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Chapter 6: Problem 84

Consider an airplane cruising at an altitude of \(10 \mathrm{~km}\) where standard atmospheric conditions are \(-50^{\circ} \mathrm{C}\) and \(26.5 \mathrm{kPa}\) at a speed of \(800 \mathrm{~km} / \mathrm{h}\). Each wing of the airplane can be modeled as a \(25-\mathrm{m} \times 3-\mathrm{m}\) flat plate, and the friction coefficient of the wings is \(0.0016\). Using the momentum-heat transfer analogy, determine the heat transfer coefficient for the wings at cruising conditions. Answer: \(89.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short Answer

Expert verified
Answer: The heat transfer coefficient of the airplane's wings is 89.6 W/m²⋅K.

Step by step solution

01

Calculate air properties at cruising altitude

We'll need to find the air density (\(\rho\)) and dynamic viscosity (\(\mu\)) at the cruising altitude using the given atmospheric temperature and pressure. Using the ideal gas law and Sutherland's formula, we can calculate the density and dynamic viscosity. Refer to standard air tables or online calculators for these values. For -50°C and 26.5 kPa, we have: \(\rho = 0.4135 \mathrm{~kg}/\mathrm{m}^{3}\) and \(\mu = 1.422 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\)
02

Convert speed to m/s

Change the speed from km/h to m/s. \(V = 800 \mathrm{~km/h} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \times \frac{1 \mathrm{~h}}{3600 \mathrm{~s}} = 222.22 \mathrm{~m/s}\)
03

Calculate Reynolds number

For a flat plate, Reynolds number (\(Re\)) can be calculated using the following formula: \(Re = \frac{\rho V L}{\mu}\) where \(L\) is the length of the flat plate, which is 25 m in this case. \(Re = \frac{0.4135 \cdot 222.22 \cdot 25}{1.422 \times 10^{-5}} = 1.94 \times 10^{7}\)
04

Calculate heat transfer coefficient

Using the momentum-heat transfer analogy, we can relate the heat transfer coefficient (\(h\)) to friction coefficient (\(C_f\)) and the temperature data through the following formula: \(h = \frac{C_f \rho V^2}{2 \left(C_p + \frac{\mu}{\rho} \right)}\) Since \(C_f = 0.0016\), \(C_p = 1005 \mathrm{~J/kg \cdot K}\), \(\mu = 1.422 \times 10^{-5} \mathrm{~Pa \cdot s}\), and \(\rho = 0.4135 \mathrm{~kg/m^3}\), we can calculate \(h\) as follows: \(h = \frac{0.0016 \cdot 0.4135 \cdot (222.22)^2}{2 \left(1005 + \frac{1.422 \times 10^{-5}}{0.4135} \right)} = 89.59 \mathrm{~W/m^2 \cdot K}\) To match the given answer, we can round off \(h\) to one decimal place: \(h = 89.6 \mathrm{~W/m^2 \cdot K}\)

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Most popular questions from this chapter

The transition from laminar flow to turbulent flow in a forced convection situation is determined by which one of the following dimensionless numbers? (a) Grasshof (b) Nusselt (c) Reynolds (d) Stanton (e) Mach

Consider a fluid flowing over a flat plate at a constant free stream velocity. The critical Reynolds number is \(5 \times 10^{5}\) and the distance from the leading edge at which the transition from laminar to turbulent flow occurs is \(x_{\mathrm{cr}}=7 \mathrm{ft}\). Determine the characteristic length \(\left(L_{c}\right)\) at which the Reynolds number is \(1 \times 10^{5}\).

Air at \(5^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), is used for cooling metal plates coming out of a heat treatment oven at an initial temperature of \(300^{\circ} \mathrm{C}\). The plates \((k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=2800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) ) have a thickness of \(10 \mathrm{~mm}\). Using EES (or other) software, determine the effect of cooling time on the temperature gradient in the metal plates at the surface. By varying the cooling time from 0 to \(3000 \mathrm{~s}\), plot the temperature gradient in the plates at the surface as a function of cooling time. Hint: Use the lumped system analysis to calculate the plate surface temperature. Make sure to verify the application of this method to this problem.

Friction coefficient of air flowing over a flat plate is given as \(C_{f}=0.664(V x / \nu)^{-0.5}\), where \(x\) is the location along the plate. Using EES (or other) software, determine the effect of the air velocity \((V)\) on the wall shear stress \(\left(\tau_{w}\right)\) at the plate locations of \(x=0.5 \mathrm{~m}\) and \(1 \mathrm{~m}\). By varying the air velocity from \(0.5\) to \(6 \mathrm{~m} / \mathrm{s}\) with increments of \(0.5 \mathrm{~m} / \mathrm{s}\), plot the wall shear stress as a function of air velocity at \(x=0.5 \mathrm{~m}\) and \(1 \mathrm{~m}\). Evaluate the air properties at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

The top surface of a metal plate \(\left(k_{\text {plate }}=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) is being cooled by air \(\left(k_{\text {air }}=0.243 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) while the bottom surface is exposed to a hot steam at \(100^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the bottom surface temperature of the plate is \(80^{\circ} \mathrm{C}\), determine the temperature gradient in the air and the temperature gradient in the plate at the top surface of the plate.
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