# Pressure drop in a pipe due to cooling

I’m trying to better my understanding of the thermodynamics and momentum balance of pipe flows. The following situation, however, is making me scratch my head and I’ve found no help in my books.

Consider a pipe of constant section in which an ideal gas is flowing with negligible friction. Somewhere along this pipe there is a cooler, and Tin > Tout. Assume the cooler is frictionless as well and of constant section equal to that of the pipe.

Will there be a pressure drop in the pipe due to the cooling?

Edit: I have done some more research on the topic and it seems that what I’m considering is a case of a so-called Rayleigh flow. I tried to solve it this way:

$p = \rho r T$ (equation of state) $p + \rho v^2 = Cst$ (conservation of momentum) $\rho v = Cst$ (conservation of mass)

There are 3 unknowns: p, $\rho$ and v. I assume the temperature profile in the flow is known.

Solving this for p gives me a second degree polynomial with two solutions for p. I am not sure yet how to discriminate between the two.

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Is there some reason that this isn't as applying the ideal gas law? –  Mark Eichenlaub Feb 11 '11 at 11:54
The pressure should equalize. Won't there be an increase in the density instead? –  Peter Shor Feb 11 '11 at 13:37
Indeed I think the density will increase rather than the pressure decrease. I was looking for a rigorous proof of this answer. What bugs me is that if density increase, conservation of mass requires that velocity decreases. But in this case how is momentum conserved? What is the force braking the fluid, if pressure is constant? –  Whelp Feb 11 '11 at 13:53
think about what temperature is and how you could decrease it without changing the momentum of gas molecules. –  gigacyan Feb 11 '11 at 15:29
It might depend on the mechanism that you use to refrigerate the gas. Are you imagining using some kind of freeze ray that miraculously stops the gas molecules from moving so fast? Such a thing isn't completely ridiculous -- look up laser cooling. In this case, I think the pressure really might decrease. On the other hand, if you cool the gas by cooling the outside of the pipe, then can you still claim to have negligible friction? –  Peter Shor Feb 13 '11 at 3:41

I think you were on the right path in your original post. For one, let's at least talk about the boundary conditions. I'll rewrite your equations with a more useful construct for the "constant".

$$\rho v = \rho_{in} v_{in}$$

$$p + \rho v^2 = p_{in} + \rho_{in} v_{in}^2$$

And state of course

$$p=\rho r T$$

The $v$ will only have one direction and will never change sign, so no need to be dealing with vector math. The unknowns are fully stated as $v$, $T$, and $\rho$, which can be a function of the position down the pipe, $x$, but I suggest that the easy way is to find an answer in terms of temperature instead.

Solving the equations you can arrive at the following in order to define $\rho$.

$$0=\rho^2 r T - (p_{in} +\rho_{in} v_{in}^2) \rho + (\rho_{in} v_{in})^2$$

Now stop and think about this equation for a little bit, and consider the part of $-b \pm \sqrt{b^2-4ac}$, and that the term $ac$ is positive, and $-b$ is positive. That means that the only way for this to be positive is for the sign to be positive, or for us to take the largest root. Forget entropy arguments, in the world I live in, density is clearly positive. We have now defined $\rho(T)$ as the positive root for the preceding equation. The rest follows easily.

$$v=\frac{\rho_{in} v_{in}}{\rho(T)}$$ $$p = \rho(T) r T$$

Take the above, and plug in $T_{out}$ and you're finished. For those of you who will yell at me for not being 100% explicit, here you go:

$$\rho(T) = \frac{p_{in} +\rho_{in} v_{in}^2 + \sqrt{(p_{in}+\rho_{in}v_{in}^2)^2-4 r T (\rho_{in} v_{in})^2}}{2 r T}$$

Just to recap, you should have known quantities for $p_{in}$, $T_{in}$, and $v_{in}$. If you don't, then you need to figure out what the question is. It follows that $\rho_{in} = p_{in}/(r T_{in})$, and these all go into the above equations. Then if you know how much it was cooled by the time it reaches the end of the pipe, plug in that for $T$ and the above equations are the answer to your problem. You may have the information in some different form, like knowing the length of $x$ and having basically $dT/dx$, although that quantity might have to come after applying different physical laws. Whatever, you have what I think is an agreeable solution giving a $T$, the rest is up to whatever it is you need.

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Solving this for p gives me a second degree polynomial with two solutions for p. I am not sure yet how to discriminate between the two.

Guess that might be ruled out by the second law:

$\displaystyle{\frac{\partial}{\partial x} \left [\rho v s \right] + \frac{q}{T} \geqslant 0}$ (if you don't take into account heat conductivity)

It is indeed depends on the temperature profile, which you assume to be known.

$\displaystyle{\rho s = n k \left[ \frac{5}{2} + \ln \left[\frac{T^{3/2}}{n} \right] + \text{const} \right]}$

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I'm lost, where did your equation for $ps$ come from? –  AlanSE Jun 14 '11 at 15:25
Which one? The second I thought to be the entropy of an ideal gas. It's not $ps$, its $\rho s$. The first one is the 2nd law of thermodynamics $\frac{\partial \rho s}{\partial t} + \text{div} \boldsymbol{j}_s = p_s$, $p_s ⩾ 0$. –  Yrogirg Jun 14 '11 at 15:57
oh ok, yes, $\rho s$, and you are just writing an expression for the entropy of an ideal gas. Yes, this could be applied to any solution of the mass, momentum, energy equations. Ultimately though, $dx=-C dT$, where $C$ is positive, implying $d/dT (\rho v s)\le0$, implying $d/dT v(T) \ge 0$. With a falling temperature you would require a falling velocity, which sure, would be true. Even after you establish this the polynomial form the questioner had involves $T$ in more than one term I think. You could still carry out the math to the end. –  AlanSE Jun 14 '11 at 16:14
I've just understood that I haven't take into account the flux of the entropy from the heater or cooler (I'll fix this). And now I actually doubt that the second law will help to rule out one of the solutions. I'll have to think more and to remember something from gas dynamics course (which I didn't like at all). –  Yrogirg Jun 14 '11 at 17:29

If the gas does not accumulate or condense in the pipe, during a given time frame $\Delta t$ there should be the same amount of substance entering the pipe and exiting the pipe.

If the gas is ideal, we then have $\dfrac{P_{in}V_{in}}{T_{in}}=\dfrac{P_{out}V_{out}}{T_{out}}$, so either pressure drops, or outgoing volume of gas per unit of time is less than incoming volume of gas per unit of time.

I have the impression that the answer to this question depends on the boundary conditions:

• if the gas is in contact with a fixed-pressure environment, e.g. it exits into an atmosphere at the same pressure as the incoming gas, then volume will decrease, so the effect of the cooling will be to increase the gas density.

• if the pipe is long enough before being in contact with another environment, while keeping a fixed section, then conservation of momentum does indicate that the incoming amount of substance will keep the same velocity (cooling will decrease the overall kinetic energy of the gas molecules, but not their overall momentum). With constant section $S$ and constant velocity $v=\dfrac{\Delta L}{\Delta t}$, the same volume $\Delta V=S\Delta L$ will be coming in and going out of the cooling zone in the time $\Delta t$, so it's the pressure that must decrease.

• depending on the length of the pipe after the cooler and before contact with a fixed pressure (and non-constant section) environment, there may be a combination of both effets through a gradient of pressure.

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cooling will decrease the overall kinetic energy of the gas molecules, but not their overall momentum - could you explain that, please? –  gigacyan Feb 11 '11 at 15:30
If you see a flying hot body, its velocity does not depend on the body temperature so cooling down means reduction of internal energy, not the kinetic energy as a whole. But cooling down decreases the gas pressure. On the other hand, the pressure difference makes work on accelerating gas. Without friction, I am afraid, there is no stationary solution: there is no equilibrium of forces. –  Vladimir Kalitvianski Feb 11 '11 at 16:01
How does a constant velocity coexist with a pressure gradient? Wont the pressure gradient accelerate the gas? –  Peter Shor Feb 11 '11 at 16:47
Constant velocity does not coexist with the pressure gradient without wall friction. So one has an accelerating flow with the pressure gradient. In other words, there will be pressure drop in the pipe if you want some non zero flow. –  Vladimir Kalitvianski Feb 11 '11 at 17:10