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When a solid object moves through a fluid drag is produced. Does this drag produce heat?

I believe drag should produce heat as it is the friction between fluid and surface of object. Is this true or not?

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  • $\begingroup$ yeah! it is true $\endgroup$ – Wolphram jonny Nov 6 '14 at 5:04
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Actually drag is NOT completely the friction between the object and fluid. Sometimes there can be almost no friction, but still highly significant drag. In a non viscous fluid, there is no friction between the object and the fluid, but there IS still drag. The phenomenon of ram pressure transfers momentum losslessly between the object and the fluid without any friction. I discuss these ideas more fully in my answer to the question "Where does the energy from a parachute go?".

In a low viscosity fluid, the energised fluid will slowly heat owing to friction with itself, not with the dragged body.

So yes, ultimately all the energy transferred to the fluid from the dragged body will wind up as heat, as it must from the first law of thermodynamics, but this is often not through direct friction.

In blunt ended spacecraft atmosphere re-entry, the effect is even more dramatic. The blunt end squashes the air, which cannot get out of the way and the air's temperature rises adiabatically just as a bike pump gets hotter if you hold your finger over the output hole and squash the plunger in suddenly. Very little friction is involved and, as long as the spacecraft can withstand the adiabatic temperature rise, it rides on a cushion of air which thrusts the atmosphere further from the aircraft out of the way. The drag is ram pressure, and the frictional heat (which would be extremely dangerous) is only generate by the fluid dragging on itself through viscous shear, at a good safe distance from the spacecraft. See the links in my other answer for my answer to the question "Where does the energy from a parachute go?".

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The first law of thermodynamics states, that $E_{int}=W+Q$.

Meaning, any work done by (or on) an object results in changing its internal energy (temperature). If, of course, $Q$ doesn't balance it out (this is when $Q=-W$). You need to have a well insulated object (when $Q=0$) and you'll definitely know, that work will result in temperature change.

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