The chrome trigger is conductive, so will not build up a charge, and has the potential of being grounded. To achieve the optimum force out of a can of duster, spray short three to five-second blast at room temperature. When an aerosol duster is sprayed continually, it acts as an efficient heat sink and will cool down.
This lowering of the can temperature also lowers the internal pressure of the can, which will greatly decrease the generated force. For more information on finding the best aerosol duster for your application, contact Chemtronics at askchemtronics chemtronics. Be the first to receive email alerts on special offers, new products, and more delivered right to your in-box. Check often for exclusive offers, contests, product alerts and more.
All rights reserved. Privacy Statement Accessibility. Aerosol dusters contain one of these propellants: HFCa 1,1,1,2-Tetrafluoroethane, CAS — This is the same material generally used in modern air conditioning systems. It is nonflammable, making it popular in professional or industrial applications because it can be sprayed on energized circuits. The negative of HFCa is its environmental impact.
The trade-off is safety, since concentrated HFCa is flammable, so could light if exposed to an ignition source e. The risks are highest when spraying in an enclosed area, where vapors could collect, or if turning the can and expelling liquid.
HFOze trans- 1,3,3,3-Tetrafluoropropene, CAS — With the greater focus on global warming, new technology has become available for blowing off dust and dirt.
Because it is newer technology that has not been widely adopted by larger volume applications like air conditioning, it is relatively expensive compared to the other alternatives. The pressure outside the nozzle is somewhat above one atmosphere, so that gas does spray away from the nozzle, but compared to one atmosphere this does not present much of a ratio, so the adiabatic expansion as the gas moves away from the nozzle is not able to account for much of the observed cooling at the can.
So let's consider what happens inside the nozzle. Here the process is, to good approximation, a pressure drop in a constriction without heat exchange.
This process is called the Joule-Thomas process. It is a well-studied process much used for cooling and liquifying gases. I tried to find values for this coefficient for butane online. I got some numbers but was unsure, so for added confidence I used the van der Waals model. It's not a perfectly accurate model but pretty good.
So we conclude that the observed cooling 10 K or more is not primarily owing to the Joule-Thomson process in the nozzle. So far we have found that the answer is not adiabatic expansion after the nozzle, nor is it Joule-Thomson isenthalpic expansion in the nozzle.
So we have to look inside the can for our answer. I will first treat the case where there is liquid and vapour in the can, and then the case where only gas is involved. With liquid in the can, let's suppose in the first instance that there is not enough time for significant heat flow into the walls of the can.
In this case we have evaporative cooling. When some vapour escapes through the nozzle, the pressure in the can falls, and consequently some liquid evaporates. Both liquid and vapour then cool. For a long time I found this cooling puzzling from a thermodynamic point of view.
From a molecular point of view it is straightforward if hard to calculate : the faster molecules preferentially move out of the liquid, and on their journey they slow down because they are escaping from attractive forces in the liquid. But how to calculate this using thermodynamic properties? Eventually the system reaches a dynamic equilibrium with pressure approximately 1 atm outside the nozzle, pressure somewhere between 1 and 2 atm inside the can, and temperature somewhere between the temperature where the vapour pressure is 1 atm and the temperature where the vapour pressue is 2 atm.
This is the heat you would have to provide to cause unit mass of substance to change from liquid to gas in conditions of constant pressure and temperature. This is the case we are used to when we boil water in a kettle. But evaporative cooling is a different process. No heat is provided, but we do something which lowers the pressure e. It came from the rest of the system as internal energy moved from the rest of the system to this part.
I am here just giving some rough feel for the sorts of numbers involved. As I already said, the temperature will not fall indefinitely; it reaches a new equilibrium; the purpose of this rough calculation was merely to show that the energy movements are consistent. The summary of the above discussion is that if there is liquid in the can then the temperature drop is primarily owing to evaporative cooling of that liquid, and the vapour, because the latent heat of vapourization has to be provided by the contents of the can in the absence of heat flow from outside.
We already established that there is only a modest cooling after the gas has left the nozzle and makes its way into the room.
Let's look inside the can again. To understand the effect of a leak in a can of gas, imagine a thin membrane dividing the gas which is about to escape from the gas which will remain. As the gas escapes this membrane moves and the gas within it expands.
That expansion is, to good approximation, adiabatic. To prove this we need to claim that there is no heat transfer across this membrane. There will be no heat transfer if the gas on either side of the membrane is at the same temperature. If you think it is not, then allow me to add another membrane further down, dividing the gas which will remain into two halves.
This gas is all simply expanding so there is no reason for temperature gradients within it. But this argument will apply no matter where we put the membrane. We conclude that the part of the gas which remains in the can simply expands adiabatically to fill the can.
So now the initial calculation which I did, describing adiabatic expansion, is the right calculation, but one must understand that the process is happening right in the can! So no wonder the can gets cold! Here is another intuition for this. As it moves through the nozzle, the gas that gets expelled is being worked on by the gas that gets left behind, giving it energy, and the gas left behind loses energy.
If there were a big hole the gas would rush out very quickly. In the case of a narrow nozzle it is prevented from getting up to very high speed.
It that case it makes its way out into the surrounding atmosphere at a similar pressure to the ambient pressure and it uses up its extra energy pushing that atmosphere back to make room for itself.
Ron Maimon is basically correct when he attributes the drop in temperature to the work being done. Note that the gas would not come out of the can if the external pressure was the same or greater than the internal pressure in the can. As to the applicability of the ideal gas law, that depends on the uniformity of the system the can of gas.
The pressure is less at the nozzle than in the bulk of the gas, but that difference disappears in roughly the time it takes a sound wave to make a couple of trips through the can. If the pressure gradient is substantial, the system is not uniform, and we are in the realm of hydrodynamics and not thermodynamics. When gas molecules rush out from the can, in the can they were tightly packed, but in the room they will be now loosely packed and have longer flight distances before bouncing to other gas molecules.
In the can and shortly after being released, there are more molecules of sprayed gas in the cubic centimeter than air molecules in the surrounding air per cm3 pressure higher , but with lower molecule flying speed times higher molecule count same temp. Gas cold. Total sum average speed is less and temperature cooler than surrounding air. Its simpler to think what happens when compressed and why it heats, then deduce what happens, when it expands and why it cools.
The best and simple answer- All the K. E of the gas molecules is lost in getting out of tight nozzle and expanding. At last u are left with lowK. E gas molecules.
The contents of a spray can are in mostly liquid form both the product to be sprayed and the propellant. There is some head space above the liquid which is essentially the propellant in gas phase. As the mixture is released from the container, the volume of liquid decreases, with a corresponding increase in head space. To maintain equilibrium, some of the liquid propellant evaporates, which requires heat, so the temperature drops slightly.
Note that the mixture product and propellant leaving the can remains in liquid form until it passes through an orifice at which point the pressure drops, and the propellant evaporates, drawing heat from its surroundings. This has no effect on the temperature of the can because it has effectively already left it.
The can gets colder because of evaporation inside the can to maintain equilibrium between liquid and gas phases of the propellant, not because of what may be going on at the nozzle. Many comments pointed out that the actual process can be significantly non-isentropic and irreversible. My earlier treatment using the first law assuming an isentropic process provides the maximum cooling case for gas only in the can no liquid.
This estimate can be off, as pointed out in a good comment by bigjosh about release from a bicycle tire. I think a more realistic evaluation should consider the following. The rate of gas flow out the opening depends on the discharge coefficient ratio actual to maximum gas flow ; for a machined nozzle the discharge coefficient can be 0. For the case of the bicycle tire, the coefficient of discharge through the valve may be very high.
Also the tire tube volume decreases with gas loss, doing work on and raising the temperature of the gas in the tube. This question can be answered using the first law of thermodynamics for an open system. The following example is based on an example in the text Elements of Thermodynamics and Heat Transfer, by Obert and Young. The reverse is also true. When a gas expands, it gets cooler. These two ideas are the basis of your refrigerator and air conditioner. A gas is squeezed by a device called a compressor.
This makes the gas quite hot. The hot gas moves through metal coils that let the heat move from the gas to the surrounding air. That is why warm air comes from under your refrigerator. Then the gas is moved to the area that you want chilled. There, it is allowed to expand. That causes it to absorb heat from its surroundings, leaving things there much cooler. But expanding gas is only part of the cooling.
Shake the spray can. You should feel something sloshing around inside. Check fill weight — Retail duster can come in a variety of fill weights the content of the material in a can including 3. This metric can only be accurately judged by the label since the size of the spray can may not vary.
Duster Doesn't Hurt The Ozone Layer There is a misconception based on old information that dusters hurt the ozone layer.
High spray force — Each type of Techspray has the potential to produce a certain amount of force. Techspray offers duster products with even higher spray force by controlling output through both the valving and the sprayer.
Under high pressure, the propellant is mostly liquid, with the empty part of the can filled with vapor. When you spray an aerosol duster, the vapors of the propellant are expelled. As a side note, this also explains why it is a bad idea to shake up a can of duster before use, as is the habit of using spray paint.
If you shake the duster, the liquid propellant is more likely to sputter out as you start spraying. These components allow you to reuse the sprayer and purchase refill cans.
The advantage of the chrome sprayer is more precise control over spray force and ESD electrostatic discharge control. The standard plastic sprayer has the potential to build up a static charge as you spray. The chrome trigger is conductive, so it will not build up a charge, and it has the potential to be grounded. Shop air — In an industrial setting, dropping an airline is very common.
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