University Curriculum Development for Decentralized Wastewater Management

Transcription

1 Page i University Curriculum Development for Decentralized Wastewater Management Hydraulics II: Energy Module Text Paul Trotta, P.E., Ph.D. Justin Ramsey, P.E. Chad Cooper September 2004

2 Page ii NDWRCDP Disclaimer This work was supported by the National Decentralized Water Resources Capacity Development Project (NDWRCDP) with funding provided by the U.S. Environmental Protection Agency through a Cooperative Agreement (EPA No. CR ) with Washington University in St. Louis. These materials have not been reviewed by the U.S. Environmental Protection Agency. These materials have been reviewed by representatives of the NDWRCDP. The contents of these materials do not necessarily reflect the views and policies of the NDWRCDP, Washington University, or the U.S. Environmental Protection Agency, nor does the mention of trade names or commercial products constitute their endorsement or recommendation for use. CIDWT/University Disclaimer These materials are the collective effort of individuals from academic, regulatory, and private sectors of the onsite/decentralized wastewater industry. These materials have been peer-reviewed and represent the current state of knowledge/science in this field. They were developed through a series of writing and review meetings with the goal of formulating a consensus on the materials presented. These materials do not necessarily reflect the views and policies of University of Arkansas, and/or the Consortium of Institutes for Decentralized Wastewater Treatment (CIDWT). The mention of trade names or commercial products does not constitute an endorsement or recommendation for use from these individuals or entities, nor does it constitute criticism for similar ones not mentioned.

3 Page iii Citation of Materials Trotta, P.D., and J.O. Ramsey Hydraulics II: Energy Text. in (M.A. Gross and N.E. Deal, eds.) University Curriculum Development for Decentralized Wastewater Management. National Decentralized Water Resources Capacity Development Project. University of Arkansas, Fayetteville, AR.

4 Page 1 I. Hydraulic Energy A. The Hydraulic Energy of Water Just like with your car, it takes energy to speed a car up from a standstill, additional energy to move it up a hill and yet more energy to overcome all the friction of wind, road and parts moving over each other. Water also needs energy to speed it up and lift it to higher elevations and overcome friction as well. In addition water requires energy to pressurize it. A cubic foot of water has more energy if it is under pressure than it has if it isn t. Water possesses energy of interest in common hydraulic in three forms: a. Elevation b. Pressure c. Velocity 1. Elevation Head The energy of elevation is the simplest to understand. Carry a bucket of water up a flight of stairs and the work you have done in part goes into potential energy of the water. Similarly, if a pump discharges water at a higher elevation than the water level in its supply tank it is providing that water with elevation head or potential energy. Consider the situation illustrated in Figure 1 where a pump discharges to a Wisconsin mound. The distance that the pump lifts the water imparts an elevation head to the water. This is also called the static lift. Figure 1 Elevation Head or Static Lift Elementary physics states work equals force times distance. Therefore, if water weighs 62.4 lbs/cu.ft. {1000 kg/m 3 } and you haul a cubic foot of it up a 10 ft {3 m} ladder, you have added 624 ft.- lbs {29,870 N - m} of potential energy to the water. Power relates to the rate of doing work or in this case how fast you haul it up the ladder. If you hauled it up the ladder in 5 seconds you would be doing 624 ft-lbs/5 seconds or 125 ft-lbs/second {497 N-m/s}. A horsepower (HP) is defined as 550 ft-lbs per second so you would be

5 Page 2 working at about ¼ HP {0.187 kw}. Ignoring friction and inefficiencies a ¼ HP pump could do about the same job. For simplicity, we describe the energy we just imparted to the water not as 624 ft-lbs but simply 10 feet of head meaning the amount of energy that a unit of water would have if lifted 10 feet. Hydraulic engineers often refer to this as static lift and generalize the elevation increase as Z. Figure 2 illustrates this use of feet as a convenient energy term. Figure 2 Elevation Head 2. Pressure Head Water under high pressure has more energy than water under low pressure. Although water is considered incompressible, water under pressure is stressed by the pressure and to a small extent the resultant strain is compressing the water and squeezing the bonds and fields in and around the water molecules. Like a bunch of stiff springs, the water absorbs the energy into the springs, which then are pushing back against the container and the surrounding water. The energy stored in the water springs is distributed over the mass of the water being squeezed. Therefore, to get the energy per unit of water, you must divide the pressure by the density of water. Pressure Head, therefore, is defined as Pressure/Density. For example, water being squeezed by a pressure of 10 pounds per square inch (psi) is being squeezed at an overall pressure of 144 x 10 pounds per square foot. (There are 144 square inches in a square foot; 12 x 12 = 144). Dividing 1440 pounds per square foot by water s density of 62.4 lbs/cu.ft (on Earth) results in: 1440 lbs/ft 2 / 62.4lbs/ft 3 = 23 feet of pressure head

6 Page 3 This is just what we would see if we calculated the pressure less than 23 feet of water. So once again, we use a convenient equivalent term for the pressure energy, we use feet of head. Figure 3 illustrates how the pressure head adds to the total energy of a unit volume of water at an arbitrary elevation z. Figure 3 Pressure Head 3. Velocity Head A water balloon resting against the side of your head doesn t have as much energy to give up as a thrown water balloon that just hit the side of your head does. The difference in energy was not due to the elevation of the balloon when it hit, your head. (In both cases, your head was at the same elevation, at least initially.) The difference in energy was not due to the pressure on the balloon. Atmospheric pressure was basically the same around both balloons. Actually, the speeding balloon has a different pressure distribution around it and if it is spinning, the differences in pressure from one side to the other could actually cause it to curve like a thrown baseball. The difference in energy between the two balloons is due to the velocity of the balloon. Physicists have shown that the energy of velocity turns into energy of elevation as an object rises. Throw a balloon upward and the height it attains is related to the initial velocity with which it leaves your hand. The velocity (or kinetic) energy of the balloon gets turned into elevation (or potential) energy as it rises and slows down. The potential energy turns back into velocity energy as it falls back down. If it were not for air friction a bullet shot straight up would come down with exactly the same velocity (and energy) that it left the barrel with. As illustrated in Figure 4, the same situation exists for water that can be visualized as a bunch of balloons being launched up and coming back down. Since the velocity energy at the beginning of the water balloon s flight is the same as the potential energy, it has at the top of its flight it is convenient, again, to measure that energy by the elevation that the balloon (or water) attains.

7 Page 4 Figure 4 Velocity Head The numerical term that is used for the velocity energy is V 2 /2g where g is the gravitational constant 32 ft/sec 2 {9.8 m/s 2 }. Mathematically, the units work out nicely; V is velocity which is measured in ft/sec; g is gravity which is measured in ft /sec 2. Multiplying V by itself (V x V = V 2 ) results in units of ft 2 /sec 2. Dividing one by the other, the measuring unit of the term, V 2 /2g, comes out to be, once again, feet (of head). 4. Conservation of Hydraulic Energy Energy is always conserved. It may turn into heat or sound or be used to break something apart, but it never truly goes away. Hydraulic energy is the same. It may change back and forth between the three forms; elevation energy, pressure energy and velocity energy and some of it may be lost to friction (heat) or sound (noise) but it is still all there. In hydraulics we are most concerned about elevation head or energy, pressure head or energy and velocity head or energy. If the additive sum of these does not remain constant as a fluid travels through a system, then, there must either be energy being added (by a pump for example), or being taken away (by friction or by a generator s turbine). Therefore, if we account for all these terms, than, we can say that between any two points in a system, point 1 and point 2, energy must be conserved for: P 1 /density + V 2 1 /2g + Z 1 + Energy Added (pump head) = P 2 /density + V 2 2 /2g + Z 2 + Head Losses Initial Hydraulic Energy Final Hydraulic Energy Pressure + Velocity + Elevation 1 + Pump Head Added Equals Pressure + Velocity + Elevation 2 + Friction Losses Or, simplifying the terms (and letting Z stand for elevation head, V stand for velocity head, P stand for pressure head, H pump stand for Pump Head, and H loss#1 stand for friction loss at a point) we can write: P@1 + V@ H pump = P@2 + V@2 + Z@ 2 + H loss#1

8 Page 5 This is known as the Energy Equation and it facilitates the solution of most of the hydraulic problems that need to be solved in on-site systems. The system shown in Figure 5 illustrates the application of this concept. Nine points (1 thru 10) are identified in the sketch illustrating nine points from the water surface in the first storage tank to the water surface in the second (higher) tank. Figure 5 Energy between Points Considering each two points in turn and the energy equation the following statements can be made. 1. The energy between point 1 and point 2 ARE EQUAL. There has been a decrease in elevation head but there has been an equal increase in the pressure head. Z@1 = P@2 + Z@ 2 2. The energy between point 2 and point 3 ARE EQUAL. There has been no change in elevation, but as the water accelerates toward the pipe entrance there is an increase in velocity. Therefore, the pressure head at 3 will be lower than the pressure head at 2 but the elevation head will have increased the same amount. There is no energy loss because there is little if any friction dissipating any energy as the flow moves toward the pipe entrance. P@2 = P@3 + V@3 3. The energy between point 3 and point 4 ARE NOT EQUAL. The elevation heads are the same and the velocity heads are the same (between just outside and just inside the pipe) but there is a friction

9 Page 6 effect as the water passes around the pipe entrance into the pipe. This is called an entrance loss. To make equality we must add the head loss due to friction back to the equation. P@3 = P@4 + H loss-entrance 4. The energy between point 4 and point 5 ARE NOT EQUAL. The elevation heads are different; since the pipe diameter is the same, the velocities heads are the same. (The water can t slow down because it is incompressible and there is no extra place for the incoming water to go. Continuity dictates that the flow in must equal the flow out so if the pipe diameters equal the same the velocities must be the same.) However, there is also pipe friction between these two points so the energy must decrease accordingly. P@4 + Z@ 4 = P@5 + Z@ 5 + H loss-pipe 5. The energy between point 5 and point 6 on either side of the pump ARE NOT EQUAL. Despite energy losses in the pump due to friction and turbulence the pump adds energy even though the elevation head remains the same on both side of the pump and the velocity head remains the same on both sides of the pump. The only way that the energy balance can be maintained is for the pressure head to go up exactly as much as the pump head provided. The only way for equality to be represented is to subtract the head added by the pump to create equality with the original condition at point 4. P@5 = P@6 - H pump 6. The energy between point 6 and point 7 ARE NOT EQUAL. The elevation head has changed and the velocity head has not changed. There is friction between points 6 and point 7 that reduces the total energy that is reflected in a change in pressure. The change in elevation also reduces the pressure, but this is not a loss due to the corresponding gain in elevation head. This is the same situation as was encountered between points 4 and 5. P@6 + Z@ 6 = P@7 + Z@ 7 + H loss-pipe 7. The energy between point 7 and point 8 ARE NOT EQUAL. Although the elevation heads are the same and the velocity heads are the same (and cancel on both sides) as the water escapes there is another friction loss at the exit. P@7 = P@8 + H loss-expansion

10 Page 7 8. The energy between point 8 and point 9 ARE NOT EQUAL. The velocity head that still existed at the pipe exit is dissipated in turbulence within the tank, but the pressure heads are equivalent since both points are at the same elevation in the tank. The two losses mention are generally considered as part of the overall same loss of energy as the flow escapes into the second tank P@8 = P@9 + H loss-velocity 9. The energy between point 9 and point 10 ARE EQUAL. Elevation head has increased while pressure head has decreased. The velocity is the same at point 9 and point 10 and there is no further friction loss between these two points. Z@9 + P@9 = + Z@ 10 This somewhat tedious exercise should demonstrate the simple fact that if we can account for all energy increases (from the pump) and all energy decreases (from friction, inlet and outlet losses), we should be able to relate the initial energy at the free water surface of the first tank to the final energy at the free water surface of the second tank. To do this we will have to add back all the losses and subtract the head added by the pump. Summarizing the above equations we see: Z@1 = Z@ 10 + H loss-friction, entrance and exit - H pump 5. Orifice Flow In onsite systems, the final discharge back into the environment is often through the use of a simple orifice or hole that squirts the treated effluent into the air or into the ground. Figure 5 illustrates this situation. The treated effluent is being used to irrigate a field through the use of sprinkler heads that make use of orifices.

11 Page 8 Figure 6 Spray Irrigation using Orifice Flow (Lenning) In addition, there are many situations where orifices are used to squirt the effluent into or over some treatment media. The traditional trickling filter which has been the basis of a new wave of attached growth biological filters makes use of orifices to not only distribute the effluent over the media, but also to drive the distributor arm around through the jet action of the individual orifices. Figure 7 illustrates this use of orifice flow. The individual doesn t only orifices spray the effluent over the media, but also result in a jet action which by Newton s law results in a force in the opposite direction of the jet which pushes the spray arms around their central pivot. Figure 7 Orifice Flow for Distribution and Drive A useful first situation to study the energy principle is an orifice drilled into the side of a large tank. Figure 8 illustrates a tank full of water up to the level, Z, with a small hole

12 Page 9 penetrating the side of the tank at its bottom. The discharge of the water relates to the pressure at the hole and the hole s configuration. Figure 8 Orifice Flow The energy at the surface of the tank consists only of the elevation head because the gage pressure is zero (atmospheric) and the velocity is zero. The energy at the exit of the flow consists only of the velocity head because the gage pressure is also zero (atmospheric) and the elevation head is now zero. Since there is little if any friction loss and no pump between these two points, we don t have to consider those terms either. Therefore, the elevation head at the top, Z, must equal the velocity head at the orifice, V 2 /2g. Z = V 2 /2g Using a little algebra allows us to solve for V. V = (Z x 2g) 1/2 The velocity V is never exactly equal to the value predicted by the above equation due to frictional losses at the orifice which are related to the thickness of the material through which the orifice is drilled and the exact shape and quality of the drilled hole. This causes the introduction of a correction factor, k, which is empirically determined (measured) with various values presented in most applied hydraulics books. Therefore, adding a correction factor to the equation for the velocity V results in: V = k x (Z x 2g) 1/2 Using a little more algebra allows us to solve for the discharge if we know the diameter of the orifice. Recognizing that the discharge will equal the velocity, V, times the area, A, enables

13 Page 10 us to put an equation together for the discharge of an orifice. The area of the orifice A orifice can be found by using the equation for the area of a circle: A orifice = Π x (D orifice ) 2 / 4 This equation is quite useful in onsite wastewater hydraulic design situations since the onsite designer is often required to discharge treated or partially treated wastewater into either treatment media (intermittent sand filter, textile filters, peat filters etc) or into disposal/dispersal trenches or chambers. Combining the two equations from above yields a useful equation for the discharge from an orifice. The discharge from an orifice with a diameter of D (ft) is related to the head Z (ft) by the following equations: Q orifice = A x V Q orifice = Π x (D orifice ) 2 ½ / 4 x k x (Z x 2g) Or Simplifying: Q orifice = C x D 2 orifice x Z 1/2 1/2 Where C = k x Π / 4 x (2g) The value of C is a constant based upon the value of Π, the gravity constant, g, and the other numerical constants including k. Various simple formulas are available for simplifying this equation for relatively standard situations. B. Momentum Issues In addition to energy considerations, the hydraulic engineer engaged in the design of onsite systems must consider two additional issues that relate to the momentum of moving water as well as the energy of moving water. Newton defined momentum as the mass multiplied by the velocity, M x V. Most engineers are familiar with the famous equation F = M x A, which relates force (F) to the mass of an object (M) and any acceleration (A), which it is experiencing. Actually Newton s equation was originally written to relate a force (F) to the change of momentum in a small interval of time. Newton s equation in this form is F = d (M x V)/dt. Viewed this way Newton s equation clearly shows that a force is needed to change the velocity of an object and its momentum and that the faster you change the velocity the greater the needed force. Consequently, there are two situations where this effect must be considered; the water hammer effect and thrust blocking. 1. Water Hammer Rapidly closing or opening a valve causes a pressure transient in pipelines which is known as water hammer. Rapid or instantaneous valve closure can result in pressures

14 Page 11 well over the steady state values, while valve opening can cause seriously low pressures, possibly so low that the flowing liquid vaporizes inside the pipe. Water hammer effects require an understanding of momentum effects. Suppose that we have water flowing in a pipe, and suddenly close a valve. The water at the front is brought to an immediate halt; the water hammer effect is caused by rebound waves created in a pipe full of liquid when the valve is closed too quickly. The sudden closure forces the liquid column to stop moving more quickly than it wants to, or when a valve is opened or a pump is started up too quickly, forcing the liquid column to start moving more quickly than it wants to. In either situation, the rebound wave travels up and down the pipe through the liquid, banging against each end of the pipe. Figure 9 Energy Grade Line during Water Hammer This banging can be heard as water hammer. Try it at home - turn on your tap, then turn it off very quickly. You should hear a bang, and maybe even several. If you turn the tap off more slowly, it should stay quiet, as the liquid in the pipes slows down more gradually. The hammer occurs because an entire column of water is being stopped quickly. The force exerted by the closed valve that has interrupted the forward motion of the water rapidly decelerates the entire mass of the water column. This sends shock waves through the pipe. This is similar to a train wreck, instead of slowing to a stop, it hits into a mountainside. The back of the train continues forward even though the front cannot go anywhere. Since the water flow is restricted inside the pipe, a shock wave of incompressible water travels back down the pipe. Assume the gate valve in the above diagram is suddenly closed. The liquid near the valve will stop moving instantaneously while the water further from the valve will still be flowing with its initial velocity and pressure. As the water in motion slams into the idle

15 Page 12 water it will compress and build pressure. The increased pressure will begin at the valve but travel back towards the inlet of the pipe. When the pressure wave finally reaches the inlet of the pipe all water in the pipe will be at rest, but will be under an excess pressure. As the pressure wave travels to the pipe inlet a transient hydraulic grade line parallel to the original hydraulic grade line will be produced with a height of p h/ γ where p h is the water hammer pressure. This is represented schematically in Figure 9. The pressure at the inlet of the pipe cannot exceed the pressure do to the height of the water, therefore, when the pressure wave arrives at the inlet the pressure instantly drops to the pressure produced by the height of the water. Remember the entire pipe is now under pressure so the liquid is compressed and the pipe walls are probably stretched. Therefore, the water starts to flow back into the reservoir and a wave of lessening pressure will work its way up the pipe towards the valve. As soon as the wave reaches, the valve the entire mass of liquid in the pipe will be under normal pressure produced by the height of the water in the reservoir. However, due to the momentum of the water, excess water will continue to flow into the reservoir producing a drop in pressure in the pipe up to the valve. This will cause a wave of pressure to travel back up the pipe towards the valve. There will be a series of pressure waves traveling back and forth over the length of the pipe. Due to the fluid friction, the fluctuation in pressure will gradually head toward the pressure for the no flow condition. A designer must be aware of the water hammer effects because the pressure generated by the initial water hammer can exceed the pressure capability of the pipe and cause it to rupture. Pipes can be protected from the effects of high water hammer pressure through the use of slow closing valves. Unfortunately, onsite systems often use solenoid valves that shut off the water flow almost instantaneously. The use of an automatic relief valves or a surge tank that permit water to escape when the pressure exceeds a preset value may have an advantage. Figure 10 Solenoid Valve

16 Page 13 A simple surge tank is a vertical standpipe connected to the pipeline (Figure 11). Steady flow in the pipe produces a pressure or water level of z 1. When the valve is suddenly closed, water rises in the surge tank up to z 2. The water surface in the vertical section of pipe will fluctuate up and down until it reaches the steady state caused by the dampening effects of the fluid friction. Figure 11 Surge Line Example Problem Given: Pipe length (L) = 200 feet Pipe Diameter (D) = 2 inches The initial Flow Rate (Q) = 30 gpm Celerity of pressure for the pipe (cp) = 3000 fps Find : a) The water hammer pressure for instantaneous valve closure b) Water hammer pressure if the valve is closed in 4.0 seconds Solution: a) V = Q/A = (30 gpm * 1 ft 3 /7.481 gal * 1 min/60 sec)/((3.14*(2/12) 2 )/4 = 3 fps p h = ρ * cp * V = 1.94 lbf sec 2 /ft 4 * 3000 ft/sec * 3 ft/sec = 17,460 lb/ft 2 *1 ft 2 /144 in 2 = 121 psi b)p h = ((2*L)/cp/closing time) * p h = 2*200 ft/4 sec / 3000 ft/sec * 121 psi = 4 psi