Chapter 2, Problem 2.36P

To determine

To find:

The angle of the tilt of the pipe.

Angle of the tilt is as follows:

O = 24.99°=0.4362 rad.

L = 2.13m

SG of oil = 0.8

SG of water = 1.0

Height of the oil, = 50cm

Height of the Water, = 50CM

Lets assume,

The bottom line as reference line,

Point 1 lies at bottom line in the tank,

°Cw = density of water = 997 kg/m³

°Co = density of oil,

H = vertical height of water level in the tube = $\text{Lsin}\theta$

g = acceleration due to gravity.

So, the total pressure at point 1 will be as follows:

P₁ $=o*g*ho+w*g*hw$

P₁ $=So*w*g*ho+w*g*hw$

Putting the values:

$P_1=0.8\times 997\times 9.8\times 0.5+997\times 9.8\times 0.5$

$P_1=8793.54N/m^{2 }$ …………. (1)

Total pressure at bottom of the tank is the cause for the rise of the water level in the tube as shown in the given diagram:

So, we have: -

$P_1=w*g*H$ = $w*g*\text{Lsin}\theta$ = $997\ast 9.8*2.13*\sin \theta$ …………. (2),

From Eqⁿ (1) & (2), we have:

$\sin \theta =0.4225$

$\theta ={\sin }^{-1}(0.4225)$

O = 24.99 degree=0.4362 rad.