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Chapter 25: Problem 14

A louvre in a \(0.25 \mathrm{~m}\) diameter extract fan and duct system is used to control air flow rates. If the flow rate is \(2.0 \mathrm{~m}^{3} \mathrm{~s}^{-1}\) with the louvre angle \(\theta\) set to \(60^{\circ}\) open, calculate: (a) the louvre separation loss coefficient at this setting and the pressure drop across the louvre, (b) the reduction in flow rate if the louvre is shut down to \(30^{\circ}\) open if the pressure drop across the louvre remains constant. Assume air density to be \(1.2 \mathrm{~kg} \mathrm{~m}^{-3}\) and that the relationship between louvre loss coefficient \(K\) and louvre open angle \(\theta\) is $$ \log _{\mathrm{e}} K=2 \times(90-\theta) / 180-1.25 $$ where \(\theta\) is measured in degrees.

Short Answer

Expert verified
(a) Louvre loss coefficient: 0.4, Pressure drop: 397.18 Pa; (b) New flow rate: 1.68 m鲁/s.

Step by step solution

01

Calculate the Louvre Loss Coefficient for 胃 = 60掳

Using the provided formula \( \log_{\mathrm{e}} K = 2 \times (90 - \theta) / 180 - 1.25 \).Substitute \( \theta = 60 \) degrees:\[ \log_{\mathrm{e}} K = 2 \times (90 - 60) / 180 - 1.25 = 2 \times 30 / 180 - 1.25 = \frac{1}{3} - 1.25 \]Calculate \( \log_{\mathrm{e}} K \):\[ \log_{\mathrm{e}} K = -0.9167 \]Solve for \( K \):\[ K = e^{-0.9167} \approx 0.4 \]
02

Calculate the Air Velocity in the Duct

The cross-sectional area of the duct can be calculated as a circle with a diameter of \( 0.25 \text{ m} \):\[ \text{Area} = \pi \times \left(\frac{0.25}{2}\right)^2 = 0.0491 \text{ m}^2 \]Using the flow rate \( Q = 2.0 \text{ m}^3/\text{s} \):Velocity \( v \) is given by:\[ v = \frac{Q}{\text{Area}} = \frac{2.0}{0.0491} = 40.73 \text{ m/s} \]
03

Calculate the Pressure Drop Across the Louvre

The pressure drop \( \Delta P \) is given by the formula:\[ \Delta P = \frac{1}{2} \times \rho \times v^2 \times K \]Using \( \rho = 1.2 \text{ kg/m}^3 \), \( v = 40.73 \text{ m/s} \), and \( K = 0.4 \):\[ \Delta P = 0.5 \times 1.2 \times (40.73)^2 \times 0.4 \]\[ \Delta P = 397.18 \text{ Pa} \]
04

Calculate the Louvre Loss Coefficient for 胃 = 30掳

Using the same formula as before with \( \theta = 30^\circ \):\[ \log_{\mathrm{e}} K = 2 \times (90 - 30) / 180 - 1.25 = \frac{2}{3} - 1.25 \]\[ \log_{\mathrm{e}} K = -0.5833 \]Solve for \( K \):\[ K = e^{-0.5833} \approx 0.558 \]
05

Calculate the New Flow Rate with 胃 = 30掳 Open

Since \( \Delta P \) remains the same at \( 397.18 \text{ Pa} \), use the formula:\[ \Delta P = \frac{1}{2} \times \rho \times v^2 \times K \]Solve for new air velocity \( v' \) with \( K = 0.558 \):\[ 397.18 = 0.5 \times 1.2 \times (v')^2 \times 0.558 \]\[ (v')^2 = \frac{397.18}{0.5 \times 1.2 \times 0.558} \quad \Rightarrow \quad v' = 34.28 \text{ m/s} \]The new flow rate \( Q' \) is:\[ Q' = v' \times \text{Area} = 34.28 \times 0.0491 = 1.68 \text{ m}^3/\text{s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Drop Calculation
Understanding how to calculate pressure drop is vital in analyzing air flow through a duct system with a louvre. Pressure drop refers to the reduction in air pressure as air flows through the system. It is important because it affects the efficiency and performance of the ventilation system. Here, pressure drop is modeled using the equation: \[\Delta P = \frac{1}{2} \times \rho \times v^2 \times K\]where:
  • \(\Delta P\) is the pressure drop in pascals (Pa)
  • \(\rho\) is the air density, given as 1.2 kg/m鲁
  • \(v\) is the air velocity in m/s
  • \(K\) is the louvre loss coefficient
First, we calculate the air velocity, as it is influenced by the flow rate \(Q = 2.0 \, \text{m}^3/\text{s}\) and the duct cross-section area. The calculated air velocity then helps us to determine the pressure drop using the given formula. By substituting the air density and louvre loss coefficient into the equation, you can derive the pressure drop value across the louvre at a specific louvre angle.
Flow Rate Reduction
Flow rate reduction explains how much the volume of air passing through the louvre decreases when the louvre is adjusted. In our analysis, we consider how closing the louvre from an angle of 60掳 to 30掳 impacts this flow. The flow rate is directly tied to air velocity and duct cross-sectional area. When the louvre angle changes, the loss coefficient \(K\) varies, affecting the velocity needed to maintain a certain pressure drop. If \(\Delta P\) remains constant, the formula to find new velocity \(v'\) is still:\[\Delta P = \frac{1}{2} \times \rho \times (v')^2 \times K\]By substituting the new \(K\) at 30掳, we can solve for the new velocity \(v'\). Finally, the flow rate \(Q'\) (in \(\text{m}^3/\text{s}\)) can be deduced using the formula:\[Q' = v' \times \text{Area}\]This calculation provides the decreased flow rate, showing the impact of partially closing the louvre on the air flow in the system.
Louvre Loss Coefficient
The louvre loss coefficient \(K\) is a crucial factor in understanding how a louvre affects airflow. This dimensionless number represents the resistance the louvre adds to airflow. The coefficient is determined by the formula:\[\log_{e} K = 2 \times \frac{(90-\theta)}{180} - 1.25\]where \(\theta\) is the open angle of the louvre. This relationship shows how \(K\) changes with different positions of the louvre. For example, more closed angles result in higher values of \(K\), demonstrating more significant flow resistance. By calculating \(K\) for different angles, we can predict how those settings affect the pressure drop and flow rate. In practical terms, it helps in adjusting the airflow system for optimal performance by balancing pressure drop with desired flow rate, ensuring energy efficiency and effective ventilation.

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Most popular questions from this chapter

The characteristics of two rotodynamic pumps at constant speed are as follows: ???? A: \(\begin{array}{cllll}Q\left(\mathrm{~m}^{3} \mathrm{~s}^{-1}\right) & 0 & 0.006 & 0.012 & 0.018 \\ H(\mathrm{~m}) & 22.6 & 21.9 & 20.3 & 17.7 \\\ \eta(\text { per cent }) & 0 & 32 & 74 & 86 \\ Q\left(\mathrm{~m}^{3} \mathrm{~s}^{-1}\right) & 0.024 & 0.030 & 0.036 \\ H(\mathrm{~m}) & 14.2 & 9.7 & 3.9 & \\ \eta(\text { per cent }) & 85 & 66 & 28 & \\ \text { Pump B: } & & & & \\ Q\left(\mathrm{~m}^{3} \mathrm{~s}^{-1}\right) & 0 & 0.006 & 0.012 & 0.018 \\ H(\mathrm{~m}) & 16.2 & 13.6 & 11.9 & 11.6 \\ \eta(\text { per cent }) & 0 & 14 & 34 & 60 \\ Q\left(\mathrm{~m}^{3} \mathrm{~s}^{-1}\right) & 0.024 & 0.030 & 0.036 \\ H(\mathrm{~m}) & 10.7 & 9.0 & 6.4 & \\ \eta(\text { per cent }) & 80 & 80 & 60 & \end{array}\) One of the above pumps is required to lift water continuously through \(3.2 \mathrm{~m}\) of vertical lift and the pipe to be used is \(21 \mathrm{~m}\) long, \(10 \mathrm{~cm}\) in diameter, and the friction coefficient is \(0.005\). Select the more suitable pump for this duty and justify your selection. What power input will be required by the selected pump?

Five operatives are employed in an aircraft hangar, volume \(5000 \mathrm{~m}^{3}\), to spray a camouflage scheme. An extract fan system removes air at a rate of \(4 \mathrm{~m}^{3} \mathrm{~s}^{-1}\), with the hangar remaining at atmospheric pressure. Paint is sprayed continuously by each operative at a steady rate of 15 litres \(\mathrm{h}^{-1}\), the paint having the following specification: $$ \begin{array}{ll} \text { Density } & 1.2 \mathrm{~kg} 1^{-1} \\ \text { Solvent content } & 25 \text { per cent by mass } \\ \text { Drying time (i.e. solvent } & \\ \quad \text { evaporation time) } & \text { assumed instantaneous } \\ \text { Specific volume } & \\ \quad \text { solvent vapour } & 0.5 \mathrm{~m}^{3} \mathrm{~kg}^{-1} \\ \text { Lower explosive limit for } & \\ \text { solvent vapour in air } & 2.0 \text { per cent by volume. } \end{array} $$ (a) Determine the maximum duration that the operatives may work before the contamination level exceeds 1 per cent of the lower explosive limit. (b) Once the process has been stopped, determine the time necessary for the contamination level to fall to 1 per cent of the lower explosive limit with the ventilation fans operating at a reduced rate of extraction of \(2 \mathrm{~m}^{3} \mathrm{~s}^{-1}\).

A ventilation duct includes a fan and control louvre. Determine the pressure rise across the fan if, at a louvre setting of \(35^{\circ}\) to the horizontal, the ventilation rate is 6 changes per hour. The room has a volume of \(300 \mathrm{~m}^{3}\) and is held at \(250 \mathrm{~N} \mathrm{~m}^{-2}\) above atmosphere. The duct is \(0.25 \mathrm{~m}\) in diameter and has a \(15 \mathrm{~m}\) overall length. Entry and exit losses are equivalent to \(33.3\) pipe diameters and the louvre has a loss coefficient defined below, \(\begin{array}{llllll}K & 2.0 & 3.2 & 6.4 & 12.8 & 30.6 \\ \text { Louvre angle } & 0 & 20 & 40 & 60 & 80\end{array}\) Take air density to be \(1.2 \mathrm{~kg} \mathrm{~m}^{-3}\) and the duct friction factor as \(0.009\)

A centrifugal pump is used to circulate water in a closed loop experimental rig, consisting of: two vertical pipes, one \(4 \mathrm{~m}\) long and the other \(3 \mathrm{~m}\) long; two horizontal pipes, each \(1.3 \mathrm{~m}\) long; three \(90^{\circ}\) bends; and a vertical 'working section' \(1 \mathrm{~m}\) long. The pump is situated in one of the two low-level corners of the circuit. The pipes and the bends are \(7.5 \mathrm{~cm}\) in diameter and the working section has a cross-sectional area of \(125 \mathrm{~cm}^{2}\). The friction factor for all pipes is \(0.006\) and the loss in each bend may be taken as \(0.1 v^{2} / 2 \boldsymbol{g}\), where \(v\) is the mean velocity in metres per second. The loss in the 'working section' may be taken as equivalent to a frictional loss in a \(1 \mathrm{~m}\) long pipe of \(7.5 \mathrm{~cm}\) diameter Determine the mean velocity in the 'working section' if the pump characteristic is as follows: $$ \begin{array}{lllllll} Q\left(\mathrm{~m}^{3} \mathrm{~s}^{-1}\right) & 0 & 0.006 & 0.012 & 0.018 & 0.024 & 0.027 \\ H(\mathrm{~m}) & 3.20 & 3.13 & 2.90 & 2.42 & 1.62 & 0.98 \end{array} $$

The characteristics of a centrifugal pump at constant speed are as follows: \(\begin{array}{lllll}Q\left(\mathrm{~m}^{3} \mathrm{~s}^{-1}\right) & 0 & 0.012 & 0.018 & 0.024 \\ H(\mathrm{~m}) & 22.6 & 21.3 & 19.4 & 16.2 \\\ \eta(\text { per cent }) & 0 & 74 & 86 & 85 \\ Q\left(\mathrm{~m}^{3} \mathrm{~s}^{-1}\right) & 0.030 & 0.036 & 0.042 & \\ H(\mathrm{~m}) & 11.6 & 6.5 & 0.6 & \\ \eta(\text { per cent }) & 70 & 46 & 8 & \end{array}\) The pump is used to lift water over a vertical distance of \(6.5 \mathrm{~m}\) by means of a \(10 \mathrm{~cm}\) diameter pipe, \(65 \mathrm{~m}\) long, for which the friction coefficient \(f=0.005\). (a) Determine the rate of flow and the power supplied to the pump. (b) If it is required to increase the rate of flow, and this may be achieved only by the addition of a second, identical pump (running at the same speed), investigate whether it should be connected in series or in parallel with the original pump. Justify your answer by determining the increased rate of flow and power consumed by both pumps.
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