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Chapter 9: Problem 4

To assess the efficacy of different liquids for cooling by natural convection, it is convenient to introduce a fure of merit, \(F_{N}\), which combines the influence of all pertinent fluid properties on the convection coefficient. If the Nusselt number is governed by an expression of the form, \(N u_{L} \sim R a^{n}\), obtain the corresponding relationship between \(F_{N}\) and the fluid properties. For a representative value of \(n=0.33\), calculate values of \(F_{N}\) for air \((k=0.026\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}, \quad \beta=0.0035 \mathrm{~K}^{-1}, \quad \nu=1.5 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \quad P r=\) \(0.70)\), water \(\left(k=0.600 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \beta=2.7 \times 10^{-4} \mathrm{~K}^{-1}\right.\), \(\nu=10^{-6} \mathrm{~m}^{2} / \mathrm{s}, \quad P r=5.0\) ), and a dielectric liquid \(\left(k=0.064 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad \beta=0.0014 \mathrm{~K}^{-1}, \quad \nu=10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right.\), \(P r=25)\). What fluid is the most effective cooling agent?

Short Answer

Expert verified
The figure of merit, \(F_N\), can be obtained using the expression: \[F_N = \left(\frac{g\beta (T_s - T_\infty )L^{3}k}{\rho c_p \nu^3}\right)^\frac{1}{n}\] Calculating the values of \(F_N\) for air, water, and dielectric liquid using the given fluid properties and \(n = 0.33\), we find that water has the highest value of \(F_N\), making it the most effective cooling agent among the three fluids.

Step by step solution

01

Recall the definitions of the Nusselt number, Grashof number, and Prandtl number

The Nusselt number, \(Nu_L\), is a dimensionless number that relates the convection heat transfer to the conduction heat transfer. It is given by: \(Nu_L = \frac{hL}{k}\), where \(h\) is the convection heat transfer coefficient, \(L\) is the characteristic length, and \(k\) is the thermal conductivity. The Grashof number, \(Gr\), is a measure of the influence of natural convection in a fluid and is given by: \(Gr = \frac{g\beta (T_s - T_\infty)L^3}{\nu^2}\), where \(g\) is the gravitational acceleration, \(\beta\) is the thermal expansion coefficient, \(T_s\) and \(T_\infty\) are the surface and ambient temperatures, and \(\nu\) is the kinematic viscosity. The Prandtl number, \(Pr\), is a dimensionless number that evaluates the relative importance of momentum and thermal diffusivity and is given by: \(Pr = \frac{\nu}{\alpha}\), where \(\alpha\) is the thermal diffusivity.
02

Determine the relationship between figure of merit F_N and fluid properties

From the expression for \(Nu_L\), \(Nu_L \sim Ra^n\), we can replace \(Ra\) by \(GrPr\): \(Nu_L^{\frac{1}{n}} \sim (GrPr)\) Using the equations for \(Gr\) and \(Pr\), including the relationship of \(\alpha=\frac{k}{\rho c_p}\), we can write F_N as a function of the fluid properties: \[F_N = \frac{hL}{k} = Nu_L = \left(\frac{g\beta (T_s - T_\infty )L^{3}k}{\rho c_p \nu^3}\right)^\frac{1}{n}\]
03

Calculate the values of F_N for air, water, and dielectric liquid

Using the given values for the fluid properties and the representative value of \(n = 0.33\), we can calculate the figure of merit, \(F_N\), for each fluid: Air: \( F_{N,A} = \left(\frac{9.81 \times 0.0035 \times L^3 \times 0.026}{1.5 \times 10^{-5}^3 \times 0.70}\right)^{0.33} \) Water: \( F_{N,W} = \left(\frac{9.81 \times 2.7 \times 10^{-4} \times L^3 \times 0.600}{10^{-6}^3 \times 5.00}\right)^{0.33} \) Dielectric liquid: \( F_{N,D} = \left(\frac{9.81 \times 0.0014 \times L^3 \times 0.064}{10^{-6}^3 \times 25}\right)^{0.33} \)
04

Determine the most effective cooling agent

To determine the most effective cooling agent, we must compare the figures of merit, F_N, for each fluid. We notice that all the expressions have a common term, \(9.81 \times L^3\), which drops out when comparing them. Thus, we can compare the remaining terms. The fluid with the highest \(F_N\) will be the most effective cooling agent. Comparing the expressions for \(F_{N,A}\), \(F_{N,W}\), and \(F_{N,D}\), we find that water has the highest value of \(F_N\), and therefore, water is the most effective cooling agent among the three fluids.

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Key Concepts

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

Nusselt Number
The Nusselt number (Nu_L) is a crucial concept in heat transfer, especially when discussing convection processes. Imagine it as a way to measure how good a fluid is at carrying heat away through convection compared to conduction. A higher Nusselt number means more effective convective heat transfer. You can calculate it using the formula:
  • \(Nu_L = \frac{hL}{k}\)
Here, \(h\) is the convection heat transfer coefficient, indicating how quickly heat is transferred by convection. \(L\) is the characteristic length, a parameter often related to the size of the physical system. Lastly, \(k\) represents the thermal conductivity of the fluid, which tells you how well the fluid conducts heat by itself.
The Nusselt number can vary based on different fluid dynamics and thermal conditions. Understanding it helps in designing systems where heat transfer efficiency is crucial, such as in coolers, radiators, and electronic devices.
Grashof Number
The Grashof number (Gr) helps us understand the role of buoyancy-driven flow in natural convection. It quantifies the effect of buoyancy compared to viscous forces in a fluid. Essentially, it measures how likely fluid is to flow spontaneously due to temperature differences. The formula for calculating the Grashof number is:
  • \(Gr = \frac{g\beta (T_s - T_\infty)L^3}{u^2}\)
In this equation:
  • \(g\) is the gravitational acceleration.
  • \(\beta\) stands for the thermal expansion coefficient, showing how much a fluid expands with temperature increase.
  • \(T_s\) and \(T_\infty\) are the surface and ambient temperatures, respectively.
  • \(u\) is the kinematic viscosity, which relates to how easily the fluid flows.
A large Grashof number indicates that the buoyancy forces are significant, leading to strong convective currents. This is essential in systems where passive fluid motion, like air around a hot object, is used to transfer heat.
Prandtl Number
The Prandtl number (Pr) is another vital dimensionless number that links the momentum and thermal diffusivities of a fluid. It essentially bridges the gap between how momentum and heat diffuse through a fluid. The Prandtl number is calculated as:
  • \(Pr = \frac{u}{\alpha}\)
Where:
  • \(u\) is the kinematic viscosity.
  • \(\alpha\) is the thermal diffusivity, representing the rate at which heat diffuses through the material.
The Prandtl number helps predict the characteristics of boundary layer development in convection processes. When the Prandtl number is high, it means heat diffuses slower than momentum, leading to a thicker thermal boundary layer compared to the velocity boundary layer. In practical applications, such as engineering heat exchangers, knowing the Prandtl number allows one to tailor the design for optimal heat transfer efficiency, ensuring that systems can be cooled or heated efficiently.

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Most popular questions from this chapter

9.27 The vertical rear window of an automobile is of thickness \(L=8 \mathrm{~mm}\) and height \(H=0.5 \mathrm{~m}\) and contains fine-meshed heating wires that can induce nearly uniform volumetric heating, \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). (a) Consider steady-state conditions for which the interior surface of the window is exposed to quiescent air at \(10^{\circ} \mathrm{C}\), while the exterior surface is exposed to air at \(-10^{\circ} \mathrm{C}\) moving in parallel flow over the surface with a velocity of \(20 \mathrm{~m} / \mathrm{s}\). Determine the volumetric heating rate needed to maintain the interior window surface at \(T_{s, i}=15^{\circ} \mathrm{C}\). (b) The interior and exterior window temperatures, \(T_{s, i}\) and \(T_{s, \rho}\), depend on the compartment and ambient temperatures, \(T_{\infty, i}\) and \(T_{\infty, p}\), as well as on the velocity \(u_{\infty}\) of air flowing over the exterior surface and the volumetric heating rate \(\dot{q}\). Subject to the constraint that \(T_{s, i}\) is to be maintained at \(15^{\circ} \mathrm{C}\), we wish to develop guidelines for varying the heating rate in response to changes in \(T_{\infty,}, T_{\infty \rho,}\), and/or \(u_{\infty}\). If \(T_{\infty, i}\) is maintained at \(10^{\circ} \mathrm{C}\), how will \(\dot{q}\) and \(T_{s, \rho}\) vary with \(T_{\infty, o}\) for \(-25 \leq T_{\infty,,} \leq 5^{\circ} \mathrm{C}\) and \(u_{\infty}=10,20\), and \(30 \mathrm{~m} / \mathrm{s}\) ? If a constant vehicle speed is maintained, such that \(u_{\infty}=30 \mathrm{~m} / \mathrm{s}\), how will \(\dot{q}\) and \(T_{s, o}\) vary with \(T_{\infty, i}\) for \(5 \leq T_{\infty, i} \leq 20^{\circ} \mathrm{C}\) and \(T_{\infty, o}=-25,-10\), and \(5^{\circ} \mathrm{C}\) ?

According to experimental results for parallel airflow over a uniform temperature, heated vertical plate, the effect of free convection on the heat transfer convection coefficient will be \(5 \%\) when \(G r_{L} / R e_{L}^{2}=0.08\). Consider a heated vertical plate \(0.3 \mathrm{~m}\) long, maintained at a surface temperature of \(60^{\circ} \mathrm{C}\) in atmospheric air at \(25^{\circ} \mathrm{C}\). What is the minimum vertical velocity required of the airflow such that free convection effects will be less than \(5 \%\) of the heat transfer rate?

Consider a 2-mm-diameter sphere immersed in a fluid at \(300 \mathrm{~K}\) and \(1 \mathrm{~atm}\). (a) If the fluid around the sphere is quiescent and extensive, show that the conduction limit of heat transfer from the sphere can be expressed as \(N u_{D, \text { cond }}=2\). Hint: Begin with the expression for the thermal resistance of a hollow sphere, Equation \(3.41\), letting \(r_{2} \rightarrow \infty\), and then expressing the result in terms of the Nusselt number. (b) Considering free convection, at what surface temperature will the Nusselt number be twice that for the conduction limit? Consider air and water as the fluids. (c) Considering forced convection, at what velocity will the Nusselt number be twice that for the conduction limit? Consider air and water as the fluids.

A biological fluid moves at a flow rate of \(\dot{m}=0.02 \mathrm{~kg} / \mathrm{s}\) through a coiled, thin-walled, 5 -mm-diameter tube submerged in a large water bath maintained at \(50^{\circ} \mathrm{C}\). The fluid enters the tube at \(25^{\circ} \mathrm{C}\). (a) Estimate the length of the tube and the number of coil turns required to provide an exit temperature of \(T_{m, o}=38^{\circ} \mathrm{C}\) for the biological fluid. Assume that the water bath is an extensive, quiescent medium, that the coiled tube approximates a horizontal tube, and that the biological fluid has the thermophysical properties of water. (b) The flow rate through the tube is controlled by a pump that experiences throughput variations of approximately \(\pm 10 \%\) at any one setting. This condition is of concern to the project engineer because the corresponding variation of the exit temperature of the biological fluid could influence the downstream process. What variation would you expect in \(T_{m, o}\) for a \(\pm 10 \%\) change in \(\dot{m}\) ?

A household oven door of \(0.5-\mathrm{m}\) height and \(0.7-\mathrm{m}\) width reaches an average surface temperature of \(32^{\circ} \mathrm{C}\) during operation. Estimate the heat loss to the room with ambient air at \(22^{\circ} \mathrm{C}\). If the door has an emissivity of \(1.0\) and the surroundings are also at \(22^{\circ} \mathrm{C}\), comment on the heat loss by free convection relative to that by radiation.
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