Pressure Problems
Pressure Problems
(OP)
I've worked myself into a bit of a loop here.
A question came up from a younger colleague today that at first seemed a bit silly but after running a model I might see his point too. Basically, the question was if my pump discharge piping immediately sizes up say from 2" to 20" and I sized the line for pump discharge pressure, could I over pressure the line from almost all of my velocity energy being converted into static energy?
What's curious is that I ran a model with a 2" discharge, 3 inches long, into a 20" line for 10ft then back to a 2" into a tank. The software is telling me that my pump TDH is 40.38 feet, discharge pressure of 19.43 psig, in the 2" discharge piping I only have 4 psig of static head pressure. This seems to be telling me that a pressure gauge on this short section of discharge piping would only read 4 psig. The static pressure then jumps up to 19 psig in the 20" line. So it seems to be that discharge pressure gauges on pumps can be incredibly misleading in cases such as mine? That the TDH converted into pressure is total developed pressure also, but not necessarily static pressure or what I'll see at the discharge of the pump on a gauge? (Is this correct?)
Though converting the TDH into pressure still gives me the correct design data and I'm in no risk of over pressuring anything downstream, it does seem that my discharge pressure gauge readings need to be read in the context of my piping diameter, as my downstream pressure gauges could actually show an INCREASE in pressure if my diameter were to increase.
Am I understanding this correctly?
A question came up from a younger colleague today that at first seemed a bit silly but after running a model I might see his point too. Basically, the question was if my pump discharge piping immediately sizes up say from 2" to 20" and I sized the line for pump discharge pressure, could I over pressure the line from almost all of my velocity energy being converted into static energy?
What's curious is that I ran a model with a 2" discharge, 3 inches long, into a 20" line for 10ft then back to a 2" into a tank. The software is telling me that my pump TDH is 40.38 feet, discharge pressure of 19.43 psig, in the 2" discharge piping I only have 4 psig of static head pressure. This seems to be telling me that a pressure gauge on this short section of discharge piping would only read 4 psig. The static pressure then jumps up to 19 psig in the 20" line. So it seems to be that discharge pressure gauges on pumps can be incredibly misleading in cases such as mine? That the TDH converted into pressure is total developed pressure also, but not necessarily static pressure or what I'll see at the discharge of the pump on a gauge? (Is this correct?)
Though converting the TDH into pressure still gives me the correct design data and I'm in no risk of over pressuring anything downstream, it does seem that my discharge pressure gauge readings need to be read in the context of my piping diameter, as my downstream pressure gauges could actually show an INCREASE in pressure if my diameter were to increase.
Am I understanding this correctly?





RE: Pressure Problems
RE: Pressure Problems
RE: Pressure Problems
RE: Pressure Problems
RE: Pressure Problems
RE: Pressure Problems
If you properly design your discharge line to handle the pump's TDH in terms of pressure (i.e. TDH converted 100% to pressure head), you will not overpressure in steady-state operation. Regardless of flow velocity, line diameter, "static pressure", etc. Energy is energy.
RE: Pressure Problems
The total energy in the pipeline is the sum of the static, dynamic and hydrostatic pressures. The static pressure is the pressure as measured moving with the fluid that can be measured with a pressure gauge. The dynamic energy depends on the velocity head and will be relatively small. The hydrostatic pressures is the pressure due to the force of gravity
The velocity head can not be measured with a pressure gauge. You need a pitot tube to measure velocity head.
However, the hydrostatic pressure will depend on the elevation. If the pipeline is relatively level, the hydrostatic pressure will be constant. If there is significant elevation difference, then there will be differences in pressure due to the elevation. For those applications, the elevation head will have to be added to the static pressure.
RE: Pressure Problems
I agree with most of the above, but there is one point that you need to keep in mind. At relatively low velocities, dynamic pressure is generally (and rightfully) ignored and Bernoulli's relationship whereby
static pressure + dynamic pressure + hydrostatic pressure = constant
is true. At higher velocities, we call "dynamic pressure" "impact pressure" and
static + impact + hydrostatic ≠ constant
and weird things happen inside shock waves.
Dynamic pressure is ½*ρ*v2 (in U.S. units you need to throw in a gc to get to lbf from lbm). Do that calculation in the 2-inch and if the dynamic pressure is less than about 3% of the static pressure, then the pressure gauge on the 2-inch is useful. If it is more than 3% of the static pressure then you'll need to move the pressure gauge to the 20-inch because you really cannot draw any conclusions about one from the other.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
RE: Pressure Problems
Bernoulli's equation also has some restrictions in its applicability that you fail to note, they are:
• Flow is steady;
• Density is constant (which also means the fluid is incompressible);
• Friction losses are negligible.
• The equation relates the states at two points along a single streamline.
Because of the restrictions, the Bernoulli equation is not used for analyzing shock waves, not because weird things happen.
RE: Pressure Problems
You are actually going to start down this tedious road again? Really? OK.
There is nothing in the derivation of Bernoulli's equation that states "constant along a streamline", in fact that addition is quite incorrect and would limit the applicability of the equation to near zero. The streamline demonstration is quite useful in describing why airplanes can fly for example, but as a demonstration, not a proof. The Bernoulli equation applies to bulk properties, not streamline properties.
You seem to be claiming that Bernoulli's big breakthrough is that he defined a relationship between static pressure, dynamic pressure and hydrostatic pressure. It wasn't. The understanding that those terms adding up to total pressure predated him by many years. His breakthrough was defining the list of conditions (much longer than your abbreviated list) whereby the sum of those terms could be considered constant so you could compare one point in a real flow to another point in that same real flow.
I don't have a clue what part of my post was either wrong or circular. I'm not really sure I care.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
RE: Pressure Problems
If the Y column is total developed head and I convert this into pressure, is this pressure my total pressure (dynamic + static)? If so, why does my total developed head drop off with an increase in flowrate, it would seem that as my flow rate increases my static pressure would go down but be canceled out by an increasing dynamic pressure and my curve should be flat. Obviously this isn't so, but I'm struggling to understand why now.
RE: Pressure Problems
RE: Pressure Problems
It is "dynamic" head, not developed. Although total developed head must equal the sum of system static plus dynamic friction loss to have sustained flow.
RE: Pressure Problems
As you can see from the description above, the energy is constant along the streamline, not just "constant".
http://mysite.du.edu/~jcalvert/tech/fluids/bernoul...
RE: Pressure Problems
Pstatic + ½*ρ*vsl12+ρ*g*z = Pstatic + ½*ρ*vsl22+ρ*g*z
Since at a plane, the static pressure is the same across the pipe (the only way that this is not true is if static pressure varies across a pipe cross-section which would invalidate the location of several million pressure sensors) and density is the same at that plane, and since we've chosen a streamline where zsl1=zsl2 this equation simplifies to:
vsl12 = vsl22
But we know that the velocity of a streamline closer to the center is higher than the velocity closer to the wall (the no-flow boundary requires this result). The only way that Bernoulli works is with bulk (also called average) velocity which is the volume weighted average of every streamline.
Before you come back with another snarky and pointless rejoinder, please ask yourself "when was the last time I was able to isolate a streamline in a flow to measure its velocity?" If the answer is not "never" then you are lying to yourself. If it is "never" then you should be able to see that the streamline model of flow is only useful for allowing humans to visualize idealized flows. We don't really have access to streamline data.
The reference you provided is instructive He goes on to include some information that really requires additional discussion, but the first section is solid.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
RE: Pressure Problems
Now go back and read the reference again so you understand fully comprehend. The third form (you just looked at the first form in the refence and appeared to stop at that point) is the usual derivative of Bernoulli's equation and includes conservation of energy, includes streamlines and is easily understood by students:
"The third form of Bernoulli's Equation is derived from the conservation of energy. Bernoulli himself took an equivalent approach, although the concept of energy was not well-developed in his time. Energy balance is a favoured method of approach in engineering, and this is the usual derivation of Bernoulli's Equation in elementary work. By the use of energy concepts, the equation can be extended usefully to compressible fluids and thermodynamic processes. In the Figure, an element of fluid is transferred from one point to another in a tube with rigid boundaries. The equation of continuity for an incompressible fluid shows that the same volume of fluid Q disappears at one point and reappears at another. The imaginary pistons move with the speed of the fluid. Capital letters are used for quantities at one point, small letters for the same quantities at the second point. The energies per unit volume, made up of kinetic, potential, and pressure terms are equated. The pressure terms can also be handled as doing work on the element of fluid, which is equivalent. The virtue of this derivation is that can be extended in various directions to give important results, and that it is easily believed by students. The rigid tube can be replaced by a surface generated by streamlines, which can be shrunk down to the neighbourhood of a single streamline, which is just the second form of Bernoulli's Equation, but here derived by energy instead of by dynamics."
RE: Pressure Problems
No comment on how your partially understood concept of this equation would result in two velocities that cannot be equal appearing to be equal in your approach? Didn't think so.
David Simpson, PE
MuleShoe Engineering
In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual. Galileo Galilei, Italian Physicist
RE: Pressure Problems
That's not true bimr. Bernoulli's can easily be adapted to analyze liquid shockwave/pressure surge at unsteady flow. However, most are only familiar with the famous steady-state formula, including 98% of textbooks. Bernoulli's is just a mathematical expression for the law of energy conservation, which continues to hold true in shockwave/pressure surge applications, even if your textbook doesn't have a section on it.
RE: Pressure Problems
If you extend the Bernoulli equation by adding some terms, deleting other terms, and changing assumptions as Hugoniot did, the resulting equation may be used to analyze shock waves. However, the equation is substantially different and is no longer known as the Bernoulli equation. The equations to analyze shock waves are known as the Rankine-Hugoniot equations, not the Bernoulli equation.
RE: Pressure Problems
Pitot tubes are good tools for measuring velocity head in a small region. However, the velocity head as typically calculated for a fluid flowing in a pipe or duct is based on the average speed of the fluid along the duct/pipe axis. If you use the velocity head you find with a Pitot tube and attempt to apply it to the full flow in a pipe, you will have errors.
"If the Y column is total developed head and I convert this into pressure, is this pressure my total pressure (dynamic + static)?"
Be careful not to assume static head and static pressure are directly related. If the fluid is not moving, and in a uniform gravitational field (close enough most of the time) then they correlate nicely. If fluids are moving, interaction with velocity head and friction will play havoc in that conversion. If you understand the rest of this conversation, I shouldn't have to explain it much more.
RE: Pressure Problems
I guess what I'm now confused about is what exactly total dynamic head on the pump curve is giving me in relation to the discharge pressure of my pump at some given flow rate. If I stuck a pressure gauge on the discharge piping (as typically seen on an operating unit) of the pump and use my TDH to calculate the pressure differential across the pump, add it to the suction side pressure, will I see this number on the gauge or will I see some number lower than this number bc the gauge is only reading static pressure on the pipe wall and some of my TDH has gone into dynamic pressure?
I suppose I could just calculate it from an actual case on one of my units but I was hoping to clear up the theory first.
RE: Pressure Problems
As stated above, the Dynamic (velocity head) will be relatively small at the typical flow rates used for industrial applications and may be neglected in your calculation.
Suction Head = Total Static plus Dynamic, measured at pump inlet.
Discharge Head = Total Static plus Dynamic, measured at pump exit.
Pump Head = Discharge Head minus Suction Head plus correction for the difference in gage elevation.
Static Head is what the absolute gage reads, converted to feet of water.
Dynamic Head is the same as Velocity Head
Review the attached paper.