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Chapter 6: Problem 14

Experiments to determine the local convection heat transfer coefficient for uniform flow normal to a heated circular disk have yielded a radial Nusselt number distribution of the form $$ N u_{D}=\frac{h(r) D}{k}=N u_{o}\left[1+a\left(\frac{r}{r_{o}}\right)^{n}\right] $$ where both \(n\) and \(a\) are positive. The Nusselt number at the stagnation point is correlated in terms of the Reynolds \(\left(R e_{D}=V D / v\right)\) and Prandtl numbers $$ N u_{o}=\frac{h(r=0) D}{k}=0.814 \operatorname{Re}_{D}^{1 / 2} \mathrm{Pr}^{0.36} $$ Obtain an expression for the average Nusselt number, \(\overline{N u}_{D}=\bar{h} D / k\), corresponding to heat transfer from an isothermal disk. Typically, boundary layer development from a stagnation point yields a decaying convection coefficient with increasing distance from the stagnation point. Provide a plausible explanation for why the opposite trend is observed for the disk.

Short Answer

Expert verified
In this problem, we found the average Nusselt number \(\overline{N u}_{D}\) for heat transfer from an isothermal disk to be: $$ \overline{N u}_{D}=N u_{o}\left[1+a\frac{2}{2+n}\right] $$ The increasing trend in convection coefficient with radial distance from the stagnation point is due to the presence of the positive exponent \(n\) in the Nusselt number distribution.

Step by step solution

01

Recall the formula for the average Nusselt number

The formula for the average Nusselt number when given the Nusselt number distribution is: $$ \overline{N u}_{D}=\frac{1}{\pi r_{o}^{2}} \int_{0}^{r_{o}} 2 \pi r N u_{D} d r $$
02

Substitute the given value of \(Nu_D\) into the formula

Substitute the given expression for \(Nu_D\) into the formula for the average Nusselt number: $$ \overline{N u}_{D}=\frac{1}{\pi r_{o}^{2}} \int_{0}^{r_{o}} 2 \pi r\left(N u_{o}\left[1+a\left(\frac{r}{r_{o}}\right)^{n}\right]\right) d r $$
03

Integrate the resulting expression

Integrate the resulting expression with respect to \(r\): $$ \begin{aligned} \overline{N u}_{D}=Nu_o \frac{1}{\pi r_{o}^{2}} \int_{0}^{r_{o}} 2 \pi r\left[1+a\left(\frac{r}{r_{o}}\right)^{n}\right]d r \\ \overline{N u}_{D}=N u_{o} \left[\frac{1}{r_{o}^{2}} \int_{0}^{r_{o}} 2 \pi r d r + \frac{a}{r_{o}^{2+n}}\int_{0}^{r_{o}} 2\pi r^{1+n} d r \right] \end{aligned} $$ Evaluate the two integrals separately: $$ \begin{aligned} \int_{0}^{r_{o}} 2\pi r d r = \left [\pi r^2 \right ]_{0}^{r_{o}} = \pi r_{o}^2 \\ \int_{0}^{r_{o}} 2\pi r^{1+n} dr = \frac{2\pi}{2+n} r_{o}^{2+n} \end{aligned} $$ Now substitute the evaluated integrals back into the expression for the average Nusselt number: $$ \overline{N u}_{D}=Nu_o \left[\frac{1}{r_{o}^{2}}(\pi r_{o}^{2})+\frac{a}{r_{o}^{2+n}}(\frac{2\pi}{2+n}r_{o}^{2+n}) \right] $$ Simplify the expression: $$ \overline{N u}_{D}=N u_{o}\left[1+a\frac{2}{2+n}\right] $$
04

Discuss the trend of the convection coefficient

Typically, boundary layer development from a stagnation point yields a decaying convection coefficient with increasing distance from the stagnation point. However, in this case, the opposite trend is observed for the disk. This is due to the presence of the positive exponent \(n\) in the expression for the Nusselt number distribution. The term \(\left(\frac{r}{r_{o}}\right)^{n}\) increases with increasing radial distance from the stagnation point, causing the increase of the convection coefficient with radial distance. In conclusion, the average Nusselt number is given by the expression: $$ \overline{N u}_{D}=N u_{o}\left[1+a\frac{2}{2+n}\right] $$ The trend of the convection coefficient is due to the presence of the positive exponent \(n\) in the Nusselt number distribution, which causes an increasing trend in convection coefficient with increasing radial distance from the stagnation point.

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Key Concepts

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

Understanding the Convection Heat Transfer Coefficient
The convection heat transfer coefficient, denoted by 'h', is a critical parameter in the field of thermal engineering. It quantifies the rate of heat transfer between a surface and a fluid moving past it due to convection. This value is inherently complex as it is influenced by various factors such as fluid velocity, viscosity, and thermal conductivity.

For a circular disk in uniform flow, as described in our exercise, 'h' varies radially, indicating how well the disk can dissipate heat into the flow at different distances from its center. A fundamental method to relate the convection heat transfer coefficient to the geometry of the system and fluid properties is by using the dimensionless Nusselt number (\(Nu\_D\_D=Nu_o[1+a(r/r_o)^n]\) where 'D' is the disk diameter).

Contrary to common trends where the convection coefficient decreases with increasing distance from a stagnation point due to boundary layer growth, the given problem presents an unusual case where this coefficient increases with radial distance, as indicated by the positive exponent 'n'. This peculiar behavior might suggest that other effects, such as a change in flow structure or surface conditions, are at play, affecting the convective heat transfer in a way that deviates from the norm.
The Role of Reynolds Number
The Reynolds number (\(Re\_D\_D=VD/v\) is a dimensionless quantity that provides insight into the flow regime of a fluid over a body. It compares the inertial forces to viscous forces within the fluid flow. In our context, a higher Reynolds number denotes a flow closer to turbulent, which enhances the mixing of fluid particles and can lead to increased convective heat transfer.

The formula for the Nusselt number at the stagnation point incorporates the Reynolds number, reflecting its direct influence on the heat transfer characteristics at that point. In the context of the provided exercise, the correlation \(Nu\_o=0.814 Re\_D^{1/2} Pr^{0.36}\) implies that the local convection heat transfer coefficient at the stagnation point (\(r=0\)) is correlated with both the Reynolds number and the Prandtl number. This correlation is imperative since it determines the initial heat transfer performance of the disk before radial effects become significant.
Significance of Prandtl Number
The Prandtl number (\(Pr\) is another dimensionless number and is defined as the ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity. It provides a measure of the relative thickness of the velocity boundary layer to the thermal boundary layer. Fluids with a high Prandtl number have a thicker thermal layer compared to the velocity layer, which means heat diffuses slower than momentum.

In the exercise, the Prandtl number is used alongside the Reynolds number to predict the stagnation point Nusselt number. This stagnation Nusselt number is used as a baseline in the calculation to determine the average Nusselt number across the disk. The Prandtl number being raised to the power of 0.36 in the correlation equation suggests that it has a significant, albeit less pronounced, impact on the convection heat transfer coefficient compared to the Reynolds number.

In practice, a fluid's Prandtl number is crucial in the design of heat exchangers and in estimating heat loss in various engineering applications. The reliance on both the Reynolds and Prandtl numbers ensures that the heat transfer analysis is comprehensive, accounting for fluid flow and thermal properties.

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Most popular questions from this chapter

Consider cross flow of gas \(\mathrm{X}\) over an object having a characteristic length of \(L=0.1 \mathrm{~m}\). For a Reynolds number of \(1 \times 10^{4}\), the average heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The same object is then impregnated with liquid \(Y\) and subjected to the same flow conditions. Given the following thermophysical properties, what is the average convection mass transfer coefficient?

It is known that on clear nights the air temperature need not drop below \(0^{\circ} \mathrm{C}\) before a thin layer of water on the ground will freeze. Consider such a layer of water on a clear night for which the effective sky temperature is \(-30^{\circ} \mathrm{C}\) and the convection heat transfer coefficient due to wind motion is \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The water may be assumed to have an emissivity of \(1.0\) and to be insulated from the ground as far as conduction is concerned. (a) Neglecting evaporation, determine the lowest temperature the air can have without the water freezing. (b) For the conditions given, estimate the mass transfer coefficient for water evaporation \(h_{\mathrm{m}}(\mathrm{m} / \mathrm{s})\). (c) Accounting now for the effect of evaporation, what is the lowest temperature the air can have without the water freezing? Assume the air to be dry.

Consider the nanofluid of Example 2.2. (a) Calculate the Prandtl numbers of the base fluid and nanofluid, using information provided in the example problem. (b) For a geometry of fixed characteristic dimension \(L\), and a fixed characteristic velocity \(V\), determine the ratio of the Reynolds numbers associated with the two fluids, \(R e_{\text {wf }} / R e_{\mathrm{w}_{\mathrm{d}}-}\) Calculate the ratio of the average Nusselt numbers, \(\overline{N u}_{L, \text {, d }} / \overline{N u}_{\text {L, b }}\), that is associated with identical average heat transfer coefficients for the two fluids, \(\bar{h}_{\mathrm{mf}}=\bar{h}_{\mathrm{bd}}\). (c) The functional dependence of the average Nusselt number on the Reynolds and Prandtl numbers for a broad array of various geometries may be expressed in the general form $$ \overline{N u}_{L}=\bar{h} L / k=C R e^{w N} P r^{1 / 3} $$ where \(C\) and \(m\) are constants whose values depend on the geometry from or to which convection heat transfer occurs. Under most conditions the value of \(m\) is positive. For positive \(m\), is it possible for the base fluid to provide greater convection heat transfer rates than the nanofluid, for conditions involving a fixed geometry, the same characteristic velocities, and identical surface and ambient temperatures?

Consider airflow over a flat plate of length \(L=1 \mathrm{~m}\) under conditions for which transition occurs at \(x_{c}=0.5 \mathrm{~m}\) based on the critical Reynolds number, \(R e_{x, c}=5 \times 10^{5}\). (a) Evaluating the thermophysical properties of air at \(350 \mathrm{~K}\), determine the air velocity. (b) In the laminar and turbulent regions, the local convection coefficients are, respectively, \(h_{\text {lam }}(x)=C_{\text {lam }} x^{-05}\) and \(h_{\text {marb }}=C_{\text {marb }} x^{-0.2}\) where, at \(T=350 \mathrm{~K}, C_{\text {lum }}=8.845 \mathrm{~W} / \mathrm{m}^{3 / 2} \cdot \mathrm{K}, C_{\text {tub }}=\) \(49.75 \mathrm{~W} / \mathrm{m}^{1.8} \cdot \mathrm{K}\), and \(x\) has units of \(\mathrm{m}\). Develop an expression for the average convection coefficient, \(\bar{h}_{\mathrm{hm}}(x)\), as a function of distance from the leading edge, \(x\), for the laminar region, \(0 \leq x \leq x_{x}\). (c) Develop an expression for the average convection coefficient, \(\bar{h}_{\text {art }}(x)\), as a function of distance from the leading edge, \(x\), for the turbulent region, \(x_{c} \leq x \leq L\). (d) On the same coordinates, plot the local and average convection coefficients, \(h_{x}\) and \(\bar{h}_{x}\), respectively, as a function of \(x\) for \(0 \leq x \leq L\).

It is desired to develop a simple model for predicting the temperature-time history of a plate during the drying cycle in a dishwasher. Following the wash cycle the plate is at \(T_{p}(t)=T_{p}(0)=65^{\circ} \mathrm{C}\) and the air in the dishwasher is completely saturated \(\left(\phi_{x}=1.0\right)\) at \(T_{x}=55^{\circ} \mathrm{C}\). The values of the plate surface area \(A_{s}\), mass \(M\), and specific heat \(c\) are such that \(M c / A_{s}=1600 \mathrm{~J} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Assuming the plate is completely covered by a thin film of water and neglecting the thermal resistances of the film and plate, derive a differential equation for predicting the plate temperature as a function of time. (b) For the initial conditions \((t=0)\) estimate the change in plate temperature with time, \(d T / d t\left({ }^{\circ} \mathrm{C} / \mathrm{s}\right)\), assuming that the average heat transfer coefficient on the plate is \(3.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
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