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Chapter 4: Problem 29

Ventilation air specifications for classrooms require that at least \(8.0 \mathrm{L} / \mathrm{s}\) of fresh air be supplied for each person in the room (students and instructor). A system needs to be designed that will supply ventilation air to 6 classrooms, each with a capacity of 20 students. Air enters through a central duct, with short branches successively leaving for each classroom. Branch registers are \(200 \mathrm{mm}\) high and \(500 \mathrm{mm}\) wide. Calculate the volume flow rate and air velocity entering each room. Ventilation noise increases with air velocity. Given a supply duct \(500 \mathrm{mm}\) high, find the narrowest supply duct that will limit air velocity to a maximum of \(1.75 \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
The volume flow rate per classroom is \(0.168 m^3/s\) and the air velocity is \(1.68 m/s\). The width of the supply duct to limit air velocity to \(1.75 m/s\) is \(1.152 m\).

Step by step solution

01

Calculate Total Volume Flow Rate

The volume flow rate is given by the product of the number of people (students and teachers per classroom) and the required ventilation rate. With each classroom having 20 students and assumed 1 instructor, and the requirement being 8 L/s per person.So, total volume flow rate required, \(Q_T\) = Number of Classrooms x Capacity of Classroom x ventilation rate per person. Therefore, \(Q_T\) = 6 (classrooms) x 21(20 students + 1 teacher) x 8 L/s = 1008 L/s. Convert this to cubic meter, as standard SI unit of \(Q_T\) would be \(m^3/s\): \(Q_T\) = 1008 * \(10^{-3} m^3/s\) = 1.008 \(m^3/s\)
02

Calculate Volume Flow Rate and Air Velocity for Each Classroom

The volume flow rate per classroom (\(Q\)) is simply \(Q_T\) divided by the number of classrooms: \(Q\) = \(Q_T / 6\) = 1.008 \(m^3/s\) / 6 = 0.168 \(m^3/s\)The ventilation air enters through a \(200 mm\) high and \(500 mm\) wide register, therefore, cross-sectional area of the register windows \(A\) = height x width = \(0.2 m * 0.5 m = 0.1 m^2\)The velocity (\(v\)) of air is volume flow rate per classroom divided by the cross-sectional area, \(v = Q / A = 0.168 m^3/s / 0.1 m^2 = 1.68 m/s\).
03

Dimensions for Supply Duct

To limit the air velocity to \(1.75 m/s\), the cross-sectional area \(A_s\) for the supply duct can be determined by dividing the total volume flow rate (\(Q_T\)) with this maximum velocity: \(A_s = Q_T / v_{max}\)Substituting the given values:\(A_s = 1.008 m^3/s / 1.75 m/s = 0.576 m^2\)Given that the height of the duct is \(0.5 m\), we can find the width of the duct by dividing the area by the height:Width of the duct = \(A_s / 0.5 m = 0.576 m^2 / 0.5 m = 1.152 m\)

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