Chapter 3, Problem 3.1PP

Write the expression for computing the pressure in a fluid.

To determine

Expression of pressure in a fluid

Explanation of Solution

Given:-

Density of fluid and depth of fluid.

In a fluid, if the fluid is stationary, the state of pressure at a point may be defined using pascal's law and hydrostatic law.

According to pascal's law, In a fluid, at a particular point, pressure acts uniformly and is equal in all the directions.

   $\begin{array}{l}\text{At a point in the fluid, }\\ {\text{pressure P = Pressure in x direction, P}}_{\text{x}}{\text{ = Pressure in y direction, P}}_{\text{y}}{\text{ = Pressure in z direction, P}}_{\text{z}}\\ {\text{P= P}}_{\text{x}}{\text{ = P}}_{\text{y}}{\text{ = P}}_{\text{z}}\end{array}$ To prove the pascal's law, consider a negligibly small wedge shaped fluid element of dimensions $(dx\times dy\times 1)$ at any arbitrary location in a fluid.

  

Resolving the forces in x direction,

   $\begin{array}{l}{\text{P}}_{\text{AB}}{\text{×AB×1 - P}}_{\text{BC}}\text{×dy×1= 0}\\ {\text{P}}_{\text{AB}}{\text{×dy×1 - P}}_{\text{BC}}\text{×dy×1= 0}\\ {\text{P}}_{\text{AB}}{\text{ = P}}_{\text{BC}}\end{array}$

Resolving the pressure forces in the y direction

   $\begin{array}{l}\text{Vertical pressure force on BC + weight of fluid element - Vertical pressure frce on AC = 0}\\ {\text{P}}_{\text{BC}}\text{×dx×1 + }\left(\frac{\text{1}}{\text{2}}\text{×dy×dx×w}\right){\text{- P}}_{\text{AC}}\text{×AC×1 = 0, (As weight of fluid element is negligible)}\\ {\text{P}}_{\text{BC}}{\text{×dx×1 - P}}_{\text{AC}}\text{×dx×1 = 0}\\ {\text{P}}_{BC}{\text{ = P}}_{AC}\\ \therefore {\text{ P}}_{AB}{\text{ = P}}_{BC}{\text{ = P}}_{AC}\text{ - eq1}\end{array}$

Since the choice of fluid element is arbitrary, force on BC could be in any direction, Hence pressure at a point in all the direction is uniform and is equal in all direction.

The hydrostatic law states that the rate of increase with respect to depth in a stationary fluid is equal to the specific weight of the fluid.

   $\begin{array}{l}\frac{\text{dP}}{\text{dZ}}\text{ = ρg}\\ \text{where Z is the depth of the point in the fluid, which is positive in the downward direction}\text{. }\end{array}$

To prove this, consider a small rectangular shaped fluid element of cross sectional area $\Delta \text{A}$.

  

   $\begin{array}{l}\text{In horizontal direction, pressure forces on AC and BD are equal and opposite}\\ \text{Resolving Pressure forces in the vertical direction}\\ \text{P}\Delta \text{A + }\left(\text{w}\times \Delta \text{A}\times \Delta \text{Z}\right)\text{ -}\left(\text{ P + }\frac{\partial P}{\partial Z}\Delta Z\right)\times \Delta A\text{ = 0}\\ \text{P}\Delta \text{A + }\left(\text{w}\times \Delta \text{A}\times \Delta \text{Z}\right)\text{ -P}\Delta A\text{ - }\frac{\partial P}{\partial Z}\Delta Z\Delta A\text{ = 0}\\ \left(\text{w}\times \Delta \text{A}\times \Delta \text{Z}\right)\text{ - }\frac{\partial P}{\partial Z}\Delta Z\Delta A\text{ = 0}\\ \frac{\partial P}{\partial Z}\Delta Z\Delta A=\left(\text{w}\times \Delta \text{A}\times \Delta \text{Z}\right)\end{array}$

   $\frac{\partial P}{\partial Z}\text{ = w},\text{ rate of pressure variation along the depth is equal to the specific weight of the fluid - eq2}$

further at a given depth in a fluid, for a given density, $\text{ ρ}$.

specific weight, w= $\text{ ρ}$ g.

Hence to find the value of pressure at any depth Z, integrate both sides of eq2

   $\begin{array}{l}{\displaystyle \int \frac{\partial P}{\partial Z}}\text{ = }{\displaystyle \int \text{w}}\\ {\displaystyle \int \partial P}\text{ = }{\displaystyle \int \rho \text{g}}\times \partial Z\because \text{w = }\rho \text{g}\\ P\text{ = }\rho \text{g}Z\end{array}$

   $\begin{array}{l}\text{Pressure at a given depth= Density of fluid × acceleration due to gravity×depth of fluid at that point}\\ \text{ P = ρ×g×Z }\end{array}$ And at the depth Z, at any point, Pressure will be the same in all directions.

Conclusion:

Pressure, P= $\text{ρgZ, and is equal in all the directions}$