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. Consider the system shown in the figure below consisting of compressed gas in a container with a hole.
There is mass flow out the system to the atmosphere which is at constant pressure Pout. As mass leaves the container, the pressure and temperature of the gas inside the container, Pin t and Tin t change with time, t. There is no heat added and no work done on the gas in the container.
To simplify the evaluation for discussion here, assume the process is isentropic; that is, the specific entropy s is constant. Using the first law, after time t. Given the initial pressure and temperature of gas in the container, the specific enthalpy, s, can be calculated.
The temperature of the gas out the hole is that for the state s, Pout. Also, once the pressure inside the container reduces to Pout, the gas inside the container is at state s, Pout and has the same temperature as the gas out the hole. For example, assume the gas inside the container is dry air initially at 0. For this state, s is 6. For the air out the hole Pout is 0. So assuming an isentropic process the temperature of the gas out the hole is K - 55 F which is also the temperature of the gas inside the contained once it reaches atmospheric pressure.
The gas inside the contained cools substantially as it ejects gas. To calculate the rate of depressurization, an evaluation of the area of the hole and the velocity out the hole is required; the flow out the hole may be choked flow while the container pressure is sufficiently high.
The Joule-Thompson coefficient is defined as the partial derivative of temperature with respect to pressure at constant enthalpy. See a good thermodynamics text such as one by Obert, or Sonntag and Van Wylen. Joule-Thompson expansion is typically explained using a throttling process in which the change in fluid velocity is negligible. As discussed above, a pure gas cools off for this process. The above are not correct reasons for cooling of gas.
The gas flowing from a high pressure inside to outside through a small nozzle or mouth. When the gas suddenly releases to the outside, there is not any energy available; the pressure reduction adjusted with volume rather than temperature in this case.
So the cooled air temperature remains the same. Their energies form a Gaussian distribution about the temperature of the gas there. It is a sorting mechanism, selecting only certain molecules. The collection of these molecules will have, more or less, a collective exit velocity. As they exit, these molecules do work against the molecules in the atmosphere. By the Furst Law, they will be cooler than the temperature inside the can.
Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Why does the gas get cold when I spray it? Ask Question. Asked 10 years, 2 months ago. Active 9 months ago. Viewed k times. Improve this question. Ken D Ken D 1 1 gold badge 4 4 silver badges 6 6 bronze badges.
Boiling is a cooling process. I wonder how much of this post went into informing that video. Add a comment. Think of it as a portable replacement for an air compressor line. This accessory can be used to blow cookie crumbs out from a keyboard, clear out computer vents, and even open up tight apertures in an SMT surface mount technology stencil in PCB assembly.
If we missed your specific question, please leave us a comment below or contact us and we'll get it answered. As a matter of fact, inhaling too many of these vapors can lead to negative health effects. Compressed Air Duster is most commonly filled with the following propellants:.
Customers looking for the lowest price often look to consumer retailers. Consumer duster is designed to be as cheap as possible, sometimes imported from outside the US.
There is a misconception based on old information that dusters hurt the ozone layer. To achieve the optimum force from a can of duster, spray a short three to five-second blast at room temperature.
When an aerosol duster is sprayed continuously, 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. As the above details demonstrate, care needs to be taken even when choosing an everyday consumable like a duster. Reach out to us or your distributor for help deciding which duster is right for your application.
Americas English Spanish. My account Login View Cart 0. Compressed Air Duster is most commonly filled with the following propellants: HFCa — Nonflammable, most common for industrial applications when spraying energized circuits because of the risk of a spark lighting a flammable material.
It is flammable, and will ignite when concentrated i. It is non-flammable and is almost indistinguishable from HFCa when sprayed, but unfortunately at a much higher price.
CO 2 - It's not as common because the spray force is not consistent. When there is the possibility of sparks or flames e. Each type propellant used in aerosol dusters have potential to produce a specific amount of force. Chemtronics offers aerosol dusters with even higher spray force by controlling output through the valve and spray head.
These provide an extra boost to push out heavier particles and anywhere more aggressive cleaning is required. Under high pressure, the propellant is mostly liquid, with the empty part of the call filled with vapor. When you spray an aerosol duster, the vapors of the propellant are expelled. It also sprays with a very high force to dislodge stubborn contamination. Aerosol dusters predominantly come in a steel can with a plastic trigger sprayer.
It is possible for the sprayer to become charged from either handling or from the propellant being discharged, so may be a concern for cleaning sensitive electronics.
The sprayer can be reused with refill cans that are sold separately. The advantage of the chrome sprayer is more precise spray control and ESD electrostatic discharge control. 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.
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