Consider a quiescent inviscid perfect gas to the right of an ideal piston in a long duct with constant cross-section. At t = t1, the piston located at x = cot1 accelerates to the left until it reaches x = 0, then its speed remains constant. The piston’s trajectory, xp(t), is such that the C+

Question

Consider a quiescent inviscid perfect gas to the right of an ideal piston in a long duct with constant cross-section. At [latex]t=t_1[/latex], the piston located at [latex]x=c_ot_1[/latex] accelerates to the left until it reaches x = 0, then its speed remains constant. The piston’s trajectory, [latex]x_p(t)[/latex], is such that the [latex]C_+[/latex] characteristics for the time the piston is accelerating all pass through the origin of coordinates (see Figure 15.18). Determine the gas velocity u(x,t) and sound speed c(x,t) in the gas to the right of the piston.

Accepted Answer

The piston moves to the left, so the gas that feels the piston motion expands and moves to the left as well; thus, u(x,t) will be negative (or zero) throughout the flow field. The method of characteristics can be used to determine u(x,t) and c(x,t) in all three regions of the flow field.
The first region, [latex]x>c_ot_1[/latex] and [latex]t The second region, [latex]x\geq x_p(t),x>0[/latex], and [latex]t>x/c_o[/latex], is more interesting. Here both u and c vary but the [latex]I_\pm[/latex] invariants are constant along the [latex]C_\pm[/latex] characteristics. For this region, all the C− characteristics (not shown on Figure 15.18) originate on x-axis where u = 0 and [latex]c=c_o[/latex], so the I_ invariant implies:

[latex]I_{-}=u(x, t)-\frac{2 c(x, t)}{\gamma-1}=u(x, 0)-\frac{2 c(x, 0)}{\gamma-1}=-\frac{2 c_o}{\gamma-1}[/latex].

In the second region, all the [latex]C_{+}[/latex] characteristics originate on the moving piston, so the [latex]I_{+}[/latex] invariant implies:

[latex]I_{+}=u(x, t)+\frac{2 c(x, t)}{\gamma-1}=\dot{x}_p( au)+\frac{2 c\left(x_p( au), au ight)}{\gamma-1} \equiv I_{+}( au)[/latex],

where τ is the time when the [latex]C_{+}[/latex] characteristic of interest touches the piston trajectory.
With these evaluations of [latex]I_{\pm}[/latex], (15.58) and (15.59) imply:

[latex]u(x, t)=\frac{I_{+}\left(x_a, 0 ight)+I_{-}\left(x_b, 0 ight)}{2}\quad ext{ and }\quad c(x, t)=\frac{\gamma-1}{4}\left(I_{+}\left(x_a, 0 ight)-I_{-}\left(x_b, 0 ight) ight)[/latex]. (15.58, 15.59)

[latex]u_{+}(x, t)=\frac{1}{2}\left(I_{+}( au)-\frac{2 c_o}{\gamma-1} ight) \quad ext { and } \quad c_{+}(x, t)=\frac{\gamma-1}{4}\left(I_{+}( au)+\frac{2 c_o}{\gamma-1} ight)[/latex].

where the subscript “+” indicates a result that applies along a [latex]C_{+}[/latex] characteristics. Thus, u(x,t) and c(x,t) are both constant along each [latex]C_{+}[/latex] characteristics (but different for different [latex]C_{+}[/latex] characteristics). Evaluating the equation for [latex]u_{+}[/latex] on the piston implies:

[latex]u_{+}\left(x_p( au), au ight)=\dot{x}_p( au)=\frac{1}{2}\left(I_{+}( au)-\frac{2 c_o}{\gamma-1} ight) \quad ext { so } \quad I_{+}( au)=2 \dot{x}_p( au)+\frac{2 c_o}{\gamma-1}[/latex].

Using this value of [latex]I_+( au)[/latex], the sound speed on the [latex]C_{+}[/latex] characteristic is:

[latex]c_{+}=c_o+\frac{\gamma-1}{2} \dot{x}_p( au)[/latex].

So, (15.57) can be integrated to find the equation for the [latex]C_{+}[/latex] characteristics:

d x / d t=u ± c. (15.57)

[latex]x=\int\left(\frac{d x}{d t} ight) d t=\left(u_{+}+c_{+} ight) t=\left(\dot{x}_p( au)+c_o+\frac{\gamma-1}{2} \dot{x}_P( au) ight) t=\left(c_o+\frac{\gamma+1}{2} \dot{x}_p( au) ight)t[/latex],

where the constant of integration is zero because the [latex]C_{+}[/latex] characteristic passes through the origin.
Now, invert this last equation to solve for [latex]\dot{x}_p( au)[/latex] in term of x and t, and substitute this result into the prior results for [latex]u_{+}[/latex]and [latex]c_{+}[/latex] to find:

[latex]u(x, t)=\dot{x}_p( au)=\frac{2}{\gamma+1}\left(\frac{x}{t}-c_o ight) \quad ext { and }\quad c(x, t)=c_o+\frac{\gamma-1}{2} \dot{x}_p( au)=\frac{2}{\gamma+1}c_o+\frac{\gamma-1}{\gamma+1} \frac{x}{t}[/latex].

where the “+” subscripts have been dropped from u and c because the parameter τ that specifies a particular [latex]C_{+}[/latex] characteristic has been eliminated in favor of (x,t)-variables that apply to the whole region.
The third region x < 0, lies at the upper left of Figure 15.18. Here the flow velocity and sound speed are constant, [latex]-u=c=2 c_o /(\gamma+1)[/latex], because the piston speed is constant. These values for u and c can be obtained by evaluating the above equations at x = 0, and are indicated in Figure 15.18, too. In this region the [latex]C_{+}[/latex] characteristics are vertical lines.
Interestingly, the solution given here is also valid for piston motion to the right, leading to compression of the gas, for a finite period of time. The situation is shown in Figure 15.19. The piston is stationary for [latex]t/t_1<-1[/latex], after which it accelerates to the right until it reaches speed [latex]c_{o}[/latex] (at [latex]t/t_1=-0.3349[/latex]), and then continues at constant speed. The main difference between this gas-compression piston motion and the gas expansion piston motion shown on Figure 15.18 is the formation of a shock wave beyond the location where the characteristics converge. Here, the homentropic assumption is violated, so a shock wave and entropy discontinuity must be fitted into the solution (see Exercise 15.21).

Question 15.7: Consider a quiescent inviscid perfect gas to the right of an...

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