After reading this answer: http://www.answers.com/Q/What_does_1714_mean_in_hydraulics, I still do not have a good idea of what 1714 represents. In fact, in the equations I was working with I saw a constant 0.0005834 and it took me until just now to realize that it is a another way to say 1/1714.

So what is this magical 1714, what does it represent, why is it needed?

In short I am looking for "1714 represents ... " in plain English that a 9th grader can understand, and yet be correct and true to its purpose and place in life to satisfy a seasoned hydraulic engineer.


It's just an artifact of using different units for the same types of quantities in the same equation.

Suppose we have an ideal pump that puts a force $F$ on a fluid to have it move at a steady velocity $v$. The power required to do this is $$ P = Fv. $$ Since pressure $p$ is force per unit area, then a flow with cross sectional area $A$ has $F = Ap$. At the same time, the volumetric flow rate is $q = Av$, so $v = q/A$. Thus we have $$ P = pq. $$

This formula is right without modification, but of course you should be consistent when you plug in dimensional quantities. The units involved are \begin{align} P & \sim \mathrm{(mass)} \cdot \mathrm{(length)}^2 \cdot \mathrm{(time)}^{-3} \\ p & \sim \mathrm{(mass)} \cdot \mathrm{(length)}^{-1} \cdot \mathrm{(time)}^{-2} \\ q & \sim \mathrm{(length)}^3 \cdot \mathrm{(time)}^{-1} \end{align} If you choose a single base unit each for mass, length, and time, and you measure $P$, $p$, and $q$ according to the above products of those base units with no additional factors, then there will be no $1714$. This is the case, for example, in SI units, where $P$ is measured in watts, $p$ in pascals, and $q$ in cubic meters per second.

Unfortunately, hydraulic engineers once used (still use? apparently?) hodgepodge units. In this case we can write $$ \frac{P}{1\ \mathrm{horspower}} = k \left(\frac{p}{1\ \mathrm{lb./in.^2}}\right) \left(\frac{q}{1\ \mathrm{gal./min.}}\right), $$ where $$ k = \frac{(1\ \mathrm{lb./in.^2})(1\ \mathrm{gal./min.})}{1\ \mathrm{horsepower}}. $$

Now \begin{align} 1\ \mathrm{horsepower} & \approx 745.7\ \mathrm{W} = 745.7\ \mathrm{kg\,m^2/s^3}, \\ 1\ \mathrm{lb./in.^2} & \approx \frac{4.448\ \mathrm{N}}{(2.54\times10^{-2}\ \mathrm{m})^2} \approx 6.895\times10^3\ \mathrm{kg/m\,s^2}, \text{ and} \\ 1\ \mathrm{gal./min.} & \approx \frac{3.785\ \mathrm{m^3}}{60\ \mathrm{s}} \approx 6.309\times10^{-5}\ \mathrm{m^3/s}, \\ \end{align} so $$ k \approx \frac{(6.895\times10^3\ \mathrm{kg/m\,s^2})(6.309\times10^{-5}\ \mathrm{m^3/s})}{745.7\ \mathrm{kg\,m^2/s^3}} \approx 5.834\times10^{-4} \approx \frac{1}{1714}. $$ That is, our equation $P = pq$ is equivalent to "power in horsepower divided by pressure in psi and flow rate in gpm is 1/1714."

  • $\begingroup$ question: I am reading and rereading the answers, but one thing puzzles me still that I cannot yet put into words, and maybe it's already in the answer, but -- if we have P = pq all in SI units, where there is no need for k, and then we put parenthesis around each P, p, and q, and convert them each to English/non-SI units individually, then such conversion will be equivalent, and the equation will (should) still hold, in which case we won't need the factor k, right? Where does k go in that case? oh .. Or is it the case that Power in SI has different units than that of non-SI? $\endgroup$ – Dennis Jan 5 '15 at 20:30
  • $\begingroup$ I think I may be getting it ... to give a very crude (non-reality based) example in SI power is measured using mass/length/time but in non-SI someone decided to use mass/length only, so to adjust for lack or add-on of physical quantities on the sides of equation the factor k was introduced. $\endgroup$ – Dennis Jan 5 '15 at 20:33
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    $\begingroup$ @Dennis The problem is that $1\ \mathrm{HP}$ is not the same magnitude as $(1\ \mathrm{psi})(1\ \mathrm{gpm})$, even though they both have dimensions of $\mathrm{(mass)(length)^2/(time)^3}$. You could use Imperial units such that $k = 1$; one such way is to measure pressure in psi, flow rate in cubic inches per second, and power in inch-pounds per second. $\endgroup$ – user10851 Jan 5 '15 at 20:41
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    $\begingroup$ The same thing occurs in any unit system. You could measure lengths in meters on one side of the equation and in light years on the other side, and then the dimensionless (but not equal to 1) quantity $(1\ \mathrm{ly})/(1\ \mathrm{m})\approx9.5\times10^{15}$ will probably appear somewhere in the equation. $\endgroup$ – user10851 Jan 5 '15 at 20:44
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    $\begingroup$ Historical side note: Watt had to define the unit "horsepower" because he priced his engines based on how much mines would save by replacing horses with his engines. $\endgroup$ – slebetman Jan 6 '15 at 3:56

1 horse power = 33000 foot-pounds per minute (by definition)

1 US gallon = 231 cubic inch (by definition)

1 psi = 1 pound per square inch (by definition)

In the equation

$$HP = k \Delta P F$$

where $F$ = flow rate in gallons per minute, $\Delta P$ is pressure difference in psi, and $HP$ is power in HP, you need a conversion factor. Doing everything in inches:

$$\frac{HP}{33000*12 *inch-pounds/min} = k\frac{\Delta P}{pounds/in^2}\frac{flow}{231 inch^3/min}$$

from which it follows that

$$k = \frac{231}{33000*12}\approx\frac{1}{1714}$$

In other words, "in plain English that a 9th grader can understand, and yet be correct and true to its purpose and place in life to satisfy seasoned physicists":

1714 represents the numerical scale factor needed to obtain pump power in HP given pressure in units of psi and flow rate in gallons per minute. It is not an exact number - only approximate."

Note - I had to hold my nose a bit while writing this answer, as working with non-SI units does not come naturally to me. But I think that it's a fair question - and NASA put a man on the moon with this system of units. No SI-based operation ever put a man on the moon. So here's to you, NASA!

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    $\begingroup$ Apollo's computers used SI internally, actually. $\endgroup$ – Félix Saparelli Jan 5 '15 at 19:44
  • $\begingroup$ @FélixSaparelli I was thinking about this report - the log of Apollo 11's flight; it's written in US imperial units. In the minds of the clever people doing the work, these were the units used. I can't comment on the "computers" on board the space ship. $\endgroup$ – Floris Jan 5 '15 at 19:52
  • $\begingroup$ wow very nice.. so the "exact" value of "1714" by definition is actually 1714 and 2/7. Aside from that, my confusion was mostly due to that when doing things in general, like converting liters into gallons, or cubic meters into cubic inches, there is no need for a factor. But in this case there is a need for a factor. That I have had a hard time understanding. $\endgroup$ – Dennis Jan 5 '15 at 20:08
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    $\begingroup$ Yes - I think it is indeed exactly $1714\frac27$. Any time you are converting units you need a factor (about 4 liters to a gallon - that's a factor). Sometimes the factor is "easy": kilogram to gram: times 1000. Sometimes it is less obvious. You might amuse yourself by determining the units of this "conversion factor". $\endgroup$ – Floris Jan 5 '15 at 21:09
  • $\begingroup$ And a mixture of SI and Imperial units put the Mars Climate Orbiter where? Who knows... $\endgroup$ – DJohnM Jan 5 '15 at 23:53

So what is this magical 1714, what does it represent, why is it needed?

Some people use archaic units. That's all it means.

Suppose you measured pressure in pascals, or newtons per square meter, flow rate in cubic meters per second, and power in watts. With these units, power = pressure * flow rate. There is no scale factor.

A similar thing happens to Newton's second law. The sane thing to do is to express force in newtons, mass in kilograms, and acceleration in meters/second2. This results in the very nice form of Newton's second law, $F=ma$. Those who one insist on using customary units, with force in pounds force, mass in pounds mass, acceleration in feet/second2 have to deal with the uglier $F=kma$, where $k=1/32.174049$. That 1/32.174049 is just a consequence of using inconsistent units. The exact same applies to your magical 1714.

  • $\begingroup$ Archaic systems causing confusion again. That's pretty much all there is to say about it. $\endgroup$ – Mast Jan 5 '15 at 21:45

protected by Qmechanic Jan 6 '15 at 12:27

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