Question

how much fluid required to raise the pressure in thin cylinder of length 3 m internal diameter 0.7 m an the wall thickness 12 mm by 0.7 bar E=210 GN/m2 and v=0.3 for a material of cylinder and k=2.1 GN/m2 for fluid

Answer 1

Given parameters are,

internal diameter= 0.7m

length of cylinder= 3m

wall thickness= 12mm= 0.012m

Pressure to raise= 0.7 bar= 70 MN/m²

poisson's ratio= v= 0.3

youngsmodulous = E = 210GN/m

Bulkmodulous 2.1GN/1 m

So for thin cylinders, two kind of stresses are induced

1) circumferential stress and 2) longitudnal stress

so, circumferential stress is given as

on = pd/21

Oh 70 * 103 +0.7m/2 * 0.012

on = 2.0416N/m

Likewise , longitudnal stress is given by

$\sigma _l=pd/4t$

$\sigma _l=70*10^3*0.7m/4*0.012m$

$\sigma _l= 1.02 GN/m^2$

Similarly, as stresses develop , the following effect that is strain will also come into existence.

so, there are two kinds of strains develop in thin cylinder.

1) Circumferential strain and 2) longitudnal strain

so, circumferential strain is given as $\epsilon _h= 1/E (\sigma _l-v\sigma _h)$

so, $\epsilon _h= (2.041-1.02*0.3)*10^6/210*10^6$

$\epsilon _h= 0.00826m$

Similarly calculating longitudnal strain,

$\epsilon _l= 1/E(\sigma _h-\upsilon \sigma _l)$

$\epsilon _h= (1.02- 0.3*2.041)*10^6/210*10^6$

$\epsilon _h= 0.00200$

so, now change in volume due to pressure is given by,

$\delta v= pd.V(5-4\upsilon )/4tE$

where V= original volume given as $=\pi r^2l=$

so, $\delta v= 70*10^3*0.7*2.70(5-4*0.3)/4*0.012*210*10^6$

$\delta v= 50*10^-^3m^3$

Finally , Bulk Modulous is given as $K= pV/\delta v$

from here we will calculate the required volume of fluid due to change in pressure given as

$V=k\delta v/p$

Hence , it is the required answer.