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Chapter 7: Problem 38

Air at 1 atm and \(20^{\circ} \mathrm{C}\) is flowing over the top surface of a \(0.5-\mathrm{m}\)-long thin flat plate. The air stream velocity is \(50 \mathrm{~m} / \mathrm{s}\) and the plate is maintained at a constant surface temperature of \(180^{\circ} \mathrm{C}\). Determine \((a)\) the average friction coefficient, \((b)\) the average convection heat transfer coefficient, and (c) repeat part (b) using the modified Reynolds analogy.

Short Answer

Expert verified
Given: - Air temperature: \(20^{\circ} \mathrm{C}\) - Length \(x\) of the flat plate: \(0.5\) m - Free stream velocity \(u_\infty\): \(10\) m/s - Air properties at \(20^{\circ} \mathrm{C}\): - Density (\(\rho\)) = 1.205 kg/m³ - Dynamic viscosity (\(\mu\)) = 1.81 x \(10^{-5}\) kg/m.s - Specific heat capacity (\(c_p\)) = 1007 J/kg.K - Thermal conductivity (\(k\)) = 0.0262 W/m.K Steps: 1. Calculate Reynolds number and Prandtl number: \(Re_x = \frac{1.205 \times 10 \times 0.5}{1.81 \times 10^{-5}} = 332486.2\) \(Pr = \frac{1.81 \times 10^{-5} \times 1007}{0.0262} = 0.692\) 2. Calculate average friction coefficient: \(C_f = \frac{0.664}{\sqrt{332486.2}} = 0.00114\) 3. Calculate average convection heat transfer coefficient: \(Nu_x = 0.664 \times 332486.2^{1/2} \times 0.692^{1/3} \Rightarrow h_x = \frac{Nu_x \times k}{x} = 31.52\) W/m².K 4. Calculate average convection heat transfer coefficient using modified Reynolds analogy: \(Nu_x = 0.5 \times 0.00114 \times 332486.2 \times 0.692 \Rightarrow h_x = \frac{Nu_x \times k}{x} = 32.78\) W/m².K Results: a) The average friction coefficient is \(0.00114\). b) The average convection heat transfer coefficient calculated using Nusselt number is \(31.52\) W/m².K. c) The average convection heat transfer coefficient calculated using modified Reynolds analogy is \(32.78\) W/m².K.

Step by step solution

01

Calculate Reynolds number and Prandtl number

First, we need to find the Reynolds number and Prandtl number for air at \(20^{\circ} \mathrm{C}\). The properties of air at \(20^{\circ} \mathrm{C}\) are as follows: Density (\(\rho\)) = 1.205 kg/m³ Dynamic viscosity (\(\mu\)) = 1.81 x \(10^{-5}\) kg/m.s Specific heat capacity (\(c_p\)) = 1007 J/kg.K Thermal conductivity (\(k\)) = 0.0262 W/m.K Reynolds number is given by: \(Re_x = \frac{\rho u_\infty x}{\mu}\) Prandtl number is given by: \(Pr = \frac{\mu c_p}{k}\)
02

Calculate average friction coefficient (a)

Using the Blasius equation for friction coefficient, we can calculate the average friction coefficient: \(C_f = \frac{0.664}{\sqrt{Re_x}}\) Calculate \(Re_x\) and \(C_f\) for air at \(x = 0.5\) m.
03

Calculate average convection heat transfer coefficient (b)

Next, we will calculate the average convection heat transfer coefficient using the Nusselt number: \(Nu_x = \frac{h_x x}{k} = 0.664 Re_x^{1/2} Pr^{1/3}\) Solve for \(h_x\).
04

Calculate average convection heat transfer coefficient using modified Reynolds analogy (c)

Finally, use the modified Reynolds analogy to find the average convection heat transfer coefficient: \(Nu_x = \frac{C_fx}{2} Re_x Pr\) Solve for \(h_x\).

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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 (Re) is a dimensionless quantity used in fluid mechanics to predict flow patterns in different fluid flow situations. It compares the relative importance of inertial effects to viscous effects and is defined by the equation:
\[Re_x = \frac{\rho u_{\infty} x}{\mu}\]
where:\
    \
  • \(\rho\) is the density of the fluid (in kg/m³),\
    • \
  • \(u_{\infty}\) is the flow velocity (in m/s),\
    • \
  • \(x\) is the characteristic length, in this case, the length of the plate (in meters), and\
    • \
  • \(\mu\) is the dynamic viscosity of the fluid (in kg/m⋅s).\
    • \

The higher the Reynolds number, the more turbulent the flow is likely to be; lower values indicate laminar flow. For a thin flat plate in an airstream, like in the above example, determining the Reynolds number helps us analyze the flow behavior: whether it's laminar or turbulent, which is essential for accurately calculating the friction coefficient and the heat transfer coefficient.
Reynolds number is crucial for predicting the onset of turbulence, and thus, its calculation forms the foundation for many convective heat transfer coefficient estimations.
Prandtl number
The Prandtl number (Pr) is another dimensionless number in fluid dynamics that is used to characterize the relative thickness of the velocity boundary layer to the thermal boundary layer. It is determined using the following equation:
\[Pr = \frac{\mu c_p}{k}\]
where:\
    \
  • \(\mu\) represents the dynamic viscosity of the fluid (in kg/m⋅s),\
    • \
  • \(c_p\) is the specific heat capacity at constant pressure (in J/kg⋅K), and\
    • \
  • \(k\) is the thermal conductivity of the fluid (in W/m⋅K).\
    • \

The Prandtl number indicates if the heat diffuses quickly or slowly compared to momentum (velocity). In the situation of air flowing over a plate, as per the given exercise, a higher Prandtl number would mean a slower thermal diffusion relative to velocity. This number affects the convection heat transfer calculations because it determines the thickness of the thermal boundary layer, which is integral to heat dissipation from the plate into the air.
Friction coefficient
The friction coefficient (Cf) is a key concept in determining the resistance that a fluid flow encounters when in contact with a surface. For the flow over a flat plate, it is estimated using the Blasius equation:
\[C_f = \frac{0.664}{\sqrt{Re_x}}\]
This equation presents the friction coefficient as inversely proportional to the square root of the Reynolds number (Re). The friction coefficient is indicative of the shear stress exerted by the flowing fluid on the plate's surface and is instrumental in predicting the frictional force experienced by the plate.
The friction coefficient affects not only the mechanical analysis of fluid-structure interactions but also thermal analysis when fluids are involved. This is demonstrated by the modified Reynolds analogy used in the exercise, which relates the friction coefficient to the convection heat transfer coefficient for the calculation of heat transfer between the air and the plate. A proper understanding of the friction coefficient aids in optimizing systems for better fluid flow and heat transfer characteristics.

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Most popular questions from this chapter

Consider a refrigeration truck traveling at \(55 \mathrm{mph}\) at a location where the air temperature is \(80^{\circ} \mathrm{F}\). The refrigerated compartment of the truck can be considered to be a 9-ft-wide, 8-ft-high, and 20 -ft-long rectangular box. The refrigeration system of the truck can provide 3 tons of refrigeration (i.e., it can remove heat at a rate of \(600 \mathrm{Btu} / \mathrm{min}\) ). The outer surface of the truck is coated with a low-emissivity material, and thus radiation heat transfer is very small. Determine the average temperature of the outer surface of the refrigeration compartment of the truck if the refrigeration system is observed to be operating at half the capacity. Assume the air flow over the entire outer surface to be turbulent and the heat transfer coefficient at the front and rear surfaces to be equal to that on side surfaces. For air properties evaluations assume a film temperature of \(80^{\circ} \mathrm{F}\). Is this a good assumption?

Define frontal area of a body subjected to external flow. When is it appropriate to use the frontal area in drag and lift calculations?

Air \((\operatorname{Pr}=0.7, k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) at \(200^{\circ} \mathrm{C}\) flows across 2-cm-diameter tubes whose surface temperature is \(50^{\circ} \mathrm{C}\) with a Reynolds number of 8000 . The Churchill and Bernstein convective heat transfer correlation for the average Nusselt number in this situation is $$ \mathrm{Nu}=0.3+\frac{0.62 \mathrm{Re}^{0.5} \mathrm{Pr}^{0.33}}{\left[1+(0.4 / \mathrm{Pr})^{0.67}\right]^{0.25}} $$ (a) \(8.5 \mathrm{~kW} / \mathrm{m}^{2}\) (b) \(9.7 \mathrm{~kW} / \mathrm{m}^{2}\) (c) \(10.5 \mathrm{~kW} / \mathrm{m}^{2}\) (d) \(12.2 \mathrm{~kW} / \mathrm{m}^{2}\) (e) \(13.9 \mathrm{~kW} / \mathrm{m}^{2}\)

A 5-m-long strip of sheet metal is being transported on a conveyor at a velocity of \(5 \mathrm{~m} / \mathrm{s}\), while the coating on the upper surface is being cured by infrared lamps. The coating on the upper surface of the metal strip has an absorptivity of \(0.6\) and an emissivity of \(0.7\), while the surrounding ambient air temperature is \(25^{\circ} \mathrm{C}\). Radiation heat transfer occurs only on the upper surface, while convection heat transfer occurs on both upper and lower surfaces of the sheet metal. If the infrared lamps supply a constant heat flux of \(5000 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the sheet metal. Evaluate the properties of air at \(80^{\circ} \mathrm{C}\).

Liquid mercury at \(250^{\circ} \mathrm{C}\) is flowing with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\) in parallel over a \(0.1-\mathrm{m}\)-long flat plate where there is an unheated starting length of \(5 \mathrm{~cm}\). The heated section of the flat plate is maintained at a constant temperature of \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at the trailing edge, \((b)\) the average convection heat transfer coefficient for the heated section, and \((c)\) the rate of heat transfer per unit width for the heated section.
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