Question

A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the container the pressure is 119 kPa . Assume Pat = 101 kPa
A) What is the depth of the fuild?

B) Find the pressure at the bottom of the container after an additional 2.35×10−3 m3 of this fluid is added to the container. Assume that no fluid spills out of the container.

Answer 1

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

$Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h$

After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

$h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}$

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

$Pb' - Pat = d \times g \times (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000$

Pb' = 122 KPa