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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

The roof of a refrigerated truck compartment is of composite construction, consisting of a layer of foamed urethane insulation \(\left(t_{2}=50 \mathrm{~mm}, k_{i}=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) sandwiched between aluminum alloy panels \(\left(t_{1}=5 \mathrm{~mm}\right.\), \(\left.k_{p}=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\). The length and width of the roof are \(L=10 \mathrm{~m}\) and W \(=3.5 \mathrm{~m}\), respectively, and the temperature of the inner surface is \(T_{s, i}=-10^{\circ} \mathrm{C}\). Consider conditions for which the truck is moving at a speed of \(V=105 \mathrm{~km} / \mathrm{h}\), the air temperature is \(T_{\infty}=32^{\circ} \mathrm{C}\), and the solar irradiation is \(G_{S}=750 \mathrm{~W} / \mathrm{m}^{2}\). Turbulent flow may be assumed over the entire length of the roof. (a) For equivalent values of the solar absorptivity and the emissivity of the outer surface \(\left(\alpha_{S}=\varepsilon=0.5\right)\), estimate the average temperature \(T_{s, o}\) of the outer surface. What is the corresponding heat load imposed on the refrigeration system? (b) A special finish \(\left(\alpha_{S}=0.15, \varepsilon=0.8\right)\) may be applied to the outer surface. What effect would such an application have on the surface temperature and the heat load? (c) If, with \(\alpha_{S}=\varepsilon=0.5\), the roof is not insulated \(\left(t_{2}=0\right)\), what are the corresponding values of the surface temperature and the heat load?

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?

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 \(25 \mathrm{~m} / \mathrm{s}\).

Consider atmospheric air at \(25^{\circ} \mathrm{C}\) and a velocity of \(25 \mathrm{~m} / \mathrm{s}\) flowing over both surfaces of a 1 - \(\mathrm{m}\)-long flat plate that is maintained at \(125^{\circ} \mathrm{C}\). Determine the rate of heat transfer per unit width from the plate for values of the critical Reynolds number corresponding to \(10^{5}\), \(5 \times 10^{5}\), and \(10^{6}\).

Copper spheres of \(20-\mathrm{mm}\) diameter are quenched by being dropped into a tank of water that is maintained at \(280 \mathrm{~K}\). The spheres may be assumed to reach the terminal velocity on impact and to drop freely through the water. Estimate the terminal velocity by equating the drag and gravitational forces acting on the sphere. What is the approximate height of the water tank needed to cool the spheres from an initial temperature of \(360 \mathrm{~K}\) to a center temperature of \(320 \mathrm{~K}\) ?
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