectional area, and dividing the result by 2 . Do you agree? Explain.

Short Answer

Expert verified
Answer: Yes, the volume flow rate in a circular pipe with laminar flow can be determined using the mentioned method, as the derived proposed formula (Q') is found to be equal to the Hagen-Poiseuille equation (Q).

Step by step solution

01

Hagen-Poiseuille Equation

The Hagen-Poiseuille equation is given by: \[Q = \frac{\pi R^4 \Delta P}{8 \eta L}\] Where: - Q is the volume flow rate - R is the radius of the pipe - ΔP is the pressure difference along the pipe - η is the dynamic viscosity of the fluid - L is the length of the pipe over which pressure difference is applied Now, let us analyze the proposition given in the problem. It states that the volume flow rate can be determined by measuring the centerline velocity (v), multiplying it by the cross-sectional area (A), and dividing the result by 2. Mathematically, this can be expressed as:
02

Proposed Formula

\[Q' = \frac{v \times A}{2}\] Let's recall that for a circular pipe, cross-sectional area A is given by:
03

Cross-sectional Area

\[A = \pi R^2\] We want to compare both formulas, Q and Q', to determine if they are equivalent or can be used interchangeably. To do this, we need to find the relationship between the centerline velocity (v) and the Hagen-Poiseuille equation. From the Hagen-Poiseuille equation, we can calculate the velocity distribution along the radius (u):
04

Velocity Distribution

\[u(r) = \frac{R^2}{4 \eta} \frac{\Delta P}{L}(1 - \frac{r^2}{R^2})\] Since the centerline velocity (v) corresponds to r = 0, we can find v by plugging r = 0 into the velocity distribution equation:
05

Centerline Velocity

\[v = \frac{R^2}{4 \eta} \frac{\Delta P}{L}(1 - 0)\] Now, let's apply the proposed formula and plug in the expression for v and A:
06

Apply the Proposed Formula

\[Q' = \frac{\frac{R^2 \Delta P}{4 \eta L} \times \pi R^2}{2}\] \[Q' = \frac{\pi R^4 \Delta P}{8 \eta L}\] Comparing the proposed formula (Q') and the Hagen-Poiseuille equation (Q), we find that:
07

Comparison

\[Q = Q'\] So, based on our analysis, we agree that the volume flow rate in a circular pipe with laminar flow can be determined by measuring the velocity at the centerline in the fully developed region, multiplying it by the cross-sectional area, and dividing the result by 2.

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Key Concepts

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

Hagen-Poiseuille Equation
The Hagen-Poiseuille equation is a fundamental principle used to describe laminar flow in circular pipes. It provides a mathematical relationship between the volume flow rate and various properties of the pipe and fluid, such as the radius of the pipe, the pressure difference, the dynamic viscosity, and the length of the pipe. The equation is expressed as:\[Q = \frac{\pi R^4 \Delta P}{8 \eta L}\]Here, \(Q\) stands for the volume flow rate, which quantifies how much fluid passes through the pipe per unit of time. The radius \(R\) is critical since the flow rate is proportional to the fourth power of the radius, highlighting that small changes in \(R\) can significantly affect \(Q\).
The pressure difference \(\Delta P\) is the driving force behind the flow, and \(\eta\) denotes the dynamic viscosity, reflecting how "thick" or "sticky" the fluid is. Finally, \(L\) is the length of the pipe where this pressure difference applies.
  • The equation assumes a steady, incompressible, and Newtonian fluid.
  • It is applicable in situations where flow is fully developed and maintains a laminar, rather than turbulent, pattern.
  • This makes it very useful in contexts like water systems, blood flow in capillaries, and various engineering applications.
Centerline Velocity
The centerline velocity in a pipe with laminar flow is the maximum velocity of the fluid. This occurs in the middle of the pipe, where the effect of the viscous forces is minimized. To find the centerline velocity, one can use the velocity distribution formula derived from the Hagen-Poiseuille equation:\[u (r) = \frac{R^2}{4\eta} \frac{\Delta P}{L} \left(1 - \frac{r^2}{R^2}\right)\]For the centerline, we set \(r = 0\), thus simplifying the expression to:\[ v = \frac{R^2 \Delta P}{4 \eta L} \]This equation shows the dependence of the velocity on the pipe radius, fluid properties, and pressure difference. Knowing the centerline velocity is crucial because it forms the basis for determining the average velocity across the pipe's cross-section.
  • The centerline velocity helps in finding the average velocity, which is half of the centerline velocity in a fully developed, laminar flow.
  • It is vital for measurements and calculations concerning flow through pipelines and similar channels.
  • Understanding the velocity distribution is key in predicting how different fluids behave under given conditions.
Volume Flow Rate Measurement
Measuring the volume flow rate in a pipe is essential for numerous practical applications, from engineering systems to biological processes. The volume flow rate \(Q\) indicates the volume of fluid moving through a pipe per unit time and is pivotal for designing systems that convey fluids efficiently. A common method to approximate the flow rate in a circular pipe with laminar flow involves using the centerline velocity.
The proposed approach involves multiplying the centerline velocity by the pipe's cross-sectional area and dividing by 2:\[ Q' = \frac{v \times A}{2} \]Here, \(A = \pi R^2\) is the cross-sectional area of the pipe. By using the centerline velocity in conjunction with the area, and factoring by 2, this formula can be used to derive the true volume flow rate, reflecting the average velocity across the pipe's section.
  • This approach aligns with the Hagen-Poiseuille equation under laminar conditions.
  • Accurate flow rate measurements ensure the systems are efficient and meet required specifications.
  • Errors can lead to undersized or oversized systems, affecting performance and costs.

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Most popular questions from this chapter

Combustion gases passing through a 3-cm-internaldiameter circular tube are used to vaporize waste water at atmospheric pressure. Hot gases enter the tube at \(115 \mathrm{kPa}\) and \(250^{\circ} \mathrm{C}\) at a mean velocity of \(5 \mathrm{~m} / \mathrm{s}\), and leave at \(150^{\circ} \mathrm{C}\). If the average heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the inner surface temperature of the tube is \(110^{\circ} \mathrm{C}\), determine \((a)\) the tube length and (b) the rate of evaporation of water.

A \(15-\mathrm{cm} \times 20\)-cm printed circuit board whose components are not allowed to come into direct contact with air for reliability reasons is to be cooled by passing cool air through a 20 -cm-long channel of rectangular cross section \(0.2 \mathrm{~cm} \times 14 \mathrm{~cm}\) drilled into the board. The heat generated by the electronic components is conducted across the thin layer of the board to the channel, where it is removed by air that enters the channel at \(15^{\circ} \mathrm{C}\). The heat flux at the top surface of the channel can be considered to be uniform, and heat transfer through other surfaces is negligible. If the velocity of the air at the inlet of the channel is not to exceed \(4 \mathrm{~m} / \mathrm{s}\) and the surface temperature of the channel is to remain under \(50^{\circ} \mathrm{C}\), determine the maximum total power of the electronic components that can safely be mounted on this circuit board. As a first approximation, assume flow is fully developed in the channel. Evaluate properties of air at a bulk mean temperature of \(25^{\circ} \mathrm{C}\). Is this a good assumption?

Water enters a 5-mm-diameter and 13-m-long tube at \(45^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). The tube is maintained at a constant temperature of \(8^{\circ} \mathrm{C}\). The exit temperature of water is (a) \(4.4^{\circ} \mathrm{C}\) (b) \(8.9^{\circ} \mathrm{C}\) (c) \(10.6^{\circ} \mathrm{C}\) (d) \(12.0^{\circ} \mathrm{C}\) (e) \(14.1^{\circ} \mathrm{C}\) (For water, use \(k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6.14, v=0.894 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=997 \mathrm{~kg} / \mathrm{m}^{3}\) )

In the fully developed region of flow in a circular tube, will the velocity profile change in the flow direction? How about the temperature profile?

A computer cooled by a fan contains eight printed circuit boards (PCBs), each dissipating \(10 \mathrm{~W}\) of power. The height of the PCBs is \(12 \mathrm{~cm}\) and the length is \(18 \mathrm{~cm}\). The clearance between the tips of the components on the \(P C B\) and the back surface of the adjacent \(P C B\) is \(0.3 \mathrm{~cm}\). The cooling air is supplied by a 10 -W fan mounted at the inlet. If the temperature rise of air as it flows through the case of the computer is not to exceed \(10^{\circ} \mathrm{C}\), determine (a) the flow rate of the air that the fan needs to deliver, \((b)\) the fraction of the temperature rise of air that is due to the heat generated by the fan and its motor, and ( \(c\) ) the highest allowable inlet air temperature if the surface temperature of the components is not to exceed \(70^{\circ} \mathrm{C}\) anywhere in the system. As a first approximation, assume flow is fully developed in the channel. Evaluate properties of air at a bulk mean temperature of \(25^{\circ} \mathrm{C}\). Is this a good assumption?

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