No more math is included than is required to understand the concepts presented. Chemical Engineering Fluid Mechanics. No more math is included than is required. Incompressible Flow. The most teachable book on incompressible flow— now fully revised, updated, and expanded Incompressible Flow, Fourth Edition is the updated and revised edition of Ronald Panton's classic text.
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Explains how fundamental principles underlying the behaviour of fluids are applied systematically to the solution of practical engineering problems. Current information and state-of-the-art anaytical methods are offered, and the work provides early coverage of dimensional analysis and scale-up. Introduction to Chemical Engineering Fluid Mechanics.
Presents the fundamentals of chemical engineering fluid mechanics with an emphasis on valid and practical approximations in modeling.
Such knowledge is especially valuable in the biochemical, chemical, energy, fermentation, materials, mining, petroleum, pharmaceuticals,. Physical and Chemical Equilibrium for Chemical Engineers. Suitable for undergraduates, postgraduates and professionals, this is a comprehensive text on physical and chemical equilibrium. Common, cheap ones are not locally calibrated, but simply used with the factory design values. In the factory they are not individually calibrated because each one coming down the production line is practically the same at the next.
For precise work, pressure gauges are calibrated this way. This is a manometer, and if bubbles are present, they make the average density on one side of the manometer different from that on the other, leading to false readings. If all the bubbles are excluded, then this is an excellent way to get two elevations equal to each other.
The horizontal part of the device plays no role. Physically, if the two fluids have exactly the same density, then they will not form a clean and simple interface. Most likely they will emulsify. Here P is not linearly proportional to radius, so this is complicated.
If we assume that over the size of this particle we can take an average radius, and calculate the "equivalent gravity" for that location, then we can use Archimedes' principle as follows.
Work in gage pressure. Two parabolas are the same. If one solves it as a manometer, from the top of the funnel to the nozzle of the jet, one will see that there is a pressure difference equal to the difference in height of the liquid in the two bottles times g times density of water - density of air.
This is clearly not a steady-state device; as it flows the levels become equal and then the flow stops. The boundaries are not fixed in space, nor are they fixed in size, but they are readily identifiable. The point is that there is one main flow out, the carbon dioxide in the exhaust, which equals the negative accumulation of carbon in the fuel, and then a lot of other minor terms.
That was supposed to reshape the beaches and in other ways make the canyon more like its pristine state. Those who organized the flood pronounced it a great success. This has the paradoxical consequence that friction, which causes the pressure to fall, causes the velocity to increase. This type of flow is discussed in Ch.
It plays no role. The answer would be the same if the initial pressure were any value, as long as the pressure was low enough for the ideal gas law to be appropriate. All real vacuum systems have leaks whose flow rate is more or less proportional to the pressure difference from outside to in assuming laminar flow in small flow passages but this comparison shows that replacing Eq.
One can look up the answers on Fig 3. Its analogs also appear in heat and mass transfer. Then, a randomly-selected liter contains 3. The deep oceans mix very slowly.
The Red Sea does not mix much with the rest of the world ocean, except by evaporation and rainfall. The perfect mixing assumption is much better for the atmosphere than for the ocean in the preceding problem.. Air resistance complicates all of this, see Prob. This appears paradoxical, but the ball is simply converting one kind of energy into another. Work was done on the ball as the airplane lifted it from the ground.
This seems odd, but the water is merely transferring work from the pump to the piston, rack and car. Here the dWn. Solving for vin dmin dWn. If we had worked it in absolute, then we would have needed to include a term for driving back the atmosphere as the car, rack and piston were driven up. But it shows how one would proceed if one did not have access to a suitable steam table. Here the students must use the fact, stated at the beginning of this set of problems, that for a perfect gas u and h are functions of T alone, and not on the pressure.
Unfortunately the adiabatic assumption cannot be realized, even approximately if one tries this in the laboratory. Even in the few seconds it takes to fill such a container the amount of heat transferred from the gas makes the observed temperature much less than one calculates this way.
The amount of heat transferred is small, but the mass of gas from which it is transferred is also small so their ratio is substantial. In spite of that this is a classic textbook exercise, which is repeated here. Using the solution to problem 4. Ask them why. Some will figure out that the heating value is for the fuel plus the oxygen needed to burn it, taken from the air, and not included in the weight of the fuel. High explosives include their oxidizer in their weight.
They do not release large amounts of energy per pound, compared to hydrocarbons. What they do is release it very quickly, much faster than ordinary combustion reactions.
You can estimate the pressure for this from the ideal gas law, finding amazing values. Fats are roughly CH2 n. Carbohydrates are roughly CH2O n. This simplifies the chemistry a little, but not much. You might ask your students what parts of plants have fats. Some will know that the seeds have fats, the leaves and stems practically zero. Then some will figure out why. The assignment of a seed is to find a suitable place, put down roots and put up leaves before it can begin to make its own food.
It is easier to store the energy for that as a fat than as a carbohydrate. Our bodies make fats out of carbohydrates so that we can store them for future use in case of famine. Thermodynamics explains a lot of basic biology! Note that the metal walls of the calorimeter are outside the system. The value here is between that for fats and for carbohydrates. This latter is not quite correct, because particles are blown off the sun by solar storms, but their contribution to the energy balance of the sun is small compared to the outward energy flux due to radiation.
What nature does is independent of how we think about it. We may show that they do not violate the first law by making an energy balance for each. These all violate the second law. They all violate common sense. If you doubt that, put a baseball on a table and watch it, waiting to see it spontaneously jump to a higher elevation and cool.
Be patient! For 1 psig, this factor is P1 If one is not going to use the approach in Ch. The ask about a can of gasoline. Torricelli's Eq. See Chapter 6. Apparently the safety doors divide the boat's hull into some number 7 on the Titanic? So if the hull is punctured, as in this problem, and the doors close properly, one compartment will fill with water up to the level of the water outside, but the other compartments will remain dry. Apparently the impact with the iceberg opened the hull to more compartments than the ship could survive, even with the safety doors closed.
Presumably if the compartments were sealed at the top, then even that accident would have been survivable, but it is much easier to provide closed doors on passageways parallel to the axis of the boat, in which there is little traffic, than on vertical passageways stairs, elevators, etc.
However it is correct. The reason is the very low density of the helium, which makes the difference in atmospheric pressure seem like a large driving force. These devices most often send a signal which is proportional to the pressure difference.
If this is shown directly on an indicator or on a chart, the desired information, the flow rate, is proportional to the square of the signal. One solution to this problem is to have chart paper with the markings corresponding to the square of the signal. The chart manufacturers refer to these as "square root" charts, meaning that in reading them one is automatically extracting the square root of the signal.
Moving blade or moving cup anemometers are most often used for low-speed gas flows, e. The students have certainly seen these in weather stations. On the dashboard of an airplane private or commercial is an "indicated air speed" dial. The lift is directly proportional to the indicated air speed, so this is also effectively a lift indicator, see Secs.
You can ask the students why commercial airliners fly as high as they can. For a given weight, there is only one indicated air speed at which they can fly steadily in level flight. As the air density goes down, the corresponding absolute velocity goes up. So by going high, they go faster.
This means fewer hours between takeoff and landing. That means less fuel used, fewer hours to pay the pilots and crews for, and happier customers who do not like to sit too long in an aircraft. Using the methods in this book, we find for that pressure difference lbf 0.
We can get other values by ratio, e. The upper one, on log-log coordinates is probably the more useful, because the fractional uncertainty in the Q, cfs reading is practically the same over the whole range. The lower one, on arithmetic coordinates is probably easier for non-technical people to use, and would probably be selected if Delta P, psig 10 Here I have chosen the pressure drop as the independent variable, 8 because that is the observational instrument reading. From Fig 5.
See Eq 5. This explains why this type of carburetor was the practically-exclusive choice of automobile engine designers for about 80 years. With a very simple device, one gets a practically constant air-fuel ratio, independent of the throttle setting.
The rest of the carburetor was devoted to those situations in which one wanted some other air-fuel ratio, mostly cold starting and acceleration, for which one wants a lower air fuel ratio " rich ratio".
The air density falls while the fuel density does not, so the air fuel ratio would become 0. High altitude conversion kits smaller diameter jets are available to deal with this problem. Rich combustion leads to increased emissions of CO and hydrocarbons; there are special air pollution rules for autos at high elevations. Demands for higher fuel economy and lower emissions are causing the carburetor to be replaced by the fuel injector.
In a way it is sad; the basic carburetor is a really clever, simple, self-regulating device. The density is proportional to the molecular weight. For the values shown, the predicted jet velocity in the burners will be the same for propane and natural gas.
Because of the higher density of propane, the diameter of the jets will normally be reduced, to maintain a constant heat input. But the velocities of the individual gas jets are held the same, to get comparable burner aerodynamics.
For 0. This is a simple scale change for each curve on Fig. Solving Eq. Either gives the same result. Clearly, the greater the depth, the higher the speed at which the propeller can turn without cavitation, so this is not as severe a problem for submarines submerged as it is for surface ships. However the noise from propeller cavitation is a serious problem for submarines, because it reveals the position of the submarine to acoustic detectors.
Submarines which do not wish to be detected operate with their propellers turning slower than their minimum cavitation speed. As in that example take 1 at the upper fluid gasoline surface and 2 at the outlet jet. Take 2a at the interface between the gasoline-water interface.
If we replaced the 20 m of gasoline with So this is really the same as Ex. Then one repeats the whole derivation in Ex 5. One can also compute the time from when the interface has passed the exit to when the surface is 1 m above the exit, finding the same time as for water, because in this period the density of the flowing fluid is the same as that of gasoline. See Prob. I have the device described in that problem. I regularly assign the problem, then run the demonstration.
One can estimate well, down to an interface one or two diameters above the nozzle, but not lower. The easiest way to work the problem is to conceptually convert the 10 ft of gasoline to 7. Then this is the same as Ex.
The final pressure is This gives the answer to part c ; the 5-fold expansion of the gas lowers its absolute pressure by a factor of 5, which would produce a vacuum and stop the flow if the initial pressure were 20 psig. Instead we proceed by a spreadsheet numerical integration. The original spreadsheet carries more digits than will fit on this table. One may test the stability of this solution, by rerunning it with smaller height increments. For increments of 0. Both round to 6.
Then this becomes the same as Ex. Following Ex. Much of the time is spent in the last inch, where the assumption that the diameter of the nozzle is negligible compared to the height of the fluid above it becomes poor see Sec 5.
Some students like that, others don't. We then square both sides and differentiate w. Then using the same ideas as in Ex 5. In this problem two Vs appear, the volume of the can and the instantaneous velocity.
Following the nomenclature, both are italic. When a V is the volume of the can, it has a "can" subscript. Try it, you'll be impressed! However it is easy to solve this on a spreadsheet. The solution is shown below. Most of the spreadsheet carries more significant figures than are shown here. For the plug flow model the agreement is V 1.
We solve V0 3. That happens here. In making up this solution I found two errors in [7]. I hope the version here is error free. For me it is a fine demonstration of unsteady-state flow.
They see all sorts of interesting combustion-related issues in it. I give this problem as an introduction to a safety lecture. Propane is by far the most dangerous of the commonly-used fuels. If there is a large leak of natural gas, buoyancy will take it up away from people and away from ignition sources.
If there is leak of any liquid fuel gasoline, diesel, heating oil it will fall on the ground and flow downhill. But the ground will absorb some, and ditches, dikes or low spots will trap some or all of it. But released propane forms a gas heavier than air, which flows downhill, and flows over ditches, dikes and low spots, looking for an ignition source, and then burning at the elevation where people are.
Thus by BE, the pressure must be rising steadily in the radial direction. In the center hole the pressure must be higher than atmospheric, in order to give the gas its initial velocity and to overcome the frictional effect of the entrance into the channel between the cardboard and the spool. Thus the figure is as sketched below. This is a very simple, portable, cheap, dramatic demonstration to use in class. Since the pressure increases by 1. It must be clear that at low values of the pressure ratio the three curves are practically identical.
As far as I know there are still ferry boats that use this system to stay above the water, regularly crossing English Channel. There are two put up each winter within five miles of my house. I wish I remembered where. As the following calculation shows, by simple B. However choosing 1. It is often in laminar flow in laboratory or analytical sized equipment. The corresponding viscometer for very high viscosity fluids replaces gravity with a pump, and attempts with cooling to hold the whole apparatus and fluid isothermal.
All serious viscometry is done inside constant temperature baths, see Fig. A1 shows that this low a viscosity is rarely encountered in liquids, so that this kind of viscometer can be applied to most liquids. Cryogenic liquids, however, have very low viscosities, so there might be a laminar-turbulent transition problem using this particular viscometer on them.
Saybolt Seconds Universal, SSU which is the standard unit of viscosity for high boiling petroleum fractions. The x component force depends on the speed of the train, and how often the balls are thrown back and forth, but not on their velocity. This shows how the 4f appears naturally in the Fanning friction factor. They don't. Assign this problem, after you have discussed laminar and turbulent flow, and you will be appalled at how few of the students can solve it.
After they have struggled with it, they will be embarrassed when you show them how trivial it is. Maybe that way they will learn about the different relation between pressure drop and velocity in laminar and turbulent flow! This is very similar to that shown in Table 6.
One enters at gpm, reads horizontally to the zero viscosity boundary which corresponds to the flat part of the curves on Fig. This is less than the available 28, so a 3 inch pipe would be satisfactory.
The most likely reason for the lower roughness in Fig 6. The value for the roughness of steel pipe Table 6. The first three columns are the same as those in Table 6. The bottom 4 rows are in addition to what is shown in Table 6. Variable First guess Solution Prob. Looking at the original spreadsheet we see it went from 0. The bottom 2 rows are in addition to what is shown in Table 6.
If there were no change in f then quadrupling the volumetric flow rate would cause the required diameter to double. As shown here it increases by a factor of 1. With the same pressure drop 1. This is only an 0. The result is in the laminar flow region, so we could solve directly from Pouisueille's equation, lbf 5. The density is easily measured in simple pyncnometers, to much greater accuracy than it could possibly be measured by any kind of flow experiment.
The viscosity is measured in laminar flow experiments, as shown in Ex. Furthermore, in a flow experiment we would presumably choose as independent variables the choice of fluid, the pipe diameter and roughness and the velocity.
We would measure the pressure drop, and compute the friction factor. If we were at a high Reynolds number, i. The friction factor would also be independent of the density which is part of the Reynolds number , although the calculation of the friction factor from the observed pressure drop would require us to know the density in advance.
Q and 6. I doubt that this has an analytical solution; it would be tolerably easy numerically, but the procedure shown in Ex. For Ex. X and 6. Again, I doubt that this has an analytical solution; it would be tolerably easy numerically, but the procedure shown in Ex. First you should point out that Fig. It has a scale ratio of about 1. If one re-plotted it on normal log-log paper, one would see that it has three families of parallel straight lines, the laminar flow lines with slope 1, the turbulent flow lines with slope about 0.
The from Fig. We draw a line of proper slope through it completes the zero viscosity boundary. We draw a line with slope 1 through it, for the laminar region, and a line with slope 0. Then we complete the laminar region, by drawing lines parallel to the first line, for various viscosities. In the laminar region, at constant flow rate the pressure drop is proportional to the viscosity, so, for example, the 20 cs line is parallel to the 40 cs line, but shifted to the left by a factor of two, and the 80 cs line is parallel to the 40 cs line, but shifted to the right by a factor of two.
We must guess a value of the kinematic viscosity for which the line will fall between the transition and zero viscosity lines. From Fig. Then for s. We could compute similar lines for other kinematic viscosities if we wished.
The problem does not ask the students to compute the lower section which corrects for differences in density. You might bring that up in class discussion. The lines there would have slope 1 on an ordinary piece of log-log paper. The values calculated above are all for s. If, for example, the true s. This is obvious for turbulent flow, where we set the density to For laminar flow it is a bit more subtle.
In making up the laminar part of the plot we used the kinematic viscosity, with an assumed s. Thus, in this formulation the laminar pressure gradient is also proportional to the specific gravity. Here I have used as the transition Reynolds number. The authors of Fig.
But this problem shows that one could. For this value we can compute the pressure gradients for laminar and turbulent equations, finding 0.
The value in the table is 0. A few of the other smallest velocity values are also laminar. All the other values are for turbulent flow. These match Fig. Comparing them to the vales from Eq. For increasing flow rates this ratio diminished. For velocities greater than 9. As a further comparison I ran the friction factors from Eq. Clearly even at this size pipe, we do not have smooth tube behavior. This is tolerable agreement. The main reason for this is that in turbulent flow, the volumetric flow rate is proportional to D5.
By visual interpolation the required pipe diameter is about 0. If the friction factors are the same, we would expect the pressure gradients to be proportional to the densities. Thus, this is only an approximate way of estimating the behavior of hydrogen.
For gases with properties more like those of air it works better. All of these give practically the same answer. To find the slopes I read the 4 inch line as beginning at 15 and ending at The curvature seems to increase with increasing pipe size. These do not flatten out until higher values of the velocity, so the curves correspond to a value of f which decreases slowly with increasing Q, giving the curves shown. From the caption for Fig. So this matches Table 6.
The computations are easy on a spreadsheet. The plot, covering the range from 3 to 12 ft is shown below. I don't know which is best, other than to follow Lapple's suggestion in the text. Another peculiarity of this comparison is that the estimate by the "K-factor" method is uninfluenced by a change in fluid viscosity, as long as the flow is turbulent.
But the equivalent length method is sensitive to changes in viscosity. I reran the spreadsheet that generated the values for the above plot, taking the fluid viscosity as 1 cP instead of the 50 cP in those examples. That raised the Reynolds numbers by a factor of 50, reducing the computed f's by almost a factor of 2. That reduced the estimated delta P by roughly a factor of 2, going from In this case the curve on the above plot ends up nearly horizontal; the computed value for 24 inches is You might ask your students what a check valve is.
Normally there will be one visible in the vicinity of your building, as part of the lawn sprinklers, which you can point out. We can simply look up the two pipe terms, in Table A. For the expansion term D 2. Then we read Fig. From Table A. The solution is a trial and error, in which one guesses f and then does a suitable trial and error.
The three terms in the denominator under the radical have values 1, 0. Each contributes to the answer. In a lecture room with a sink one can fill the tank with a piece of tygon tubing, and let the students watch the results. Then we do the calculations, then run the test. The above value is for galvanized pipe, as shown in the problem. They use the value for steel pipe, getting a much lower value. As it was used more and more and the pipe rusted the time climbed slowly to 85 sec.
Cleaning it with a wire brush got it back to its original value. The fundamental cussedness of inanimate objects This works well, one can see several oscillations in the flow rate.
One can sketch Fig. True leakage paths are certainly more complex than the uniform annulus assumed here, but this description of the leakage path is generally correct. You might ask bright students how much error is introduced that way? I checked by comparing the slit solution to that for an annulus with inside diameter 0.
The computed flow for the slit is 1. The other values are taken from Ex.
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