Consider adiabatic compressible flow of a perfect gas in a round duct with constant diameter D, and interior skin friction coefficient, Cƒ (the Fanning friction factor). Starting from (15.25), use the definition of the Mach number, u = Mc, and (15.28) to derive an equation for dM²/dx, where x is

Question

Consider adiabatic compressible flow of a perfect gas in a round duct with constant diameter D, and interior skin friction coefficient, $C_f$ (the Fanning friction factor). Starting from (15.25), use the definition of the Mach number, u = Mc, and (15.28) to derive an equation for dM²/dx, where x is the distance along the duct in the direction of flow.

$\dot{m}\frac{d u}{d x}=-\frac{d}{d x}(p A)+p \frac{d A}{d x}-F_f=-A \frac{d p}{d x}-F_f$ , (15.25)

$\frac{T_0}{T}=1+\frac{u^2}{2 c_{ p } T}=1+\frac{\gamma-1}{2} \frac{u^2}{\gamma R T}=1+\frac{\gamma-1}{2} M^2$ , (15.28)

[latex]\dot{m}\frac{d u}{d x}=-\frac{d}{d x}(p A)+p \frac{d A}{d x}-F_f=-A \frac{d p}{d x}-F_f[/latex], (15.25)

[latex]\frac{T_0}{T}=1+\frac{u^2}{2 c_{ p } T}=1+\frac{\gamma-1}{2} \frac{u^2}{\gamma R T}=1+\frac{\gamma-1}{2} M^2[/latex], (15.28)

Accepted Answer

The perimeter friction per unit length in the duct is [latex]F_f=\pi D(1 / 2) ho u^2 C_f[/latex], so (15.25) becomes:

[latex]\dot{m} \frac{d u}{d x}= ho u A \frac{d u}{d x}=-A \frac{d p}{d x}-\frac{\pi D}{2} ho u^2C_f[/latex],

where the first equality follows from (15.18). Here, A = πD²/4 is constant so the equation can be simplified:

[latex] ho_1 u_1 A_1= ho_2 u_2 A_2, \quad ext { or } \quad ho u A=\dot{m}=const[/latex]., (15.18)

[latex] ho u\frac{d u}{d x}=-\frac{d p}{d x}-\frac{2 C_f}{D} ho u^2 ext {, and rearranged: } \frac{1}{u^2} \frac{d u^2}{d x}=-\frac{2}{ ho u^2} \frac{d p}{d x}-\frac{4 C_f}{D}[/latex].

First, convert the u²-derivative into one involving the Mach number. Start with (15.28), note that [latex]T_0[/latex] will be constant for adiabatic flow, multiply by c², and differentiate:

[latex]\frac{T_0}{T}=\frac{c_0^2}{c^2}=1+\frac{\gamma-1}{2} M^2 ext {, which implies: } c_0^2=c^2+\frac{\gamma-1}{2} u^2 \quad ext { or } \quad 0=\frac{d c^2}{d x}+\frac{\gamma-1}{2} \frac{d u^2}{d x}[/latex].

Now differentiate u² = M²c², substitute in the differentiated result from (15.28), and isolate (1/u²)(du²/dx):

[latex]\frac{d u^2}{d x}=c^2 \frac{d M^2}{d x}+M^2 \frac{d c^2}{d x}=c^2 \frac{d M^2}{d x}-M^2 \frac{\gamma-1}{2} \frac{d u^2}{d x}, \quad ext { or } \quad \frac{1}{u^2} \frac{d u^2}{d x}=\left(1+\frac{\gamma-1}{2} M^2 ight)^{-1} \frac{1}{M^2} \frac{d M^2}{d x}[/latex].

To similarly convert the pressure gradient term in the rearranged version of (15.18) to one involving only u and M, use (15.1h) to eliminate p in favor of c and ρ:

[latex]Speed of Sound: c=\sqrt{\gamma R T}=\sqrt{\gamma p / ho}[/latex], (15.1h)

[latex]\frac{2}{ ho u^2} \frac{d p}{d x}=\frac{2}{ ho u^2} \frac{d}{d x}\left(\frac{ ho c^2}{\gamma} ight)=\frac{2}{\gamma u^2} \frac{d c^2}{d x}+\frac{2}{\gamma M^2} \frac{1}{ ho} \frac{d ho}{d x}[/latex].

Substitute for dc²/dx from above, and note that (1/ρ)(dρ/dx) = −(1/u)(du/dx) for constant-area flow where ρu = constant. Thus, the pressure gradient term becomes:

[latex]-\frac{2}{ ho u^2} \frac{d p}{d x}=\frac{\gamma-1}{\gamma u^2} \frac{d u^2}{d x}+\frac{1}{\gamma M^2} \frac{1}{u^2} \frac{d u^2}{d x}=\left[\frac{(\gamma-1) M^2+1}{\gamma M^2} ight] \frac{1}{u^2} \frac{d u^2}{d x}[/latex].

Using this result and that from (15.28), (1/u²)(du²/dx) can be eliminated from the rearranged version of the momentum equation (15.18) so that − after some algebra − it takes the final form:

[latex]\frac{4 C_f}{D}=\left[\frac{1-M^2}{\gamma M^4} ight]\left(1+\frac{\gamma-1}{2} M^2 ight)^{-1} \frac{d M^2}{d x}[/latex],

The left-side term will always be positive. Thus, when M < 1, dM²/dx must be positive and the flow’s Mach number must increase toward unity. And, when M > 1, dM²/dx must be negative and the flow’s Mach number must decrease toward unity. Therefore, this equation supports the contention that the effect friction on a compressible flow is to drive the Mach number toward unity. Interestingly, this equation can be integrated, too (see Thompson 1972, p. 299).

Question 15.6: Consider adiabatic compressible flow of a perfect gas in a r...

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