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Kerbin Dallas Multipass

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Posts posted by Kerbin Dallas Multipass

    Who will solve this first ?

    in Science & Spaceflight

    Posted

    Now work out how much energy is in 1 L of water flowing at .5 m/s. Now work out how much power it takes to supply that much energy to 500 L of water every second.

    You seem to be missing the part where the pump power is the energy output per second, which is equal to the energy imparted to each 500 L water.

    I imagine it is significantly less than 1 MW?

    Who will solve this first ?

    in Science & Spaceflight

    Posted

    Guys, how about I just show him every single step, because I don't think he quite gets what the units have to do with it.

    500 L/s

    10^6 W

    Watch me do magic here.

    1000000 W = 1000000 J/s

    1000000 J/s / 500 L/s

    Okay, take the numbers: 2000

    Now take the units: J/s / L/s. Let's simplify. J/s * s/L = J/L

    2000 J/L. That's how much energy that is delivered to each liter. This is because the amount of work depends solely on how much water is going through at once, and how much power is supplied, as that determines how much work is supplied to each liter. Notably, this ratio is determined by the size of the hole through which the water exits, however, knowing how big the hole is is not required if you already know the throughput and the power (because then you know their ratio already, and the only thing that matters is that ratio).

    When you play with infinities, things break down because you hit a mathematically singularity here, and things don't work right at all then. Just know that larger holes require less power to force the same throughput as smaller holes.

    Moving on,

    1 Liter of water is also 1 kg of water, due to its density.

    So, 2000 J/kg.

    dU = m * g * h. Or dU/m = g * h

    Assume 1 kg of water, it doesn't matter, because it instantly simplifies out. 2000J/kg * 1 kg/1 kg = 9.81 * h.

    h = 203.87.

    Thank you for explaining that. It makes sense. And it answers why both, the flow rate and the power are described in the task.

    But is it realistic? As I illustrated above, pumping 500l/s through a 1m² pipe will not get us anywhere near a 200m fountain, independent of the power at hand. I think that confused me.

    If I were to just write down "Since we're pumping water, it depends on the nozzle"... would that be wrong? (I see it's not the desired answer, but isn't that true?)

    Who will solve this first ?

    in Science & Spaceflight

    Posted · Edited by Kerbin Dallas Multipass

    The pump has a power that it supplies to the water. Obviously the speed that the water is accelerate to with this given power depends on how many litres the power is supplied to.

    No.

    (charlimit makes me post this, too.)

    Who will solve this first ?

    in Science & Spaceflight

    Posted

    I guess I'm nitpicking here.

    The question "What height does the water reach" should be "what maximum height can the water reach".

    If that is the actual question then there is no point in mentioning the 500l/s

    If I force 500l/s through an infinitely small hole i theoretically get infinite speed. If I don't have infinite power, I obviously can not do that (the power at hand is stated).

    So what is the 500l/s mention for? is that a distraction?

    Who will solve this first ?

    in Science & Spaceflight

    Posted

    No, the volume flow is .5m3/s

    Since the height is 200m (assuming g=10m/s like the rest of the thread), velocity has to be 63 m/s.

    For .5m3/s that means the area of the muzzle is 0.0079m2 and that puts the diameter at 10cm if I'm right.

    Passing .5m³ through a 1m² opening in 1 second gives me a flow speed of .5m/s.

    The 200m height is not specified in the question, neither is the nozzle size.

    - - - Updated - - -

    Attach a 100KW pump to that nozzle, and you'll get much more than 500l/s of flow.

    Yea but why is the 500l/s specified in the question?

    Dawn at Ceres Thread

    in Science & Spaceflight

    Posted · Edited by Kerbin Dallas Multipass

    WP states that the entire project was cancelled numerous times, So maybe the tech on board is just a bit old. Also consider that 1 MPixel doesn't even sound so bad regarding that Hubble had just 2.56 MP.

    This is what the cameras look like btw:

    Kamera_0e71cfe92c.jpg

    (notice all the space tape)

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