A pressure vessel has an internal volume of 0.5 `m^3` at atmospheric pressure. It is desired to test vessel at 3000 bar by pumping water into it. The estimated variation in change of empty volume of container due to pressurization to 3000 bar is 0.6 percent. Calculate mass of water to be pumped into vessel to attain desired pressure level given bulk modulus of water as 2000 Mpa.

SOLUTION:

Given,

Internal volume of pressure vessel at atmospheric pressure, `V_1=0.5 m^3`

Initial atmospheric pressure, `P_1=1.01325` bar

We know, the mass of water required to fill the vessel at atmospheric pressure is,

`m_1=rho_w*V_1=1000**0.5=500 kg`

Final pressure, `P_2=3000` bar

Change in volume of the container`(dV)/V=0.6%=0.006`

Bulk modulus of water, `K=2000`MPa

Now,

The additional quantity of water needed for pressurization is attained due to increase in the vessel volume and compression of water under pressure. 

`m_w=rho_w*v`

Taking natural logarithm on both sides, we get,

`lnm=ln (rho*v)`

or, `ln m=ln rho +ln v`

Differentiating both sides, we get,

`(dm)/m=(drho)/(rho)+(dv)/v`

Now, we know,

Bulk modulus, `K=(dP)/((drho)/rho)`

or,`(drho)/rho=10^5/(2000**10^....Show More

SOLUTION:

Given,

Internal volume of pressure vessel at atmospheric pressure, `V_1=0.5 m^3`

Initial atmospheric pressure, `P_1=1.01325` bar

We know, the mass of water required to fill the vessel at atmospheric pressure is,

`m_1=rho_w*V_1=1000**0.5=500 kg`

Final pressure, `P_2=3000` bar

Change in volume of the container`(dV)/V=0.6%=0.006`

Bulk modulus of water, `K=2000`MPa

Now,

The additional quantity of water needed for pressurization is attained due to increase in the vessel volume and compression of water under pressure. 

`m_w=rho_w*v`

Taking natural logarithm on both sides, we get,

`lnm=ln (rho*v)`

or, `ln m=ln rho +ln v`

Differentiating both sides, we get,

`(dm)/m=(drho)/(rho)+(dv)/v`

Now, we know,

Bulk modulus, `K=(dP)/((drho)/rho)`

or,`(drho)/rho=10^5/(2000**10^6)*(3000-1.01325)`

`=0.15`

Thus,

`(dm)/m=0.15+0.6/100=0.156`

Thus, the .ass of water to be  pumped in the vessel after filling is,

`dm=0.156*m_1=0.156**500=78 kg`