Red_CCF said:
My bad I was thinking about a water-air system (not pure water system) for some reason. From the steam table the equilibrium vapor pressure you are referring to is 3.17kPa?
Yes.
I ran into a little problem with this one.
Since the table in my combustion book is referenced at standard state the sensible component is zero and I get h(25C, 1atm) = hf = -241845kJ/kmol.
I was taught to approximate liquid water enthalpy at a specific temperature to the enthalpy of saturated liquid at the same temperature. In this case the enthapies of the liquid is pressure independent and the change in enthalpies from 25C 1atm to 25C equilibrium pressure is zero; I'm not sure if there's a better approach than this. The only thing left is the hfg = 2409.8kJ/kg = 43412.5kJ/kmol. However, the reference state for this value is that for liquid water at 0 C, 1atm and I'm not sure how I would convert it to standard state reference.
The heats for formation table on the internet gives:
Water vapor in hypothetical state of 25 C and 1 atm: -241.818 kJ/mol
Liquid water at 25C and 1 atm: -285.8 kJ/mol
The heat of vaporization of water at 25C is 43.99 kJ/mol
If one also wishes to include the change in enthalpy in dropping the pressure of liquid water from 1 atm to the equilibrium vapor pressure at 25C, one would use:
ΔH=V(1-αT)ΔP
where V is the specific volume of liquid water at 25C, α is the coefficient of volumetric thermal expansion of liquid water at 25C, and ΔP is the pressure difference between the equilibrium vapor pressure and 1 atm. See what magnitude this term contributes, compared to the heat of vaporization.
Sorry the k was a typo and should be a 1.
After mutiplying both sides by dx/dt I get:
P_{Ef}\frac{(P_{Ei}-P_{Ef})}{C}e^{\frac{-kt}{C}}=P_{Ei}\frac{(P_{Ei}-P_{Ef})}{C}e^{\frac{-kt}{C}}-\frac{(P_{Ei}-P_{Ef})^2}{C}e^{\frac{-2kt}{C}}-\frac{(P_{Ei}-P_{Ef})^2}{C}(e^{\frac{-kt}{C}}-e^{\frac{-2kt}{C}})
Integrating from t = 0 to t = infinity, without simplification
P_{Ef}\frac{(P_{Ei}-P_{Ef})}{k}=P_{Ei}\frac{(P_{Ei}-P_{Ef})}{k}-\frac{(P_{Ei}-P_{Ef})^2}{2k}-\frac{(P_{Ei}-P_{Ef})^2}{2k}
From the above it looks like the damper and spring does equal amount of work. What is the significance of the sign difference between the damper/spring and the initial/final pressures?
In the first term on the right hand side, if we write
P_{Ei}=\frac{P_{Ei}+P_{Ef}}{2}+\frac{P_{Ei}-P_{Ef}}{2}
We obtain:
P_{Ef}\frac{(P_{Ei}-P_{Ef})}{k}=\frac{P_{Ei}^2-P_{Ef}^2}{2k}-\frac{(P_{Ei}-P_{Ef})^2}{2k}
From this, it follows that:
Total work done by surroundings to compress spring and damper = P_{Ef}\frac{(P_{Ef}-P_{Ei})}{k}
Work done by surroundings to compress spring = \frac{P_{Ef}^2-P_{Ei}^2}{2k}
Work done by surroundings to compress damper = \frac{(P_{Ef}-P_{Ei})^2}{2k}
Note that the work done by the surroundings to compress the spring is just the elastic energy stored in the spring, and this is analogous to the reversible P-V work required to compress the gas. The work done by the surroundings to compress the damper is positive definite, and this is analogous to the viscous energy dissipation.
Next, we are going to see how these results are affected if we do the compression in a two-step process, rather than the single large step just completed.
Step 1: Compress the gas from pressure P
Ei to pressure (P
Ei+P
Ef)/2, allowing the system to re-equilibrate at the new pressure
Step 2: Compress the gas from pressure (P
Ei+P
Ef)/2 to pressure P
Ef, allowing the system to re-equilibrate at the new pressure
I want you to determine the following for each of these Steps:
(a) The total work done by the surroundings to compress the spring and damper
(b) The work done by the surroundings to compress the spring
(c) The work done by the surroundings to compress the damper
(You don't need to do the integrations again to obtain these results. You can use the algebraic results from the final equation describing the work terms in the single step process).
Then, I'd like you to determine the sum of these quantities over the combination of the two steps. We can then compare the results in going from P
Ei to P
Ef in a two step process with going directly from P
Ei to P
Ef in a single step process. The results, particularly with respect to the work to compress the damper will be very revealing.
Chet
