In the derivation of Bernoulli's equation
$$W_{nc} = (P_1 – P_2)V $$
$$W_{nc} = \Delta KE + \Delta PE $$
$$(P_1 – P_2)V = 1/2mv_2^2 – 1/2mv_1^2 + mgh_2 – mgh_1 $$
$$P_1V + 1/2mv_1^2 + mgh_1 = P_2V + 1/2mv_2^2 + mgh_2$$
$$P_1 + 1/2\rho v_1^2 + \rho gh_1 = P_2 + 1/2\rho v_2^2 + \rho gh_2$$
I have read that this is just conservation of energy, but how is energy conserved even though a non conservative force ((P1-P2)V) is applied on the system? Won't just the existence of the non conservative work not make the energy of the system conserved? Bernoulli's equation is definitely mathematically true(and very useful), but I'm finding it hard to accept that 'energy is conserved'.
Best Answer
It depends on the energies you are considering.
You're right in the "introductory mechanics" sense, energy is conserved when $\Delta E=\Delta K+\Delta U=0$ for a system.
However, in this case the work is being done by the force(s) associated with the pressure. So one can include this in a change in total "energy" of the system. Then we have a conserved quantity: $$\Delta E=\Delta K+\Delta U-(P_1-P_2)V=0$$
This quantity is conserved because the work done by the fluid pressure goes into changing its kinetic and potential energy.
Of course this means that the claim that Bernoulli's principle is equivalent to energy conservation is not entirely true, but one can still fudge the wording around a bit and people will usually still know what you mean by it.