California State University, Northridge



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|College of Engineering and Computer Science

Mechanical Engineering Department

Mechanical Engineering 390

Fluid Mechanics | |

| |Spring 2008 Number: 11971 Instructor: Larry Caretto |

Solutions to Exercise Seven – Control Volume Analysis of Momentum and Energy

1 A 100-ft-wide river with a flow rate of 2400 ft3/s flows over a rock pile as shown in the figure at the right. Determine the direction of flow and the head loss associated with the flow across the rock pile. (Problem 5.87 and Figure P5.87 copied from Munson et al., Fluid Mechanics text.)

We know that the head loss must be positive so we can assume a flow direction and compute the head loss. If the head loss is negative, we have assumed the incorrect direction.

Start by assuming that the flow goes from point 1 to point 2 (from right to left) and apply the energy equation to a control volume bounded by the top and bottom of the river and vertical lines at section 1 and section 2. Note that our equation has to have point 1 as the assumed inlet and point 2 as the assumed outlet.

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Along the top of the river, there is essentially no difference in pressure between two liquid elevations because the air density is negligible. Thus, we have p1 = p2 = atmospheric pressure. There is no shaft work so hs = 0. We can find the velocities from the stated flow rate and the areas.

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Rearranging the energy equation to solve for head loss with p2 – p1 = hs = 0 and substituting the velocities just found and the elevations shown in the diagram gives the head loss shown below.

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Since this head loss is positive our original assumption that the river flows from right to left is correct, and the head loss, hL = 0.32 ft.

If a ¾ hp motor is required to produce a 24-in stream of air having a velocity of 40 ft/s as shown in the figure at the right, estimate (a) the efficiency of the fan and (b) the thrust of the supporting member on the conduit enclosing the fan. (Problem 5.88 and figure taken from Munson, Fluid Mechanics.)

This problem is similar to example 5.24 starting on page 239 of the text. In that example the following equation is used to provide “a reasonable estimate of the efficiency”.

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We can find the numerator of this efficiency definition from the energy equation.[1]

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For this fan, po = pi = 0 because both are atmospheric pressure, and zo = zi, so the pressure and elevation terms vanish. In addition we can assume that the air entering the duct from the outer atmosphere has such a low velocity that Vi2 ................
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