Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In flow across tube banks, why is the Reynolds number based on the maximum velocity instead of the uniform approach velocity?

Short Answer

Expert verified
Answer: The Reynolds number in flow across tube banks is based on the maximum velocity because it provides a more accurate representation of the flow behavior in these complex flow environments, accounting for the increased shear stress and flow disturbances in high-velocity regions. This allows for a better prediction of the transition from laminar to turbulent flow and a more accurate analysis of heat transfer and pressure drop.

Step by step solution

01

Understanding the Reynolds number

The Reynolds number is a dimensionless quantity that is used to predict flow patterns in fluid dynamics. It is calculated using the formula Re = ρvL/μ, where ρ is the fluid density, v is the fluid velocity, L is the characteristic length (such as pipe diameter or tube spacing), and μ is fluid viscosity. The Reynolds number helps to determine if the flow is laminar or turbulent, which is essential for analyzing fluid flow behavior, heat transfer, and pressure drop.
02

Maximum velocity in tube banks

In tube banks, the fluid flow is complicated due to the presence of multiple tubes and the interaction between the fluid and tube surfaces. The fluid velocity is not uniform across the entire flow area because of the presence of tubes. The maximum velocity occurs in the region where the flow area is the smallest, typically between two closely spaced tubes. This region of highest velocity will have a significant impact on the overall flow characteristics, such as the heat transfer rate and pressure drop.
03

Why use maximum velocity in the Reynolds number calculation

By using the maximum velocity instead of the uniform approach velocity, we get a more accurate representation of the flow behavior in tube banks. Using the maximum velocity accounts for the increased shear stress and flow disturbances in these high-velocity regions, leading to a better prediction of the transition from laminar to turbulent flow and a more accurate analysis of the heat transfer and pressure drop. In tube bank applications, understanding the flow behavior in regions of high velocity is essential for an accurate analysis of the overall performance. To summarize, the Reynolds number in flow across tube banks is based on the maximum velocity because it provides a more accurate representation of the flow behavior in these complex flow environments, accounting for the increased shear stress and flow disturbances in high-velocity regions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Air at \(20^{\circ} \mathrm{C}\) flows over a 4-m-long and 3 -m-wide surface of a plate whose temperature is \(80^{\circ} \mathrm{C}\) with a velocity of $5 \mathrm{~m} / \mathrm{s}$. The rate of heat transfer from the surface is (a) \(7383 \mathrm{~W}\) (b) \(8985 \mathrm{~W}\) (c) \(11,231 \mathrm{~W}\) (d) 14,672 W (e) 20,402 W (For air, use $k=0.02735 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7228, \nu=1.798 \times\( \)\left.10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)$

In a piece of cryogenic equipment, two metal plates are connected by a long ASTM A437 B4B stainless steel bolt. Cold gas, at \(-70^{\circ} \mathrm{C}\), flows between the plates and across the cylindrical bolt. The gas has a thermal conductivity of \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), a kinematic viscosity of \(9.3 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), and a Prandtl number of \(0.74\). The diameter of the bolt is \(9.5 \mathrm{~mm}\), and the length of the bolt exposed to the gas is \(10 \mathrm{~cm}\). The minimum temperature suitable for the ASTM A437 B4B stainless steel bolt is \(-30^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-2M). The temperature of the bolt is maintained by a heating mechanism capable of providing heat at \(15 \mathrm{~W}\). Determine the maximum velocity that the gas can achieve without cooling the bolt below the minimum suitable temperature of \(-30^{\circ} \mathrm{C}\).

Hot water vapor flows in parallel over the upper surface of a 1-m-long plate. The velocity of the water vapor is \(10 \mathrm{~m} / \mathrm{s}\) at a temperature of \(450^{\circ} \mathrm{C}\). A coppersilicon (ASTM B98) bolt is embedded in the plate at midlength. The maximum use temperature for the ASTM B98 copper-silicon bolt is \(149^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-2M). To devise a cooling mechanism to keep the bolt from getting above the maximum use temperature, it becomes necessary to determine the local heat flux at the location where the bolt is embedded. If the plate surface is kept at the maximum use temperature of the bolt, what is the local heat flux from the hot water vapor at the location of the bolt?

Consider laminar flow over a flat plate. Will the friction coefficient change with distance from the leading edge? How about the heat transfer coefficient?

The local atmospheric pressure in Denver, Colorado (elevation $1610 \mathrm{~m}\( ), is \)83.4 \mathrm{kPa}$. Air at this pressure and at \(30^{\circ} \mathrm{C}\) flows with a velocity of \(6 \mathrm{~m} / \mathrm{s}\) over a \(2.5-\mathrm{m} \times 8-\mathrm{m}\) flat plate whose temperature is \(120^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the plate if the air flows parallel to the \((a) 8-\mathrm{m}-\) long side and \((b)\) the \(2.5\)-m side.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free