It is pressure difference that causes fluid motion.

@ACuriousMind It was helpful, thanks.

@mb28025 Edited, thanks. I am making a point about an arbitrary basis, whose kets are linearly independent and not necessarily orthogonal or normalized. If you limit yourself to bases obtained from each other by unitary transformations then obviously the inner product is preserved, by definition of a unitary transformation -- not an interesting case for me.

@Prahar Assuming an orthogonal set of kets, the problem of normalization still remains. Please see my comment to Er Jio's answer.

Please see my comment to Er Jio's answer.

Your answer partly clears my confusion. There are still two questions however. (1) Say the basis $\{|a\rangle, |b\rangle\}$ is orthogonal but otherwise arbitrary. We still have the conclusion $\langle a|a \rangle =1=\langle b|b \rangle$ when the components are written in its own basis. This conclusion is violated when we switch basis (see my question for details). (2) How do we even decide if $\{|a\rangle, |b\rangle\}$ are orthogonal? In geometry, to know whether two vectors $u,v$ are orthogonal we have the explicit recipe $g_{ij}u^iv^j$ where $g$ is the metric tensor. What is it in QM?

@Mazura I don't think so. See the comments to joseph h's answer.

Thermodynamics is a field plagued with bad textbooks. I would recommend that you try H.B. Callen's book instead of trying to disentangle what your current book is saying. If you are an engineer then you may like the one by Van Wylen and Sontagg.

It means nothing special. You are only calculating the loss due to bends, fittings etc. and that loss can be any multiple of dynamic/velocity pressure. Larger the loss (larger the $C$) larger the pressure difference needed to maintain a particular flow rate.