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Basics of Fluid Mechanics, 2014a

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11.3. SPEED OF SOUND 379<br />

to be continuous. In the control volume it is convenient to look at a control volume<br />

which is attached to a pressure pulse (see Figure 11.2). Applying the mass balance<br />

yields<br />

or when the higher term dU dρ is neglected yields<br />

ρc=(ρ + dρ)(c − dU) (11.1)<br />

ρdU = cdρ=⇒ dU = cdρ<br />

ρ<br />

(11.2)<br />

From the energy equation (Bernoulli’s equation), assuming isentropic flow and neglecting<br />

the gravity results<br />

(c − dU) 2 − c 2<br />

+ dP 2 ρ<br />

=0 (11.3)<br />

neglecting second term (dU 2 ) yield<br />

−cdU + dP ρ<br />

=0 (11.4)<br />

Substituting the expression for dU from equation (11.2) into equation (11.4) yields<br />

Sound Speed<br />

( ) dρ<br />

c 2 = dP ρ ρ =⇒ c2 = dP<br />

dρ<br />

(11.5)<br />

An expression is needed to represent the right hand side <strong>of</strong> equation (11.5). For an ideal<br />

gas, P is a function <strong>of</strong> two independent variables. Here, it is considered that P = P (ρ, s)<br />

where s is the entropy. The full differential <strong>of</strong> the pressure can be expressed as follows:<br />

dP = ∂P<br />

∂ρ<br />

∣ dρ + ∂P<br />

s<br />

∂s ∣ ds (11.6)<br />

ρ<br />

In the derivations for the speed <strong>of</strong> sound it was assumed that the flow is isentropic,<br />

therefore it can be written<br />

dP<br />

dρ = ∂P<br />

∂ρ ∣ (11.7)<br />

s<br />

Note that the equation (11.5) can be obtained by utilizing the momentum equation<br />

instead <strong>of</strong> the energy equation.<br />

Example 11.1:<br />

Demonstrate that equation (11.5) can be derived from the momentum equation.

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