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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/m鲁 - 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/m虏.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/m虏.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/m虏.K. c) The average convection heat transfer coefficient calculated using modified Reynolds analogy is \(32.78\) W/m虏.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/m鲁 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/m鲁),\
    • \
  • \(u_{\infty}\) 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鈰卻).\
    • \

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鈰卻),\
    • \
  • \(c_p\) is the specific heat capacity at constant pressure (in J/kg鈰匥), and\
    • \
  • \(k\) is the thermal conductivity of the fluid (in W/m鈰匥).\
    • \

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

The local atmospheric pressure in Denver, Colorado (elevation \(1610 \mathrm{~m}\) ), is \(83.4 \mathrm{kPa}\). Air at this pressure and \(20^{\circ} \mathrm{C}\) flows with a velocity of \(8 \mathrm{~m} / \mathrm{s}\) over a \(1.5 \mathrm{~m} \times 6 \mathrm{~m}\) flat plate whose temperature is \(140^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the plate if the air flows parallel to the \((a)\)-m-long side and \((b)\) the \(1.5 \mathrm{~m}\) side.

During a plant visit, it was noticed that a 12-m-long section of a \(10-\mathrm{cm}\)-diameter steam pipe is completely exposed to the ambient air. The temperature measurements indicate that the average temperature of the outer surface of the steam pipe is \(75^{\circ} \mathrm{C}\) when the ambient temperature is \(5^{\circ} \mathrm{C}\). There are also light winds in the area at \(10 \mathrm{~km} / \mathrm{h}\). The emissivity of the outer surface of the pipe is \(0.8\), and the average temperature of the surfaces surrounding the pipe, including the sky, is estimated to be \(0^{\circ} \mathrm{C}\). Determine the amount of heat lost from the steam during a 10 -h-long work day. Steam is supplied by a gas-fired steam generator that has an efficiency of 80 percent, and the plant pays \(\$ 1.05 /\) therm of natural gas. If the pipe is insulated and 90 percent of the heat loss is saved, determine the amount of money this facility will save a year as a result of insulating the steam pipes. Assume the plant operates every day of the year for \(10 \mathrm{~h}\). State your assumptions.

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by the following correlation $$ \mathrm{Nu}_{x}=0.035 \mathrm{Re}_{x}^{0.8} \operatorname{Pr}^{1 / 3} $$ Determine the ratio of the average convection heat transfer coefficient \((h)\) over the entire plate length to the local convection heat transfer coefficient \(\left(h_{x}\right)\) at \(x=L\).

In flow over blunt bodies such as a cylinder, how does the pressure drag differ from the friction drag?

Kitchen water at \(10^{\circ} \mathrm{C}\) flows over a 10 -cm-diameter pipe with a velocity of \(1.1 \mathrm{~m} / \mathrm{s}\). Geothermal water enters the pipe at \(90^{\circ} \mathrm{C}\) at a rate of \(1.25 \mathrm{~kg} / \mathrm{s}\). For calculation purposes, the surface temperature of the pipe may be assumed to be \(70^{\circ} \mathrm{C}\). If the geothermal water is to leave the pipe at \(50^{\circ} \mathrm{C}\), the required length of the pipe is (a) \(1.1 \mathrm{~m}\) (b) \(1.8 \mathrm{~m}\) (c) \(2.9 \mathrm{~m}\) (d) \(4.3 \mathrm{~m}\) (e) \(7.6 \mathrm{~m}\) (For both water streams, use \(k=0.631 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=4.32\), \(\left.\nu=0.658 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\)
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