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Chapter 3: Problem 52

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 \(A=πr^2\), where \(r\) is the radius of the circle, which is half of the diameter, so \(A_2=π(3ft/2)^2\).
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 \( v_{square} = \sqrt {(2*(p_{atm}-p_{box})/ρ}\) and using \(Q = A_{square}v_{square}\) gives \(Q\). Substituting \(Q\) into \(A_{circle}v_{circle}\) gives \(v_{circle}\).

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