Applied Fluid Mechanics (7th Edition)
Applied Fluid Mechanics (7th Edition)
7th Edition
ISBN: 9780132558921
Author: Robert L. Mott, Joseph A. Untener
Publisher: PEARSON
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Textbook Question
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Chapter 3, Problem 3.1PP

Write the expression for computing the pressure in a fluid.

Expert Solution & Answer
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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.

   At a point in the fluid, pressure P = Pressure in x direction, Px = Pressure in y direction, Py = Pressure in z direction, PzP= Px = Py = Pz To prove the pascal's law, consider a negligibly small wedge shaped fluid element of dimensions (dx×dy×1) at any arbitrary location in a fluid.

  Applied Fluid Mechanics (7th Edition), Chapter 3, Problem 3.1PP , additional homework tip  1

Resolving the forces in x direction,

   PAB×AB×1 - PBC×dy×1= 0PAB×dy×1 - PBC×dy×1= 0PAB = PBC 

Resolving the pressure forces in the y direction

   Vertical pressure force on BC + weight of fluid element - Vertical pressure frce on AC = 0PBC×dx×1 + (12×dy×dx×w)- PAC×AC×1 = 0, (As weight of fluid element is negligible)PBC×dx×1 - PAC×dx×1 = 0PBC = PAC PAB = PBC = PAC                                                                                                 - eq1

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.

   dPdZ = ρgwhere Z is the depth of the point in the fluid, which is positive in the downward direction

To prove this, consider a small rectangular shaped fluid element of cross sectional area ΔA.

  Applied Fluid Mechanics (7th Edition), Chapter 3, Problem 3.1PP , additional homework tip  2

   In horizontal direction, pressure forces on AC and BD are equal and oppositeResolving Pressure forces in the vertical directionPΔA + (w×ΔA×ΔZ) -( P + PZΔZ)×ΔA = 0PΔA + (w×ΔA×ΔZ) -PΔA  - PZΔZΔA = 0(w×ΔA×ΔZ)  - PZΔZΔA = 0PZΔZΔA = (w×ΔA×ΔZ)

   PZ = w, 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,  ρ.

specific weight, w=  ρ g.

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

   PZ = wP = ρg×Z                                                  w = ρgP = ρgZ

   Pressure at a given depth= Density of fluid × acceleration due to gravity×depth of fluid at that point                                 P  = ρ×g×Z  And at the depth Z, at any point, Pressure will be the same in all directions.

Conclusion:

Pressure, P= ρgZ, and is equal in all the directions

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Chapter 3 Solutions

Applied Fluid Mechanics (7th Edition)

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