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My problem involves an object completely immersed in a fluid, moving at some speed, using a certain amount of power to sustain that speed. It could be an airplane or a torpedo, among other things.

(This problem originally arose from trying to calculate certain things about ships and torpedoes, so I choose the torpedo.)

The torpedo has mass m, speed v, and power P being applied by the engine.

What I want to know is, if m is doubled, but v is the same, how does P change? I cannot find a formula for this anywhere.

(The torpedo also has exactly the same shape, so drag from water is the same.)

In other words: How much more power is needed to sustain the same speed of an object that's twice as heavy, all else being equal?


If we need concrete numbers, I'll just choose these: m = 1,000 kg, v = 10 m/s. Just let C and A be 1. Therefore drag = 50 kN and power = 500 kW.

Formulas I know:

$F_D = \frac{1}{2} \rho v^2 CA$

$P = \frac{1}{2} \rho v^3CA$

These formulas relate drag force and power to fluid density, speed, a coefficient of drag, and cross-section area. They do not involve mass (m) anywhere.

I'm confused because there are two ways of thinking about this, that I can see, that conflict with each other. On the one hand, Kinetic Energy is just half the mass times v squared. So to sustain the KE (against drag) of a doubled weight, it seems twice the power is necessary.

On the other hand, drag force in my problem is the same, so the power to overcome that force should be the same. But it doesn't make sense that you can get twice the weight "for free". It seems there must be more power needed to push it.

So I don't get it. Is there some kind of summed formula combining two ways of measuring power? I also know that power multiplied by time gives you a unit equivalent to energy. Not sure if this is a clue or not.

Since speed is the same, Reynolds number is the same, so hopefully we should be able to avoid "complex fluid dynamics things."

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  • $\begingroup$ There is power required to maintain your kinetic energy, which depends on mass. The fluid effects, such as drag, only depend on the shape and size of your object. The mass of your object does not influence the flow field. $\endgroup$ Dec 1, 2018 at 21:35

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On the one hand, Kinetic Energy is just half the mass times v squared. So to sustain the KE of a doubled weight, it seems twice the power is necessary.

That's the most common misconception about Newtonian mechanics, by far. In Newtonian mechanics, in the absence of any forces, objects will just continue moving with the same velocity. But there's a stubbornly persistent misconception that it "costs force" to keep the momentum the same. People will say that an object stops moving because it "runs out of force".

You're doing the same thing, but instead of mixing up momentum with force you're mixing up energy with power. Having high energy doesn't require high power. A rock standing on a high ledge has a lot of gravitational potential energy, and that doesn't cost any power at all. It just sits there.

Similarly, it costs more energy to get a heavier object moving, but once it is moving, that doesn't matter. The power needed to keep it going is simply $P = Fv$, and if the velocities $v$ are the same, and the drag forces $F$ are the same, then the powers are the same.

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  • $\begingroup$ In Newtonian mechanics, in the absence of any forces, objects will just continue moving with the same velocity. Right, I understand the first law. What I meant was sustaining the KE against the drag force, which I just figured needed some power to pump in energy to the KE to increase it, whereas drag pumps in energy to decrease it. I should make an edit. $\endgroup$
    – DrZ214
    Dec 1, 2018 at 21:41
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    $\begingroup$ @DrZ214 The drag force is associated with a negative power $Fv = -\rho v^3 CA/2$. To keep it going at the same speed, you need to put in positive power $+\rho v^3 CA/2$. $\endgroup$
    – knzhou
    Dec 1, 2018 at 21:42
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These formulas relate drag force and power to fluid density, speed, a coefficient of drag, and cross-section area. They do not involve mass (m) anywhere.

They don't because they don't need to.

As the torpedoes travel at constant speed $v$, there is no net force acting on them (2nd Law of Newton): the drag force exerted by the water equals the trust delivered by the motor, in both cases.

Since as the drag force does not depend on the mass of the torpedo, only on the shape of the torpedo, the viscosity of the water and the torpedo's speed $v$, the power is invariant of the torpedo mass.

Mass comes into pay to determine the amount of energy needed to accelerate the torpedoes from $0$ to $v$ but that is not the situation here: there is no acceleration in play.

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When calculating the power required to keep the torpedo at the same speed, you're actually calculating how the kinetic energy is changing over time due to the drag force. This quantity doesn't have any dependence on the total kinetic energy of the torpedo, in the same way that the lines $10-x$ and $100-x$ have the same slope. It only depends on the force. Since the force to be overcome is the same, the required power is the same.

What might be confusing you is the following: it will indeed take roughly twice as much energy to accelerate the torpedo from rest to the speed v if the mass was doubled. But once it is already at that speed, the amount of power required to keep the torpedo moving at speed v is the same.

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  • $\begingroup$ *it will indeed take twice as much energy to accelerate the torpedo from rest to the speed v if the mass was doubled. * Only if you disregard the drag. $\endgroup$
    – Gert
    Dec 1, 2018 at 21:53
  • $\begingroup$ @Gert That is correct. The real ratio depends on what exactly the accelerating forces are in each case, and it's only exactly twice if the acceleration is the same in each case. $\endgroup$ Dec 1, 2018 at 21:56

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