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March 22, 2016, 16:17 |
Mass-Weighted Average Specific Heat
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#1 |
New Member
Join Date: Nov 2015
Posts: 16
Rep Power: 10 |
Hi All,
I am currently looking into the properties of my mixture and under the specific heat portion, it is currently selected to "Mixing Law" and there is no option for a "Mass-weighted average". Is "Mixing Law" the same as a "Mass-Weighted Average"? If not, what is the equation for the "Mixing Law" option. Thank you all! |
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March 25, 2016, 07:45 |
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#2 |
Member
Join Date: Oct 2011
Posts: 34
Rep Power: 14 |
mixing law for specific heat is mole fraction based. sigma(y_i * cp_i). where cp_i is by default from NASA polynomials in the units of Joul/mole/k.
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March 25, 2016, 10:33 |
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#3 |
New Member
Join Date: Nov 2015
Posts: 16
Rep Power: 10 |
Hi chittipo,
Thank you for the reply! I just have a couple of questions. 1. In my material properties settings, the units for specific heats are J/kg/K. Is this wrong and is it suppose to be in mole? 2. In the equation you state, sigma(y_i * cp_i). Is Y mass fraction of the ith species? Thank you for the help! |
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March 25, 2016, 11:40 |
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#4 |
New Member
adam
Join Date: Mar 2015
Posts: 18
Rep Power: 11 |
1- i don't think it's wrong, since Cphas the unit of enthalpy/temperature....specific-->mass.
2-Y is normally the mass fraction yeah. hope it helped. |
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March 25, 2016, 12:52 |
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#5 |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,674
Rep Power: 65 |
specific heat capacity has units of J/kg/K
with moles, J/mol/K, this should be referred to as molar heat capacity. Unfortunately many people will call this specific heat, I have no clue why they do this because specific means per unit mass. I can understand getting heat capacity mixed up, but specific heat should never be confused with molar heat. Mixing law for specific heat capacity should be based on mass fraction and mixing law for molar heat capacity should be based on mole fractions. |
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March 25, 2016, 13:32 |
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#6 |
New Member
Join Date: Nov 2015
Posts: 16
Rep Power: 10 |
LuckyTran and poutivar,
Thank you for the reply. I believe this answers my questions! Thank you again for the help. |
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