Pressure of fluid on bottom of container

[faoltaem]

A cylindrical container with a cross sectional area of 65.2 cm^{2} holds a fluid of density 806kg/m^{3}. At the bottom of the container the pressure is 116 kPa
a) What is the depth of the fluid?
b) Find the pressure at the bottom of the container after an additional 2.05 \times 10^{-3} m^{3} of this fluid is added to the container. Assume no fluid spills out of the container.


a) A = 65.2 cm^{2}
\rho = 806 kg/m^{3}
P(bottom) = 116 kPa = 1.16 \times 10^{5} Pa

closed manometer \rightarrow P = \rhogh

h = \frac{P}{\rho g} = \frac{1.16 \times 10^{5}}{806 \times 9.81}
= 14.67 m


i'm not really sure how to do part (b), this is all that i could come up with

P1 + \frac{1}{2}\rhov1^{2} + \rhogh = constant
P2 + \frac{1}{2}\rhov2^{2} + \rhogh = constant
v2 = v1 + 2.05 \times 10^{-3}

P1 + \frac{1}{2}\rhov1^{2} + \rhogh = P2 + \frac{1}{2}\rhov2^{2} + \rhogh

P1 + \frac{1}{2}\rhov1^{2} = P2 + \frac{1}{2}\rhov2^{2}

P1 + \frac{1}{2}\rhov1^{2} = P2 + \rho(v1 + 2.05 \times 10^{-3})^{2}

P2 = P1 + \rhov1^{2} - \rho(v1 + 2.05 \times 10^{-3})^{2}

= 116 + 403v1^{2} - 403(v1 + 2.05 \times 10^{-3})^{2}


would someone be able to tell me if (a) is correct and possibly a better way to solve (b)
thanks

 

[learningphysics]

Find out how much more depth is added to the fluid... You know the volume added... you know the cross sectional area of the cylinder... you should be able to get the new height and hence the new pressure.

 

[faoltaem]

new pressure

ok so
v = 2.05 x 10^{-3} m^{3}
CSA = 65.2 cm2 = 6.52 x 10^{-3} m^{3}

h = \frac{v}{CSA} =0.3144m

P = \rhogh
= 806 x 9.81 x (14.67 + 0.314)
= 806 x 9.81 x 14.985
= 118 486.05 Pa
= 1.18 x 10^{5} Pa = 118 kPa

thankyou

 

[learningphysics]

ok so
v = 2.05 x 10^{-3} m^{3}
CSA = 65.2 cm2 = 6.52 x 10^{-3} m^{3}

h = \frac{v}{CSA} =0.3144m

P = \rhogh
= 806 x 9.81 x (14.67 + 0.314)
= 806 x 9.81 x 14.985
= 118 486.05 Pa
= 1.18 x 10^{5} Pa = 118 kPa

thankyou

looks good. no prob. you're welcome.