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1. ### Questions Regarding the Phase Diagram

How does the squeezing effect reduce the melting point? I think I'm reading the phase diagram wrong. Looking at the phase diagram here, if we start at the triple point with no air in the head space, and we add air until the head space pressure is 1atm, the vapour pressure should remain...
2. ### Questions Regarding the Phase Diagram

Taking a step back, this whole time I was under the (wrong) impression that for Case 2 (triple point, head space, no air), if I add air (ideal gas) until the total pressure in the head space is 1atm, the equilibrium state will not change; I did not realize that it actually does decrease by 0.01C...
3. ### Questions Regarding the Phase Diagram

So regardless of whether head space is present, the melting point when there is 1atm of pressure acting on the liquid (head space or not) is 0C, even if we are dealing with ideal gases in the head space? My steam tables actually stop at 0.01C and I haven't been able to track down a water...
4. ### Questions Regarding the Phase Diagram

Sorry, I am still confused about why there is a difference in melting point between Cases 1 and 3. I used the example here and the vapor pressure is calculated to be 0.00596bar (assuming heat of sublimation and vaporization is close enough). Thank you
5. ### Questions Regarding the Phase Diagram

Thank you, this summarizes the cases I was confused about perfectly. My original question is basically how and why freezing point vary between Case 1 and 3, since the system in both cases are under 1atm of pressure albeit the way the pressure is applied is different. By vapour pressure of...
6. ### Questions Regarding the Phase Diagram

For the case of a pressure of 1atm applied from a piston, in the phase diagram in my OP, I believe the melting point is 0C and 1atm. In the case of a fixed volume container, the melting/triple point is at 0.01C with the vapour pressure at 0.00611bar. Essentially I see that the water is at two...
7. ### Questions Regarding the Phase Diagram

Sorry, I was thinking of a different case (1atm pressure applied from by a piston); in that case it would be closer to 0C than 0.01C? Another way of asking my original question would be, why do we need a significantly higher total (air) pressure/force than an applied piston pressure to obtain...
8. ### Questions Regarding the Phase Diagram

With regards to the first statement, I'm confused on what is meant by the same state. In the phase diagram for 1atm of pressure applied by the piston the freezing point is 0C (exactly) but for the case of cooling in an isolated fixed volume box with 1atm of air, the triple point/freezing point...
9. ### Questions Regarding Effect of Nucleation in Phase Change

1/2. I'm basically wondering if nucleation is a requirement for phase change. For instance, when liquid water is at equilibrium with its vapour, does individual molecules simply enter/leave the interface, or does nuclei have to form to induce the phase change? 3. Thermodynamically, I'm...
10. ### Questions Regarding the Phase Diagram

I phrased that poorly, it was meant as a question not a statement. I'm was thinking, from the liquid's perspective, if I have air at 1atm, or a piston applying a pressure at 1atm, the liquid "feels" the same force, but the equilibrium state is very different. I get that the liquid has no idea...
11. ### Questions Regarding the Phase Diagram

I can't get my head around why ideal gas has no effect on equilibrium state or in a sense, hold the water down. The pressure on the liquid (and in the head space) applied by the air at 1atm has a drastically different effect than 1atm applied by a piston even though the pressure felt by the...
12. ### Questions Regarding the Phase Diagram

In the above examples I assumed that air was an ideal gas and that changing the air pressure did not affect the phase equilibrium (if our volume was constant adding air wouldn't have changed anything). Base on the link you recommended, when real gas effects are considered, increasing the total...
13. ### Questions Regarding Effect of Nucleation in Phase Change

1. Is nucleation a phenomenon that occurs in all phase change (freezing/melting, evaporation/condensation)? 2. I've always read evaporation/condensation described as a liquid-vapour interface phenomenon (water molecules going entering-leaving the interface at equal rates in equilibrium). If...
14. ### Questions Regarding the Phase Diagram

I think I will increase pressure to 50atm: the mass of liquid water in the bottom half: 0.499kg the partial pressure of water vapor in the top half: 70.14kPa the partial pressure of air in the top half: 4994.86kPa the mass of water vapor in the top half: 5.2mg the total mass of water in the...
15. ### Straightforward and Rigorous Rocket Thrust Derivation

I'm not totally sure what you mean but check out Mechanics and Thermodynamics of Propulsion by Hill and Peterson Chp 10.
16. ### Questions Regarding the Phase Diagram

I think now I will vary the pressure to say 10atm (i think we are still in the ideal gas range?) at 90C. New volume is 0.065L the mass of liquid water in the bottom half: 0.499kg the partial pressure of water vapor in the top half: 70.14kPa the partial pressure of air in the top half: 942.86kPa...
17. ### Questions Regarding the Phase Diagram

I was thinking 90C The new vapour/air volume is: V = mRT/P = 0.000588*0.287*363.15/31.36 = 1.95L the mass of liquid water in the bottom half: 0.499kg the partial pressure of water vapor in the top half: 70.14kPa the partial pressure of air in the top half: 31.16kPa the mass of water vapor in...
18. ### Questions Regarding the Phase Diagram

Actually I will do 0.01C (right at the triple point) since that's the lowest temperature my book's steam table goes to. I'm going to assume that we only have liquid and vapour here (I don't have the properties for solid but I think density at least is close to the liquid) The new vapour/air...
19. ### Questions Regarding the Phase Diagram

the mass of liquid water in the bottom half: 0.499kg (assuming it is 50% by volume) the partial pressure of water vapor in the top half: 2.339 kPa the partial pressure of air in the top half: 98.96 kPa (assuming 1atm = 101.3kPa) the mass of water vapor in the top half: 8.65mg the total mass of...
20. ### Questions Regarding the Phase Diagram

That is essentially what I'm looking for. However, in this setup, we cannot independently control the vapour pressure since the piston will raise the partial pressure of the vapour by the same proportion as the air pressure. Is there another possible setup where air pressure and vapour pressure...
21. ### Questions Regarding the Phase Diagram

So in the above derivation we are essentially looking at something like a closed piston-cylinder assembly where we can apply pressure differently? How does the analysis change if we have a closed, fixed volume container with a cup of water inside, and total pressure is varied by changing the...
22. ### Questions Regarding the Phase Diagram

Is the P used here the vapour pressure or the total pressure? Thank you
23. ### Questions Regarding the Phase Diagram

Hi Chet Apologies, it look me a little while to look at this. I'm confused about how the above method can be used to predict the melting temperature change. Assuming that my mixture of liquid and vapour is already at equilibrium at some total pressure (vapour is saturated), if I increase the...
24. ### Questions Regarding the Phase Diagram

For an ideal gas mixture, this is what I found (I couldn't figure out how to do equations in this new format): u_i = g_i + RTln(y_i*p/p_ref) Taking the derivative with respect to T and p while holding one of them constant should give the change in chemical potential with respect to each...
25. ### Questions Regarding the Phase Diagram

What if the air pressure is exaggerated to the order of 100MPa? What I'm trying to get at is, on what order of magnitude does the air pressure have to increase to get to the freezing point in Case 1 (1atm constant applied pressure) or about a 0.01C decrease? Had this scenario occurred and I...
26. ### Questions Regarding the Phase Diagram

Hello I’m wondering if my understanding of how different processes relate to the phase diagram here is correct. 1) If a piston cylinder assembly maintains a constant 1atm pressure on some water and the water is cooled, we would be going horizontally along the red line from 373.15K to 273.15K...
27. ### Question about Auto-Ignition Temperature

I have a couple of questions 1. Are all of these quantities empirical (i.e. we can't calculate the Flash Point or AIT etc.)? 2. For Flash and Flame Point, why is the ignition temperature irrelevant, as a higher ignition temperature would add more energy to the system? 3. When we do tests to...
28. ### Question about Auto-Ignition Temperature

Hello I was wondering about explanation of the auto ignition temperature (AIT). From the definition in ASTM E659: What I get from the last sentence is that fuel-air reaction at some level occurs at any temperature, but the number of particles reacting is not large enough to produce enough...
29. ### Mechanisms of Supercooling and Supersaturation

I have a couple of questions: The homogeneous freezing temperature of water is listed at -42C. However, from the equations formulated to find critical radius here, I did not see any factors which restricts the homogeneous nucleation temperature to a certain value. How is the homogeneous...
30. ### Question about Flow between Parallel Plates

The ODE looked like it was setup for the combined process but I wasn't 100% sure. So if I am isolating for the second step as below: \frac{dx}{dt}+\frac{k}{C}x=A\frac{(P_{Ef}-P_{Ei})}{2C} After substituting x_2(0) = -A\frac{(P_{Ef}-P_{Ei})}{2k}e^{-4} the x2(t) I get is: x_2(t) =...
31. ### Question about Flow between Parallel Plates

I'm confused about why the RHS of the ODE is not divided by 2 since the initial pressure is (PEi + PEf )/2 and the final pressure is PEf. Also, is x(t) and Δx(t) equivalent? Is initial condition equal to x1 (4C/k)? What I am confused by is why the initial position has a (1 - e-4) term as...
32. ### Question about Flow between Parallel Plates

Was the total work I calculated correct? I actually had a typo and the extra damper work should have been \frac{A^2e^{-4}(P_{Ef}^2-P_{Ei}^2)}{4k} which is still different. The original total work when each step takes to t = infinity was: \frac{3P_{Ef}^2-P_{Ei}^2-2P_{Ef}P_{Ei}}{4k} and I...
33. ### Question about Flow between Parallel Plates

I got the pressure terms mixed up and forgot that the pressure terms are applied pressure and not related to the cylinder gas pressure. For total work I got: W = \frac{A^2}{4k}[(e^{-4}-3)P_{Ef}^2-(e^{-4}-1)P_{Ei}^2+2P_{Ei}P_{Ef}] and the extra damper work should be...
34. ### Question about Flow between Parallel Plates

From previous results for n = 2 where both stages go from t = 0 to infinity W_{spring} = \frac{P_{Ef}^2-P_{Ei}^2}{2k} W_{damper} = \frac{(P_{Ef}-P_{Ei})^2}{4k} For the new first step from t = 0 to 4C/k and second step for t = 0 to infinity (assuming we go to equilibrium in the second step...
35. ### Question about Flow between Parallel Plates

So if we perform n = finite cycles at t = 4C/k, we would not get the desired final pressure until the last step (which is on purposely made bigger?) but in the limit of n→∞, since we are performing ∞ steps at t = 4C/k the "loss" at each step is made up for? With regards to C and k, in a real...
36. ### Question about Flow between Parallel Plates

Is this because x(t) and hence dx/dt is linearly proportional to the step size PEf - PEi, so the process is faster with larger step sizes? What is the actual penalty of not doing the process to full completion (t = ∞)? The only one I can think of is that the final desired pressure at each...
37. ### Question about Flow between Parallel Plates

Does the number of discrete steps have any bearing on the time it takes to reach 98% displacement? Here it appears that regardless of step size the time to reach 98% displacement is constant at 4C/k, which I find a big strange since intuitively I thought that decreasing the step size would...
38. ### Question about Flow between Parallel Plates

I was able to get the x(t) equation but unable to get the form for the work equation. For the work equation, are we solving: -W=\int{C(\frac{dx}{dt})^2dt}+\int{kx\frac{dx}{dt}dt} where we integrate t from 0 to infinity? If so I probably made some arithmatic errors during...
39. ### Question about Flow between Parallel Plates

I got myself mixed up a bit due to the time variable. From the original ODE AP_{ext}(t)=AP_{ext}(0)-C\frac{dx}{dt}-kx for P_{ext}=P_{Ei}+(P_{Ef}-P_{Ei})\frac{t}{τ} substitution of the second into the first equation gives (P_{Ei}-P_{Ef})\frac{t}{τ}=C\frac{dx}{dt}+kx The notation is a bit...
40. ### Question about Flow between Parallel Plates

Is the main difference in quasistatic process with a continuous force variation Pext(t) is that it is a direction function of time unlike a sequence of discrete steps? In the case of continuous forces, what is considered analogous to a "step"? What were some of my mis-interpretations...
41. ### Question about Flow between Parallel Plates

The way quasistatic processes was taught to me from digging through my notes from awhile back is similar to what is in the attached diagram (it neglects viscous effects). In Case 1 if M is pushed onto the piston from its platform, a work of Mzg (or lost energy) is needed to bring the piston back...
42. ### Question about Flow between Parallel Plates

I thought quasistatic process is the step-by-step process we solved for but at infintiessimal step-sizes. Does increasing pressure in dP incrment count as a discrete steps? From my recollection of first year calculus, continuous function y(x) is one where infinitessimal dx variation causes...
43. ### Question about Flow between Parallel Plates

Looking at this process qualitatively, is the quasistatic process essentially: apply a infinitessimal pressure increase -> piston compresses until interface/external pressure balances -> wait for equilibrium -> repeat? With regards to the piston kinematics (if massless and frictionless), the...
44. ### Question about Flow between Parallel Plates

I think each step is identical, so the below would apply to both steps: a) -W_{irrev}=RT\left(\frac{\sqrt{P_{Ef}}}{\sqrt{P_{Ei}}}-1\right) b) -W_{rev}=RT\ln{\left(\frac{\sqrt{P_{Ef}}}{\sqrt{P_{Ei}}}\right)} c) -W_{visc}=RT\frac{(\ln{(P_{Ef})}-\ln{(P_{Ei})})^2}{8} The entire process...
45. ### Question about Flow between Parallel Plates

For this case, would it be W = ∫pdV = ∫RT/VdV = RTln(Vf/Vi), where Vf/Vi = PEi/PEf The difference between the two work terms would be Wnon-quasi - Wquasi = RT[1-PEf/PEi - ln(PEi/PEf)]. I expect the difference to be positive, but am not sure how to show this in the expression above. Thank...
46. ### Question about Flow between Parallel Plates

The initial volume would be Vi = nRT/PEi and the final volume is Vf = nRT/PEf for both cases. So W = PΔV = nRT(1-PEf/PEi)? Thanks
47. ### Question about Flow between Parallel Plates

I had some trouble figuring this out. Do I integrate ∫APEfdx with the quasistatic case being C = 0? After substituting dx = (PEi-PEf)/C*ekt/Cdt, if C = 0 then the solution is undefined? I think the first law (Q = W) should be used somewhere but could not out figure how. Thanks
48. ### Question about Flow between Parallel Plates

So both -285.5kJ/mol and 1.89kJ/mol are calculated by subtracting the same "absolute enthalpy" of H2O (at 25C) with different reference state enthalpies (liquid water at 0C and H2/O2 absolute enthalpies at 25C)? Does this mean that to convert the enthalpy from one reference state to the other we...
49. ### Question about Flow between Parallel Plates

From my steam tables the enthalpy of saturated liquid at 25C is 104.9kJ/kg = 1.89kJ/mol. Is this value equivalent to the heat of formation of liquid water -285.8kJ/mol relative to standard state (i.e. they have the same "absolute" enthalpy) if pressure effect on liquid enthalpy is neglected...
50. ### Question about Flow between Parallel Plates

From the above equation (using alpha at 20oC): ΔH=V(1-αT)ΔP = 1.0029*10^{-3}(1-207x10^{-6}*25)*(101.3-3.17)= 0.0979kJ/kg =0.00176kJ/mol I wasn't sure how to deal with the αT term since α is given per degree of temperature change (so 1K = 1C) while I used the absolute temperature, although...