Another Thermodynamics Problem

[Charles Link]

@Delta² and @Chestermiller Please see my Edited last part of post 25, where I have once again made a couple additional changes and inputs. I don't know that Chestermiller is taking $P$ to be constant, but there are a couple of items that I still think need to be looked at very carefully. I think there are a couple of underlying assumptions in @Chestermiller 's analysis, but I am quite impressed by it, in any case.

 

[Delta2]

@Delta² and @Chestermiller Please see my Edited last part of post 26, where I have once again made a couple additional changes and inputs. I don't know that Chestermiller is taking $P$ to be constant, but there are a couple of items that I still think need to be looked at very carefully. I think there are a couple of underlying assumptions in @Chestermiller 's analysis, but I am quite impressed by it, in any case.

He was taking $P$ to be constant in his approach in post #12. In his new approach , he doesn't make such an assumption.

 

[Charles Link]

He was taking $P$ to be constant in his approach in post #12. In his new approach , he doesn't make such an assumption.

In his latest approach, the one difficulty I am having with the introduction of pressure $P$ is that it seems there needs to be some obstruction either by multiple collisions, or introducing a closed container with a small hole in it like I am doing in the last part of post 25. Otherwise, there doesn't seem to be a mechanism to cause any pressure to occur. If the walls of the container are supplying this reverse pressure by slowing some of the particles, then there could be some momentum loss in the x-direction that needs to be computed there.$\\$ Pressure would imply a force from particles moving in the minus-x direction, and there is for the most part, no particle motion in the minus x-direction, unless some additional assumptions and/or additional boundaries are applied. $\\$ Edit: I do think the assumption of many collisions between the particles (in order to create a pressure $P$) may be inherent in @Chestermiller 's calculations. In any case, if this is assumed, the equations of @Chestermiller 's derivation look to be correct.

 

[Chestermiller]

@Chestermiller it is obvious from this

$F=PA=\frac{\phi^2 RT}{\mu PA}$

that the pressure $P$ in the boundary cannot be considered constant , since in the most general case $\phi=\phi(t)=-\frac{dm}{dt}$ will be time varying.

How do we know that dm/dt won't be constant. Apparently it will if P is constant. And there is nothing in the other equation coupling the force F with dV/dt to prevent this.

 

[Delta2]

How do we know that dm/dt won't be constant. Apparently it will if P is constant. And there is nothing in the other equation coupling the force F with dV/dt to prevent this.

Well to be honest I am not an expert in sublimation process, but I think that at start dm/dt going to be large, since the mass that is sublimating is larger in volume and surface boundary, and afterwards , dm/dt will get smaller and smaller as the volume and surface boundary shrinks due to loss of mass...What do you think?

EDIT: OK, if the sublimation is done in such a way that the layer of the sublimating substance has always surface A, then dm/dt probably will be constant.

But where exactly your approach at post #12 goes wrong if it isn't because of the non constant pressure?

 

[Charles Link]

To add to my comments on the last part of post 33, it seems these particles would emerge with a wide distribution of velocities, and thereby, it would seem that there could be a lot of collisions occurring between the particles, making the emerging particles behave in all directions as a gas of temperature $T$ and pressure $P$.

 

[Delta2]

To add to my comments on the last part of post 33, it seems these particles would emerge with a wide distribution of velocities, and thereby, it would seem that there could be a lot of collisions occurring between the particles, making the emerging particles behave in all directions as a gas of temperature $T$ and pressure $P$.

Yes I think that's the idea. In an infinitesimal volume dV of surface A and thickness dx just above the sublimating layer we can view the particles to behave as an ideal gas of moles dn, pressure P and temperature T and if we apply the ideal gas law, we get $P=\frac{dn}{dV}RT=\frac{dm}{\mu dV}RT=\frac{\rho}{\mu}RT$

 

[Chestermiller]

Well to be honest I am not an expert in sublimation process, but I think that at start dm/dt going to be large, since the mass that is sublimating is larger in volume and surface boundary, and afterwards , dm/dt will get smaller and smaller as the volume and surface boundary shrinks due to loss of mass...What do you think?

EDIT: OK, if the sublimation is done in such a way that the layer of the sublimating substance has always surface A, then dm/dt probably will be constant.

But where exactly your approach at post #12 goes wrong if it isn't because of the non constant pressure?

My error was in assuming that mechanical energy is conserved. It's one thing to say that momentum is always conserved (which it is), but quite another to say that mechanical energy is conserved. There is heat added to cause the solid to evaporate and to ultimately cause the increase in kinetic energy of the cup. So the mechanical energy equation can't balance. Here's an example:
From the development I presented, we have $$F=\phi v$$where $\phi=-dm/dt$ and v is the discharge velocity of the vapor at the phase boundary: $v=\sqrt{\frac{RT}{\mu}}$. So the rate of doing work to force the vapor out of the control volume is obtained by multiplying the force by the velocity:
$$\dot{W}=Fv=\left(-\frac{dm}{dt}\right)\frac{RT}{\mu}=-Pv_m\frac{dn}{dt}$$where $v_m$ is the molar volume at T and P, and n is the number of moles of solid remaining. This is basically equivalent to the relationship for the work done in discharging the vapor derived in post #12. I equated this to the rate of change of kinetic energy of the cup $mV\frac{dV}{dt}$. But that is not what I get if I just multiply the momentum balance by v. There I obtain:
$$\dot{W}=Fv=mv\frac{dV}{dt}=mV\frac{dV}{dt}\left(\frac{v}{V}\right)$$This is clearly inconsistent with mechanically energy being balanced. So, the error was assuming that it was.

 

[Chestermiller]

It's even worse than I made out in my previous post. If I had done the mechanical energy balance on the cup correctly, I would have started with the force balance on the cup and multiplied by cup velocity to obtain:
$$FV=\left(-\frac{dm}{dt}\right)vV=MV\frac{dV}{dt}$$But, the momentum balance gave us $$V=\frac{(m(0)-m(t))}{M}v$$Substituting this on the left hand side of the equation would have yielded: $$\left(-m\frac{dm}{dt}\right)\frac{v^2}{M}=MV\frac{dV}{dt}$$Multiplying both sides by M and integrating with respect to t would have yielded$$(m(0)-m(t))^2v^2=M^2V^2$$This is the same relationship as obtained from the force balance.

 

[Charles Link]

It's even worse than I made out in my previous post. If I had done the mechanical energy balance on the cup correctly, I would have started with the force balance on the cup and multiplied by cup velocity to obtain:
$$FV=\left(-\frac{dm}{dt}\right)vV=MV\frac{dV}{dt}$$But, the momentum balance gave us $$V=\frac{(m(0)-m(t))}{M}v$$Substituting this on the left hand side of the equation would have yielded: $$\left(-m\frac{dm}{dt}\right)\frac{v^2}{M}=MV\frac{dV}{dt}$$Multiplying both sides by M and integrating with respect to t would have yielded$$(m(0)-m(t))^2v^2=M^2V^2$$This is the same relationship as obtained from the force balance.

The simple momentum balance seems to work for this one, but I like your derivation of post 24. Otherwise, the momentum balance requires a calculation or estimate of $\bar{v}_x$, and your derivation of post 24 comes up with that result.