MDME: MANUFACTURING, DESIGN, MECHANICAL ENGINEERING 

GASES


A perfect (or ideal) gas is assumed. This is accurate enough if the pressure is not too high, nor the gas close to liquifaction point (saturation).
Calculations with such gases obey;
1. The perfect gas law pV = mRT
2. The 1st law of Thermo U2-U1= Q-W (everything does!)
See table for various gases processes that determine how we treat Q=mcT, and W=pV 

SmartBoard Notes:  Gases.pdf     Gases.one

Perfect Gas

The perfect gas equation;

pV = mRT


Where;
p = pressure (Pa) absolute!
V = volume (m3)
m = mass of gas (kg)
R = 8.314472 / M  (J/kgK)  where M = relative molecular mass (no units)
T = temperature (K) Kelvin!



Notes:
Kelvin (K) = Celsius (oC) + 273.15
Standard Pressure is defined as one atm. or 760.0 mm Hg or 101.325 kPa.
Standard Temperature is 20oC
STP = Standard Temperature and Pressure

General Gas Equation (Also called Combined Gas Equation)
For before-and-after processes on a constant mass of gas, using the perfect gas pV/T = mR = constant, so;


Summary of Special Gas Processes 

Just use this table for all your gas problems. From Kinsky p113.



The polytropic index varies according to the type of process, as shown in the table below.

Variation of polytropic index n
Polytropic
index
Relation Effects
n < 0  — An explosion occurs
n = 0 pV0 = p
(constant)
Isobaric: Equivalent to an isobaric process (constant pressure)
n = 1 pV = NkT
(constant)
Isothermal: Equivalent to an isothermal process (constant temperature)
1 < n < γ  — Polytropic: A quasi-adiabatic process such as in an internal combustion engine during expansion, or in vapor compression refrigeration during compression
n = γ  — Adiabatic: γ =   is the adiabatic index, yielding an adiabatic process (no heat transferred)
 — Equivalent to an isochoric process (constant volume)

Perfect Gas Simulator 


Tim Lovett 2010

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See also: NASA Gas Lab Animations

The simple processes;

CONSTANT VOLUME (ISOCHORIC)
(Gay-Lussac's Law 1809)

Constant Volume
Modify Temperature
Watch Pressure change

CONSTANT PRESSURE (ISOBARIC)
(Charles' Law 1787)

Constant Pressure
Modify Temperature
Watch Volume change

CONSTANT TEMPERATURE  (ISOTHERMAL)
(Isothermal: Slow enough to allow complete heat transfer)

Constant Temperature
Modify Volume
Watch Pressure change


Example:


This 1.25L PET (Polyethylene terephthalate) bottle was pumped up to 150 PSI.

 

How much energy was released when it exploded?

Since this gas process happened so quickly (faster than 1/240th of a second), there is no time for heat transfer.
The gas process is adiabatic, so n = γ. (Where, for air, γ = 1.4)

Work =
       
We need to find the pressure and volumes in the first and second states.
p1 = 150 PSI = 1034.214 kPa = 1034214 Pa
p2 = atmospheric = 101300 Pa
V1 = 1.25 L = 0.00125 m3
V2 = ?

Use the relation: pVn= constant.
In other words: p1V1n= p2V2n
Therefore we can find V2;
p2V2n = p1V1n
V2n = (p1V1n) / p2
V2 = ((p1V1n) / p2)^(1/n)
      = ((1034214 * 0.001251.4) / 101300) ^ (1/1.4)
      = 0.006571 m3

This is the volume to which the air expanded after the bottle burst. It expands only 5.25 times, even though the pressure was 10.2 times. Had it been an isothermal process (constant temperature) it would have to draw heat in during expansion, making it expand 10.2 times. Why is this final volume less when adiabatic? Because as air expands it gets colder, which makes the volume smaller.


Now put these figures back into the Work equation:
Work =  (p1V1 - p2V2) / (n-1)
        =  (1034214 * 0.00125 - 101300 * 0.006571) / (1.4-1)
        = 1567.813 Joules (1.5 kJ)

How does this compare to the energy of a bullet?

Example muzzle energy levels of different types of firearms (Wikipedia)
Firearm (except listed air guns)CaliberMuzzle energy
JoulesWhft-lbs
air gun spring.177200.0115
air gun mag spring.22300.0122
air gun PCP.2240+0.0130+
pistol.22LR1590.04117
pistol9 mm5190.14383
pistol.45 ACP5640.16416
pistol.357 Magnum8730.24640
pistol10mm1,0570.29775
1.25L PET bottle 1 MPa 1,500    
rifle5.56×45 mm1,7960.51,325
rifle7.62×39 mm2,0700.581,527
rifle7.62 × 51 mm3,7991.062,802
rifle.338 Lapua Magnum6,6341.844,893
heavy.50 BMG15,0374.1811,091
heavy14.5 × 114 mm32,0008.8923,744

The PET bottle released more energy than a 10mm pistol, and a bit less than a 5.56mm rifle. Glad it didn't pop while I was holding it.

What was the temperature of the air after the explosion?

For an adiabatic expansion,

Assming T1 = 25 oC (298 K)

T2 = ( (p2 / p1) ^ (n-1/n) ) * T1

= ( (101300/1034214) ^ (1.4 - 1/1.4) ) * 298

= 60.58 K (-212.4 oC)


What was the mass of air in the pressurized PET bottle?

pV = mRT

m = pV / RT = 1034214*0.00125 / (287 * 298) = 0.01512 kg (15 grams)

Compare this to atmospheric pressure for 1.25L.

1 cubic meter of air at atmospheric pressure and 25 oC will have a mass of...

m = (101300*1) / (287 * 298) = 1.1844 kg

So for 1.25 L, the mass is: 1.1844 *1.25/1000 = 0.0014805 kg (1.5 grams)

As expected, the compressed air is 10 times the mass of atmospheric air.

 

 



Questions:

Homework Assignment: Kinksy
Do all questions; Chapter 5: Gases
5.1 to 5.25 (page 114-117) Perfect Gas Laws
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