(Redirected from Drain waste vent)
Soil stack is connected to the sewer at the bottom and vented at the top, while each plumbing fixture has its own trap.
In modern plumbing, a drain-waste-vent (or DWV) is part of a system that allows air to enter a plumbing system to maintain proper air pressure to enable the removal of sewage and greywater from a dwelling. Waste is produced at fixtures such as toilets, sinks, and showers. As the water runs down, proper venting is required to avoid a vacuum from being created. As the water runs down air must be allowed into the waste pipe either through a roof vent, or the 'drain waste vent.' (or DWV)
- 5Island fixture vent
Overview[edit]
DWV systems maintain neutral air pressure in the drains, allowing free flow of water and sewage down drains and through waste pipes by gravity. It is critical that a sufficient downward slope be maintained throughout, to keep liquids and entrained solids flowing freely towards the main drain from the building. In relatively rare situations, a downward slope out of a building to the sewer cannot be created, and a special collection pit and grinding lift 'sewage ejector' pump are needed. By contrast, potable water supply systems operate under pressure to distribute water up through buildings, and do not require a continuous downward slope in their piping.
Every fixture is required to have an internal or external trap; double trapping is prohibited by plumbing codes due to its susceptibility to clogging. Every plumbing fixture must also have an attached vent. The top of stacks must be vented too, via a stack vent, which is sometimes called a stink pipe.[1]
All plumbing waste fixtures use traps to prevent sewer gases from leaking into the house. Through traps, all fixtures are connected to waste lines, which in turn take the waste to a 'soil stack', or 'soil vent pipe'. At the building drain system's lowest point, the drain-waste vent is attached, and rises (usually inside a wall) to and out of the roof. Waste exits from the building through the building's main drain and flows through a sewage line, which leads to a septic system or a public sewer. Cesspits are generally prohibited in developed areas.
The venting system, or plumbing vents, consists of a number of pipes leading from waste pipes to the outdoors, usually through the roof. Vents provide a means to release sewer gases outside instead of inside the house. Vents also admit oxygen to the waste system to allow aerobic sewage digestion, and to discourage noxious anaerobic decomposition.[further explanation needed] Vents provide a way to equalize the pressure on both sides of a trap, thereby allowing the trap to hold the water which is needed to maintain effectiveness of the trap, and avoiding 'trap suckout' which otherwise might occur.
Operation[edit]
A sewer pipe is normally at neutral air pressure compared to the surrounding atmosphere. When a column of waste water flows through a pipe, it compresses air ahead of it in the pipe, creating a positive pressure that must be released so it does not push back on the waste stream and downstream trap water seals. As the column of water passes, air must freely flow in behind the waste stream, or negative pressure results. The extent of these pressure fluctuations is determined by the fluid volume of the waste discharge.
Excessive negative air pressure, behind a 'slug' of water that is draining, can siphon water from traps at plumbing fixtures. Generally, a toilet outlet has the shortest trap seal, making it most vulnerable to being emptied by induced siphonage. An empty trap can allow noxious sewer gases to enter a building.
On the other hand, if the air pressure within the drain becomes suddenly higher than ambient, this positive transient could cause waste water to be pushed into the fixture, breaking the trap seal, with serious hygiene and health consequences if too forceful. Tall buildings of three or more stories are particularly susceptible to this problem. Vent stacks are installed in parallel to waste stacks to allow proper venting in tall buildings.
External venting[edit]
Most residential building drainage systems in North America are vented directly through the building roofs. The DWV pipe is typically ABS or PVC DWV-rated plastic pipe equipped with a flashing at the roof penetration to prevent rainwater from entering the buildings. Older homes may use copper, iron, lead or clay pipes, in rough order of increasing antiquity.
Under many older building codes, a vent stack (a pipe leading to the main roof vent) is required to be within a 5-foot (1.5 m) radius of the draining fixture it serves (sink, toilet, shower stall, etc.). To allow only one vent stack, and thus one roof penetration as permitted by local building code, sub-vents may be tied together inside the building and exit via a common vent stack. One additional requirement for a vent stack connection occurs when there are very long horizontal drain runs with very little slope to the run. Adding a vent connection within the run will aid flow, and when used with a cleanout allows for better serviceability of the long run.
A blocked vent is a relatively common problem caused by anything from leaves, to dead animals, to ice dams in very cold weather, or a horizontal section of the venting system, sloped the wrong way and filled with water from rain or condensation. Symptoms range from bubbles in the toilet bowl[citation needed] when it is flushed, to slow drainage,[citation needed] and all the way to siphoned (empty) traps which allow sewer gases to enter the building.
When a fixture trap is venting properly, a 'sucking' sound can often be heard as the fixture vigorously empties out during normal operation. This phenomenon is harmless, and is different from 'trap suckout' induced by pressure variations caused by wastewater movement elsewhere in the system, which is not supposed to allow interactions from one fixture to another. Toilets are a special case, since they are usually designed to self-siphon to ensure complete evacuation of their contents; they are then automatically refilled by a special valve mechanism.[citation needed]
Internal venting[edit]
Mechanical vents (also called cheater vents[2]) come in two types: Air admittance valves and check vents, the latter being a vent with a check valve.
Air admittance valves (AAVs, or commonly referred to in the UK as Durgo valves and in the US as Studor vents and Sure-Vent®) are negative-pressure-activated, one-way mechanical valves, used in a plumbing or drainage venting system to eliminate the need for conventional pipe venting and roof penetrations. A discharge of wastewater causes the AAV to open, releasing the vacuum and allowing air to enter the plumbing vent pipe for proper pressure equalization.
Since AAVs will only operate under negative pressure situations, they are not suitable for all venting applications, such as venting a sump, where positive pressures are created when the sump fills. Also, where positive drainage pressures are found in larger buildings or multi-story buildings, an air admittance valve could be used in conjunction with a positive pressure reduction device such as the PAPA positive air pressure attenuator, to provide a complete venting solution for more-complicated drainage venting systems.
Using AAVs can significantly reduce the amount of venting materials needed in a plumbing system, increase plumbing labor efficiency, allow greater flexibility in the layout of plumbing fixtures, and reduce long-term roof maintenance problems associated with conventional vent stack roofing penetrations.
While some state and local building departments prohibit AAVs, the International Residential and International Plumbing Codes allow it to be used in place of a vent through the roof. AAV's are certified to reliably open and close a minimum of 500,000 times, (approximately 30 years of use) with no release of sewer gas; some manufacturers claim their units are tested for up to 1.5 million cycles, or at least 80 years of use. AAVs have been effectively used in Europe for more than two decades.[when?] US manufacturers offer warranties that range from 1 year to 'lifetime'.
Island fixture vent[edit]
Island fixture vent for under-cabinet waste plumbing
An island fixture vent, sometimes colloquially called a 'Chicago Loop', “Boston loop” or 'Bow Vent', is an alternate way of venting the trap installed on an under counter island sink or other similar applications where a conventional vertical vent stack or air admittance valve is not feasible or allowed.
As with all drains, ventilation must be provided to allow the flowing waste water to displace the sewer gas in the drain, and then to allow air (or some other fluid) to fill the vacuum which would otherwise form as the water flows down the pipe.
An island fixture vent provides an elegant solution for this necessity: when the drain is opened, water displaces the sewer gas up to the sanitary tee, the water flows downward while sewer gas is displaced upward and toward the vent. The vent can also provide air to fill any vacuum created.
The key to a functional island fixture vent is that the top elbow must be at least as high as the 'flood level' (the peak possible drain water level in the sink). This ensures that the vent never becomes waterlogged.
Cost[edit]
The cost of installation is high because of the number of elbows and small pieces of pipe required. The largest cost outlay with modern plastic drain pipes is labor. Use of street elbows is helpful.
Alternately if moving sink to an island sink, install the P-trap below the floor of the island and vent off the top of the drain. Attach toward the trap and reverse 180 degrees so any water in the vent flows down the drain. Slope drain down, slope vent up, and attach to existing vent from previous existing fixture that is now abandoned. Patch previously existing drain to become vent. In Canada, the national plumbing code requires that the minimum trap arm be at least 2 times the pipe diameter, (e.g., 1.25 inch pipe needs a 2.5-inch trap arm, 1.5 pipe needs a 3-inch trap arm, etc.) and that the vent pipe be one size larger than the drain that it serves, also a cleanout is required on both the vent and the drain. The reason for this is in the event of a plugged sink, the waste water will back up and go down the vent, possibly plugging the vent (as it is under the countertop), and a clean-out would permit the cleaning of the pipes.
Fittings[edit]
Drainage and venting systems require not only pipe, but also many specialized fittings which add considerably to their cost of construction. Special access fittings such as 'clean-outs' enhance the long-term maintainability of the systems, and are required by most plumbing codes.
See also[edit]
References[edit]
- ^Gee, Steve (30 July 2007). 'Cat burglar's crude escape'. The Daily Telegraph. Retrieved 20 March 2015.
- ^Saltzman, Reuben (21 November 2012). 'Illegal Plumbing Products in Minnesota'. Star Tribune.
Mechanical vents are not allowed in Minnesota. These are often referred to as cheater vents, and they come in two varieties - an air admittance valve and a check vent.
Further reading[edit]
- Fink, Justin (16 September 2015). 'Drain-Waste-Vent Systems'. Fine Homebuilding. Taunton Press. 154: 18–19. Retrieved 25 September 2015.
External links[edit]
Retrieved from 'https://en.wikipedia.org/w/index.php?title=Drain-waste-vent_system&oldid=903688459'
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In nonideal fluid dynamics, the Hagen–Poiseuille equation, also known as the Hagen–Poiseuille law, Poiseuille law or Poiseuille equation, is a physical law that gives the pressure drop in an incompressible and Newtonian fluid in laminar flow flowing through a long cylindrical pipe of constant cross section. It can be successfully applied to air flow in lungalveoli, or the flow through a drinking straw or through a hypodermic needle. It was experimentally derived independently by Jean Léonard Marie Poiseuille in 1838[1] and Gotthilf Heinrich Ludwig Hagen,[2] and published by Poiseuille in 1840–41 and 1846.[1]
The assumptions of the equation are that the fluid is incompressible and Newtonian; the flow is laminar through a pipe of constant circular cross-section that is substantially longer than its diameter; and there is no acceleration of fluid in the pipe. For velocities and pipe diameters above a threshold, actual fluid flow is not laminar but turbulent, leading to larger pressure drops than calculated by the Hagen–Poiseuille equation.
- 3Derivation
Equation[edit]
In standard fluid-kinetics notation:[3][4][5]
where:
- ΔP is the pressure difference between the two ends,
- L is the length of pipe,
- μ is the dynamic viscosity,
- Q is the volumetric flow rate,
- R is the pipe radius.
The equation does not hold close to the pipe entrance.[6]:3
The equation fails in the limit of low viscosity, wide and/or short pipe. Low viscosity or a wide pipe may result in turbulent flow, making it necessary to use more complex models, such as Darcy–Weisbach equation. If the pipe is too short, the Hagen–Poiseuille equation may result in unphysically high flow rates; the flow is bounded by Bernoulli's principle, under less restrictive conditions, by
Relation to Darcy–Weisbach[edit]
Normally, Hagen–Poiseuille flow implies not just the relation for the pressure drop, above, but also the full solution for the laminar flow profile, which is parabolic. However, the result for the pressure drop can be extended to turbulent flow by inferring an effective turbulent viscosity in the case of turbulent flow, even though the flow profile in turbulent flow is strictly speaking not actually parabolic. In both cases, laminar or turbulent, the pressure drop is related to the stress at the wall, which determines the so-called friction factor. The wall stress can be determined phenomenologically by the Darcy–Weisbach equation in the field of hydraulics, given a relationship for the friction factor in terms of the Reynolds number. In the case of laminar flow, for a circular cross section:
where Re is the Reynolds number, ρ is the fluid density, and v is the mean flow velocity, which is half the maximal flow velocity in the case of laminar flow. It proves more useful to define the Reynolds number in terms of the mean flow velocity because this quantity remains well defined even in the case of turbulent flow, whereas the maximal flow velocity may not be, or in any case, it may be difficult to infer. In this form the law approximates the Darcy friction factor, the energy (head) loss factor, friction loss factor or Darcy (friction) factorΛ in the laminar flow at very low velocities in cylindrical tube. The theoretical derivation of a slightly different form of the law was made independently by Wiedman in 1856 and Neumann and E. Hagenbach in 1858 (1859, 1860). Hagenbach was the first who called this law the Poiseuille's law.
The law is also very important in hemorheology and hemodynamics, both fields of physiology.[7]
Poiseuille's law was later in 1891 extended to turbulent flow by L. R. Wilberforce, based on Hagenbach's work.
Derivation[edit]
The Hagen–Poiseuille equation can be derived from the Navier–Stokes equations. Although more lengthy than directly using the Navier–Stokes equations, an alternative method of deriving the Hagen–Poiseuille equation is as follows.
Liquid flow through a pipe[edit]
a) A tube showing the imaginary lamina. b) A cross section of the tube shows the lamina moving at different speeds. Those closest to the edge of the tube are moving slowly while those near the center are moving quickly.
Assume the liquid exhibits laminar flow. Laminar flow in a round pipe prescribes that there are a bunch of circular layers (lamina) of liquid, each having a velocity determined only by their radial distance from the center of the tube. Also assume the center is moving fastest while the liquid touching the walls of the tube is stationary (due to the no-slip condition).
To figure out the motion of the liquid, all forces acting on each lamina must be known:
- The pressure force pushing the liquid through the tube is the change in pressure multiplied by the area: F = −A ΔP. This force is in the direction of the motion of the liquid. The negative sign comes from the conventional way we define ΔP = Pend − Ptop < 0.
- Viscosity effects will pull from the faster lamina immediately closer to the center of the tube.
- Viscosity effects will drag from the slower lamina immediately closer to the walls of the tube.
Viscosity[edit]
Two fluids moving past each other in the x direction. The liquid on top is moving faster and will be pulled in the negative direction by the bottom liquid while the bottom liquid will be pulled in the positive direction by the top liquid.
When two layers of liquid in contact with each other move at different speeds, there will be a shear force between them. This force is proportional to the area of contact A, the velocity gradient perpendicular to the direction of flow Δvx/Δy, and a proportionality constant (viscosity) and is given by
The negative sign is in there because we are concerned with the faster moving liquid (top in figure), which is being slowed by the slower liquid (bottom in figure). By Newton's third law of motion, the force on the slower liquid is equal and opposite (no negative sign) to the force on the faster liquid. This equation assumes that the area of contact is so large that we can ignore any effects from the edges and that the fluids behave as Newtonian fluids.
Faster lamina[edit]
Assume that we are figuring out the force on the lamina with radiusr. From the equation above, we need to know the area of contact and the velocity gradient. Think of the lamina as a ring of radius r, thickness dr, and length Δx. The area of contact between the lamina and the faster one is simply the area of the inside of the cylinder: A = 2πr Δx. We don't know the exact form for the velocity of the liquid within the tube yet, but we do know (from our assumption above) that it is dependent on the radius. Therefore, the velocity gradient is the change of the velocity with respect to the change in the radius at the intersection of these two laminae. That intersection is at a radius of r. So, considering that this force will be positive with respect to the movement of the liquid (but the derivative of the velocity is negative), the final form of the equation becomes
where the vertical bar and subscript r following the derivative indicates that it should be taken at a radius of r.
Slower lamina[edit]
Next let's find the force of drag from the slower lamina. We need to calculate the same values that we did for the force from the faster lamina. In this case, the area of contact is at r + dr instead of r. Also, we need to remember that this force opposes the direction of movement of the liquid and will therefore be negative (and that the derivative of the velocity is negative).
Putting it all together[edit]
To find the solution for the flow of a laminar layer through a tube, we need to make one last assumption. There is no acceleration of liquid in the pipe, and by Newton's first law, there is no net force. If there is no net force then we can add all of the forces together to get zero
or
First, to get everything happening at the same point, use the first two terms of a Taylor series expansion of the velocity gradient:
The expression is valid for all laminae. Grouping like terms and dropping the vertical bar since all derivatives are assumed to be at radius r,
Finally, put this expression in the form of a differential equation, dropping the term quadratic in dr.
It can be seen that both sides of the equations are negative: there is a drop of pressure along the tube (left side) and both first and second derivatives of the velocity are negative (velocity has a maximum value at the center of the tube, where r = 0). Using the product rule, the equation may be rearranged to:
The right-hand side is the radial term of the Laplace operator ∇2, so this differential equation is a special case of the Poisson equation. It is subject to the following boundary conditions:
— 'no-slip' boundary condition at the wall | |
— axial symmetry. |
Axial symmetry means that the velocity v(r) is maximum at the center of the tube, therefore the first derivative dv/dr is zero at r = 0.
The differential equation can be integrated to:
To find A and B, we use the boundary conditions.
First, the symmetry boundary condition indicates:
A solution possible only if A = 0. Next the no-slip boundary condition is applied to the remaining equation:
so therefore
Now we have a formula for the velocity of liquid moving through the tube as a function of the distance from the center of the tube
or, at the center of the tube where the liquid is moving fastest (r = 0) with R being the radius of the tube,
Poiseuille's law[edit]
To get the total volume that flows through the tube, we need to add up the contributions from each lamina. To calculate the flow through each lamina, we multiply the velocity (from above) and the area of the lamina.
Finally, we integrate over all lamina via the radius variable r.
Startup of Poiseuille flow in a pipe[8][edit]
When a constant pressure gradient is applied between two ends of a long pipe, the flow will not immediately obtain Poiseuille profile, rather it develops through time and reaches the Poiseuille profile at steady state. The Navier-Stokes equations reduce to
with initial and boundary conditions,
The velocity distribution is given by
where is the Bessel function of the first kind of order zero and are the positive roots of this function and is the Bessel function of the first kind of order one. As , Poiseuille solution is recovered.
Poiseuille flow in annular section[9][edit]
Poiseuille flow in annular section
If is the inner cylinder radii and is the outer cylinder radii, with applied pressure gradient between the two ends , the velocity distribution and the volume flux through the annular pipe are
When , the original problem is recovered.
Plane Poiseuille flow[edit]
Plane Poiseuille flow
Plane Poiseuille flow is flow created between two infinitely long parallel plates, separated by a distance with a constant pressure gradient is applied in the direction of flow. The flow is essentially unidirectional because of infinite length. The Navier-Stokes equations reduce to
with no-slip condition on both walls
Therefore, the velocity distribution and the volume flow rate per unit length are
Poiseuille flow through some non-circular cross-sections[10][edit]
Joseph Boussinesq[11] derived the velocity profile and volume flow rate in 1868 for rectangular channel and tubes of equilateral triangular cross-section and for elliptical cross-section. Joseph Proudman[12] derived the same for isosceles triangles in 1914. Let be the constant pressure gradient acting in direction parallel to the motion.
The velocity and the volume flow rate in a rectangular channel of height and width are
The velocity and the volume flow rate of tube with equilateral triangular cross-section of side length are
The velocity and the volume flow rate in the right-angled isosceles triangle are
The velocity distribution for tubes of elliptical cross-section with semi-axis and is[8]
Here, when , Poiseuille flow for circular pipe is recovered and when , plane Poiseuille flow is recovered.
Poiseuille flow through arbitrary cross-section[edit]
The flow through arbitrary cross-section satisfies the condition that on the walls. The governing equation reduces to[13]
If we introduce a new dependent variable as
then it is easy to see that the problem reduces to that integrating a Laplace equation
satisfying the condition
on the wall.
Poiseuille's equation for compressible fluids[edit]
For a compressible fluid in a tube the volumetric flow rate and the linear velocity are not constant along the tube. The flow is usually expressed at outlet pressure. As fluid is compressed or expands, work is done and the fluid is heated or cooled. This means that the flow rate depends on the heat transfer to and from the fluid. For an ideal gas in the isothermal case, where the temperature of the fluid is permitted to equilibrate with its surroundings, and when the pressure difference between ends of the pipe is small, the volumetric flow rate at the pipe outlet is given by
where:
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- Pi is inlet pressure
- Po is outlet pressure
- L is the length of tube
- is the viscosity
- R is the radius
- V is the volume of the fluid at outlet pressure
- v is the velocity of the fluid at outlet pressure
This equation can be seen as Poiseuille's law with an extra correction factor Pi + Po/2Po expressing the average pressure relative to the outlet pressure.
Electrical circuits analogy[edit]
Electricity was originally understood to be a kind of fluid. This hydraulic analogy is still conceptually useful for understanding circuits. This analogy is also used to study the frequency response of fluid-mechanical networks using circuit tools, in which case the fluid network is termed a hydraulic circuit.
Poiseuille's law corresponds to Ohm's law for electrical circuits, V = IR.
Since the net force acting on the fluid is equal to
where S = πr2, i.e. ΔF = πr2 ΔP, then from Poiseuille's law
it follows that
- .
For electrical circuits, let n be the concentration of free charged particles (in m−3) and let q* be the charge of each particle (in coulombs). (For electrons, q* = e = 1.6×10−19 C.)
Then nQ is the number of particles in the volume Q, and nQq* is their total charge. This is the charge that flows through the cross section per unit time, i.e. the currentI. Therefore, I = nQq*. Consequently, Q = I/nq*, and
But ΔF = Eq, where q is the total charge in the volume of the tube. The volume of the tube is equal to πr2L, so the number of charged particles in this volume is equal to nπr2L, and their total charge is
Now,
Since the voltageV = EL, we get
This is exactly Ohm's law, where the resistanceR = V/I is described by the formula
- .
It follows that the resistance R is proportional to the length L of the resistor, which is true. However, it also follows thatthe resistance R is inversely proportional to the fourth power of the radius r, i.e. the resistance R is inversely proportional to the second power of the cross section area S = πr2 of the resistor, which is wrong according to the electrical analogy.
The correct relation is
where ρ is the specific resistance; i.e. the resistance R is inversely proportional to the cross section area S of the resistor.[14]
The reason why Poiseuille's law leads to a wrong formula for the resistance R is the difference between the fluid flow and the electric current. Electron gas is inviscid, so its velocity does not depend on the distance to the walls of the conductor. The resistance is due to the interaction between the flowing electrons and the atoms of the conductor. Therefore, Poiseuille's law and the hydraulic analogy are useful only within certain limits when applied to electricity.
Both Ohm's law and Poiseuille's law illustrate transport phenomena.
Medical applications – intravenous access and fluid delivery[edit]
The Hagen–Poiseuille equation is useful in determining the flow rate of intravenous fluids that may be achieved using various sizes of peripheral and central cannulas. The equation states that flow rate is proportional to the radius to the fourth power, meaning that a small increase in the internal diameter of the cannula yields a significant increase in flow rate of IV fluids. The radius of IV cannulas is typically measured in 'gauge', which is inversely proportional to the radius. Peripheral IV cannulas are typically available as (from large to small) 14G, 16G, 18G, 20G, 22G. As an example, the flow of a 14G cannula is typically twice that of a 16G, and ten times that of a 20G. It also states that flow is inversely proportional to length, meaning that longer lines have lower flow rates. This is important to remember as in an emergency, many clinicians favor shorter, larger catheters compared to longer, narrower catheters. While of less clinical importance, the change in pressure can be used to speed up flow rate by pressurizing the bag of fluid, squeezing the bag, or hanging the bag higher from the level of the cannula. It is also useful to understand that viscous fluids will flow slower (e.g. in blood transfusion).
See also[edit]
Pipe Flow Pressure Drop
Notes[edit]
- ^ abSutera, Salvatore P.; Skalak, Richard (1993). 'The History of Poiseuille's Law'. Annual Review of Fluid Mechanics. 25: 1–19. Bibcode:1993AnRFM.25..1S. doi:10.1146/annurev.fl.25.010193.000245.
- ^István Szabó, ;;Geschichte der mechanischen Prinzipien und ihrer wichtigsten Anwendungen, Basel: Birkhäuser Verlag, 1979.
- ^Kirby, B. J. (2010). Micro- and Nanoscale Fluid Mechanics: Transport in Microfluidic Devices. Cambridge University Press. ISBN978-0-521-11903-0.
- ^Bruus, H. (2007). Theoretical Microfluidics.
- ^'Poiseuille and his law'(PDF). pdfs.semanticscholar.org.
- ^Vogel, Steven (1981). Life in Moving Fluids: The Physical Biology of Flow. PWS Kent Publishers. ISBN0871507498.
- ^Determinants of blood vessel resistance.
- ^ abBatchelor, George Keith. An introduction to fluid dynamics. Cambridge university press, 2000.
- ^Rosenhead, Louis, ed. Laminar boundary layers. Clarendon Press, 1963.
- ^Drazin, Philip G., and Norman Riley. The Navier–Stokes equations: a classification of flows and exact solutions. No. 334. Cambridge University Press, 2006.
- ^Boussinesq, Joseph. 'Mémoire sur l’influence des Frottements dans les Mouvements Réguliers des Fluids.' J. Math. Pures Appl 13.2 (1868): 377-424.
- ^Proudman, J. 'IV. Notes on the motion of viscous liquids in channels.' The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science 28.163 (1914): 30-36.
- ^Samuel Newby Curle and H.J. Davies. Modern Fluid Dynamics. Volume 1, Incompressible Flow, Van Nostrand Reinhold Inc (1971)
- ^Fütterer, C. et al. 'Injection and flow control system for microchannels' Lab-on-a-Chip (2004): 351-356.
References[edit]
- Sutera, S. P.; Skalak, R. (1993). 'The history of Poiseuille's law'. Annual Review of Fluid Mechanics. 25: 1–19. Bibcode:1993AnRFM.25..1S. doi:10.1146/annurev.fl.25.010193.000245.
- Pfitzner, J (1976). 'Poiseuille and his law'. Anaesthesia. 31 (2) (published Mar 1976). pp. 273–5. doi:10.1111/j.1365-2044.1976.tb11804.x. PMID779509.
- Bennett, C. O.; Myers, J. E. (1962). 'Momentum, Heat, and Mass Transfer'. McGraw-Hill.
External links[edit]
Retrieved from 'https://en.wikipedia.org/w/index.php?title=Hagen–Poiseuille_equation&oldid=896923610'
Posted by2 years ago
Archived
First of all, apologies I can't get some of the LaTeX to work.
For my games computing dissertation module I'm making a game to simulate water from from reservoir, to water purification, to pump, to water tower to villages. Within the game there will be pipes underground that can change in thickness and this is where my problems are occurring.
constants set: [; g = 9.8 ;], [; rho = 998 ;] (apparently that's water density at 20*C). Pipes change in diameter, but they're made up of pipe segmented, all pipe lengths are 5m (if that's relevant).
I dont think this is the issue, but here's where the values came from if it is the problem: To test the change in pipe diameter, i'm making the change a little bit after the water tower. The standheight is 10m, and then the water height inside slowly fills. The pressure from the water tower was calculated as: around 98,000 N/m2 (slowly increasing as it fills) using [; g rho h ;]. Output velocity as 14m/s (slowly increasing as it fills) using [; sqrt{2gh} ;]
SO.
Pipe1 has a Diameter: 2m (Area: [; pi m{2} ;] ), Pressure: 98 000N/m2 and Velocity: 14m/s. Pipe2 (next pipe along) has a Diameter: 1m (Area: 0.7853m2). I need to calculate the pressure and velocity for P2.
To do this I used flow rate equation: [; v_1 A_1 = v_2 A_2 ;]
Used this to find v2: [; v_2 = frac{v_1 A_1}{A_2} = frac{14*pi}{0.7853} = 56m/s ;]
Next I used the Bernoulli equation (where the height part has been taken away as both heights are the same): [; P_1 + frac{1}{2} rho {v_1}{2} = P_2 + frac{1}{2} rho {v_2}{2} ;]
Rearranged for P2: [; P_2 = P_1 + frac{rho}{2}({v_1}{2} - {v_2}{2}) = 98000 + frac{998}{2}({14}{2} - {56}{2}) = -1,466,668 ;]
So yeah basically using that I've got a negative pressure, where have I gone wrong? This is intented to be a game to simulate the basics of physics, I understand that in reality it's much more complicated and I'm no physicist, just trying to model it from these basic equations. Any Help, really stuck on this?
Bonus Question: So if after pipe2 there's another pipe, pipe3, with diameter 2m, then surely there'll be less water in pipe3 as pipe2 acted as a bottleneck? If less water comes through, then would I model the equations from the diameter of the pipe, or the diameter of the water within the pipe? (could be calculated from volumeContained/5 = areaOfWater). And so how does that all work exactly?? Have I misunderstood how things work? If i'm wrong and the same amount of water can fit into the smaller pipe, just at a faster speed, then whats the point in having big pipes? That's why I'm thinking it must be less water passing through.
tl;dr; Tried using the flowrate equation with Bernoulli equation to find velocity and pressure from one pipe to another then the diameter changes, however using those equations I get a negative pressure.
18 comments
Negative Pressure In Pipe Flow Calculator
1. The problem statement, all variables and given/known data
(This is more of a discussion question)
My question is, if we have a pipe between two reservoir A and reservoir B (Height of A > height of B) , then if flow in the pipe is to happen from A to B, what sign would the pressure sign be at any point in the pipe?
2. Relevant equations
Energy Equation:
$$ frac{P_1}{gamma}+h_1+frac{v_1^2}{2g} = frac{P_2}{gamma}+h_2+frac{v_2^2}{2g} + h_f +h_L$$
3. The attempt at a solution
I think it makes sense that the pressure at any point in the pipe would be negative - because if we analyse the absolute pressures, we have atmospheric pressure pushing down on reservoir A and if the pressure along the pipe is negative, then $$P_{absolute, pipe} < P_{atm}$$ which implies flow from reservoir A to B.
Can someone please provide me with some discussion / their point of view?
Thank you.
(This is more of a discussion question)
My question is, if we have a pipe between two reservoir A and reservoir B (Height of A > height of B) , then if flow in the pipe is to happen from A to B, what sign would the pressure sign be at any point in the pipe?
2. Relevant equations
Energy Equation:
$$ frac{P_1}{gamma}+h_1+frac{v_1^2}{2g} = frac{P_2}{gamma}+h_2+frac{v_2^2}{2g} + h_f +h_L$$
3. The attempt at a solution
I think it makes sense that the pressure at any point in the pipe would be negative - because if we analyse the absolute pressures, we have atmospheric pressure pushing down on reservoir A and if the pressure along the pipe is negative, then $$P_{absolute, pipe} < P_{atm}$$ which implies flow from reservoir A to B.
Can someone please provide me with some discussion / their point of view?
Thank you.