Calling all thermo whiz's...
Calling all thermo whiz's...
(OP)
Here's the problem statement:
A 0.5m^3 rigid tank has equal volumes of saturated refrigerant vapor and refrigerant liquid at 312 deg K. Additional refrigerant is then slowly introduced into the tank until the total mass of refrigerant (liquid and vapor) is 400 kg. Some vapor is bled off to maintain the original temperature and pressure. After adding the refrigerant, the refridgerant is saturated at 312 deg K and 0.9334 MPa. Under these saturated conditions, vf=0.000795 m^3/kg and vg=0.01872 m^3/kg.
And the question:
What was the mass of refrigerant added to the tank?
If someone can explain the solution, it would be much appreciated...
A 0.5m^3 rigid tank has equal volumes of saturated refrigerant vapor and refrigerant liquid at 312 deg K. Additional refrigerant is then slowly introduced into the tank until the total mass of refrigerant (liquid and vapor) is 400 kg. Some vapor is bled off to maintain the original temperature and pressure. After adding the refrigerant, the refridgerant is saturated at 312 deg K and 0.9334 MPa. Under these saturated conditions, vf=0.000795 m^3/kg and vg=0.01872 m^3/kg.
And the question:
What was the mass of refrigerant added to the tank?
If someone can explain the solution, it would be much appreciated...





RE: Calling all thermo whiz's...
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RE: Calling all thermo whiz's...
My solution, which assumes a negligible amount is "bled off", is 72 kg. If my memory serves me right, and it has been 7+ years since I thought about thermodynamics... when something is done slowly, like mass added to a rigid tank, you assume that there is no temperature change of the system (the temperature change I speak of would be due to the compression of the vapor in the tank). If there is no temperature change of the system, then there is no pressure rise (can't have one without the other, right?). If there is no pressure rise, then adding mass without changing volume just pushes you closer to the saturated liquid point (vf) WHEN YOU ARE UNDER THE VAPOR DOME.
The solution indicates that one must first calculate the volume that is bled off. This is done by taking the final volume of the vapor (which we known from solving for the quality using the final mass and volume, or .5m^3/400kg) and then subtracting from the initial volume of the vapor:
0.25m^3 - 0.19m^3 = 0.06m^3
Then the solution goes on to use the 0.06m^3 to find the resultant mass of vapor that was bled off.
What I'm asking is, what is the change in vapor volume due to (i.e. the 0.25m^3 to 0.19m^3)? Is it truly "bled off"? Or, is this the result of the quality of the mixture being pushed more toward the saturated liquid point (vf)???
RE: Calling all thermo whiz's...
Good luck,
Latexman
Technically, the glass is always full - 1/2 air and 1/2 water.
RE: Calling all thermo whiz's...
What am I missing here???
RE: Calling all thermo whiz's...
Good luck,
Latexman
Technically, the glass is always full - 1/2 air and 1/2 water.
RE: Calling all thermo whiz's...
[0.25 m^3 - 0.19 m^3]/0.01872 m^3/kg= 3.2kg
RE: Calling all thermo whiz's...
It's a convoluted problem as written.
So, let's ASSuME the refrigerant added is 312 deg K and 0.9334 MPa too.
400 kg - 0.25/0.000795 - 0.25/0.01872 = 72.18 kg added IF the refrigerant added is 312 deg K and 0.9334 MPa too.
Since we know some refrigerant was bled off, it must be greater than 72.18 kg.
75 kg is the only viable answer.
Good luck,
Latexman
Technically, the glass is always full - 1/2 air and 1/2 water.
RE: Calling all thermo whiz's...
I'm going to report it as errata: it is misleading at best, bad analysis at worse.
RE: Calling all thermo whiz's...
Good luck,
Latexman
Technically, the glass is always full - 1/2 air and 1/2 water.
RE: Calling all thermo whiz's...
Logic dictates that there must be more than 72 kg added to the tank if some of it was bled off, but science does not dictate that it was exactly 3 kg that was bled off, thus the book's solution to the problem is at fault.
RE: Calling all thermo whiz's...
Good luck,
Latexman
Technically, the glass is always full - 1/2 air and 1/2 water.
RE: Calling all thermo whiz's...
Prove it. Prove that the book's solution is incorrect (or define "at fault" in that context) with only the given information.
Can't do it? Well now who is incorrect, you or the book?
You should learn the difference between should, would and could.
If you only pick the answers that "should" be correct, nobody will care what your piece of paper says when something doesn't work out during testing due to erroneous assumptions.
If you only pick the answers that "would" be correct "if only..." then you are just wasting peoples' time.
Sometimes in the real world, you have to pick the answer that "could" be correct, because it works, and you can't prove that it is incorrect.
Should, would, and/or could you agree?
RE: Calling all thermo whiz's...
As that cubic metre leaves as vapour to make room for the liquid out goes 1/0.01872 kilograms = 53.42
The net effect is to add 1204.44 to the trapped mass for every 1257.86 of inflow.
So to add the 72.18 noted upthread requires an inflow of 72.18 * 1257.86/1204.44 = 75.38
RE: Calling all thermo whiz's...
total volume 0.5 m3
liquid density 1257.861635 kg/m3
gas density 53.41880342 kg/m3
intial vol each phase 0.25 m3
mass gas initially 13.35470085 kg
mass liquid initially 314.4654088 kg
total mass initially 327.8201097 kg
final mass 400 kg
v_tot = v_gas + v_liq
m_tot = (dens gas)*v_gas + (dens liq) * v_liq
v_gas = 0.5 - v_liq
final conditions
0.309860689 v_liq
0.190139311 v_gas
0.059860689 displaced vgas
3.197686386 displaced mgas
added mass is 400 - 327.82 + 3.20 = 75.38kg
RE: Calling all thermo whiz's...
I don't think that's the point of the problem?
What I'm asking is, what is the change in vapor volume due to (i.e. the 0.25m^3 to 0.19m^3)? Is it truly "bled off"? Or, is this the result of the quality of the mixture being pushed more toward the saturated liquid point (vf)???
Why would the quality (or vapor pressure) change if pressure and temperature are held constant?
RE: Calling all thermo whiz's...