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."