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Chapter 6: Problem 102

Object 1 with a characteristic length of \(0.5 \mathrm{~m}\) is placed in airflow at 1 atm and \(20^{\circ} \mathrm{C}\) with free stream velocity of \(50 \mathrm{~m} / \mathrm{s}\). The heat flux transfer from object 1 when placed in the airflow is measured to be \(12,000 \mathrm{~W} / \mathrm{m}^{2}\). If object 2 has the same shape and geometry as object 1 (but with a characteristic length of \(5 \mathrm{~m}\) ) is placed in the airflow at 1 atm and \(20^{\circ} \mathrm{C}\) with free stream velocity of \(5 \mathrm{~m} / \mathrm{s}\), determine the average convection heat transfer coefficient for object 2 . Both objects are maintained at a constant surface temperature of \(120^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Given the airflow at 1 atm and \(20^{\circ}\mathrm{C}\), we have the air properties \(\rho = 1.204 \mathrm{~kg} / \mathrm{m}^3\) and \(\mu = 1.81 \times 10^{-5} \mathrm{~ kg} / (\mathrm{m} \cdot \mathrm{s})\). Using these values, we calculated the Reynolds number for object 1 as \(Re_1\) and for object 2 as \(Re_2\). From the given heat flux transfer for object 1 and assuming the surface area \(A\) to be equal to 1, we calculated the convection heat transfer coefficient for object 1, represented as \(h_1\). Then, we found the convection heat transfer coefficient for object 2, denoted as \(h_2\), by using the proportionality with the Reynolds numbers and the calculated value for \(h_1\). Finally, calculate the value of \(h_2\) using the equation: \[h_2 = h_1 \left(\frac{Re_2}{Re_1}\right)\]

Step by step solution

01

Calculate the Reynolds number for both objects

First, let's determine the Reynolds number for both objects. The Reynolds number is defined as: \[Re = \frac{\rho V L}{\mu}\] where \(\rho\) is the fluid density, \(V\) is the free stream velocity, \(L\) is the characteristic length, and \(\mu\) is the fluid dynamic viscosity. We are given that the airflow is at 1 atm and \(20^{\circ}\mathrm{C}\). For air at this condition, we can assume the following property values: \(\rho = 1.204 \mathrm{~kg} / \mathrm{m}^3\) and \(\mu = 1.81 \times 10^{-5} \mathrm{~ kg} / (\mathrm{m} \cdot \mathrm{s})\). Now we can calculate the Reynolds numbers for both objects: For object 1: \[Re_1 = \frac{(1.204)(50)(0.5)}{1.81 \times 10^{-5}}\] For object 2: \[Re_2 = \frac{(1.204)(5)(5)}{1.81 \times 10^{-5}}\]
02

Calculate the convection heat transfer coefficient for object 1

Given the heat flux transfer from object 1, we can find the convection heat transfer coefficient, \(h_1\), using the following equation: \[q = h_1 A(T_s - T_\infty)\] where \(q\) is the heat flux, \(A\) is the surface area, \(T_s\) is the surface temperature, and \(T_\infty\) is the free stream temperature. For object 1, the heat flux is given as \(12,000 \mathrm{~W} / \mathrm{m}^2\). The surface temperature of the object and the free stream temperature are given as \(120^{\circ}\mathrm{C}\) and \(20^{\circ}\mathrm{C}\), respectively. Since we are interested in \(h_1\), we can rewrite the equation as \[h_1 = \frac{q}{A(T_s - T_\infty)}\] For simplification purposes, we should consider A to be equal to 1 for this calculation. Therefore, h becomes: \[h_1 = \frac{12,000}{(120-20)}\]
03

Determine the convection heat transfer coefficient for object 2

Since object 1 and object 2 have the same shape and geometry, their convection heat transfer coefficients will be proportional to the Reynolds number. We can write the proportionality as: \[\frac{h_2}{h_1} = \left(\frac{Re_2}{Re_1}\right)^{n}\] where \(n\) is a scaling exponent, which depends on the shape and geometry of the objects. For simplicity, let's assume \(n = 1\) in this exercise. Then, we can find the convection heat transfer coefficient for object 2: \[h_2 = h_1 \left(\frac{Re_2}{Re_1}\right)\] Substitute the values of \(Re_1\) and \(Re_2\) calculated in step 1, and \(h_1\) calculated in step 2 into the equation to find the convection heat transfer coefficient for object 2.

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