Mixing chamber with multiple inlets

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Discussion Overview

The discussion revolves around a problem involving a mixing chamber with multiple inlets, focusing on the conservation of mass and energy principles in thermodynamics. Participants analyze the enthalpy and mass flow rates of fluids entering and exiting the chamber, with specific reference to the properties of R-134a refrigerant.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that mass conservation implies m1 + m2 = m3, while energy conservation leads to m1h1 + m2h2 = m3h3.
  • One participant expresses confusion about how to incorporate velocity into the calculations without knowing the area of the pipe.
  • Another participant questions the enthalpy values, noting that the outlet enthalpy appears lower than the inlet enthalpy for one of the inlets, which seems inconsistent given the states of the fluids.
  • There is a discussion about the validity of the enthalpy values for R-134a, with one participant correcting another's interpretation of the saturation properties at given pressures.
  • Some participants express uncertainty about the relevance of the velocity information provided in the problem, suggesting it may not be necessary for solving the equations.
  • A later reply clarifies that the process through the valve is a throttling process, which conserves enthalpy.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the relevance of the velocity in the calculations, and there is disagreement regarding the correct enthalpy values for the saturated liquid and vapor states of R-134a. The discussion remains unresolved regarding the implications of these discrepancies.

Contextual Notes

Participants note limitations in the problem, such as the unclear relationship between velocity and mass flow without the area of the pipe, and the potential confusion caused by the provided enthalpy values and states of the refrigerant.

canadiansmith
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Hi I was wondering if someone could help me with this question. Question 4 uploaded in picture.
Ok, so far this is what I have found

Pipe 1:
m1=2kg/s
P1=1000kpa
T1=100 C
h1=334.82kJ/kg

Pipe 2:
sat-liquid
T2=60 C
P2= 1000kpa
h2=291.36KJ/kg

Outlet:
sat vapor
P3=1000kpa
V=20 m/s
h3=267.97 kJ/kg
v3= 0.0202 m^3/kg

and I know that the mass in = mass out
so m1 + m2 = mout
My problem is that I don't know what to do with the velocity because I do not have area to use formula Vout=VA

Also I believe to solve the problem the formula m1*h1 + m2*h2 = m3*h3
but I have tried a variation of this and it did not work.
 

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Enthalpy is conserved for each input, right. That's because dH = dQ + Vdp, dQ = 0, dp = 0. So you have m1h1 + m2h2 = m3h3. You also have m1 + m2 = m3. In other words, you've already stated all the equations you need.

Two equations, 2 unknowns (m2 ans m3). Outflow = m3*V.

I'm not claiming total knowledge here - does anyone see any flaws in this?
 
rude man said:
Enthalpy is conserved for each input, right. That's because dH = dQ + Vdp, dQ = 0, dp = 0. So you have m1h1 + m2h2 = m3h3. You also have m1 + m2 = m3. In other words, you've already stated all the equations you need.

Two equations, 2 unknowns (m2 ans m3). Outflow = m3*V.

I'm not claiming total knowledge here - does anyone see any flaws in this?

I have tied to use that method which would be
m1h1 + m2h2 = (m1 + m2)h3
but then m2 = -5.716 and this is wrong
I don't now how to relate the velocity into the equation. You said that outflow = m3*V right? what would outflow be? wouldn't the units be like kgm/s^2 or something like that?
 
My apologies. I did not think when I wrote that outflow = m3*V. Outflow should be just m3.

Something bothers me. Your enthalpy at the outlet is less than the enthalpy at inlet 2. That sounds weird. At 2 we have saturated liquid, and at 3 we have saturated vapor , so h3 must be > h2, yet you have h3 < h2. h3 looks OK since its temp. is 100 vs. 60 for the other two, so its h should be the highest of all.

I don't know what to do with V, if it's a red herring or needed. Must think more.

EDIT:

To conserve energy, looks like h3 should be reduced by ρV^2 where ρ = density at outlet.

BTW your insert is really hard to read, at least on my display.
 
Last edited:
Actually I believe the h3 value is correct because r-134a at 1000kpa the saturation temp is only 39.39 C and the sat vapor value in the table hg= 267.97. The numbers are much less intuitive because it is not water. Still confused by this velocity given. It feels like It is a necessary piece of info but without area of pipe or diameter I can't seem to do anything with it.
 
According to my table, saturated liquid enthalpy of r134a at 1 MPa is 105.29 kg/kJ, not 291.36. The table agrees with your T = 39.39C and sat. vapor h of 267.97.

http://energy.sdsu.edu/testhome/Test/solve/basics/tables/tablesPC/pSatR134a.html

I don't see how inlet 2 can have T = 60C and p = 1 Mpa. There's something fishy about that. As you can see from the table, and as I believed a priori, there is one and only one temperature associated with saturated r134a at 1 MPa, and that is 39.39C. I wonder if there is something tricky about the statement "after the valve", though I sure can't see it.
 
Sorry is this better?
 

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Oh my bad you are very correct. I read the table wrong. I believe the pressure at inlet 2 is so high due to the valve.
 
hmm. I'm not sure. Like there shouldn't be. min to valve should = mout
 
  • #10
canadiansmith said:
Sorry is this better?

Much! Thanks.
 
  • #11
Ok I got it. So h3 will be 267.97 because it is sat vapor at 1Mpa and h2 will be 137.42 which is the sat-liquid value for 60 C at a pressure of 1681.3 Kpa. And because the hin = hout of a valve this can be used. Plug all this in the the formula we talked about and a got the right answer. Thanks for your help and for your time.
 
  • #12
You absolutely got it. I was just writing to point out that the valve made it a throttling process with h conserved. Good going!

PS so V was never used?
 
  • #13
rude man said:
You absolutely got it. I was just writing to point out that the valve made it a throttling process with h conserved. Good going!

PS so V was never used?

Nope. I guess it was to throw you off as well as the pressure at p2 after the valve was never used. I hate extra info that that. Thanks again
 

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