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Chapter 6: Problem 34

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?

Short Answer

Expert verified
We calculated the Prandtl numbers for both the base fluid and nanofluid as \(Pr_{bf} = 5.71\) and \(Pr_{nf} = 5.90\). The ratio of the Reynolds numbers is 1.131, and the ratio of the average Nusselt numbers is 1. Comparing the convection heat transfer rates, we found that the ratio of the Nusselt numbers for the nanofluid and the base fluid is \(\frac{\overline{Nu}_{nf}}{\overline{Nu}_{bf}} = \frac{1.131^m \cdot (5.90)^{1/3}}{(5.71)^{1/3}}\), where \(m\) is a positive constant. Since this ratio is greater than 1, it is not possible for the base fluid to provide a greater convection heat transfer rate than the nanofluid under the given conditions.

Step by step solution

01

(a) Calculate Prandtl numbers

The Prandtl number is defined as the ratio of the fluid's kinematic viscosity to its thermal diffusivity: \( Pr = \frac{\nu}{\alpha} \) Using the values from Example 2.2, calculate the Prandtl numbers of the base fluid and nanofluid. Base fluid: \( \nu_{bf} = 8.4 \times 10^{-7} \: m^2/s, \alpha_{bf} = 1.47 \times 10^{-7} \: m^2/s \) Nanofluid: \( \nu_{nf} = 9.50 \times 10^{-7} \: m^2/s, \alpha_{nf} = 1.61 \times 10^{-7} \: m^2/s \) Prandtl number for base fluid: \( Pr_{bf} = \frac{\nu_{bf}}{\alpha_{bf}} = \frac{8.4 \times 10^{-7}}{1.47 \times 10^{-7}} = 5.71 \) Prandtl number for nanofluid: \( Pr_{nf} = \frac{\nu_{nf}}{\alpha_{nf}} = \frac{9.50 \times 10^{-7}}{1.61 \times 10^{-7}} = 5.90 \)
02

(b) Determine the ratio of Reynolds numbers and average Nusselt numbers

The Reynolds number is defined as the ratio of the inertial forces to the viscous forces in the fluid flow and is calculated using the formula: \( Re = \frac{VL}{\nu} \) Let's determine the ratio of the Reynolds numbers for the two fluids: \( \frac{Re_{wf}}{Re_{wd}} = \frac{\frac{VL}{\nu_{bf}}}{\frac{VL}{\nu_{nf}}} \) Simplifying the equation, we get: \( \frac{Re_{wf}}{Re_{wd}} = \frac{\nu_{nf}}{\nu_{bf}} = \frac{9.50 \times 10^{-7}}{8.4 \times 10^{-7}} = 1.131 \) Now we need to calculate the ratio of the average Nusselt numbers: \( \frac{\overline{Nu}_{L,d}}{\overline{Nu}_{L,b}} = \frac{\bar{h}_{mf}}{\bar{h}_{bd}} \) Since \(\bar{h}_{mf}=\bar{h}_{bd}\), this ratio is equal to 1.
03

(c) Analyze convection heat transfer rates

For the general functional dependence of the average Nusselt number on the Reynolds and Prandtl numbers, we have: \( \overline{Nu}_L = CRe^m Pr^{1/3} \) In order to answer the question, let's compare the base fluid and nanofluid Nusselt numbers: \( \frac{\overline{Nu}_{nf}}{\overline{Nu}_{bf}} = \frac{C(Re_{nf})^m (Pr_{nf})^{1/3}}{C(Re_{bf})^m (Pr_{bf})^{1/3}} = \frac{(Re_{nf})^m (Pr_{nf})^{1/3}}{(Re_{bf})^m (Pr_{bf})^{1/3}} \) We previously found that \(\frac{Re_{nf}}{Re_{bf}} = 1.131\), and we have the Prandtl numbers for both fluids. Therefore: \( \frac{\overline{Nu}_{nf}}{\overline{Nu}_{bf}} = \frac{1.131^m \cdot (5.90)^{1/3}}{(5.71)^{1/3}} \) For a positive value of \(m\), the term \(1.131^m\) will always be greater than 1. Since the Prandtl number ratio is also greater than 1, the overall ratio of \(\frac{\overline{Nu}_{nf}}{\overline{Nu}_{bf}}\) will be greater than 1 as well. This means that, under the given conditions, it is not possible for the base fluid to provide a greater convection heat transfer rate than the nanofluid.

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Key Concepts

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

Prandtl number
The Prandtl number (Pr) is an essential concept in fluid dynamics and heat transfer. It represents the ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity. In simpler terms, it tells us how momentum and heat are transported through a fluid.
  • Kinematic viscosity (\( u \)) is the fluid's resistance to flow and movement.
  • Thermal diffusivity (\( \alpha \)) indicates how quickly heat spreads through a fluid.
To calculate the Prandtl number for a fluid, use the formula:\[Pr = \frac{u}{\alpha}\]A higher Prandtl number means the fluid can transfer momentum more quickly than heat, often seen in oils. Conversely, a low Prandtl number indicates heat diffuses faster than momentum, typical for liquid metals. Understanding the Prandtl number helps predict how a fluid will behave in thermal environments.
Reynolds number
The Reynolds number (Re) is a key factor in determining the flow regime of a fluid, whether laminar or turbulent. It measures the ratio of inertial forces to viscous forces and gives insight into the movement and flow characteristics of fluid.
  • Inertial forces are related to the fluid's velocity and density.
  • Viscous forces are linked to its viscosity.
You can calculate the Reynolds number using:\[Re = \frac{VL}{u}\]where \( V \) is the velocity, \( L \) is the characteristic length, and \( u \) is the kinematic viscosity. A low Reynolds number (less than 2000) suggests laminar flow, where the fluid moves in parallel layers. A high Reynolds number (greater than 4000) indicates turbulent flow, characterized by mixing and chaotic fluctuations. The Reynolds number helps engineers design systems that optimize flow conditions.
Nusselt number
The Nusselt number (Nu) is a dimensionless value that measures the enhancement of heat transfer through convection compared to pure conduction. It essentially tells us how effective a heat transfer process is in a given fluid situation.
  • If Nu = 1, heat transfer is purely conductive.
  • If Nu > 1, convection plays a significant role.
This number is calculated by:\[\overline{Nu}_L = \frac{\bar{h} L}{k}\]where \( \bar{h} \) is the average heat transfer coefficient, \( L \) is the characteristic length, and \( k \) is the thermal conductivity of the fluid. High Nusselt numbers indicate strong convective heat transfer, which is desirable in heating or cooling applications. Understanding this concept helps in evaluating and improving heat exchanger designs.
Convection heat transfer
Convection heat transfer is the process of heat movement caused by the bulk motion of fluid. It combines conduction and advection processes, significantly impacting the efficiency of heating or cooling systems.
  • Natural convection occurs due to buoyancy forces, driven by temperature differences within the fluid.
  • Forced convection involves external forces like fans or pumps to enhance heat transfer.
The rate of convection heat transfer can be described using:\[Q = \bar{h}A(T_s - T_ ext{fluid})\]where \( Q \) is the heat transfer rate, \( \bar{h} \) is the heat transfer coefficient, \( A \) is the surface area, and \( T_s \) and \( T_ ext{fluid} \) are the temperatures of the surface and fluid, respectively. Effective convection heat transfer is crucial in numerous applications, from radiators to climate control systems.
Thermal properties
Thermal properties are characteristics of a material that define how it transfers heat. Understanding these properties helps us predict how materials will behave in various thermal environments.
  • Thermal conductivity (\( k \)) is a measure of a material's ability to conduct heat. High thermal conductivity means heat moves easily through the material.
  • Specific heat capacity (\( c_p \)) indicates how much heat energy is needed to change a material's temperature by a certain amount.
  • Thermal diffusivity is the rate at which temperature spreads through a material, calculated by \( \alpha = \frac{k}{\rho c_p} \) where \( \rho \) is the density.
These properties are crucial for designing thermal management systems and selecting materials suitable for specific heat transfer applications, such as insulators or conductors.

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Most popular questions from this chapter

The heat transfer rate per unit width (normal to the page) from a longitudinal section, \(x_{2}-x_{1}\), can be expressed as \(q_{12}^{\prime}=\bar{h}_{12}\left(x_{2}-x_{1}\right)\left(T_{s}-T_{\infty}\right)\), where \(\bar{h}_{12}\) is the average coefficient for the section of length \(\left(x_{2}-x_{1}\right)\). Consider laminar flow over a flat plate with a uniform temperature \(T_{s}\). The spatial variation of the local convection coefficient is of the form \(h_{x}=C x^{-1 / 2}\), where \(C\) is a constant. (a) Beginning with the convection rate equation in the form \(d q^{\prime}=h_{s} d x\left(T_{s}-T_{x}\right)\), derive an expression for \(\bar{h}_{12}\) in terms of \(C, x_{1}\), and \(x_{2}\). (b) Derive an expression for \(\bar{h}_{12}\) in terms of \(x_{1}, x_{2}\), and the average coefficients \(\bar{h}_{1}\) and \(\bar{h}_{2}\), corresponding to lengths \(x_{1}\) and \(x_{2}\), respectively.

A streamlined strut supporting a bearing housing is exposed to a hot airflow from an engine exhaust. It is necessary to run experiments to determine the average convection heat transfer coefficient \(\bar{h}\) from the air to the strut in order to be able to cool the strut to the desired surface temperature \(T_{x}\). It is decided to run mass transfer experiments on an object of the same shape and to obtain the desired heat transfer results by using the heat and mass transfer analogy. The mass transfer experiments were conducted using a half-size model strut constructed from naphthalene exposed to an airstream at \(27^{\circ} \mathrm{C}\). Mass transfer measurements yielded these results: \begin{tabular}{rr} \hline \multicolumn{1}{c}{\(\boldsymbol{\boldsymbol { e } _ { \boldsymbol { L } }}\)} & \(\overline{\boldsymbol{S h}}_{\boldsymbol{L}}\) \\ \hline 60,000 & 282 \\ 120,000 & 491 \\ 144,000 & 568 \\ 288,000 & 989 \\ \hline \end{tabular} (a) Using the mass transfer experimental results, determine the coefficients \(C\) and \(m\) for a correlation of the form \(\overline{S h}_{L}=C R e_{L}^{m} S c^{1 / 3}\). (b) Determine the average convection heat transfer coefficient \(\bar{h}\) for the full-sized strut, \(L_{H}=60 \mathrm{~mm}\), when exposed to a free stream airflow with \(V=60 \mathrm{~m} / \mathrm{s}\), \(T_{\infty}=184^{\circ} \mathrm{C}\), and \(p_{\infty}=1 \mathrm{~atm}\) when \(T_{s}=70^{\circ} \mathrm{C}\). (c) The surface area of the strut can be expressed as \(A_{s}=2.2 L_{H} \cdot l\), where \(l\) is the length normal to the page. For the conditions of part (b), what is the change in the rate of heat transfer to the strut if the characteristic length \(L_{H}\) is doubled?

Photosynthesis, as it occurs in the leaves of a green plant, involves the transport of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) from the atmosphere to the chloroplasts of the leaves. The rate of photosynthesis may be quantified in terms of the rate of \(\mathrm{CO}_{2}\) assimilation by the chloroplasts. This assimilation is strongly influenced by \(\mathrm{CO}_{2}\) transfer through the boundary layer that develops on the leaf surface. Under conditions for which the density of \(\mathrm{CO}_{2}\) is \(6 \times 10^{-4} \mathrm{~kg} / \mathrm{m}^{3}\) in the air and \(5 \times 10^{-4} \mathrm{~kg} / \mathrm{m}^{3}\) at the leaf surface and the convection mass transfer coefficient is \(10^{-2} \mathrm{~m} / \mathrm{s}\), what is the rate of photosynthesis in terms of kilograms of \(\mathrm{CO}_{2}\) assimilated per unit time and area of leaf surface?

On a summer day the air temperature is \(27^{\circ} \mathrm{C}\) and the relative humidity is \(30 \%\). Water evaporates from the surface of a lake at a rate of \(0.10 \mathrm{~kg} / \mathrm{h}\) per square meter of water surface area. The temperature of the water is also \(27^{\circ} \mathrm{C}\). Determine the value of the convection mass transfer coefficient. 6.53 It is observed that a 230 -mm-diameter pan of water at \(23^{\circ} \mathrm{C}\) has a mass loss rate of \(1.5 \times 10^{-5} \mathrm{~kg} / \mathrm{s}\) when the ambient air is dry and at \(23^{\circ} \mathrm{C}\). (a) Determine the convection mass transfer coefficient for this situation. (b) Estimate the evaporation mass loss rate when the ambient air has a relative humidity of \(50 \%\). (c) Estimate the evaporation mass loss rate when the water and ambient air temperatures are \(47^{\circ} \mathrm{C}\), assuming that the convection mass transfer coefficient remains unchanged and the ambient air is dry.

To a good approximation, the dynamic viscosity \(\mu\), the thermal conductivity \(k\), and the specific heat \(c_{p}\) are independent of pressure. In what manner do the kinematic viscosity \(v\) and thermal diffusivity \(\alpha\) vary with pressure for an incompressible liquid and an ideal gas? Determine \(\alpha\) of air at \(350 \mathrm{~K}\) for pressures of 1,5 , and \(10 \mathrm{~atm}\). Assuming a transition Reynolds number of \(5 \times 10^{5}\), determine the distance from the leading edge of a flat plate at which transition will occur for air at \(350 \mathrm{~K}\) at pressures of 1,5 , and 10 atm with \(u_{s}=2 \mathrm{~m} / \mathrm{s}\).
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