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Chapter 7: Problem 111

Air (1 atm, \(\left.5^{\circ} \mathrm{C}\right)\) with free stream velocity of \(2 \mathrm{~m} / \mathrm{s}\) flows in parallel to a stationary thin \(1 \mathrm{~m} \times 1 \mathrm{~m}\) flat plate over the top and bottom surfaces. The flat plate has a uniform surface temperature of \(35^{\circ} \mathrm{C}\). Determine \((a)\) the average friction coefficient, \((b)\) the average convection heat transfer coefficient, and \((c)\) the average convection heat transfer coefficient using the modified Reynolds analogy and compare with the result obtained in \((b)\).

Short Answer

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Question: Determine the average friction coefficient, average convection heat transfer coefficient, and average convection heat transfer coefficient using the modified Reynolds analogy for an air flow parallel to a stationary flat plate at 1 atm, 0.5°C, and 2 m/s velocity. The length of the flat plate is 1 meter. Compare the results of the average convection heat transfer coefficient calculated using both methods. Solution: Step 1: The properties of air at 1 atm and 0.5°C are: Density (ρ) = 1.292 kg/m³ Dynamic viscosity (μ) = 1.731 x 10⁻⁵ kg/ms Thermal conductivity (k) = 0.0244 W/mK Specific heat (c_p) = 1006 J/kgK Step 2: Calculate the Reynolds number: Re = (2 m/s × 1.292 kg/m³ × 1 m) / (1.731 x 10⁻⁵ kg/ms) ≈ 149118 Step 3: Calculate the average friction coefficient Cf: Cf = 0.074 × Re^(-0.2) ≈ 0.00867 Step 4: Calculate the average convection heat transfer coefficient h_avg: Prandtl number (Pr) = (1006 J/kgK × 1.731 x 10⁻⁵ kg/ms) / 0.0244 W/mK ≈ 0.713 Nusselt number (Nu) = 0.0366 × Re^(0.8) × Pr^(1/3) ≈ 75.02 h_avg = (0.0244 W/mK × 75.02) / 1 m ≈ 1.83 W/m²K Step 5: Calculate the average convection heat transfer coefficient using the modified Reynolds analogy h_avg-mod: F_Pr = Pr^(2/3) - 1 ≈ -0.096 h_avg-mod = (0.00867 × (-0.096) × 1.292 kg/m³ × (2 m/s)²) / 2 ≈ -0.109 W/m²K Comparison: The average convection heat transfer coefficient calculated using the standard method (h_avg) is 1.83 W/m²K, while the average convection heat transfer coefficient calculated using the modified Reynolds analogy (h_avg-mod) is -0.109 W/m²K. The negative value indicates that the modified Reynolds analogy does not provide an accurate prediction of the heat transfer coefficient in this case.

Step by step solution

01

Find the properties of air at the given condition

First, we need to find out the properties of air at the given temperature, pressure, and velocity. Here are the properties of air at 1 atm and 0.5°C: Density (\(\rho\)) = 1.292 kg/m³ Dynamic viscosity (\(\mu\)) = 1.731 x 10⁻⁵ kg/ms Thermal conductivity (k) = 0.0244 W/mK Specific heat (\(c_p\)) = 1006 J/kgK
02

Calculate the Reynolds number

Next, calculate the Reynolds number (\(Re\)) for the air flowing over the flat plate using the following formula: \(Re = \dfrac{U\rho L}{\mu}\) Where: \(U\) = 2 m/s (velocity of air) \(\rho\) = 1.292 kg/m³ (density of air) \(L\) = 1 m (length of flat plate) \(\mu\) = 1.731 x 10⁻⁵ kg/ms (dynamic viscosity of air)
03

Calculate the average friction coefficient \(C_f\)

For flow over a flat plate with a Reynolds number (\(Re\)) greater than 1000, the average friction coefficient (\(C_f\)) can be calculated using the Blasius Equation: \(C_f = 0.074 \times Re^{-0.2}\) Substitute the calculated Reynolds number in the equation and determine \(C_f\).
04

Calculate the average convection heat transfer coefficient \(h_{avg}\)

To calculate the average convection heat transfer coefficient, \(h_{avg}\), the Nusselt number, \(Nu\), needs to be calculated first. The Nusselt number can be found using the following correlation: \(Nu = 0.0366 \times Re^{0.8} \times Pr^{1 / 3}\) Where: \(Pr\) = Prandtl number Calculate the Prandtl number (\(Pr\)) using the properties of air: \(Pr = \dfrac{c_p\mu}{k}\) Now, substitute the obtained Reynolds number, Prandtl number, and Nusselt number in the correlation. \(h_{avg} = \dfrac{k \times Nu}{L}\)
05

Calculate the average convection heat transfer coefficient using the modified Reynolds analogy \(h_{avg-mod}\)

The modified Reynolds analogy states that the ratio of the average convection heat transfer coefficient (\(h_{avg-mod}\)) to the friction coefficient (\(C_f\)) depends on the Prandtl number (\(Pr\)): \(h_{avg-mod} = \dfrac{C_tF_Pr \rho U^2}{2}\) Where: \(F_Pr = Pr^{(2/3)} - 1\) Insert the known values and calculate \(h_{avg-mod}\). Now, compare the results of the average convection heat transfer coefficient calculated in step 4 and step 5.

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Most popular questions from this chapter

In an experiment, the temperature of a hot air stream is to be measured by a thermocouple with a spherical junction. Due to the nature of this experiment, the response time of the thermocouple to register 99 percent of the initial temperature difference must be within \(5 \mathrm{~s}\). The properties of the thermocouple junction are \(k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\rho=8500 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=320 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The hot air has a free stream velocity and temperature of \(3 \mathrm{~m} / \mathrm{s}\) and \(140^{\circ} \mathrm{C}\), respectively. If the initial temperature of the thermocouple junction is \(20^{\circ} \mathrm{C}\), determine the thermocouple junction diameter that would satisfy the required response time of \(5 \mathrm{~s}\). Hint: Use the lumped system analysis to determine the time required for the thermocouple to register 99 percent of the initial temperature difference (verify application of this method to this problem).

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?

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 \(x=L\).

Consider laminar flow over a flat plate. Will the friction coefficient change with distance from the leading edge? How about the heat transfer coefficient?

Jakob (1949) suggests the following correlation be used for square tubes in a liquid cross-flow situation: $$ \mathrm{Nu}=0.102 \mathrm{Re}^{0.675} \mathrm{Pr}^{1 / 3} $$ Water \((k=0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6)\) at \(50^{\circ} \mathrm{C}\) flows across a \(1-\mathrm{cm}\) square tube with a Reynolds number of 10,000 and surface temperature of \(75^{\circ} \mathrm{C}\). If the tube is \(2 \mathrm{~m}\) long, the rate of heat transfer between the tube and water is (a) \(6.0 \mathrm{~kW}\) (b) \(8.2 \mathrm{~kW}\) (c) \(11.3 \mathrm{~kW}\) (d) \(15.7 \mathrm{~kW}\) (e) \(18.1 \mathrm{~kW}\)
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