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Forces on the Fluid Element

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I came across this in Fluid Mechanics book by Prof F.M.White. Here he gave an expression for the Surface Pressure acting over a fluid element. I have great trouble understanding the expression. I would be grateful if somebody can help me understand the mathematical idea behind it.

Thank You!

Jenkins
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The net force in the $x$-direction due to pressure (in excess of the reference pressure $p$) acting on the circular cross-section on the extreme right is clearly $-dp \,(A +dA)$ -- the product of the excess pressure $dp$, assumed constant here, times the cross-sectional area $A + dA$.

If $\theta $ is the angle between the lateral surface and the $x$-axis, then the lateral surface area is $dA/\sin \theta$. The excess pressure can be assumed to vary linearly in this approximation for small $dA$ from $0$ on the extreme left to $dp$ on the extreme right. The average excess pressure is then $dp/2$ and the net force in the $x$-direction is given by

$$dF_{x, \,\text{side}}= \frac{1}{2} \ dp \cdot \frac{dA}{\sin \theta} \cdot \sin \theta = \frac{1}{2} \ dp \ dA,$$

where the last multiplication by $\sin \theta $ is used to obtain the component of the lateral force in the $x$-direction.

Adding together we get

$$dF_x = \frac{1}{2} \ dp \ dA- dp\,(A + dA) = -A \,dp - \frac{1}{2} \, dp \, dA$$

Assuming that the second term on the RHS is negligible with respect to the first term, since it involves a product of small quantities $dp \cdot dA$, we get $dF_x \approx - A \, dp$.

RRL
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  • Hi, perhaps I found an answer at a my question but I would like have the opinion of an person with more experience than me: so could I ask your opinion about my this answer, please? Excuse me for the bother. – Antonio Maria Di Mauro Sep 04 '20 at 22:31
  • Sorry to trouble you sir, but can you elaborate this line, "If θ is the angle between the lateral surface and the x-axis, then the lateral surface area is dA/sinθ." I mean A and A + dA are cross sectional areas of the two opposite faces of the frustum right? How can dA be related to the lateral surface area of the frustum? Even if you just mention the underlying mathematical property behind the statement, it would be extremely helpful to me. Have a nice day! – Jenkins Sep 06 '20 at 06:41
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    @Jenkins: The small and large base of the frustum have radius $R$ and $R+dR$, respectively. The lateral surface area is $S = \pi \cdot (R + (R+dR)) \cdot L$ where $L$ is the slant height. The slant height is the hypotenuse of the triangle with $dR$ as the length of the side opposite $\theta$, so $L = dR/\sin \theta$. Thus, $S = (2 \pi R + \pi R) (dR / \sin \theta) = (2\pi Rd R + \pi dR^2)/\sin \theta$. Note that $$dA = \pi(R+dR)^2 - \pi R^2 = 2 \pi R dR + \pi dR^2 $$ Hence, $S = dA / \sin \theta$. – RRL Sep 06 '20 at 16:29
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    The lateral surface area of a conical frustum is given here – RRL Sep 06 '20 at 16:37