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 Unit 12 Quiz – Do one of the following problems

Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cross section area of the throat of the converging duct is 0.05 ft2. Determine the mass flowrate through the duct if the receiver pressure is (a) 10 psia; (b) 5 psia.

This problem is similar to example 11.5 in the text. We compute the mass flow rate as the product of density times velocity times area at the throat. The density is related to the Mach number and the velocity is the Mach number times the speed of sound.

This problem is for a duct that has a converging section only. It cannot become supersonic because there is no increase in area beyond the minimum throat area. We first have to determine if the flow is choked or not. To do this we compute the critical pressure ratio, p* from the equation below, using the value of k = 1.4 for air and a stagnation pressure equal to the standard atmospheric pressure of 14.696 psia.

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If the receiver pressure is less than p*, the flow is choked and we know that the throat Mach number is one; if the receiver pressure is greater than p*, the flow is not choked; the throat pressure is the receiver pressure and we have to compute the throat Mach number from this pressure.

The initial receiver pressure of 10 psia is greater than p* = 7.763 psia so the flow is not choked; the throat pressure is 10 psia and we can compute the throat Mach number by rearranging equation 11.59 in the text to solve for the Mach number.

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Equation 11.60 gives the density at this Mach number. The density of standard air, ρ0 = 0.00238 slug/ft3 is taken from Table 1.7.

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In order to find the velocity at the throat as V = c(Ma) we have to find c which requires the temperature. This is found from equation 11.56, using the standard temperature from table 1.7, T0 = 59oF = 518.67 R.

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We can now compute the sound speed using the gas constant for air from Table 1.7: R = 1716 ft·lbf/slug·R.

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So our speed, V = c(Ma) = (1057 ft/s)(0.7625) = 805.6 ft/s.

We now have all the terms necessary to compute the mass flow rate at the throat.

[pic]0.0728 slug/s

In the second case, the receiver pressure of 5 psia is less than p* so the throat pressure is p* = 7.763 psia and the throat Mach number is 1. We use the same equations as above to find the density and temperature at the throat.

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Since the Mach number is one, the velocity at the throat is the sound speed at the temperature just found. Once we find the velocity we can compute the mass flow rate at the throat.

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[pic]0.0769 slug/s

Water flows in a rectangular channel with a flowrate per unit width of q = 1.5 m2/s and a depth of 0.5 m at section (1). The head loss between sections (1) and (2) is 0.03 m. Sketch the specific energy diagram for this flow and indicate the relative location of states (1) and (2) on this diagram. Is it possible to have a head loss of 0.06 m? Explain.

We can apply the energy equation to the diagram shown above. Here we use the surface of the liquid as the points to apply the energy equation. This means that p1 = p2 = 0. We assume that the slope is very small so that we can use the fluid depth as the elevation. This gives z1 = y1 = 0.5 ft and z2 = y2. With these simplifications our energy equation becomes.

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We replace the velocity, V, by the flow per unit width: V = Q/A = (qb)/(yb) = q/y. This gives the following result in our energy equation. Note that Q = qb is constant as the flow changes depth. If the width, b, is constant, then q = Q/b, is also constant. Thus we do not have to subscript q in the equation below.

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Although the specific energy changes in the flow as a result of head loss, the relationship between specific energy and depth does not change. At any point in the flow we must have

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We can compute the values of E for various values of y and plot the results as shown in the figure to the left. In particular, we know the value of the depth, y = 0.5 m at section (1) in the flow. Thus, the specific energy at this point can be found as

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This confirms the value that we can read from the chart for y = 0.5 m. According to the energy equation, the head loss means that the energy at station 2 is 0.03 m less than the energy at station (1). This gives E2 = 0.929 ft. This specific energy corresponds to the following a depth of 0.550 m. This is found by solving the final equation below, using a numerical root finder to find y. Both points are shown on the energy curve with the value of their depts. indicated

[pic] We see that the lower energy corresponds to a higher depth with the same flow.

If we had a head loss of 0.06 m from station 1 the energy at station (2) would be 0.959 m – 0.06 m = 0.859 m. This is only possible if the minimum energy is less than 0.859 m. To check this we can compute the minimum energy by first finding the critical depth for this energy, yc. and then applying the formula that Emin = 1.5 yc.

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Since Emin = 0.9183 m we cannot have a head loss of 0.06 m because this would make the specific energy less than its minimum value.

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