Question

Fluid Statics - Manometer Question.

Can anyone help me?

From this diagram http://imgur.com/HeceaBS , I need to derive an expression for the density of the oil in terms of a, b and c.

I have no idea how to do this and I need to learn it.
Thanks

Answer 1

We can determine the density of the oil by using the column heights a, b, and c given that we know the density of water.

Recall that a column of a substance exerts a pressure. This pressure proportional of the column height, h, and the density of the fluid in the column. It is independent on the diameter of the column. Let's see why.


The pressure exerted by a column of liquid is expressed as\[P = \rho gh\] where \(\rho\) is the density of the liquid in the column, g is gravitational acceleration, and h is the height of the column.

This is derived directly from the definition of pressure\[\rm Pressure = {\rm Force \over \rm Area} = {mg \over A}\]
Density is defined as\[\rho = {\rm mass \over volume}\]
Therefore, \[P = {\rho \cdot V \cdot g \over A}\]
Volume of a column can be expressed as \[V = A \cdot h\]
This yields, \[P = \rho g h\]

Since this manometer has stationary fluid, we know that the pressure exerted by the left hand column is equal to the pressure exerted by the right hand column.
\[P_L = P_R\]

The left hand column is composed entirely of water and has a height of "a". Therefore, \[P_L= \rho_w g a\]

The right hand column is made up of a column of water, with height "b", and a column of oil, with height "c."
\[P_R = \rho_w g b + \rho_o g c\]

Setting these equal to each other. \[\rho_w g a = \rho_w g b + \rho_o g c\]With a little Algebra magic\[\rho_w (a -b) = \rho_o c\]
Expressing in terms of the density of the oil\[\rho_o = {\rho_w (a-b) \over c}\]

Answer 2

Thank you so much! This was great help.