Explanation Given: The bulk modulus ( E V ) is 2.20 GPa . The depth of the water is 300 m . Density of the water ( ρ ) is 1000 kg/m 3 Calculation: Calculate the bulk modulus for incompressible fluid. E V = d p d V / V (I) Here, the pressure at depth 300 m is p and Volume is V . Calculate the density of water ( ρ ) using the formula. V = m ρ Here, the mass is m . Find the d V V parameter. d V V = − m / ρ 2 d ρ m / ρ = − d ρ ρ Substitute ( − d ρ ρ ) in Equation (1). E V = − d p ( d ρ / ρ ) Assume that density ( ρ ) as 1 , 000 kg / m 3 and pressure ( p ) as 0 at water surface. Calculate the pressure from 0 to p ; when density changes from 1 , 000 kg / m 3 as depth increases. E V ∫ ρ 0 ρ d ρ ρ = ∫ p 0 p d p Here, the pressure at the surface (absolute pressure) is p 0 and the density at the surface is ρ 0 . Substitute p 0 = 0 , ρ 0 = 1 , 000 kg / m 3 , and E V = 2.20 G P a in the above equation. ( 2.20 × 10 9 N / m 2 ) ∫ 1 , 000 ρ d ρ ρ = ∫ 0 p d p ( 2.20 × 10 9 ) ln ( ρ 1 , 000 ) = p ρ = 1 , 000 e p ( 2.20 × 10 9 ) (II) Calculate the unit weight of water ( γ ) using the relation. γ = ρ g Here, the acceleration due to gravity is g . Write the expression for The pressure ( p ). p = p 0 + γ z Here, the absolute pressure is p 0 , the unit weight of fluid is γ , and the vertical height is z . Substitute γ = ρ g in the above equation. p = p 0 + ( ρ g ) z Differentiate on both sides of the above equation. d p = ( ρ g ) d z d p ρ = g × d z Substitute Equation (II) for ρ and g = 9

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2, Problem 17P

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The value of pressure at a depth of 300 m when water is compressible and the value of pressure at a depth of 300 m when water is incompressible.

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Chapter 2, Problem 16P

Chapter 2, Problem 18P

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