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Figure \(P 3.52\) shows a duct for testing a centrifugal fan. Air is drawn from the atmosphere \(\left(p_{\mathrm{am}}=14.7 \mathrm{psia}, T_{\mathrm{atm}}=70^{\circ} \mathrm{F}\right) .\) The inlet box is \(4 \mathrm{ft} \times 2 \mathrm{ft}\). At section 1 , the duct is \(2.5 \mathrm{ft}\) square. At sanction 2 the duct is circular and has a diameter of \(3 \mathrm{ft}\). A water manometer in the inlet box measures a static pressure of -2.0 in. of water. Calculate the volume flow rate of air into the fan and the average fluid velocity at both sections 1 and \(2 .\) Assume constant density.

Short Answer

Expert verified
\(A_1 = 6.25 ft^2\), \(A_2 = 7.07 ft^2\), \(P_{box} = -0.0722 psia\), \(v_{square} = 107 ft/s\), \(Q = 668 ft^3/s\), \(v_{circle} = 94.4 ft/s\)

Step by step solution

01

Calculate the Area of the Two Sections

For section 1, being a square section, the area \(A_1\) will be the square of the side which is given as \(2.5 ft\), thus \(A_1 = (2.5ft)^2\). For section 2, being a circular section, we will calculate the area \(A_2\) using the formula for the area of a circle \(A=蟺r^2\), where \(r\) is the radius of the circle, which is half of the diameter, so \(A_2=蟺(3ft/2)^2\).
02

Calculate the Volume Flow Rate of Air into the Fan

The volume flow rate \(Q\) is given by the equation \(Q = Av\), where \(A\) is the cross-sectional area and \(v\) is the velocity. We know that the pressure in the inlet box is -2.0 in. of water. Converting this to feet of water, it is -2.0/12 ft of water. To convert this to psia, we use the conversion factor of 1 in. water = 0.0361 lb/in^2 = 0.0361 psia, so the pressure in the inlet box \(P_{box}\) is -2*0.0361 psia. Since the air is drawn from the atmosphere into the fan, the volume flow rate \(Q\) will be \(A_{square}v_{square} = A_{circle}v_{circle}\), where \(v_{square}\) and \(v_{circle}\) are the velocities at the square and circular sections respectively.
03

Calculate the Velocity

We can find the velocities using Bernoulli's equation, which states that for an incompressible, frictionless fluid, the sum of the pressure energy, kinetic energy and potential energy per unit volume is the same at all points along a streamline. Assuming the fluid is incompressible because we are dealing with air and its density does not change much, we can apply Bernoulli's equation which gives \( v_{square} = \sqrt {(2*(p_{atm}-p_{box})/蟻}\) and using \(Q = A_{square}v_{square}\) gives \(Q\). Substituting \(Q\) into \(A_{circle}v_{circle}\) gives \(v_{circle}\).

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Key Concepts

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

Centrifugal Fan Testing
Centrifugal fan testing involves evaluating the performance and efficiency of a fan by measuring parameters like pressure differences and flow rates. This kind of testing is crucial to ensure the fan operates efficiently and meets specified criteria. In the testing setup described, air is drawn from the atmosphere into a duct system ending with the fan.

The test setup typically includes measuring the static pressure in the inlet box, which is crucial for determining the pressure loss and understanding the energy requirements for the fan operation. The pressure is measured using instruments like a water manometer, which provides the pressure difference in inches of water. In this example, the static pressure is recorded as -2.0 inches of water, indicating a slight vacuum relative to atmospheric pressure.

By understanding these measurements, engineers can determine how well the fan will perform under various conditions.
Volume Flow Rate Calculation
Determining the volume flow rate (Q) involves finding how much air is passing through the system per unit of time. The volume flow rate is a critical factor; it tells us how effective the fan is at moving air.

To calculate this, we use the formula \[ Q = A \times v \] where \( A \) is the cross-sectional area of the duct, and \( v \) is the velocity of the air. For different sections of the duct, the areas are different. At section 1, the duct is square; at section 2, it is circular. Using the formulas for the area of a square and a circle, we can compute each.

Once the areas are known, we will apply Bernoulli鈥檚 principle to find the velocities. With velocities known, the continuity equation helps express the flow rates through the different duct sections, assuming constant density.
Bernoulli's Equation
Bernoulli鈥檚 Equation is a principle that describes the behavior of fluid flow. It is instrumental in fluid mechanics for explaining how pressure, velocity, and height relate within flowing fluid. The principle can be expressed as \[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \] where \( P \) is the pressure, \( \rho \) is the fluid density, \( v \) is the velocity, and \( g \) is the acceleration due to gravity, and \( h \) is the height.

Bernoulli鈥檚 equation assumes an incompressible, frictionless fluid and steady flow. In the fan testing context, it helps compute air velocity at different duct sections . The difference in static pressure from atmospheric pressure is used to determine the change in kinetic energy, yielding the velocity. By using Bernoulli鈥檚 equation with the differential pressures, we can deduce the velocities efficiently.
Incompressible Fluid Assumption
The assumption of incompressible fluid simplifies calculations significantly. It means that the fluid's density remains constant throughout the flow. This is a valid assumption for liquids and low-speed gas flows like air within certain limits.

In the fan testing situation, air can be modeled as incompressible because its density change is negligible at the speeds and conditions tested. This lets us use formulas like Bernoulli鈥檚 equation more straightforwardly, avoiding complex calculations needed for compressible fluids.

Assuming incompressibility also simplifies volume flow rate calculations and the application of fluid dynamics principles, ensuring straightforward solving of velocity and discharge quantities.

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Most popular questions from this chapter

Observations show that it is not possible to blow the table tennis ball from the funnel shown in Fig. \(\mathrm{P} 3.122 a\). In fact, the ball can be kept in an inverted funnel, Fig. \(P 3.122 b,\) by blowing though it. The harder one blows through the funnel, the harder the ball is held within the funnel. Try this experiment on your own. Explain this phenomenon.

Water flows through a hole in the bottom of a large, open tank with a speed of \(8 \mathrm{m} / \mathrm{s}\), Determine the depth of water in the tank. Viscous effects are negligible.

Air flows steadily along a streamline from point (1) to point (2) with negligible viscous effects. The following conditions are measured: At point \((1) z_{1}=2 \mathrm{m}\) and \(p_{1}=0 \mathrm{kFa}\); at point \((2) \mathrm{z}_{2}=10 \mathrm{m}\) \(p_{2}=20 \mathrm{N} / \mathrm{m}^{2},\) and \(V_{2}=0 .\) Determine the velocity at point (1).

A small card is placed on top of a spool as shown in Fig. \(P 3.121 .\) It is not possible to blow the card off the spool by blowing air through the hole in the center of the spool. The harder one blows, the harder the card "sticks" to the spool. In fact by blowing hard enough it is possible to keep the card against the spool with the spool turned upside down. Give this experiment a try. (Note: It may be necessary to use a thumb tack to prevent the card from sliding from the spool.) Explain this phenomenon.

Air flows smoothly over the hood of your car and up past the windshield. However, a bug in the air does not follow the same path; it becomes splattered against the windshield. Explain why this is so.

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