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Chapter 7: Problem 46

Consider the following fluids, each with a velocity of \(V=5 \mathrm{~m} / \mathrm{s}\) and a temperature of \(T_{\infty}=20^{\circ} \mathrm{C}\), in cross flow over a 10-mm-diameter cylinder maintained at \(50^{\circ} \mathrm{C}\) : atmospheric air, saturated water, and engine oil. (a) Calculate the rate of heat transfer per unit length, \(q^{\prime}\), using the Churchill-Bernstein correlation. (b) Generate a plot of \(q^{\prime}\) as a function of fluid velocity for \(0.5 \leq V \leq 10 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
In summary, for fluids atmospheric air, saturated water, and engine oil flowing over a 10-mm-diameter cylinder maintained at 50掳C, we calculated the rate of heat transfer per unit length, \(q^{\prime}\), using the Churchill-Bernstein correlation. First, fluid properties at 20掳C, Reynolds number, Prandtl number, and Nusselt number were determined. Then, \(q^{\prime}\) was calculated using the formula \(q^{\prime} = Nu \frac{k}{D} (T_w - T_{\infty})\). Lastly, a plot was generated to display the rate of heat transfer per unit length as a function of fluid velocity ranging from 0.5 m/s to 10 m/s.

Step by step solution

01

Fluid Properties

The properties of atmospheric air, saturated water, and engine oil at 20掳C can be obtained from standard tables or reliable sources. The properties we need are the density 蟻, specific heat capacity c_p, viscosity 渭, and thermal conductivity k. You can find these properties in standard engineering textbooks or online sources such as NIST Webbook or Engineering Toolbox.
02

Calculation of Reynolds Number

We will now calculate the Reynolds number for each fluid using the formula: \[ Re = \frac{\rho V D}{\mu} \] where Re is the Reynolds number, 蟻 is the fluid density, V is the fluid velocity (5 m/s), D is the diameter of the cylinder (0.01 m), and 渭 is the fluid viscosity. Calculate the Reynolds number for each fluid using their respective properties. Now, we can proceed to calculate the Nusselt number using the Churchill-Bernstein correlation.
03

Churchill-Bernstein Correlation

The Churchill-Bernstein correlation is given by: \[ Nu = 0.3 + \frac{(0.62 Re^{1/2} Pr^{1/3})}{[1 + (0.4 / Pr)^{2/3}]^{1/4}} \left[1 + \left(\frac{Re}{282000}\right)^{5/8}\right]^{4/5} \] where Nu is the Nusselt number, Re is the Reynolds number, and Pr is the Prandtl number (Pr = 渭*c_p/k, the ratio of the fluid's viscosity to its thermal conductivity). Calculate the Prandtl number for each fluid and use the correlation to find the Nusselt number.
04

Rate of Heat Transfer Per Unit Length

With the Nusselt number calculated, we can now find the rate of heat transfer per unit length, q', using the formula: \[ q^{\prime} = Nu \frac{k}{D} (T_w - T_{\infty}) \] where q' is the rate of heat transfer per unit length, Nu is the Nusselt number, k is the thermal conductivity, D is the diameter of the cylinder (0.01 m), T_w is the temperature of the cylinder (50掳C), and T鈭� is the temperature of the fluid (20掳C). Calculate the rate of heat transfer per unit length, q', for each fluid. Finally, let's generate the plot of q' as a function of fluid velocity.
05

Plotting q' vs Fluid Velocity

To generate the plot of the rate of heat transfer per unit length, q', as a function of fluid velocity (V) ranging from 0.5 m/s to 10 m/s, we will follow these steps: 1. Create a list of fluid velocities ranging from 0.5 m/s to 10 m/s (e.g., V = [0.5, 1, 1.5, ..., 10]). 2. For each fluid velocity, calculate the Reynolds number, Prandtl number, and Nusselt number using the formulas discussed above. 3. Use the q' formula to compute the rate of heat transfer per unit length for each fluid at their respective velocities. 4. Create a plot with the fluid velocities (x-axis) and q' values (y-axis) for each fluid (air, water, and oil). This plot will visually display the relationship between the rate of heat transfer per unit length and the fluid velocity for each fluid flowing over the 10-mm-diameter cylinder.

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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 quantity in fluid dynamics used to predict flow regimes in different fluid scenarios. It is derived from the balance of inertial forces to viscous forces and 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 characteristic length (in this exercise, the cylinder diameter);
  • \(\mu\) is the dynamic viscosity.
A low Reynolds number indicates laminar flow where viscous forces dominate, while a high number suggests turbulent flow with inertial forces playing a greater role.
This understanding helps in understanding how heat transfer behaves differently under various flow conditions.
When calculating the Reynolds number for different fluids like air, water, and engine oil, knowing each fluid's unique properties is essential for accurate analysis.
Nusselt Number
The Nusselt number represents the enhancement of heat transfer through a fluid as compared to pure conduction. It is another dimensionless number calculated knowing the Reynolds and Prandtl numbers, often via the Churchill-Bernstein correlation for cylindrical objects. The formula used is:\[ Nu = 0.3 + \frac{(0.62 Re^{1/2} Pr^{1/3})}{[1 + (0.4 / Pr)^{2/3}]^{1/4}} \left[1 + \left(\frac{Re}{282000}\right)^{5/8}\right]^{4/5} \]Important Highlights:
  • A high Nusselt number implies more effective convective heat transfer.
  • It bridges the connection between fluid flow and heat transfer.
  • The Nusselt number depends on flow conditions (Reynolds number) and fluid properties (Prandtl number).
Along a cylinder, the Nusselt number provides insight into how heat dissipates with different fluid velocities and properties,
helping determine the conduction and convection balance over the object's surface.
Prandtl Number
The Prandtl number is another dimensionless value that aids in characterizing fluid flow. It primarily relates the kinematic viscosity of a fluid to its thermal diffusivity:\[ Pr = \frac{\mu c_p}{k} \]where:
  • \(\mu\) is dynamic viscosity;
  • \(c_p\) is specific heat capacity;
  • \(k\) is thermal conductivity.
Key Points:
  • A low Prandtl number indicates that thermal diffusivity is dominant over momentum diffusivity, common in liquid metals.
  • A high Prandtl number signifies momentum diffusivity dominance, typical in oils.
  • The Prandtl number heavily influences the Nusselt number calculation.
In cross-flow heat transfer situations, different fluids will exhibit distinct Prandtl numbers,
which results in variation in heat transfer efficiency over the cylindrical surface.
Churchill-Bernstein Correlation
The Churchill-Bernstein correlation is a widely used empirical relation to estimate the Nusselt number for heat transfer over cylindrical bodies in cross flow. This correlation is particularly useful when dealing with varied Reynolds and Prandtl numbers. Its formula is complex yet comprehensive, adjusting for various fluid dynamic conditions effectively. Features of the Churchill-Bernstein Correlation:
  • Covers a broad range of Reynolds numbers, making it adaptable for both laminar and turbulent flows.
  • Incorporates the Prandtl number, offering flexibility in different fluid conditions.
  • Derives from experimental data, providing practical heat transfer predictions in engineering applications.
By utilizing this correlation, engineers and students can predict the Nusselt number for different fluid dynamics scenarios directly,
which aids in efficient design and analysis of heat transfer equipment and processes.
Fluid Dynamics
Fluid dynamics is the study of fluids in motion, encompassing a multitude of physical principles and laws to predict and analyze fluid behavior in various settings. Core Aspects of Fluid Dynamics:
  • Considers factors like fluid velocity, pressure, density, and temperature.
  • Employs mathematical equations like the Navier-Stokes equations to describe the motion of fluids.
  • An integral part of many engineering disciplines such as mechanical, chemical, and civil engineering.
When solving real-world problems, fluid dynamics is essential to understanding how different fluids will transfer heat when flowing across surfaces like a cylinder.
It informs decisions about designing systems in heating, ventilating, and air-conditioning (HVAC), engine cooling, and many more applications,
where efficient and effective heat transfer plays a critical role.

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Most popular questions from this chapter

A cryogenic probe is used to treat cancerous skin tissue. The probe consists of a single round jet of diameter \(D_{e}=2 \mathrm{~mm}\) that issues from a nozzle concentrically situated within a larger, enclosed cylindrical tube of outer diameter \(D_{o}=15 \mathrm{~mm}\). The wall thickness of the AISI 302 stainless steel probe is \(t=2 \mathrm{~mm}\), and the separation distance between the nozzle and the inner surface of the probe is \(H=5 \mathrm{~mm}\). Assuming the cancerous skin tissue to be a semiinfinite medium with \(k_{c}=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(T_{c}=37^{\circ} \mathrm{C}\) far from the probe location, determine the surface temperature \(T_{s}\). Neglect the contact resistance between the probe and the tissue. Cold nitrogen exits the jet at \(T_{e}=100 \mathrm{~K}, V_{e}=20 \mathrm{~m} / \mathrm{s}\). Hint: Due to the probe walls, the jet is confined and behaves as if it were one in an array such as in Figure \(7.18 c\).

Consider mass loss from a smooth wet flat plate due to forced convection at atmospheric pressure. The plate is \(0.5 \mathrm{~m}\) long and \(3 \mathrm{~m}\) wide. Dry air at \(300 \mathrm{~K}\) and a free stream velocity of \(35 \mathrm{~m} / \mathrm{s}\) flows over the surface, which is also at a temperature of \(300 \mathrm{~K}\). Estimate the average mass transfer coefficient \(\bar{h}_{m}\) and determine the water vapor mass loss rate \((\mathrm{kg} / \mathrm{s})\) from the plate.

A spherical droplet of alcohol, \(0.5 \mathrm{~mm}\) in diameter, is falling freely through quiescent air at a velocity of \(1.8 \mathrm{~m} / \mathrm{s}\). The concentration of alcohol vapor at the surface of the droplet is \(0.0573 \mathrm{~kg} / \mathrm{m}^{3}\), and the diffusion coefficient for alcohol in air is \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Neglecting radiation and assuming steady-state conditions, calculate the surface temperature of the droplet if the ambient air temperature is \(300 \mathrm{~K}\). The latent heat of vaporization is \(8.42 \times 10^{5} \mathrm{~J} / \mathrm{kg}\).

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 plain carbon steel by an airstream moving at a velocity \(u_{\infty}\) in cross flow over the top and bottom surfaces 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, and assuming a uniform sheet temperature in the direction of airflow, derive a differential equation that governs the temperature distribution, \(T(x)\), along the sheet. Consider the effects of radiation, as well as convection, and express your result in terms of the velocity, thickness, and properties of the sheet \(\left(V, \delta, \rho, c_{p}, \varepsilon\right)\), the average convection coefficient \(\bar{h}_{W}\) associated with the cross flow, and the environmental temperatures \(\left(T_{\infty}, T_{\text {sur }}\right)\). (b) Neglecting radiation, obtain a closed form solution to the foregoing equation. For \(\delta=3 \mathrm{~mm}, V=\) \(0.10 \mathrm{~m} / \mathrm{s}, L=10 \mathrm{~m}, W=1 \mathrm{~m}, u_{\infty}=20 \mathrm{~m} / \mathrm{s}, T_{\infty}=\) \(20^{\circ} \mathrm{C}\), and a sheet temperature of \(T_{i}=500^{\circ} \mathrm{C}\) at the onset of cooling, what is the outlet temperature \(T_{o}\) ? Assume a negligible effect of the sheet velocity on boundary layer development in the direction of airflow. The density and specific heat of the steel are \(\rho=7850 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=620 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while properties of the air may be taken to be \(k=0.044\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}, \nu=4.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \operatorname{Pr}=0.68\). (c) Accounting for the effects of radiation, with \(\varepsilon=\) \(0.70\) and \(T_{\text {sur }}=20^{\circ} \mathrm{C}\), numerically integrate the differential equation derived in part (a) to determine the temperature of the sheet at \(L=10 \mathrm{~m}\). Explore the effect of \(V\) on the temperature distribution along the sheet.

Packed beds of spherical particles can be sintered at high temperature to form permeable, rigid foams. A foam sheet of thickness \(t=10 \mathrm{~mm}\) is comprised of sintered bronze spheres, each of diameter \(D=0.6 \mathrm{~mm}\). The metal foam has a porosity of \(\varepsilon=0.25\), and the foam sheet fills the cross section of an \(L=40 \mathrm{~mm} \times W=40 \mathrm{~mm}\) wind tunnel. The upper and lower surfaces of the foam are at temperatures \(T_{s}=80^{\circ} \mathrm{C}\), and the two other foam edges (the front edge shown in the schematic and the corresponding back edge) are insulated. Air flows in the wind tunnel at an upstream temperature and velocity of \(T_{i}=20^{\circ} \mathrm{C}\) and \(V=10 \mathrm{~m} / \mathrm{s}\), respectively. (a) Assuming the foam is at a uniform temperature \(T_{s}\), estimate the convection heat transfer rate to the air. Do you expect the actual heat transfer rate to be equal to, less than, or greater than your estimated value? (b) Assuming one-dimensional conduction in the \(x=\) direction, use an extended surface analysis to estimate the heat transfer rate to the air. To do so, show that the effective perimeter associated with Equation \(3.70\) is \(P_{\mathrm{eff}}=A_{p, \mathrm{r}} / L\). Determine the effective thermal conductivity of the foam \(k_{\text {eff }}\) by using Equation 3.25. Do you expect the actual heat transfer rate to be equal to, less than, or greater than your estimated value?
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