Let’s assume we have an adiabatic pipe with a viscous fluid passing through. To reduce complexity, let’s further assume a constant heat capacity for the fluid. Independently from the choice of fluid model (constant density or ideal gas), the friction forces lead to a pressure drop between pipe inlet (1) and pipe outlet (2). However, when taking a look at the fluid temperature, different behaviors can be observed.
For a constant density fluid the velocity within the pipe stays constant. Thus, the energy balance reduces to the enthalpy balance between outlet an inlet: cpT1 + p1/rho = cpT2 + p2/rho. This means that a pressure drop due to pipe friction leads to an increase of fluid temperature, i.e. T2 > T1.
For ideal gas the contrary can be observed. Without further deduction of the energy balance (the given problem can be found in literature as “Fanno Flow”), the following effects are present: When moving downstreams, the velocity increases but the temperature drops, i.e. T2 < T1.
The question is: When investigating the temperature of airflow close to atmospheric pressure and low velocities (
Dear Axel,
Very interesting question, I know the ideal gas interpretation is the correct one as it incorporates the correct (thermodynamical) physics.
Falsifying the constant density case is a little tricky though.
It should be related to the fact that thermodynamics, in the form of the definition of a thermodynamic pressure, breaks down in the limit of Mach number approaching zero or when you're speaking of an "incompressible fluid."
You can see the link if you think of the formal definition of the speed of sound: a2 = dp / drho (at constant entropy).
Roughly speaking, a constant density, drho = 0, implies an infinite speed of sound.
In this limit many thermodynamical relations become nonsensical.
The pressure has a mechanical role in the problem solely and doesn't influence the thermodynamics.
I can't directly pinpoint how this can be used in your example (I would myself be very interested in seeing this), but I think the search for the incorrect assumption in the constant density case is the way to go.
Most probably you simply can no longer use the enthalpy balance.
To help you out, I can recall that Panton talks about this in his book (Incompressible Flow, 1984).
For constant property adiabatic flow, the energy balance should be Cp T1 + 0.5* u1^2 = Cp T2 + 0.5*u2^2. rho1 = rho2, A1 = A2, => u1 = u2. This leads to T1=T2. Unless, it is not adiabatic.
Dear Koen and Shiuh-Hwa, thanks for your answers.
I will try to get hold of Panton’s book, maybe this will clear things up.
Shiuh-Hwa, I think the energy balance you give is valid for ideal gas but not for a constant density flow. In the constant density case enthalpy (h=u(T)+pv) is not only a function of temperature anymore as the pv term in the equation cannot be replaced as for ideal gas (pv=RT).
Let me rephrase my question in order to may get some more input: Is the practical influence of the effects described above negligible for low Mach flows, or is the assumption to model low Mach airflow as constant density gas only applicable for isothermal calculations?
Interesting question, I thought I'd let it pass and just read some of the responses. So I did a bit of reading and learned something, thanks for that! But your second question on "practical influence" makes me think... there are a few assumptions in your first question: "adiabatic pipe", "constant heat capacity", "low Mach number" (but then air-flow up to 100 m/s). Could be that these assumptions already move the flow problem far away from any "practical" effects?
This problem can be solved at various levels, but essentially you have the coupling between the viscous flow in the pipe and the temperature field that develops due to viscous heating. Hence one needs to solve the thermal energy equation in conjunction with the linear momentum equation for the fluid. The two equations are coupled through viscous heating. A stronger coupling can occur if one accounts for temperature dependent viscosity. In that case the momentum equations cannot be solved independently. Another simplification one can adopt is to neglect convective effects in the thermal energy balance and to assume the flow field is fully developed. ( which would be the case when the thermal and momentum boundary layers are developed, small Peclet number, small Reynolds number).
Check out my report on "Computing Hysteresis Effects in Plane Poiseuille Flow with Viscous Heating" which can be downloaded from RG. Note in this problem the walls of the channel are not insulated. I also have notes on the derivation of the thermal energy equation that might be helpful in formulating your mathematical problem: see https://sites.google.com/site/ekayasolutionsteaching/classes/reaction-engineering-notes/1-conservation-principles
Let's make it simple: Suppose there is no heat generation, no work, no heat flux across the boundary. The 1st tell us that h1 + 0.5 u1^2 = h2 + 0.5 u2^2. Since u1=u2 as I claim before, h1=h2. You then check thermodynamic table, you will find that the temperature hardly change at all.
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