Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Jacaranda (Engineering) 3333Mail CodePhone: 818.677.6448
E-mail: 8348Fax: 818.677.7062
April 15homework solutionsME 390, L. S. Caretto, Spring 2008Page 1
April 15Homework Solutions
8.5Carbon 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:
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:
Substituting this expression for velocity into the Reynolds number (and the condition for turbulent flow) gives.
The dynamic viscosity of carbon dioxide at 20oC and atmospheric pressure is found in Table 1.8 as = 1.47x10-5 Ns/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
= 8.83 cm
8.6The 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 p/x as an approximation to dp/dx according to the following equation.
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) / p/x /0.00 / 520 / mm H2O/m
0.25 / -186
0.50 / 427
0.75 / -152
1.00 / 351
1.25 / -126
1.50 / 288
1.75 / -104
2.00 / 236
2.25 / -96
2.50 / 188
2.75 / -86
3.00 / 145
3.25 / -72
3.50 / 109
3.75 / -72
4.00 / 73
4.25 / -74
4.50 / 36
4.75 / -72
5.00 / 0
We see that the magnitude of the pressure gradient decreases with distance and eventually reaches a nearly constant value after x = 3.25 m. In the plotted data the straight line fit appears to match the data to a point slightly below x = 3.0 m. Thus a reasonable estimate of the entry length, to within the accuracy of the data is that the entry length is 3 meters.
The relationship between the wall shear stress and the pressure gradient in fully-developed flow can be used to compute the wall shear stress.
The average pressure gradient in the region where the gradient is nearly constant is 72.5 mm H2O/m. To get this into conventional units we have to multiply by the specific weight of water. Thus, 1 mm H2O = (0.001 m)(9800 N/m3) = 9.80 N/m2. With this conversion factor we can find the wall shear stress as follows
8.88 N/m2
8.15A fluid of density = 1000 kg/m3 and viscosity = 0.30 Ns/m2 flows steadily down a vertical 0.10-m-diameter pipe and exits as a free jet from the lower end. Determine the maximum pressure allowed in the pipe at a location 10 m above the pipe exit if the flow is to be laminar.
The maximum pressure allowed in the pipe will be the pressure that produces the largest flow rate that remains laminar. Since laminar flow requires the Reynolds number to be less than 2100, we have to satisfy the following inequaltiy.
So we have the result that V < 6.30 m/s; we need to connect this result to the presure. The relationship for pressure drop in laminar flow in a pipe that makes an angle with the horizontal is given by equation 8.11 on page 412 of the text, where p is defined as pin – pout.
Setting the inequality that V < 6.30 m/s and solving for the inlet pressure gives.
Since the exit is a free jet, pout = 0. The coordinate system for takes= 0 when the flow is in a horizontal direction to the right. A pipe that has a vertical upflow would have = 90o; a pipe, like the one here, that has a vertical downflow has = –90o. and sin() = –1. We are given the density of the fluid as 1000 kg/m3, so its specific weight, = g = (1000 kg/m3)(9.80655 m/s2)(1Ns2/kgm) = 9807 N/m3.
pin < –3.759x104Pa = –37.6 kPa
8.20Oil flows through the horizontal pipe shown in Figure P8.20 (copied from the text at the right) under laminar conditions. All sections of the pipe are the same except one. Which section of the pipe (A, B, C, D, or E) is slightly smaller diameter than the others? Explain.
For laminar flow in a horizontal pipe the pressure drop is related to the diameter by the following equation.
This equation tells us that the pressure gradient, p/l, varies as D-4 if all other terms in the equation are constant. If we assume that the temperature in the pipe is constant, the viscosity and density in the pipe will be constant, and the constant density means that the volume flow rate, Q, will be constant, even if the diameter changes, according to the continuity equation
We can compute the gradient in head measured by the piezometer tubes as a measure of the pressure gradient. This is shown in the table below.
Section / A / B / C / D / EDistance, x (ft) / 15 / 25 / 50 / 66 / 96
Head, h (in) / 60 / 56 / 46 / 39 / 26
h/x / -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.
8.22Oil of SG = 0.78 and a kinematic viscosity, = 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 p/ = p/g =[ f (l/D) V2/2 ] / (g) = 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).
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 = D2/4 = (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.
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.
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.
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.
=18.5 m
8.29Air 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.
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.)
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 / p (in H2O) / f3950 / 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