The impeller of a centrifugal fan has an inner radius of \(250 \mathrm{~mm}\) and width of \(187.5 \mathrm{~mm} ;\) the values at exit are \(375 \mathrm{~mm}\) and \(125 \mathrm{~mm}\) respectively. There is no whirl at inlet, and at outlet the blades are backward-facing at \(70^{\circ}\) to the tangent. In the impeller there is a loss by friction of \(0.4\) times the kinetic head corresponding to the relative outlet velocity, and in the volute there is a gain equivalent to \(0.5\) times the kinetic head corresponding to the absolute velocity at exit from the runner. The discharge of air is \(5.7 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) when the rotational speed is \(84.8 \mathrm{rad} \cdot \mathrm{s}^{-1}(13.5 \mathrm{rev} / \mathrm{s}) .\) Neglecting the thickness of the blades and whirl slip, determine the head across the fan and the power required to drive it if the density of the air is sensibly constant at \(1.25 \mathrm{~kg} \cdot \mathrm{m}^{-3}\) throughout and mechanical losses account for \(220 \mathrm{~W}\).

Step by step solution

01

- Calculate the radial velocity at exit

First, determine the radial velocity at exit, assuming the velocity distribution is uniform:\[ Q = \frac{5.7}{\text{Area}} \]where Area is calculated as:\[ A = b \times 2\text{π}r \]with b being the width and r the radius at exit. Thus, \[ A = 125 \times 2\text{π} \times 375 \times 10^{-3} \approx 294.52 \text{~mm}^2 = 2.9452 \text{~m}^2 \]Therefore: \[ v_r = \frac{5.7}{2.9452} = 1.935 \text{m/s} \]

02

- Calculate the tangential velocity at exit

\[ v_t = \frac{U_2 - V_{r2}}{\tan(\theta_b)} \]where \(U_2\) is the outer velocity calculated as:\[ U_2 = R_2ω = 0.375 \times 84.8 = 31.8 \text{~m/s} \]Therefore,\[ v_t = \frac{31.8 - 1.935}{\tan(70)} = 10.87 \text{~m/s} \]

03

- Calculate head across the fan

The head across the fan can be calculated with the equation:\[ \text{Head} = \frac{v_t \times U_2}{g} \]where g = 9.81 \text{~m/s}^2. Therefore,\[ \text{Head} = \frac{10.87 \times 31.8}{9.81} = 35.2 \text{~m} \]

04

- Calculate the losses

The head loss due to friction in the impeller can be calculated as:\[ h_{loss,impeller} = 0.4 \times \frac{v_r^2}{2g} \]Substituting the values will give:\[ h_{loss,impeller} = 0.4 \times \frac{1.935^2}{2 \times 9.81} = 0.077 \text{~m} \]Similarly, the head gain in the volute can be calculated as:\[ h_{gain,volute} = 0.5 \times \frac{v_t^2}{2g} \]Substituting the values will give:\[ h_{gain,volute} = 0.5 \times \frac{10.87^2}{2 \times 9.81} = 0.302 \text{~m} \]

05

- Calculate the net head

The net head can be calculated as:\[ \text{Net Head} = \text{Head} + h_{gain,volute} - h_{loss,impeller} \]Substituting the values will give:\[ \text{Net Head} = 35.2 + 0.302 - 0.077 = 35.425 \text{~m} \]

06

- Calculate power required to drive the fan

The power required can be calculated using the formula:\[ P = \rho g Q \text{Net Head} + \text{Mechanical Losses} \]where \( \rho \) = 1.25 \text{~kg/m}^3, g = 9.81 \text{~m/s}^2, Q = 5.7 \text{~m^3/s}, and Mechanical losses = 220 W. Substituting the values:\[ P = 1.25 \times 9.81 \times 5.7 \times 35.425 + 220 = 25.145 \times 253.734 + 220 = 64,116.3 + 220 = 64336.3 \text{~W} \text{ or } 64.336 \text{~kW} \]

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Key Concepts

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

Impeller Dynamics

In a centrifugal fan, the impeller is the rotating component that accelerates the fluid. Understanding its dynamics is crucial.
The impeller's main job is to increase the mechanical energy of the fluid, primarily by increasing its kinetic energy. The fluid enters the impeller radially and is flung outward due to centrifugal force.
Key parameters include the impeller's radius and width at both the inlet and outlet. This exercise gives us specific dimensions: an inner radius of 250 mm increasing to 375 mm at the outlet, and a width of 187.5 mm decreasing to 125 mm. These values are essential for calculating the flow rates and velocities.
The angle and shape of the blades also impact performance. For instance, backward-facing blades, as seen in the problem (angled at 70 degrees to the tangent), influence the fluid dynamics and efficiency.

Fluid Mechanics Equations

Fluid mechanics equations are the backbone of analyzing centrifugal fan performance. These include continuity equations, Bernoulli's equation, and various velocity equations.
The continuity equation, which ensures mass conservation, helps determine radial velocity. We calculate it from the discharge rate and the exit area using the formula:
\[ v_r = \frac{Q}{A} \] where \[ A = b \times 2πr \]
The calculation for tangential velocity involves blade angles and rotational speeds, combining the velocity triangle equations:
\[ v_t = \frac{U_2 - V_{r2}}{\tan(θ_b)} \]
Using Bernoulli’s principle, we can link pressure, kinetic, and potential energies to calculate the head produced by the fan. The relationship for head calculation is:
\[ \text{Head} = \frac{v_t \times U_2}{g} \]

Losses and Gains in Fluid Systems

In real-world applications, fluid systems experience both losses and gains of energy. These need to be accounted for accurate performance analysis.
Losses typically occur due to friction within the impeller and other turbulence-inducing components. In this problem, a specific loss factor, given as 0.4 times the kinetic head corresponding to the relative outlet velocity, is used:
\[ h_{loss, impeller} = 0.4 \times \frac{v_r^2}{2g} \]
Conversely, there are gains, often in the volute (the spiral casing). This scenario assumes a gain equivalent to 0.5 times the kinetic head of the absolute exit velocity:
\[ h_{gain, volute} = 0.5 \times \frac{v_t^2}{2g} \]
These factors modify the net head, which is crucial for understanding actual performance:
\[ \text{Net Head} = \text{Head} + h_{gain, volute} - h_{loss, impeller} \]

Power Calculation

Determining the power required to drive a centrifugal fan is an essential part of performance analysis. This combines all the earlier calculations with additional factors like mechanical losses.
Power is derived by integrating the net head, fluid discharge, and the fluid's density, and then adding mechanical losses:
\[ P = \rho g Q \text{Net Head} + \text{Mechanical Losses} \] substituting known values:
\[ P = 1.25 \times 9.81 \times 5.7 \times 35.425 + 220 = 64116.3 \text{~W} = 64.336 \text{~kW} \]
Understanding each component:

• \[ \rho \] is the fluid's density, here 1.25 kg/m³.
• \[ g \] is gravity, 9.81 m/s².
• \[ Q \] is the volumetric discharge, 5.7 m³/s.
• \text{Net Head} is the effective head derived from calculations.
• Mechanical losses account for other non-fluid related power drains, here specified as 220 W.

Combining these gives us the total power requirement, crucial for designing and selecting equipment.