# Can anyone explain calculation of net force due to pressure in a streamtube of fixed control volume and infinitesmal size

I am studying derivation of bernoulli equation from FM White's fluid mechanics and in that pressure distribution is as shown in the image 1. I am unable to understand force calculation as shown in image .

There seem to me to be three main things that appear initially confusing about the derivation. I believe they both have to do with the $$\frac{1}{2} dpdA$$ term. I think the $$-dp(A+dA)$$ term is quite self explanatory.

I think the three main points are:

(1) Why is the term of the form $$\frac{1}{2} dpdA$$

(2) Why doesn't the $$dA$$ term take into account the geometry of our streamtube (i.e. why doesn't the area use the formula for the side surface area of a frustum instead of using $$dA$$).

(3) Why doesn't the resultant force take into account the angle of deviation of the side of the streamtube when calculating the force, (i.e. why is the term $$\frac{1}{2} dpdA$$ instead of $$\frac{1}{2} dpdA \theta$$ where $$\theta$$ is the angle of deviation from the axis of symmetry of the streamtube.)

The answer to part (1) is that when we remove the constant pressure $$P$$ acting on all surfaces, we notice that we are left with a linearly increasing pressure, from the 'left' of streamtube, to the 'right'. If you are familiar with distributed loads from structural mechanics, the general formula for calculating the average force across a section with varying force, is $$\frac{1}{2}Fl$$, where $$F$$ is the force distribution and $$l$$ is the length of our section. This is analogous to how we treat the pressure, i.e. think of it as a 'pressure distribution' of sorts. The average pressure is then just $$\frac{1}{2}dp$$. Note here that we merely assume a linear increase of the pressure across the volume. (I do not know the answer for why this is a valid assumption for full disclosure.)

To answer part (2), This is simply because we are dealing with such small changes in area the actual geometry doesn't figure into the derivation. An extremely tiny change in surface area of an extremely tiny cube is exactly analogous to an extremely tiny change in area of a frustum for example.

The answer to part (3) is simply that pressure is isotropic. Thus regardless of the angle, we know the force in the direction of $$S$$ will still be $$\boldsymbol{n}p\partial S$$ where $$\boldsymbol{n}$$ is the unit vector in the specified direction, $$p$$ is the pressure, and $$\partial S$$ is the surface element. Have a look at this http://www.atm.damtp.cam.ac.uk/people/mem/FLUIDS-IB/dyn.pdf which will give a much better (and detailed) explanation than I have here.

Hopefully this covers the more odd circumstances (they were at least the most peculiar aspects of the formula to me ). Note I assumed the other term $$-dp(A+dA)$$ is straightforward; it is simply an expression of force acting on the 'right' side surface of the element.

Hope this helps somewhat.