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 |

April 15 Homework Solutions

8.5 Carbon dioxide at 20oC and a pressure of 550 kPa (abs) flows in a pipe at a rate of 0.04 N/s. Determine the maximum diameter allowed if the flow is to be turbulent.

To ensure turbulent flow, the Reynolds number must be greater than 4000. This gives:

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We are given the weight flow rate which is the mass flow rate times g. We can use this to get a relationship between the velocity and the area as follows:

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Substituting this expression for velocity into the Reynolds number (and the condition for turbulent flow) gives.

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The dynamic viscosity of carbon dioxide at 20oC and atmospheric pressure is found in Table 1.8 as μ = 1.47x10-5 N(s/m2. Since the dyanmic viscosity does not have a significant dependence on pressure, except for very large pressures, we can use this value at the given pressure of 550 kPa(abs). Substituting this value and the other given data into the inequality for diameter gives

[pic]= 8.83 cm

8.6 The pressure distribution measured along a straight horizontal portion of a 50-mm-diameter pipe attached to a tank is shown in the table below. Approximately how long is the entrance length? In the fully developed portion of the flow, what is the wall shear stress?

The data in the table below have been used to compute a finite-difference approximation to the pressure gradient. That is we compute Δp/Δx as an approximation to dp/dx according to the following equation.

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Locating the value of the estimated gradient at the mid point of the measured pressure data gives a so-called cnetral difference approximation to the deivative which is more accurate than placing the gradient at either of the end points. The gradient data are also shown in table below along with a plot of the data. The plot also shows a straight line fitted through the last few data poitns and extrpolated back to x = 0..

|x (m) |p (mm H2O) |Δp/Δx | | | | | | | |

| | | | | | | | | | |

| | | | | | | | | | |

|Distance, x (ft) |15 | |25 | |50 | |66 | |96 |

|Head, h (in) |60 | |56 | |46 | |39 | |26 |

|Δh/Δx | |-0.4 | |-0.4 | |-0.438 | |-0.433 | |

We see that the gradient is the same (-0.4 in/ft) for the differences between section A and B and section B and C; the first change occurs in section D, where the magnitude of the gradient increases. This indicates that section D is the one with the smaller diameter. To confirm this we can compute the gradient in section E to see if it is the same as the previous gradients, -0.4 in/ft.

To compute the gradient in section E, we have to start at the measurement point in section C, where the total distance, x = 50 ft. The distance to the end of section C (which the same as the start of section D) is 10 ft. (We know this because we are told that each section is 20 ft long.) Thus with the gradient of -0.4 in/ft, the head at the start of section D would be 46 in – (0.4in/ft)(10 ft) = 42 in. We can then compute the gradient in D because we know that the head is 39 in over a 6 ft length of section D. This gives the gradient in section D as (39 in – 42 in)/(6 ft) = -0.5 in/ft. This gradient would reduce the head over the remaining 14 ft section D to 39 in – (0.5 in/ft)(14 ft) = 32 in. With this head at the end of section D, which is the same as the start of section E, we can compute the gradient in section E from the measured head of 26 in, 6 ft into section E as (26 in – 32 in)/(6 ft) = -0.4 in/ft, the same as at the start of the pipe. This confirms our preliminary conclusion that section D has the smaller diameter.

8.22 Oil of SG = 0.78 and a kinematic viscosity, ν = 2.2x10-4 m2/s flows through the vertical pipe shown in Figure P8.22 (copied at the right) at a rate of 4x10-4 m3/s. Determine the manometer reading, h.

We can write the head loss in the energy equation as Δp/γ = Δp/ρg =[ f (l/D) ρV2/2 ] / (ρg) = f(l/D)V2/2g so that the energy can be written to give the following relationship between the downstream point (2) and the upstream point (1).

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We have to compute the Reynolds number as the first step in determining the friction factotr. For this pipe D = 20 mm = 0.02 m and the area = πD2/4 = π(0.02 m)2 = 0.001257 m2. The velocity is found from the volume flow rate by the usual equation and then used to compute the Reynolds number.

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Since Re < 2,100, the flow is laminar and we find the friction factor as 64/Re. With this friction factor, we can find the pressure difference from the energy equation after cancelling the equal velocity terms.

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The manometer reading can be analyzed by starting with the lower level on the right of the manometer. At this point, the pressures on both sides of the mamometer are the same and can be equated to pressures in the pipe by the manometer formula.

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From the diagram we see that the sum h1 + h2 is the length of 4 m between the pressure taps. Furthermore, the ratio of the specific weights of the oil and manometer fluid can also be written as the ratio of the specific gravity of each fluid. Thus we can solve for the unknown height, h, as follows.

[pic]=18.5 m

8.29 Air at standard conditions flows through an 8-in-diameter, 14.6-ft-long, straight duct with the velocity versus pressure drop data indicated in the table below. Determine the average friction factor over this range of data.

The usual equation for pressure drop in terms of friction factor can be solved for the friction factor as follows.

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The data shown in the table below have the velocity in ft/min and pressure drop in inches of water. To get the pressure drop in lbf/ft2 we have to multiply the pressure drop in inches of water by the specific weight of water. Doing this and inserting the necessary conversion factors gives the following calculation formula for the friction factor. (The density of air at standard conditions, 0.00238 slug/ft3, is taken from Table 1.7.)

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Applying this computational equation to all the data in the table below gives a set of data for friction factors from which we can compute the average value.

faverage = 0.0162

|V ft/min |Δp (in H2O) |f |

|3950 |0.35 |0.0161 |

|3730 |0.32 |0.0165 |

|3610 |0.30 |0.0165 |

|3430 |0.27 |0.0165 |

|3280 |0.24 |0.0160 |

|3000 |0.20 |0.0160 |

|2700 |0.16 |0.0158 |

|Average f |0.0162 |

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h2

h1

2

1

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