All gases are Newtonian fluids and exhibit viscous behavior. Bird, Stewart, and Lightfoot, Transport Phenomena, Section 1.4 MOLECULAR THEORY OF THE VISCOSITY OF GASES AT LOW DENSITY (i.e., ideal gas limit).I don't like that answer. What if there are no viscous behaviors?
I didn't say Gravitational Potential Energy. I was referring to the potential energy of changing the volume of a gas / spring etc..without potential energy effect.
I was under the impression that this was only true for perfectly inviscuous fluids (or viscous fluids in very high pressure gradients or flowing in a converging nozzle). For viscous fluids at lower pressure gradients, I thought it was flow instability (turbulence) that limits the speed at which energy can be transported? Am I wrong about this? Sorry again, not trying to nitpick, just testing my understanding.The speed at which the effect of applied pressure transmits through the gas is basically the speed of sound in the gas. There will be a delay as with any force deforming anything.
The context of the thread is a gas, which is why I introduced the speed of sound. I have no strong feelings about this though.Am I wrong about this?
The mass of the piston is important only in so far as some members thought that it might affect the total amount of work that is done by the gas between the initial and final thermodynamic equilibrium states. The analysis I have done has shown that, because of the viscous damping of the piston motion, even though the piston mass affects the amount of work done by the gas during the process, in the end it has no effect on the ultimate work done by the gas from initial to final thermodynamic equilibrium states.I didn't say Gravitational Potential Energy. I was referring to the potential energy of changing the volume of a gas / spring etc..
I don't know why the mass of the piston is of interest here, in any case. That's a bit like wanting to know the sort of battery or PSU that's connected to a theoretical circuit. We (more than often) assume the supply is constant voltage and I would say that the power for the piston is a high impedance source ( 'constant displacement' could be the name).
In this case, why didn't you include the viscous foces in the energy balance in the post where you introduced the Newton's second law and integrate it to get the energy balanceIt is dissipated by the viscous behavior of the gas. Think of the gas as if it is a pre-compressed spring in parallel with a viscous damper. The spring mimics the equilibrium P-V behavior of the gas and the damper comes into play in rapid irreversible deformations in mimicking the viscous effect. So, think of mean compressive stress in the rapidly deforming gas as being approximated by the linearized relationship $$P_{int}=\bar{\sigma}_z=P_{int,init}-C_1(V-V_{init})-C_2\frac{dV}{dt}$$where C2 is proportional to the gas viscosity. So, from our force balance on the piston, we have the linearized differential equation: $$\frac{m}{A}\frac{d^2V}{dt^2}+C_1A(V-V_{init})+C_2A\frac{dV}{dt}$$$$+
(P_0(t)-P_{int,init})A=0$$This equation describes a forced oscillation on the piston. Basically, what will happen is that the piston will oscillate about the final equilibrium position until the amplitude of its oscillation is damped to zero by viscous stresses in the gas.
The viscous forces are implicitly included in ##P_{int}## in the energy balance I introduced and in its integration. To complete the derivation, we needed to assume that at infinite time, the piston is no longer moving.In this case, why didn't you include the viscous foces in the energy balance in the post where you introduced the Newton's second law and integrate it to get the energy balance
If you take my point about the way people approach electric circuit theory then there's definitely a different attitude to the behaviour of gases. How many people even stop to think about what happens in an electrical supply when they are given a value for the supply voltage? Does the battery charge or discharge when a Potentiometer is moved?some members thought that it might affect the total amount of work that is done by the gas between the initial and final thermodynamic equilibrium states.
In the basic model, it's the piston that is being moved (by an irresistible applied force) so doesn't the piston stops moving when you choose to make it stop?The piston stops moving because its motion is damped by viscous stresses in the gas.
It's the first thing we're taught for gases at thermodynamic equilibrium. It isn't until we take a course in fluid mechanics that we learn about the irreversible deformational behavior of gases at finite deformation rates; that's when we learn about the viscous aspect of their behavior.I suppose there is one big difference in that Heat Engines involve Work Done on a piston but the behaviour of gases under pressure is the first thing that we're taught.
In homework problems in thermodynamics, there are cases where the piston is stopped manually or with a detent within the cylinders, but in most of the cases, the gas, piston, and other surroundings are allow to continue moving spontaneously until they finally re-equilibrate on their own.In the basic model, it's the piston that is being moved (by an irresistible applied force) so doesn't the piston stops moving when you choose to make it stop?
Thank you. You are very kind, and it's a pleasure interacting with you.PS I find you a terrific source of help with Thermodynamics.
yeah well - my thermodynamics extends to the basics of A level. Not surprisingly. I can easily get hold of the wrong end of the stick. But it's all grist to the mill.I'm a chemical engineer, and thermodynamics is our bread and butter.
I would like to add that it is always true in force of 3rd Newton law: the average force per unit area the gas exerts on the inside face of the piston is the same as the average force the inside face of the piston exerts on the gas (here we are assuming the gas inside the cylinder is the system).I'm going to continue what I was discussing in post #7 for an irreversible deformation of the gas. If ##\bar{\sigma_z}=P_{int}## is the average force per unit area exerted by the gas on the inside face of the piston, and we choose the gas alone as our system, then, calling the average external force per unit area exerted by the piston on the gas ##P_{ext}##, we must have that $$P_{ext}=P_{int}=\bar{\sigma_z}$$