My fire academy textbook says there is a limit to how fast water can flow through a pipe or hose. When is this limit acheived? What are the problems reaching this limit? And please describe the occurrences as you approach this limit.
I suppose there are two ways to think about this: real life (fire fighting or plumbing) conditions and theoretical conditions where you could have an infinitely strong pipe and an infinitely powerful pump.
Just by visualizing this scenario, I can tell the water will heat up tremendously from the friction and turbulence, but I don’t know if it could reach boiling point and I don’t know how this would affect the process anyway. But my theory is that there is some limited velocity the water would have and pushing any harder would be like pushing your hands together… I just don’t know what mechanism/phenomenon could generate this infintely huge resistance.
I’m not sure exactly what you’re trying to ask, but if I may reasonably interpret your post, I think I can clarify a few things.
First off, if we’re assuming an idealized pipe and fluid, then there isn’t any strict upper limit to how fast the stream can exit the pipe (none due to the pipe alone, anyway). Flow through a pipe is governed by two equations, the continuiity equation and the Bernoulli equation, as follows:
Continuity (for an incompressible fluid; in this case water):
Where V is the inlet (1) and outlet (2) velocities, A is the cross sectional area of the pipe or hose at inlet (1) and outlet (2), and Rho is the density. Basically, this is a conservation of mass equation, saying that since no mass is stored in the pipe, the mass flow in must equal the mass flow out.
Where P is pressure, V is velocity, Rho is density, g is the acceleration due to gravity, and h is the height of the inlet (1) and outlet (2). Basically, this defines the relationshop between static pressure P (the pressure in a still fluid, say, in a water main), the dynamic pressure (the pressure due to the inertia of a moving fluid), and pressure caused by gravitational potential.
Okay, so how is this related to the question of how fast water can move through a pipe? Let’s take the Bernoulli equation, and drop the first and last terms from each side. In other words, the pressure at each end of the pipe is equal, and the pipe is horizontal. This reduces to:
.5RhoV[sub]1[/sub]^2 = .5RhoV[sub]2[/sub]^2
and thus to:
V[sub]1[/sub]^2 = V[sub]2[/sub]^2
and:
V[sub]1[/sub] = V[sub]1[/sub]
So the input velocity is equal to the output velocity, at least in idealized circumstances. Similarly, if we imagine a horizontal pipe with a pressure difference and assume that the velocity of the fluid on the left hand side of the pipe (and the equation) is negligible, we get:
So we see that the velocity is driven by the difference in pressures between the inlet and outlet of the pipe. In the case of a fire hose, the inlet pressure would be the pressure in the water main under the street or connected to the hydrant, and the outlet pressure would be the atmospheric pressure. Hypothetically, the velocity could be scaled up indefinitely; in practice this isn’t possible, in part because of head loss due to irreversible processes, such as turbulence, and in part because it’s impossible for us to generate an infinite pressure difference, which is what we’d need to scale up the velocities indefinitely. More on this later; right now I have to get to class.
On review, I noticed a typo: That last V[sub]1[/sub] should be V[sub]2[/sub].
So, where was I? Oh yeah, head loss.
Head loss calculations are a little complicated, but are conceptually pretty easy. Basically, fluid flowing through a conduit (such as water through a pipe), loses head ( a measure of the energy stored in the fluid), due to internal viscous effects, scale on the walls of the pipe, valves, fittings, and so on. The calculation of head loss is broken into two parts minor and major head loss (also known as friction loss). Minor head loss concerns the losses due to valves and fittings, and major head loss concerns the losses due to the parameters of the pipe.
Here are a set of web sites that might come in handy:
The important thing to know is that the major and minor head losses rely not just on the valves and pipes, but on the square of the velocity. Here’s an example; the major (friction) loss equation most commonly used is:
h[sub]f[/sub] = f*(L/D)(V^2)/(2g)
Where h[sub]f[/sub] is the head loss, f is the friction factor (determined experimentally), L is the length of the pipe, D is the diameter of the pipe, V is the average flow velocity, and g is the acceleration due to gravity. [Note: the friction factor is a function of Reynolds number, relative roughness, and pipe diameter. We’ll come back to that.]
So, since f, L, D, and g are constants defined by the pipe and gravity, the head loss is proportional to the square of the velocity.
Using a modified form of the Bernoulli equation (I’m not going to type it in, just see the major head loss link above) that takes into account head added by pumps and major and minor head loss, and making the same assumptions as in the previous post (horizontal pipe, negligible initial flow), and assuming no head added by a pump, and no minor losses, we get:
The only problem with this is that while Rho, L, D, P[sub]1[/sub], and P[sub]2[/sub] are constants in a given situation, f is - you guessed it - a function of velocity. As mentioned above, f depends on relative roughness and pipe diameter, which are both constants, and on Reynolds number, which is dependent on velocity:
Re = V*D/v
Where Re is Reynolds number, D is diameter, V is velocity, and v is kinematic viscosity (a constant). So:
As pressure increases, so does velocity, and so does head loss, until eventually you reach a plateau, and the change in pressure difference is counteracted by the increase in the friction factor and thus the head loss. BUT, it isn’t a universal limit; rather, it depends on the system parameters involved.
One final thing, and I hope I haven’t gone on too long: the velocity of the stream exiting the hose has more to do with the volume flow rate than the velocity of the fluid inside the pipe. This is because the exiting stream is run through a nozzle, which exploits the continuity equation. By making the cross -sectional area of the exit less than that of the hose, a nozzle forces the fluid to move at a greater speed through the exit than it does within the hose.
I hope I’ve answered your question. Has any of this helped?
Gack!
You just took me almost twenty years back to Navy Nuclear Power School in lovely Orlando. I thought I had excised all of that stuff from my brain years ago in order to make room for things more relevant to my day-to-day activities.
In a nutshell: Don’t exceed 5 feet per second water velocity for general plumbing/drainage/irrigation applications. Firefighting applications probably push this number, but it’s an emergency, isn’t it?
Water turbulence in the pipe or hose causes friction, which causes pressure loss, which increases exponentially (maybe exponentially is the wrong word, but the curve does get dramatically steeper). This means you need more pumping power to try to cram a ton of water through an undersized pipe.
For example, the pressure loss in a 1" copper pipe at 10 GPM is just 3.53 PSI/100’
If you try to push 20 GPM through the same pipe you’ll lose 12.73 PSI.
30 GPM through the same pipe will cost you 26.97 PSI.
There is no magical upper limit, it just gets harder and harder.
Well, that’s some post Frix, thanks. I’ll come back to it when I have some more time to think about it. I’ll look through those sites too, Shiva.
Really, my question is simple: If you needed to push as much water as possible through a pipe, no nozzle, and your life depended on it, what problems would you run into and what would be the limitations? What would physically happen in the pipe? Theoretical, and real life answers…
Well, Firx has really put it right out there for you (provided you can look up obscure things like the friction losses in fire hoses). You should probably specify what size pipes you’re using–give us a practical maximum for firefighting–so that we don’t use outlandish boundary conditions to tell you anything is possible. To wit:
I was going to cite the ungodly flow rates found in liquid-fuel rocketry (a 3m-by-30m ICBM can empty its fuel tanks in less than three minutes) and then another example hit me: the “rainbirds” on the Space Shuttle launching pad. They reach a peak flow rate of 900,000 gallons per minute through six nozzles, with diameters of 42" (2 nozzles) and 30" (4 nozzles). A smaller system used to dampen acoustic noise and cool down the liquid engines in case of an emergency puts 2,500 gallons per minute through a 6" diameter pipe. I imagine that NASA is probably pretty close to the practical maximum–they tend to ignore the optimum in favor of the gleeful pursuit of the theoretical maximum.
Well, the short answer is yes with an if, the long answer is no with a but. You see, in the fluidic domain power is transferred not just by the flow of a fluid, but also by the pressure in the fluid. Thus the following relationship:
P = p*v
where P is power [watts], p is pressure [newtons/square meter], and v is volume flow rate [cubic meters/second]. Head is basically the energy per unit volume, so when you have head losses due to friction, you decrease the energy per unit volume; in other words, since power is just an energy flow rate, you decrease the power delivered to the exit. Since the volume flow rate is constant (thanks to the continuity equation), this translates to a decrease in pressure.
Where does that leave us? Since the energy we initially give the fluid (from, say, a pump) doesn’t all reach the exit, it has to go somewhere - thanks to the first law of thermodynamics, which states that energy is conserved (neither created nor destroyed, merely changed from one form to another). In this case, it’s probably divided between the hose and the water, with most of it going into the water. This results in a temperature increase in the water.
Now for the ‘if’. While the energy does increase the temperature of the water, water has a very high specific heat, which means that a (comparatively) lot of energy must go into the water to increase its temperature. Add this to the fact that you have a pretty high volume of water flowing through the hose, and this means that the water doesn’t increase in temperature very much. I would guess that you probably wouldn’t be able to readily detect it by touch - you’d need a fairly sensitive thermometer. Even if you can feel the difference, it shouldn’t get anywhere near boiling, unless you have some insane circumstances.
The long answer consists of all of the above (the ‘but’). In addition, it should be noted that this doesn’t really have any direct relationship to the turbulence of the water per se. If the flow were completely laminar, everything I’ve said here and in the previous posts would hold, except that you could calculate the friction factor rather than determining it experimentally (and the friction factor would have a different value, resulting in a different power loss and a different increase in temperatue). So any temperature change isn’t really due to turbulence, but exacerbated by it.
Fortunately, we rarely have to worry about this, since just about all pipe flows are completely turbulent.