When I began this series I pointed out that whenever a waterjet is going to be used both the target material and the waterjet delivery system have to be considered, if the work is to be done well. In the last four posts (set 2 in the series) I have tried to emphasize the role of cracks and flaws in the way in which water penetrates into and removes material. It is easier to see this with large-scale operations, such as in the removal of large volumes of soil, but it equally holds true in the abrasive cutting of glass. Now in set 3 the focus is going to swing back to the ways in which high-pressure waterjets are developed, particularly in the different choices of equipment that can be used. Because this series is meant to help folk understand how systems work, and through that, how to improve production and quality it will tend to shy away from putting a lot of formulae into the presentations. There is a reason that I, an academic, don’t like having students learn equations by rote. It is that it becomes, quite possible, to misremember them. If you are used to looking them up (particularly true in today’s computer world where formulae can easily be used to generate tables) then you are less likely to mis-remember the exact relationships, and to make a possible critical mistake. But, as I showed in the third post, when the tables of jet flow, horsepower and thrust were generated, there are a few, critical equations that need to be born in mind. And the one that underlies the economics of many operations is tied up in the size of the power that is available to do the work. The basic power equation itself is relatively straightforward: In the course of this small set of posts the different components that make up this circuit are going to be discussed in turn. But at the end of the first set I mentioned that in an early comparison of the relative cleaning performance of 10,000 psi waterjets of nominally equal power, and flow (10 gpm IIRC) there was a dramatic difference in the cleaning efficiency, as the Navy reported at the time.
Figure 3. Relative cleaning efficiency in areal percentage cleaned, of five competing systems in cleaning heat exchanger tubes in Navy boilers. (Tursi, T.P. Jr., & Deleece, R.J. Jr, (1975) Development of Very High Pressure Waterjet for Cleaning Naval Boiler Tubes, Naval Ship Engineering Center, Philadelphia Division, Philadelphia, PA., 1975, pp. 18.)
Why such a difference? Consider how the power changes from the time that it first enters the pump motor, and then is converted into power along the line to the target. The numbers that I am going to use may seem extreme, but they actually mirror an early experimental set-up in our laboratory, before we learned better.
A water flow of 10 gallons a minute (gpm) at a pressure of 10,000 pounds per square inch (psi) pressure will contain – using the above equation;
10,000 x 10/1714 = 58.34 horsepower (hp)
But that is the power in the water. Pumps are not 100% efficient, and so there has to be some additional power put into the pump to allow for the relative efficiency of the pump itself. For the sake of illustration let us say that the pump converts the energy at 90% efficiency. Thus the power that is supplied to the drive shaft of the pump will need to be:
58.34/0.9 = 64.8 hp
But that is still not the power that we have to supply, since that power – usually – comes from an electric power cord that feeds into a motor, which then, in turn, drives the pump shaft. That motor itself is also not 100% efficient. Let us, for the sake of discussion, say that it is 92.6% efficient. Then the electrical power supplied will be:
64.8/0.926 = 70 hp
Now, as the calculation progresses, remember that this is the power that is being paid for. And so, in the first part of the flow, the power is transformed, from electric power to water power, but at the pump.
in that earlier post that a competitor, running at a pump pressure of 45,000 psi was losing 35,000 psi of that pressure, just to overcome friction in pushing the water down through a tube that was too narrow. As a result the water coming out of the nozzle had barely enough pressure (10,000 psi) to cut into the rock.
At the same time very few people pay a lot of attention to how their nozzle fits on the end of the feed line, or how well it is made. Think of this – you have just spent $200,000 on a system, and yet, because the nozzle is a disposable part, you look around for the cheapest source you can get. You don’t size it for a good fluid fit, nor do you check how well it is machined. And yet the entire performance of your system is controlled by that small item. The difference between a very good nozzle and a standard nozzle can give as much as a factor of 10 improvement on performance – but who checks. The one you use saved you $15 relative to what you would have paid if you had bought the competing product, what a bargain – right?
There are different ways in which pumps operate and produce the high-pressure flow. With a fixed size of orifice in the nozzle and with a given pressure drop along the feed line the pressure at the nozzle will be correspondingly reduced. So that if, for example, we use a 0.063 inch diameter nozzle then the chart you developed after generating the table will show that this will carry a flow of 9.84 gpm at 10,000 psi. But let us suppose that the hose looses 20 psi per foot of length, and that the hose is 200 ft long, then the pressure drop along the hose will be 20 x 200 = 4,000 psi.
Thus the pressure of the water coming out of the hose will be only 6,000 psi. And at an orifice of 0.063 inches, the flow through the orifice will now only be 7.62 gpm. (The way in which the pressure is controlled is assumed to be through bypassing extra flow back to the reservoir through a bleed-off circuit).
Now the pump is still putting out 10 gpm at 10,000 psi, but now the flow out of the nozzle is only 7.62 gpm at 6,000 psi. The power in this jet is (7.62 x 6,000/1714) only 26.7 hp. This is only 38% of the energy going into the pump.