Compressibility of Fluids

Example 1 Estimate the increase in pressure (in psi) required to decrease a unit volume of

mercury by 0.1%.

From tables the bulk modulus for mercury is: E_V = 4.14 x 10⁶ lb/in²

E_V = - dP / (dV/V) So -dP = E_V (dV/V)

dP = - (4.14 x 10⁶ lb/in² ) (- 0.001) = 4140 psi (result)

Example 2 Natural gas at 70 ^oF and standard atmospheric pressure of 14.7 psi (abs) is compressed

isentropically to a new absolute pressure of 70 psi. Find the final density and temperature of the

gas.

For ideal gases P = ρ R T.

From tables for natural gas: R = 3.099 x 10³ ft lb / slug ^oR

At state 1: T₁ = 70 ^oF = (70 + 460 ) = 530 ^oR and P₁ = (14.7 psi)(144 in²/ft²) = 2116.8 psfa

So ρ₁ = P₁ / RT₁ = (2116 lb/ft²) / [ (3.099 x 10³ ft lb / slug ^oR) ( 530 ^oR ) = 1.288 x 10 ^-3 slugs/ft³

For an isentropic process P / ρ^K = constant where K is the ratio of specific heats for

natural gas (in this case). K = 1.31

So P₁ / ρ₁^K = P₂ / ρ₂^K Therefore ρ₂^K = (P₂ /P₁) ρ₁^K So ρ₂ = (P₂ /P₁)^1/K ρ₁

For state 2:

ρ₂ = ( 70 / 14.7)^1/1.3 ( 1.288 x 10 ^-3 ) = 4.28 x ^{10 -3} slugs/ft³ (result)

Now P₂ = ρ₂ R T₂

So T₂ = P₂ / ρ₂ R = (70 psi)( 144 in²/ft²) / 4.28 x ^{10 -3} slugs/ft³ )( 3.099 x 10³ ft lb / slug ^oR)

T₂ = 760 ^oR = 300 ^oF (result)

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