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Chapter 4: Problem 22

A university laboratory that generates \(15 \mathrm{m}^{3} / \mathrm{s}\) of air flow at design condition wishes to build a wind tunnel with variable speeds. It is proposed to build the tunnel with a sequence of three circular test sections: section 1 will have a diameter of \(1.5 \mathrm{m},\) section 2 a diameter of \(1 \mathrm{m},\) and section 3 a diameter such that the average speed is \(75 \mathrm{m} / \mathrm{s}\) (a) What will be the speeds in sections 1 and \(2 ?\) (b) What must the diameter of section 3 be to attain the desired speed at design condition?

Short Answer

Expert verified
The speeds in sections 1 and 2 will be approximately \(8.476 m/s\) and \(19.739 m/s\) respectively. The diameter of section 3 needs to be approximately \(0.505 m\) to attain the desired speed at design condition.

Step by step solution

01

Calculate the speeds in sections 1 and 2

Using the simplified continuity equation, we can find the speeds in sections 1 and 2. Remember that the area of a circle is calculated as \( A = \pi r^2 \), where \( r \) is the radius of the circle, which is half the diameter. Therefore, \( v_1 = \frac{Q}{A_1} = \frac{15 m^3/s}{\frac{\pi}{4} * (1.5 m)^2} \approx 8.476 m/s \) , and \( v_2 = \frac{Q}{A_2} = \frac{15 m^3/s}{\frac{\pi}{4} * (1 m)^2} \approx 19.739 m/s \).
02

Find the diameter of section 3

The speed in section 3 is given as \(75 m/s\). We can rearrange the simplified continuity equation to find the area of section 3: \( A_3 = \frac{Q}{v_3} = \frac{15 m^3/s}{75 m/s} = 0.2 m^2 \). Then we can calculate the diameter of section 3 by rearranging the equation for the area of a circle to solve for the radius, then doubling it to find the diameter: \( d_3 = 2 * \sqrt{\frac{A_3}{\pi}} = 2 * \sqrt{\frac{0.2 m^2}{\pi}} \approx 0.505 m \).

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Most popular questions from this chapter

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