Another Thermodynamics Problem

[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.