16100lectre39_cj

16100lectre39_cj - Derivation of Sound Wave Properties...

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Unformatted text preview: Derivation of Sound Wave Properties Sound wave propagating to right ρ ρ + dρ du p + dp a u = 0 ⇒ ρ ( x) = p Similar for u ( x), p( x) xs l Assume: ∗ Sound wave creates small disturbances in an isentropic manner. Mass dl ρ ( x)dx + ρu l − ρu 0 = 0 dt ∫0 l d ⎡ xs ( ρ + dρ )dx + ∫ ρdx ⎤ + ρu l −( ρ + dρ )du = 0 ⎢ ∫0 ⎥ xs ⎦ dt ⎣ 0 xs l d⎡ dρdx + ∫ ρdx ⎤ − ( ρ + dρ )du = 0 ⎥ 0 ⎦ ⎣ dt ⎢ ∫0 Constant time d [dρx s ] − ( ρ + dρ )du = 0 dt dx dρ s − ( ρ + dρ )du = 0 dt dρ a − ρdu − dρdu = 0 Higher order ⇒ dρa = ρdu ρ + dρ , x < x s ρ , x > xs Derivation of Sound Wave Properties Isentropic disturbances Constant entropy disturbances for perfect, ideal gases satisfy: p = const. γ ρ ⇒ p + dp p =γ γ ( ρ + dρ ) ρ p + dp ⎛ ρ + dρ ⎞ ⇒ =⎜ ⎜ρ⎟ ⎟ p ⎝ ⎠ dp ⎛ dρ ⎞ ⎟ 1+ = ⎜1 + p⎜ ρ⎟ ⎝ ⎠ dp dρ 1+ = 1+ γ p ρ dp = γ γ γp dρ ρ Conservation of Momentum dl ρudx + ( ρu 2 + p ) l − ( ρu 2 + p) 0 = 0 dt ∫0 ρdua ⇒ + − ( p + dp) = 0 p Higher order terms eliminated ρadu = dp Summarizing: adρ = ρdu Mass: γp Isentropic: dp = dρ ρ Momentum: ρadu = dp (1) (2) (3) Combining a * (1) − (3) gives: a 2 dρ = dp Then, using (2) gives: γp a 2 dρ = dρ ρ ⎛ γp ⎞ ⇒ ⎜ a 2 − ⎟dρ = 0 ⎜ ρ⎟ ⎝ ⎠ Since dρ ≠ 0 , then 16.100 2002 a2 = γp ρ . We’ve just derived the speed of sound for an 2 Derivation of Sound Wave Properties Ideal, perfect gas. ∂p s = const . ∂ρ One other thing of interest: Suppose the sound wave caused a change in pressure, dp . Then, the change in velocity is: 1 du = dp ⇒ du has same sign as dp ρa Note: without assuming ideal, perfect gas, the general result is a 2 = 16.100 2002 3 ...
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16100lectre39_cj - Derivation of Sound Wave Properties...

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