Fluid Mechanics: Fundamentals and Applications
Fluid Mechanics: Fundamentals and Applications
4th Edition
ISBN: 9781259696534
Author: Yunus A. Cengel Dr., John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 2, Problem 118P

Although liquids, in general, are hard to compress, the compressibility effect (variation in the density) may become unavoidable at the great depths in the oceans due to enormous pressure increase. At a certain depth the pressure is reported to be 100 MPa and the average coefficient of compressibility is about 2350 MPa .
(a) Taking the liquid density at the free surface to be ρ 0 = 1030 kg/m 3 obtain an analytical relation between density and pressure, and determine the density at the specified pressure. Answer: 1074 kg/m 3
(b) Use Eq. 2-13 to estimate the density for the specified pressure and compare your result with that of part (a).

Expert Solution
Check Mark
To determine

(a)

The density of water at the specified depth.

Answer to Problem 118P

The density of water at the specified depth is 1078.744kg/m3.

Explanation of Solution

Given information:

The pressure at a certain depth is 100MPa, the average coefficient of compressibility is 2350MPa and the liquid density at the free surface is ρ=1030kg/m3.

Write the expression for density at a depth.

  α=1ρ( ρ P)T...... (I)

Here, the partial derivative of density is ρ, the partial derivative of density is P and isothermal compressibility of water is α.

Write the expression for ratio of partial derivative of density with respect to pressure.

  ρP=( ρ 1 ρ P 1 P )T

Here, the pressure at the specified depth is P1, the initial pressure of water at the free surface is P, the density at the free surface is ρ and the density of the liquid at the specified depth is ρ1.

Substitute ρ1ρP1P for ρP in Equation (I).

  α=1ρ( ρ 1 ρ P 1 P )T(ρ1ρ)=αρ(P1P)ρ1=(αρ( P 1 P ))+ρ...... (II)

Substitute, 4.80×105atm1 for α, 986.9232atm for P1, 0.986923atm for P and 1030kg/m3 for ρ in Equation (II).

  ρ1=[( ( 4.80× 10 5 atm 1 )×1030 kg/ m 3 ( 986.9232atm0.986923atm ))+1030 kg/ m 3 ]=(4.80× 10 5 atm 1×1030kg/ m 3( 985.936277atm))+1030kg/m3=48.7446kg/m3+1030kg/m3=1078.744kg/m3

Conclusion:

The density of water at the specified depth is 1078.744kg/m3.

Expert Solution
Check Mark
To determine

(b)

The density of water at the specified depth and its comparison with part (a).

Answer to Problem 118P

The density of water at the specified depth is 1073.7859kg/m3.

Explanation of Solution

Write the expression for the density of water using Equation 2-13.

  k=ΔPΔρ/ρo...... (III)

Write the expression for pressure change.

  ΔP=P1Po

Write the expression for change in density.

  Δρ=ρ1ρo

Substitute P1Po for ΔP and ρ1ρo for Δρ in Equation (III).

  k=( P 1 P )( ρ 1 ρ ρ )ρ1ρ1=( P 1 P )kρ1=( P 1 P k+1)×ρ...... (IV)

Here, the coefficient of compressibility is k, the pressure at the specified depth is P1, the initial pressure of water at the free surface is P, the density at the free surface is ρ and the density of the liquid at the specified depth is ρ1.

Substitute, 2350MPa for k, 100×106Pa for P1, 100×103Pa for P and 1030kg/m3 for ρ in Equation (IV).

  ρ1=( 100MPa-100kPa 2350MPa+1)×1030kg/m3=( 100MPa( 10 6 Pa 1MPa )100kPa( 10 3 Pa 1kPa ) 2350MPa( 10 6 Pa 1MPa )+1)×1030kg/m3=(0.4251+1)×1030kg/m3=1073.7859kg/m3

The value of density obtained in part (a) that is 1078.744kg/m3 is greater than the value obtained in part (b) that is 1073.7859kg/m3.Thus, the density calculated in part (b) is more exact as the isothermal compressibility of water in part (a) is an approximate value.

Conclusion:

The density of water at the specified depth is 1073.7859kg/m3.

The density calculated in part (b) is more exact as the isothermal compressibility of water in part (a) is an approximate value.

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Chapter 2 Solutions

Fluid Mechanics: Fundamentals and Applications

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