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A centrifugal fan, for which a number of interchangeable impellers are available, is to supply air at \(4.5 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) to a ventilating duct at a head of \(100 \mathrm{~mm}\) water gauge. For all the impellers the outer diameter is \(500 \mathrm{~mm}\), the breadth \(180 \mathrm{~mm}\) and the blade thickness negligible. The fan runs at \(188.5 \mathrm{rad} \cdot \mathrm{s}^{-1}\) ( \(\left.30 \mathrm{rev} / \mathrm{s}\right)\). Assuming that the conversion of velocity head to pressure head in the volute is counterbalanced by the friction losses there and in the impeller, that there is no whirl at inlet and that the air density is constant at \(1.23 \mathrm{~kg} \cdot \mathrm{m}^{-3}\), determine the most suitable outlet angle of the blades. (Neglect whirl slip.)

Short Answer

Expert verified
The most suitable outlet blade angle is approximately 88.41 degrees.

Step by step solution

01

Calculate the flow velocity

First, determine the flow velocity (peripheral velocity) at the outer diameter of the impeller. Use the formula: \[ U = \frac{\text{Outer diameter}}{2} \times \text{angular velocity} \]Given the outer diameter, D = 0.5 m, and angular velocity, \(\omega = 188.5 \text{ rad} \cdot \text{s}^{-1} \), the flow velocity is: \[ U = \frac{0.5}{2} \times 188.5 = 47.125 \text{ m} \cdot \text{s}^{-1} \]
02

Calculate the volume flow rate

Next, use the given volume flow rate, \( Q = 4.5 \text{ m}^3 \cdot \text{s}^{-1} \), to find the radial velocity at the outlet. The radial velocity, \(V_r\), can be calculated using the formula: \[ Q = A \cdot V_r \]where A is the area at the impeller outlet. Since the breadth (b) is 0.18 m and the diameter (D) is 0.5 m, the area is: \[ A = b \cdot D = 0.18 \cdot 0.5 = 0.09 \text{ m}^2 \]The radial velocity, \(V_r\), is then:\[ V_r = \frac{Q}{A} = \frac{4.5}{0.09} = 50 \text{ m} \cdot \text{s}^{-1} \]
03

Apply Bernoulli's equation for the flow head

Using the head (pressure head), convert the given water gauge head to a velocity head. Given head, \( h = 100 \text{ mm} = 0.1 \text{ m} \). Using the density of air, \( \rho = 1.23 \text{ kg} \cdot \text{m}^{-3} \), the velocity head can be found using Bernoulli鈥檚 equation:\[ P = \rho \cdot g \cdot h \]\[ P = 1.23 \cdot 9.81 \cdot 0.1 \approx 1.20663 \, \text{ Pa} \]
04

Relate velocity head to velocities

To find the relationship between the tangential and radial velocities at the outlet, recall that the head is related to these velocities by:\[ P = \rho \left( \frac{V_{t}^2}{2} \right) \]Since the pressure head must be counterbalanced by the conversion of velocity head to pressure head, solve for the tangential velocity \( V_t \):\[ 1.20663 = 1.23 \left( \frac{V_{t}^2}{2} \right) \]\[ V_{t}^2 = \frac{1.20663 \times 2}{1.23} \approx 1.962 \] \[ V_{t} \approx \sqrt{1.962} \approx 1.4 \text{ m} \cdot \text{s}^{-1} \]
05

Determine the outlet blade angle

Finally, calculate the angle of the outlet blades, \(\beta\), using the tangential and radial components of velocity. The blade angle at the outlet can be determined from the relation:\[ \tan(\beta) = \frac{V_r}{V_t} \]\[ \tan(\beta) = \frac{50}{1.4} \approx 35.714 \]\[ \beta = \arctan(35.714) \approx 88.41^\circ \]

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Key Concepts

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

Flow Velocity
When dealing with centrifugal fans, the flow velocity at the impeller鈥檚 outer diameter is key. It measures how fast the air moves due to the fan鈥檚 rotation.
To find this, use the formula: \[ U = \frac{\text{Outer diameter}}{2} \times \text{angular velocity} \]
Here, the given outer diameter (0.5 m) and angular velocity (188.5 rad/s) yield a flow velocity of 47.125 m/s. This tells us how quickly the air is rotated by the fan blades.
Radial Velocity
Radial velocity (Vr) shows the speed of air moving from the center to the edge of the fan.
This is crucial because it affects the volume of air moved by the fan.
With a volume flow rate (Q) of 4.5 m鲁/s and an outlet area (A) of 0.09 m虏, we find Vr using: \[ V_r = \frac{Q}{A} = \frac{4.5}{0.09} = 50 \text{ m} \cdot \text{s}^{-1} \]
This calculation helps determine how efficiently the fan moves air outward from the impeller.
Bernoulli's Equation
Bernoulli's equation links pressure, velocity, and height in fluid dynamics.
In this fan problem, we convert water gauge head to velocity head.
Given a head (h) of 0.1 m and air density (蟻) of 1.23 kg/m鲁, Bernoulli鈥檚 equation is: \[ P = \rho \cdot g \cdot h \]
Substituting values, we get: \[ P = 1.23 \cdot 9.81 \cdot 0.1 \approx 1.20663 \text{ Pa} \]
This pressure difference is key to understanding the energy transformation within the fan.
Outlet Blade Angle
The outlet blade angle (尾) impacts how air exits the fan.
Using the tangential (Vt) and radial velocities (Vr), we apply: \[ \tan(\beta) = \frac{V_r}{V_t} \]
Given Vr of 50 m/s and Vt of 1.4 m/s, we find 尾: \[ \tan(\beta) = \frac{50}{1.4} \approx 35.714 \]
Solving, 尾 鈮 88.41掳.
This steep angle ensures that air leaves the fan blades effectively, maximizing the fan's performance.
Air Density
Air density (蟻) influences how much air mass is moved by the fan.
In this problem, 蟻 is given as 1.23 kg/m鲁 and used in calculations involving pressure and velocity relationships.
For instance, 蟻 was critical in Bernoulli鈥檚 equation to convert the water gauge head to velocity head.
This density impacts the fan鈥檚 ability to generate pressure and thus efficiently move air through the duct system.

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Most popular questions from this chapter

A vertical-shaft Francis turbine is to be installed in a situation where a much longer draft tube than usual must be used. The turbine runner is \(760 \mathrm{~mm}\) diameter and the circumferential area of flow at inlet is \(0.2 \mathrm{~m}^{2}\). The overall operating head is \(30 \mathrm{~m}\) and the speed \(39.27 \mathrm{rad} \cdot \mathrm{s}^{-1}(6.25 \mathrm{rev} / \mathrm{s}) .\) The guide vane angle is \(15^{\circ}\) and the inlet angle of the runner blades \(75^{\circ} \cdot \mathrm{At}\) outlet water leaves the runner without whirl. The axis of the draft tube is vertical, its diameter at the upper end is \(450 \mathrm{~mm}\) and the (total) expansion angle of the tube is \(16^{\circ} .\) For a flow rate of \(Q\left(\mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\right)\) the friction loss in the tube (of length \(l\) ) is given by \(h_{\mathrm{f}}=0.03 Q^{2} l .\) If the absolute pressure head at the top of the tube is not to fall below \(3.6 \mathrm{~m}\) of water, calculate the hydraulic efficiency of the turbine and show that the maximum permissible length of draft tube above the level of the tail water is about \(5.36 \mathrm{~m}\). (The length of the tube below tailwater level may be neglected. Atmospheric pressure \(\equiv 10.33 \mathrm{~m}\) water head.)

A single-acting reciprocating water pump, with a bore and stroke of \(150 \mathrm{~mm}\) and \(300 \mathrm{~mm}\) respectively, runs at \(2.51 \mathrm{rad} \cdot \mathrm{s}^{-1}(0.4 \mathrm{rev} / \mathrm{s}) .\) Suction and delivery pipes are each \(75 \mathrm{~mm}\) diameter. The former is \(7.5 \mathrm{~m}\) long and the suction lift is \(3 \mathrm{~m}\). There is no air vessel on the suction side. The delivery, pipe is \(300 \mathrm{~m}\) long, the outlet (at atmospheric pressure) being \(13.5 \mathrm{~m}\) above the level of the pump, and a large air vessel is connected to the delivery pipe at a point \(15 \mathrm{~m}\) from the pump. Calculate the absolute pressure head in the cylinder at beginning, middle and end of each stroke. Assume that the motion of the piston is simple harmonic, that losses at inlet and outlet of each pipe are negligible, that the slip is \(2 \%\), and that \(f\). for both pipes is constant at 0.01. (Atmospheric pressure \(10.33 \mathrm{~m}\) water head.)

The following duties are to be performed by rotodynamic pumps driven by electric synchronous motors, speed \(100 \pi / n \mathrm{rad} \cdot \mathrm{s}^{-1}(=50 / n \mathrm{rev} / \mathrm{s})\), where \(n\) is an integer: (a) \(14 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) of water against \(1.5 \mathrm{~m}\) head; \((b)\) oil (relative density \(0.80\) ) at \(11.3 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against \(70 \mathrm{kPa}\) pressure; \((c)\) water at \(5.25 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against \(5.5 \mathrm{MPa}\). Designs of pumps are available with specific speeds of \(0.20,0.60,1.20,2.83,4.0\) rad. Which design and speed should be used for each duty?

The following data refer to a Pelton wheel. Maximum overall efficiency \(79 \%\), occurring at a speed ratio of \(0.46 ; \mathrm{C}_{\mathrm{v}}\) for nozzle \(=0.97\); jet turned through \(165^{\circ}\). Assuming that the optimum speed ratio differs from \(0.5\) solely as a result of losses to windage and bearing friction which are proportional to the square of the rotational speed, obtain a formula for the optimum speed ratio and hence estimate the ratio of the relative velocity at outlet from the buckets to the relative velocity at inlet.

In a vertical-shaft inward-flow reaction turbine the sum of the pressure and kinetic heads at entrance to the spiral casing is \(120 \mathrm{~m}\) and the vertical distance between this section and the tail-race level is \(3 \mathrm{~m}\). The peripheral velocity of the runner at entry is \(30 \mathrm{~m} \cdot \mathrm{s}^{-1}\), the radial velocity of the water is constant at \(9 \mathrm{~m} \cdot \mathrm{s}^{-1}\) and discharge from the runner is without whirl. The estimated hydraulic losses are: (1) between turbine entrance and exit from the guide vanes, \(4.8 \mathrm{~m}\) (2) in the runner, \(8.8 \mathrm{~m},(3)\) in the draft tube, \(790 \mathrm{~mm}\), (4) kinetic head rejected to the tail race, \(460 \mathrm{~mm}\). Calculate the guide vane angle and the runner blade angle at inlet and the pressure heads at entry to and exit from the runner

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