- #1

decentfellow

- 130

- 1

## Homework Statement

A cylindrical vessel of height ##H## and radius ##R## contains liquid of density ##\rho##. Determine the circumferential tension at a height ##h##; also determine its maximum and minimum values.

This is the scan of the original question(solved), I couldn't understand what was done here so I tried to solve it by myself and it has frustrated me so much that I am about to cry.

## Homework Equations

## The Attempt at a Solution

So my line of thought was that what would have been the case when there was no liquid filled inside it, would there still be any kind of circumferential tension. So, what I want to ask here is that is there any kind of manufacturing stress present in the cylinder due to its curved shape. This has me stuck for quite some time.

Also, I tried attempting the question as follows:-

Consider a differential ring of height ##dh## situated a depth of ##h## as shown below:-

Now due to pressure of the liquid the force exerted on the walls of the differential element by water is given by

$$F_{\text{liq.}}=P(dh)(Rd\theta) \qquad \& \qquad P=\rho gh$$

Now, here is the part where I am having a doubt as to how to determine the tension that acts on the differential cylindrical element due to the expansion that occurs due to the pressure exerted by the liquid. I was confused b/w two cases:-

1. If there was already tension along the circumference due to the shape of the cylinder before it was filled with liquid. If so is the case then, does the expansion due to the liquid pressure add to it.

2. If there was no tension when there was no pressure due to the liquid(i.e. there was vacuum inside the cylinder), then after adding the liquid the cylinder expands due to the pressure exerted by the liquid, hence tension acts to not let the vessel expand more.

I don't know how to proceed with case-1, because I don't know how does the tension act in a cylinder which has no internal pressure due to any kind of fluid.

But with case-2, the forces that start acting on the circumference are as shown in the diagram below:-

After all the changes that can occur in the dimensions of the cylinder has occurred there will be static equilibrium, hence we get:-

$$2T\sin\left(\frac{d\theta}{2}\right)=P(dh)(Rd\theta)\\

\implies \require{cancel} T\cancel{d\theta}=P(dh)R\cancel{d\theta}\implies T=\rho Rg\;h\;dh$$

So, it seems that the circumferential tension depends on the differential height. That seems weird.

Also, doesn't the cylinder should have some thickness or else when it expands it will break immediately. Do correct me if my reasoning is wrong.

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