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Sonic Velocity Calculation

johnsin (Chemical) (OP) 
3 Apr 08 20:15 
Please let me know how to solve.
T= 732.2F (389 C) P= 3.67psia (190mmHg) MW= 128.5
I can calculate Density that is 0.591395 kg/m3 ideal gas equation is applicated .
so how can I calculate. I don't know K=Cp/Cv this fact. also there are some calculation sheet. I don't know what I have to apply any calculation sheet. Please help me.


johnsin20: For an ideal gas:c _{p}  c _{v} = R (the universal gas constant) k = c _{p} / c _{v} = c _{p}/(c _{v}  R) c _{p} and c _{v} are independent of temperature and pressure, so they are constant for a given gas. They are not universal constants, they will vary from one gas to gas another. For a real gas:c _{p} and c _{v} are dependent on the gas temperature and pressure, so they are not constant for any gas. For ideal or real gases; You cannot calculate c _{p}. It must be experimentally measured or obtained from a correlation or a graph of experimental values. Milton Beychok (Visit me at www.airdispersion.com) .


mbeychok (Chemical) For a perfect gas cp  cv = R (the universal gas constant) However, either Cp or Cv is temperature dependent.
Regards 

correction to above pv=RT perfect gas
CpCv=R
For semi perfect gas Cp or Cv is function of temp For ideal gas, Cp is generally assumed constant.


try the computer programs available from NASA, distributed by Cosmic, such as : < http://www.openchannelfoundation.org/projects/CET93_PC/> The ability to calcualte the thremodynamic propoerties of any gas is fundamental to the process uses by the types of objects that Nasa designs, and the referenced program ( or its background references ) may provide a means to obtain a value for Cp/ Cv. 

zdas04 (Mechanical) 
7 Apr 08 16:08 
To reduce confusion on mobeychok's post above
c_{p}  c_{v} = R
k= c_{p}/c[sub]v/sub] = c_{p}/ (c_{p}  R)
(the equation above had (c_{v}  R) in the denominator)
David 

zdas04: David, thanks for picking up my typo. ====================================== sailoday28: My posting very clearly said: (a) the the specific heats were constant and independent of temperature and pressure for ideal gases and (b) the specific heats were not constant and were dependent on temperature and pressure for real gases. Milton Beychok (Visit me at www.airdispersion.com) .


mbeychok (Chemical) Yes for an ideal gas Pv=RT specific heats are constant. For a semiperfect gas, Pv=RT, specific heats are dependent upon temperature.
What I am trying to emphasize is that for pv=RT, gamma is temperature dependent. For the ideal gas, other than relying on tables, etc, how does one obtain the specific heats?
Regards 

sailoday28: Quote:What I am trying to emphasize is that for pv=RT, gamma is temperature dependent. For the ideal gas, other than relying on tables, etc, how does one obtain the specific heats?
Please read my original response in this thread slowly and carefully and you will see the answer to your question. Quote:Yes for an ideal gas Pv=RT specific heats are constant. For a semiperfect gas, Pv=RT, specific heats are dependent upon temperature.
I don't understand why you are repeating information already included in my original response. Milton Beychok (Visit me at www.airdispersion.com) .


mbeychok (Chemical) Sorry, but I didn't see where you addressed a semiperfect gas, unless you are stating that pv=RT with a variable specific heat is a real gas. My understanding is that a real gas approaches pv=RT as p approaches 0.
regards 

sailoday28: Now you have really confused me. Are you saying that there are three types of gases: ideal, semiperfect and real? Or are saying that all real gases are semiperfect gases? Or what are you saying? I just searched through 5 books on thermodynamics and none of them use the terminology "semiperfect gas"? Milton Beychok (Visit me at www.airdispersion.com) .


25362 (Chemical) 
9 Apr 08 2:23 
To give you a general idea about γ:
• For a gas containing simple particles helium atoms, for example, or the electronproton mixtures that characterize matter at very high temperatures γ has the value 5/3.
• For diatomic molecules like those in air, γ takes the lower value 7/5. Since Cv = 5/2 R, and as stated by mbeychok Cp = Cv + R.
• For more complicated molecules γ is lower still; in triatomic CO_{2} or NO_{2}, for example, γ is close to 4/3.
As an example. A mixture consisting of 2.0 mol of oxygen and 1.0 mol of argon the Cv = (1/n) ΔU/ΔT would be 6.5R/3.0 mol = 2.2R.
This is the result of applying the equipartition theorem for the total internal energy of the molecules (the average energy per molecule of a system in thermodynamic equilibrium is 1/2 kT for each degree of freedom):
for oxygen [O_{2}] with 5 independent components of motion, U_{ox} = (2.0 mol)(5/2RT) = 5.0RT for monoatomic argon [Ar] U_{ar} = (1.0 mol)(3/2RT) = 1.5RT U_{total} = 6.5RT Cv = 6.5R/3.0 mol = 2.2R Cp = 3.2R Cp/Cv = 1.45
These assumptions seem to hold in the range of 250 to 750 K; outside these limits molecules show different degrees of freedom. This is explained by quantum mechanics (lack of rotation at low temperatures, and springlike vibrations at higher temperatures).


mbeychok (Chemical) Pv=RT applies to gases as they approach p=0. From that definition, one obtains the simple relation between Cp and Cv. (and only applicable at low pressure) The specific heats at low pressure are basically a function of only temperature. My experience is that ideal and perfect gas assume Cp or Cv to be constant. Semiperfect uses the fact that with pv=RT, the specific heats are temperature dependent. All the above are for Pv=RT. As to where to look I don't have my texts available at present, but suggest you try Keenan, Osbourne and others in addition to a google search. You will see that I have not dreamed up these terms. But in summary a real gas is a gas and becomes perfect or semiperfect as p approaches zero pressure.
Regards 

johnsin (Chemical) (OP) 
9 Apr 08 20:36 
Thanks everyone. I appreciate that. but you guys don't have to be angle. just calm down 

Sailoday28: As I said earlier, the c _{p} of real gases are dependent on both temperature and pressure. See: (1) Figure 34 on page 3143 of the 6th Edition of Perry's Chemical Engineers' Handbook. (2) "The Variation of the Specific Heats (c _{p}) of Oxygen, Nitrogen and Hydrogen with Pressure", E. J. Workman, Physics Review, 37, 1345  1355 (1931) (3) "Data Book on Hydrocarbons", J.B. Maxwell, D. Van Nostrand, 1950 (page 75, first sentence) Also, the female of a species is either pregnant or not pregnant. There is no such thing as being semipregnant. The same hold true for gases, they are either ideal in their behavior or they are real gases in their behavior. There is no such thing as a "semireal" or "semiideal" or "semiperfect" gas. Of course, as the pressure of a real gas approaches 0, its behavior approaches that of an ideal gas. There is no disputing that statement ... but the real gas is still a real gas until it becomes completely ideal in its behavior. There is reason to create a new term and call it a "semiperfect" no matter who coined that new term. Just call it a real gas that is approaching a pressure of zero (or approaching a compressibility factor of 1). I promise this is my last word in this thread. Milton Beychok (Visit me at www.airdispersion.com) .


Sailoday28: Pardon my typographical error: Quote:There is reason to create a new term and call it a "semiperfect" no matter who coined that new term. Just call it a real gas that is approaching a pressure of zero (or approaching a compressibility factor of 1).
should read: There is no reason to create a new term and call it a "semiperfect" no matter who coined that new term. Just call it a real gas that is approaching a pressure of zero (or approaching a compressibility factor of 1). Milton Beychok (Visit me at www.airdispersion.com) .


mbeychok (Chemical) You stateThere is no such thing as a "semireal" or "semiideal" or "semiperfect" gas.
How did you come up up with semireal, semiideal? I did not. I am not playing with words. I am not creating a new termPerhaps you should look in other thermo texts, or do the google search that I previously suggested
Also, should we now, define the perfect gas as a gas that approaches a compressibility of one?
regards
Regards 

25362 (Chemical) 
10 Apr 08 10:13 
This interesting discussion is of semantic value.
To avoid confusion let me clarify that while pregnancy is a yes or no situation, real gases are the only option.
As far as I remember semiperfect gases aren't real gases approaching perfection; on the contrary, they are actually defined as "ideal" gases deviating from perfection, in that their heat capacity [Cp] is only a function of temperature, and is independent of pressure.
Perfect or semiperfect gases are just idealizations which may be approached asymptotically but are still abstractions.
In fact, Cp and Cv of real gases at low pressures rarely depart from the ideal gas values, and are estimated just as functions of temperature, namely as if they were semiperfect gases.


25362 (Chemical) "In fact, Cp and Cv of real gases at low pressures rarely depart from the ideal gas values, and are estimated just as functions of temperature, namely as if they were semiperfect gases."
Your point is will taken. It is interesting to note, however, that for a VDW gas at all other conditions, Cv is a function of only temperature. Regards 

johnsin20:
what is the chemical description of this gas with a mole wt of 128.5 ? 

25362 (Chemical) 
10 Apr 08 13:01 
From thermodynamics the difference between molar heat capacities is: CpCv = T(∂P/∂T)_{V}(∂V/∂T)_{P} which can be estimated using an EOS. For a vdW gas, ie,, a gas that follows the equation (P+a/V ^{2})(Vb) = RT CpCv = R + 2aP/RT^{2} 

25362 (Chemical) 
10 Apr 08 13:42 
To sailoday28, of course, the result I gave for an vdW gas is a good approximation after eliminating negligible terms. Using the Berthelot EOS using the same approach: CpCv ≈ R[1 + (27/16)(P/Pc)(Tc^{3}/T^{3})] where Pc and Tc are critical properties. 

25362 (Chemical) I have no disagreement with your posting. I was just pointing out that for VDW, Cv=Cv(T). Cp and gamma are functions of two variables. Relating to the original post. Having an equation of state,Berthelot, VDW,etc Cp or Cv at low pressure, one may calculate the sound speed. 

johnsin (Chemical) (OP) 
10 Apr 08 20:12 
I'm Working on furnace company. anyways I need to calculate sonic velocity. I almost do it. there is option which is
v=81.7(p/?)^0.5 V P ? Where: ?=Density P=Absolute essure V=Sonic Velocity
the sonic velocity does not exceed 80% of sonic. The conditions at the outlet of the furnace are: ?=0.0368lb/ft^3 (0.5895kg/m^3) MW=128.5 P=3.67psia (190mmhg ) T=732.2F (389 C) Per the above conditions, the Sonic Velocity is: v=81.7*(3.67/0.0368)^0.5 V=816ft/s (249m/s)
I don't understand how calculated 81.7. tell me. now I'm studying. thank you. 

25362 (Chemical) 
11 Apr 08 0:36 
The factor is the result of units conversion in the following equation from metric to british units: v = (γP/ρ)^{0.5} v, 1 m/s = 3.281 ft/s P, 1 psia = 6,894.8 N/m ^{2}ρ, 1 lb/cf = 16.02 kg/m ^{3}γ = Cp/Cv ≈ 1.44 3.281× (1.44×6,894.8÷16.02)^{0.5} = 81.7 

Hello, all: I really must explain why I keep repeating that the specific heat of a real gas is dependent on BOTH temperature and pressure. Here is a partial copy of a Figure 314 from page 3143 of the 6th Edition of Perry's Chemical Engineers' Handbook which I believe fully confirms that fact: Milton Beychok (Visit me at www.airdispersion.com) .


25362 (Chemical) 
15 Apr 08 1:27 
I'm sure nobody denied mbeychok's statement on the dependence of real gases Cp on P and T. The only point I thought worth of attention was that the socalled semiperfect gases aren't a subfamily of real gases. They are, contrarywise, members of the ideal gas tribe, and as such only arbitrary works of fiction.


mbeychok (Chemical) In the regions where P=P(v,T) in which VDW is met Cv is a function of only temperature. Regards 

There are 2 ways to calculate the sonic velocity:
1) Vs=223*SQRT(k*T/M) where Vs is ft/s where SQRT is square root where k is ratio of specifc heats (Cp/Cv) where T is temperature in degrees Rankine where M is gas mol weight
2) Vs=68.1*SQRT(k*P/rhov) where Vs is ft/s where SQRT is square root where k is ratio of specific heat (Cp/Cv) where P is pressure in psia where rhov is vapour density in lb/ft3 

One way really. These two equations are related via the ideal gas law T/M = 1/R x P/rho. Multiplying 223 by (1/R) ^{1/2} just happens to = 68.1 Good luck, Latexman 

hacksaw (Mechanical) 
2 May 08 21:41 
we now know why the term semiperfect is not used
interesting discussion


hacksaw (Mechanical) I suggest you "google" the term and see if it not used. You should be supprised. Regards


Hi
Can some explains me the Back Pressure on PSV.
Is it the Pressure immediately after the PSV minus the upstream pressure.
Is it the Pressure drop in the downstream pipeline.
What is the corelation of Back Pressure with Static Pressure.
Thanks 



