/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 A circular pipe of 25 -mm outsid... [FREE SOLUTION] | 91影视

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A circular pipe of 25 -mm outside diameter is placed in an airstream at \(25^{\circ} \mathrm{C}\) and 1 -atm pressure. The air moves in cross flow over the pipe at \(15 \mathrm{~m} / \mathrm{s}\), while the outer surface of the pipe is maintained at \(100^{\circ} \mathrm{C}\). What is the drag force exerted on the pipe per unit length? What is the rate of heat transfer from the pipe per unit length?

Short Answer

Expert verified
The drag force exerted on the circular pipe per unit length is approximately \(1.33\, \mathrm{N/m}\), and the rate of heat transfer from the pipe per unit length is approximately \(372.99\, \mathrm{W/m}\).

Step by step solution

01

Calculate the Reynolds Number (Re)

To compute the Reynolds number, we need the airstream speed (V), pipe diameter (D), the air kinematic viscosity (谓) at 25掳C. Using this data, we can calculate the Reynolds number as follows: \( Re = \frac{VD}{谓} \) For air at 25掳C and 1 atm pressure, 谓 is approximately \( 15.52 脳 10^{-6}\, \mathrm{m^2/s} \) (you can find this value in a table or online resources). Now, we have: D = 0.025 m (25 mm diameter) V = 15 m/s (airstream speed) Now, calculate the Reynolds number: \( Re = \frac{(15\, \mathrm{m/s})(0.025\, \mathrm{m})}{15.52 脳 10^{-6}\, \mathrm{m^2/s}} \approx 24,172 \)
02

Calculate the Drag Coefficient (Cd)

To compute the drag coefficient, we can use an empirical correlation based on the Reynolds number (Re) for flow over a circular cylinder. For a Reynolds number of approximately 24,172 as computed in Step 1, the drag coefficient is Cd 鈮 0.4 (you can find this value in a table or online resources).
03

Calculate the Drag Force (Fd) per Unit Length

To compute the drag force per unit length, first, we need to calculate the dynamic pressure of the airstream. To do this, we need the density (蟻) of air at 25掳C and 1 atm pressure (approximated as \(1.184\, \mathrm{kg/m^3}\)), the airstream speed (V), and the drag coefficient (Cd) as calculated in Step 2. Next, we can calculate the dynamic pressure (q) of the airstream and drag force (Fd) per unit length: \( q = \frac{(1/2)蟻V^2} \) \( Fd = \mathrm{Cd} \cdot q \cdot d \) Now, calculate the dynamic pressure and drag force per unit length: \( q = \frac{(1/2)(1.184\, \mathrm{kg/m^3})(15\, \mathrm{m/s})^2} \approx 132.87\, \mathrm{N/m^2} \) \( Fd = (0.4) (132.87\, \mathrm{N/m^2})(0.025\, \mathrm{m}) \approx 1.33\, \mathrm{N/m} \) Therefore, the drag force exerted on the pipe per unit length is approximately 1.33 N/m.
04

Calculate the Nusselt Number (Nu) and Convective Heat Transfer Coefficient (h)

To compute the Nusselt number, we need the Prandtl number (Pr) of air at 25掳C and 1 atm pressure (approximated as 0.706), and the Reynolds number (Re) as calculated in Step 1. Using the Reynolds number, the Prandtl number, and an empirical correlation for flow over a circular cylinder, we can calculate the Nusselt number as follows: \( Nu 鈮 0.3 + \frac{(0.62\,Re^{1/2}\,Pr^{1/3})}{(1 + (0.4/Pr)^{2/3})^{1/4}}(1 + (\frac{Re}{282000})^{5/8})^{4/5} = 152.59 \) Now, compute the convective heat transfer coefficient (h) using the Nusselt number and pipe diameter (D): \( h = \frac{Nu \cdot k_{air}}{D} \) For air at 25掳C and 1 atm pressure, the thermal conductivity (k) is approximately \(0.026\, \mathrm{W/(m \cdot K)}\). Now, calculate the convective heat transfer coefficient (h): \( h = \frac{(152.59)(0.026\, \mathrm{W/(m \cdot K)})}{0.025\, \mathrm{m}} \approx 158.95\, \mathrm{W/(m^2 \cdot K)} \)
05

Calculate the Rate of Heat Transfer (q) per Unit Length

Finally, we can calculate the rate of heat transfer per unit length by multiplying the convective heat transfer coefficient (h) by the surface area per unit length (A) and by the temperature difference (螖T) between the surface temperature and the airstream temperature: \( q = h \cdot A \cdot 螖T \) For a circular pipe, the surface area per unit length (A) is the circumference of the pipe, or \( A = 蟺D \). Now, insert the known values and compute the rate of heat transfer per unit length: \( q = (158.95\, \mathrm{W/(m^2 \cdot K)}) (蟺 \cdot 0.025\, \mathrm{m})(100^{\circ} \mathrm{C} - \,25^{\circ} \mathrm{C}) \approx 372.99\, \mathrm{W/m} \) In conclusion, the drag force exerted on the pipe per unit length is approximately 1.33 N/m, and the rate of heat transfer from the pipe per unit length is approximately 372.99 W/m.

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Key Concepts

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

Reynolds Number
Understanding the Reynolds number is crucial in analyzing fluid flow over a surface, such as a circular pipe. It is a dimensionless quantity that helps us determine whether the flow is laminar or turbulent. The Reynolds number, denoted by \( Re \), depends on several factors including the velocity of the fluid \( V \), the characteristic length (or diameter of the pipe \( D \)), and the kinematic viscosity \( u \) of the fluid. The formula for the Reynolds number is given by:
  • \( Re = \frac{VD}{u} \)
A higher Reynolds number usually indicates turbulent flow, while a lower value suggests laminar flow.
For example, in the calculation above, the Reynolds number was determined to be 24,172. This suggests that the airflow over the pipe is likely turbulent. Turbulence affects calculations of drag force and heat transfer rates by altering how the fluid moves over the pipe's surface.
Nusselt Number
The Nusselt number is another dimensionless number, which connects the convective and conductive heat transfer processes. It represents the enhancement of heat transfer in a fluid as a result of convection, compared to conduction alone. Typically, calculated using the Reynolds number \( Re \) and the Prandtl number \( Pr \), it is expressed by the formula:
  • \( Nu = 0.3 + \frac{(0.62 Re^{1/2} Pr^{1/3})}{(1 + (0.4/Pr)^{2/3})^{1/4}}(1 + (\frac{Re}{282000})^{5/8})^{4/5} \)
The Nusselt number helps in finding the convective heat transfer coefficient. In our example, the Nusselt number was calculated to be approximately 152.59, indicating a significant enhancement of heat transfer through the convection over conduction.
This understanding allows engineers and scientists to design systems more effectively by knowing how well heat is being expelled from surfaces into moving fluids, whether in pipes, heat exchangers, or over complex geometries.
Convective Heat Transfer Coefficient
The convective heat transfer coefficient \( h \) is a critical factor in determining how efficiently heat is transferred from a solid surface to a fluid or vice versa. It is derived from the relation between the Nusselt number \( Nu \), the fluid's thermal conductivity \( k \), and the characteristic length or diameter \( D \). The formula is:
  • \( h = \frac{Nu \cdot k_{air}}{D} \)

Given the Nusselt number and the thermal conductivity of air, the convective heat transfer coefficient was estimated to be approximately 158.95 W/(m\(^2\cdot\)K) in our example.
A higher \( h \) value suggests a more efficient heat transfer process, ideal for applications where rapid heating or cooling is needed. Understanding this helps in designing better heat exchangers or optimizing cooling in systems like engines, electronic devices, or environmental control systems.

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Most popular questions from this chapter

The boundary layer associated with parallel flow over an isothermal plate may be tripped at any \(x\)-location by using a fine wire that is stretched across the width of the plate. Determine the value of the critical Reynolds number \(R e_{x, c, o p}\) that is associated with the optimal location of the trip wire from the leading edge that will result in maximum heat transfer from the warm plate to the cool fluid.

Highly reflective aluminum coatings may be formed on the surface of a substrate by impacting the surface with molten drops of aluminum. The droplets are discharged from an injector, proceed through an inert gas (helium), and must still be in a molten state at the time of impact. \(V=3 \mathrm{~m} / \mathrm{s}\), and \(T_{i}=1100 \mathrm{~K}\), respectively, traverse a stagnant layer of atmospheric helium that is at a temperature of \(T_{\infty}=300 \mathrm{~K}\). What is the maximum allowable thickness of the helium layer needed to ensure that the temperature of droplets impacting the substrate is greater than or equal to the melting point of aluminum \(\left(T_{f} \geq T_{\text {mp }}=933 \mathrm{~K}\right)\) ? Properties of the molten aluminum may be approximated as \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=\) \(1200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

In the production of sheet metals or plastics, it is customary to cool the material before it leaves the production process for storage or shipment to the customer. Typically, the process is continuous, with a sheet of thickness \(\delta\) and width \(W\) cooled as it transits the distance \(L\) between two rollers at a velocity \(V\). In this problem, we consider cooling of an aluminum alloy (2024-T6) by an airstream moving at a velocity \(u_{\infty}\) in counter flow over the top surface of the sheet. A turbulence promoter is used to provide turbulent boundary layer development over the entire surface. (a) By applying conservation of energy to a differential control surface of length \(d x\), which either moves with the sheet or is stationary and through which the sheet passes, derive a differential equation that governs the temperature distribution along the sheet. Because of the low emissivity of the aluminum, radiation effects may be neglected. Express your result in terms of the velocity, thickness, and properties of the sheet \(\left(V, \delta, \rho, c_{p}\right)\), the local convection coefficient \(h_{x}\) associated with the counter flow, and the air temperature. For a known temperature of the sheet \(\left(T_{i}\right)\) at the onset of cooling and a negligible effect of the sheet velocity on boundary layer development, solve the equation to obtain an expression for the outlet temperature \(T_{a}\). (b) For \(\delta=2 \mathrm{~mm}, V=0.10 \mathrm{~m} / \mathrm{s}, L=5 \mathrm{~m}, W=1 \mathrm{~m}\), \(u_{\infty}=20 \mathrm{~m} / \mathrm{s}, T_{\infty}=20^{\circ} \mathrm{C}\), and \(T_{i}=300^{\circ} \mathrm{C}\), what is the outlet temperature \(T_{a}\) ?

Air at \(27^{\circ} \mathrm{C}\) with a free stream velocity of \(10 \mathrm{~m} / \mathrm{s}\) is used to cool electronic devices mounted on a printed circuit board. Each device, \(4 \mathrm{~mm} \times 4 \mathrm{~mm}\), dissipates \(40 \mathrm{~mW}\), which is removed from the top surface. A turbulator is located at the leading edge of the board, causing the boundary layer to be turbulent. (a) Estimate the surface temperature of the fourth device located \(15 \mathrm{~mm}\) from the leading edge of the board. (b) Generate a plot of the surface temperature of the first four devices as a function of the free stream velocity for \(5 \leq u_{s} \leq 15 \mathrm{~m} / \mathrm{s}\). (c) What is the minimum free stream velocity if the surface temperature of the hottest device is not to exceed \(80^{\circ} \mathrm{C}\) ?

Dry air at atmospheric pressure and \(350 \mathrm{~K}\), with a free stream velocity of \(25 \mathrm{~m} / \mathrm{s}\), flows over a smooth, porous plate \(1 \mathrm{~m}\) long. (a) Assuming the plate to be saturated with liquid water at \(350 \mathrm{~K}\), estimate the mass rate of evaporation per unit width of the plate, \(n_{\mathrm{A}}^{\prime}(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m})\). (b) For air and liquid water temperatures of 300,325 , and \(350 \mathrm{~K}\), generate plots of \(n_{\mathrm{A}}^{\prime}\) as a function of velocity for the range from 1 to \(25 \mathrm{~m} / \mathrm{s}\).

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