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Rotational sprinkler pressure

  1. Jul 27, 2010 #1
    1. whirling lawn sprinklers

    A sprinkler head with a 1.5” dia center hub and two sprinkler jets of 1/8” dia. opening mounted on 3/8” tubing is rotating in a 9” dia. center. There is a ¾”dia hose that is supplying water at 7 gal per min to the sprinkler.

    What is the velocity out of the 1/8”dia jet and the rotational force the jet of water?

    2. Relevant equations
    P = r * g * h
    HP= psi*gpm/1714

    3. The attempt at a solution
    My attempt is missing because Im know sure of the relative equations are correct
  2. jcsd
  3. Jul 31, 2010 #2
    Trying to get a handle on the problem ... what happens if you reach into the middle of the sprinker, and hold the arm so it does not rotate? What would be the volume and velocity of water exiting the 1/8" diameter jets?

    Second query -- it seems reasonable to assume that the water is exiting the jets at right angles to the sprinker arms? Not exactly what I've seen in real sprinklers, but sounds like the problem is assuming right angle exit. Is that the case in the problem statement?
  4. Jul 31, 2010 #3
    Yes, 90 degrees for max forces.

    Yes , What would be the volume and velocity as well as the forces to hold it from turning.
  5. Jul 31, 2010 #4
    Homework, right? So, what do You think would be the volume of water coming out of one of the 1/8" diameter jets?
  6. Aug 1, 2010 #5
    well , it starts at 3/4"dia at 7gal/min and go to 3/8 dia and again to 1/8" dia
    the volume is changing so so the flow rate would too??
  7. Aug 1, 2010 #6
    Ok, so what is the velocity of the water flowing in? 7 gal/min in a 3/4" diam pipe.
  8. Aug 2, 2010 #7
    C=vf(119.8)/A 7gal/min/60sec= .116666gal/sec A = 0.785 di 2 , A=.44156
    C=(.1166gal/sec)(119.8lb/gal)/(.44156 in2)
    C= 6.18in/s
  9. Aug 2, 2010 #8
    Good, thanks! Now we're started. I'll comment on individual parts of your calculation.

    A = cross-section area of 3/4" pipe = .785 diam^2 = .44 in^2 ... ok
    (You can carry many sig figures at intermediate steps if you really want to, but not necessary to carry more than 2-3 digits, because final answers will have only 2 digits. Measurements like 7 gal/min or 3/4" diameter are at best only plus or minus 1 percent, and probably have much more slop than that.)

    Flow rate is 7 gal/min / (60 sec/min) = .1166 gal/sec ... ok

    The volume of flow figure you are using, 119.8, does not seem right. I think you meant to write down the volume of 1 gallon of water, in cubic inches. That would be 231 in^3.

    That is if you are working with US gallons. The problem statement does not say. If using imperial gallons, the volume would be higher. Ah, that is possibly where the 119.8 figure comes from. One imperial gallon is about 1.2 times one US gallon, so perhaps a conversion factor for going to imperial gallons has crept in?

    Anyway, your line #2 has a units error - lb/gal is not correct units for what you wanted to end up with in line #3. You would need (some number) of cubic inches / gallon. I'm going to assume you are working with US gallons, which are 231 cubic inches.

    So your line #2 becomes (with correction)

    C=(.1166gal/sec)(231 in3/gal)/(.44156 in2)

    Notice that the units balance: (gal/sec) * (in3/gal) / (in2) = in/sec

    Line 3 is then C = velocity of water in 3/4" hose = 61 in/sec

    You have actually got the right figures in your line 3, though off by a factor of 10 from what I calculated. Perhaps you had more written down than you wrote in your post?

    Anyway, let's move on. Think about what's going on in the sprinkler. No formulas for the moment.

    7 gal/min of water is coming in thru the hose. Where is it going? What is the flow thru each arm of the sprinkler?

    Now, using similar methods to what was done above, what is the velocity (inches/sec) of the water in the 3/8" tubes of each sprinkler arm? What is the velocity thru each of the 1/8" jet openings?
  10. Aug 2, 2010 #9
    ¾ Hose A = 0.785 di 2 , A=.785(.752) , A=.44156
    C1=.1166gal/sec(231in3/gal)/.44156in2 = 61 in/sec
    Now, 3/8 tube= .375 A = 0.785 di 2 , A=.785(.3752) , A=.1104
    C2=.1166gal/sec(231in3/gal)/.1104in2 = 272.7 in/sec
    And 1/8 tube= .125 A = 0.785 di 2 , A=.785(.1252) , A=.0123
    C3=.1166gal/sec(231in3/gal)/.0123in2 = 2189.8 in/sec

    I see the velocity is increasing a lot but is the force increasing or constant? Ho wait, there are two tubes so the velocity and forces are cut in half, so
    272.7/2=136.35in/sec & 2189.8/2=1094.9in/sec
    well, the forces related to velocity are Velocity= power/force. What would you use for power to find the forces?
  11. Aug 2, 2010 #10
    Great! I agree with your numbers.

    What I would do is forget about power, and just use F=ma. Think of the jet of a sprinkler like a rocket engine. Every second, it is throwing out some quantity of mass, at some velocity. That is an increase in momentum, ie delta mv. Delta mv is ma = applied force.

    The velocity of the water coming out the jet is being measured with respect to the jet nozzle, ie it is a moving reference frame. But that is correct. Whether the jet moves or not, or how fast it is already moving, is not related to the new force it is experiencing. Think of a rocket engine bolted to a test frame; when you fire the rocket engine, you observe a force, even if the rocket engine does not move (because it is fastened to the earth, which has lots of mass).

    This sprinkler problem is actually pretty tricky. It looks simple on the surface, but there are subtleties. You probably feel those intuitively, and the problem becomes complex in the mind. But ... the good news ... the textbook question has focused on two aspects which are straightforward, so if you ignore the subtleties of how the nozzle might rotate, you can determine the water velocity out the nozzle, and the applied force. The way I like to think of it, is if you were to hold the sprinkler so it could not rotate, and do the calculations for that situation. Like the rocket engine bolted to its test frame.
  12. Aug 2, 2010 #11
    That is an increase in momentum, ie delta mv. Delta mv is ma = applied force.

    F=mv/t or p(momentum) = mv 1 gallon of water is equal to 8.33 lbs
    F= (.1166gal/sec) 1094.9in/sec *(8.33lbs/1gal) = 1063.45lb*in/sec
    1 (pound * in) / second = 0.0115212462 m kg / s----------------12.25m*kg/sec
    at this point Im not sure, is this newtons?
  13. Aug 2, 2010 #12
    And you are there! Yes, newtons. You dropped a /sec unit in the middle of your workout,
    which should have ended up with pound*in / second^2 or m*kg/sec^2, which is newtons.

    Rearrange m*kg / sec^2 as kg * (m/sec^2) and you can see it is mass * acceleration, which equals force, which is in newtons if you are working in meters, kg, secs. 1 newton is force that causes a mass of 1 kg to be accelerated by 1 m/sec^2.

  14. Aug 2, 2010 #13
    So to stop the one single jet, I would have to apply 12.25 Newtons it seems a lot?
    or 24.5 to stop the sprinkler.

    I appreciate your help greatly
  15. Aug 3, 2010 #14
    Yes, it does seem a lot of force. 1 newton = 0.225 pound, so it would be about 11 pounds. That's what the calculation came out as, and I agree with it, but I may be wrong!

    Yesterday, just for fun, I fooled around with the garden hose and sprinkler. The problem is overstated in a few regards. My hose flow rate, with the tap on full, was only 5.1 gal/min. The sprinkler arms are not as imagined in the problem statement, either. Mine has three arms, and each arm has 7 pinhole (1.2 mm) openings - one of them along the arm, pointing upwards, producing a spray at about 80 degrees inclination to the horiziontal, and the other six at end of arm. The arm has a bend, so it points up about 20 degrees, and also backwards about 20 degrees. The sprays out of the arm end are between 20 and 40 degrees inclination to the horizontal.

    So, much of the force is being exerted downwards - it holds the sprinkler in place. And other force is directed outward. Only part of the force is causing rotation of the arms.

    Unfortunately I don't have a suitable setup to measure the sprinker forces directly.

    Still, I think 11 pounds (5.5 pounds per each of the two arms) is not reasonable. Maybe someday I'll get some more measurements from an actual sprinkler.

    I'll be interested in hearing what feedback you get from others, re the sprinkler problem.

    Best wishes,
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