Heat reduction based on compressionHeat Exchanger CalculationUnderstanding heat pump efficiencyHeat diffusion in a houseWhat is the temperature of compress air entering the atmosphere?For air conditioning, why is it 20BTU/hr/sq ft?Why does the refrigerant release heat when it is compressed into a liquid?Why do objects stop cooling down / losing heat?
Read file lines into shell line separated by space
What is this symbol: semicircles facing each other?
Why did MS-DOS applications built using Turbo Pascal fail to start with a division by zero error on faster systems?
Would the Republic of Ireland and Northern Ireland be interested in reuniting?
Remarkable applications of Dickson's lemma
Is there any method of inflicting the incapacitated condition and no other condition?
Tensorflow - logistic regrssion -oneHot Encoder - Transformed array of differt size for both train and test
Was it ever possible to target a zone?
Pythagorean triple with hypotenuse a power of 2
Why doesn't 'd /= d' throw a division by zero exception?
“T” in subscript in formulas
Thank God it's Friday, tomorrow is THE weekend. Why the definite article?
In the MCU, why does Mjölnir retain its enchantments after Ragnarok?
French abbreviation for comparing two items ("vs")
How do I request a longer than normal leave of absence period for my wedding?
Who was president of the USA?
How do I get a decreased-by-one x in a foreach loop?
The Knight's estate
Why isn't "I've" a proper response?
Compelling story with the world as a villain
What are some interesting features that are common cross-linguistically but don't exist in English?
Why in most German places is the church the tallest building?
Do they have Supervillain(s)?
Does norwegian.no airline overbook flights?
Heat reduction based on compression
Heat Exchanger CalculationUnderstanding heat pump efficiencyHeat diffusion in a houseWhat is the temperature of compress air entering the atmosphere?For air conditioning, why is it 20BTU/hr/sq ft?Why does the refrigerant release heat when it is compressed into a liquid?Why do objects stop cooling down / losing heat?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
So I was thinking about air conditioner today and how we run air across compressed freon to cool down air but why do we need freon at all why not store just compressed air. My guess is because its inefficient.
My question given a 1 cubic meter tank of air, if the room temperature at 1 atmosphere is 80 degrees. At what atmosphere would you have to have the tank so that the initial release of the air would be the temperature 70 degrees?
thermodynamics pressure air
asked 8 hours ago
johnny 5johnny 5
2341 gold badge2 silver badges7 bronze badges
$endgroup$
add a comment |
$begingroup$
So I was thinking about air conditioner today and how we run air across compressed freon to cool down air but why do we need freon at all why not store just compressed air. My guess is because its inefficient.
My question given a 1 cubic meter tank of air, if the room temperature at 1 atmosphere is 80 degrees. At what atmosphere would you have to have the tank so that the initial release of the air would be the temperature 70 degrees?
thermodynamics pressure air
asked 8 hours ago
johnny 5johnny 5
2341 gold badge2 silver badges7 bronze badges
$endgroup$
add a comment |
$begingroup$
So I was thinking about air conditioner today and how we run air across compressed freon to cool down air but why do we need freon at all why not store just compressed air. My guess is because its inefficient.
My question given a 1 cubic meter tank of air, if the room temperature at 1 atmosphere is 80 degrees. At what atmosphere would you have to have the tank so that the initial release of the air would be the temperature 70 degrees?
thermodynamics pressure air
asked 8 hours ago
johnny 5johnny 5
2341 gold badge2 silver badges7 bronze badges
$endgroup$
So I was thinking about air conditioner today and how we run air across compressed freon to cool down air but why do we need freon at all why not store just compressed air. My guess is because its inefficient.
My question given a 1 cubic meter tank of air, if the room temperature at 1 atmosphere is 80 degrees. At what atmosphere would you have to have the tank so that the initial release of the air would be the temperature 70 degrees?
thermodynamics pressure air
thermodynamics pressure air
asked 8 hours ago
johnny 5johnny 5
2341 gold badge2 silver badges7 bronze badges
asked 8 hours ago
johnny 5johnny 5
2341 gold badge2 silver badges7 bronze badges
edited 6 hours ago
johnny 5
asked 8 hours ago
johnny 5johnny 5
2341 gold badge2 silver badges7 bronze badges
asked 8 hours ago
johnny 5johnny 5
2341 gold badge2 silver badges7 bronze badges
asked 8 hours ago
johnny 5johnny 5
2341 gold badge2 silver badges7 bronze badges
2341 gold badge2 silver badges7 bronze badges
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let's first walk you through the usual refrigeration cycle:
We compress the freon, which heats it up. That's because we've put energy into in during the compression: It took work to push the piston.
Next we cool the gas back to room temperature. That's the big black coil on the back of the fridge.
Then we let it expand. The expansion cools the gas, because it does work (energy is removed) as it expands. Since it started at room temperature, it's cooler than room temperature after the expansion.
Finally, we let that cool gas absorb heat energy from whatever we want to keep cold. That warms the gas a bit, and gets us ready to start the cycle again.
If we didn't have step 2, the expansion in step 3 would just be the reverse of step 1 and get the freon back where it started. But since we did some of the cooling in step 2 (which we can do because it's hot from the compression), we end up with colder-than-room-temperature gas after 3.
Now, why do we do this with freon? The process I described above is for something like air which stays gaseous throughout. If you use freon (or sometimes ammonia or other chemicals) that will switch between gas and liquid during the process, it becomes more efficient. The physics is still basically the same, it's just that the boiling and condensing can transfer more heat energy without having to use really high pressures.
Back to a gas like air: If you want to calculate how much it chills as it expands, you're talking about "adiabatic expansion". The Wikipedia article is a good starting point. For that operation, there's a relation between the initial and final pressure and temperature:
$$P_i^1-gammaT_i^gamma = P_f^1-gammaT_f^gamma$$
Where $gamma = 1.6$ is a constant value for air. If we take the initial temperature to be roughly 80F or 300K, the final temp to be 70F or 295K and the final pressure to be $P_A = 10^5$ for atmospheric, then
$$ P_i = P_f (T_f/T_i)^gamma/(1-gamma) $$
$$ P_i = (10^5) (295/300)^1.6/(-0.6) = 106,000$$
That's a messy calculation, but because of the exponents ends up being that about a 6% increase in the pressure, when let out, will given you a 10F decrease in the temperature. In US units, that's from 14.7 PSIA to 15.6 PSAI; not all that much. Of course, a 10F drop isn't really enough to run a refrigerator that you'd really want...
answered 8 hours ago
Bob JacobsenBob Jacobsen
7,98213 silver badges24 bronze badges
$endgroup$
add a comment |
$begingroup$
Compression brings the molecules closer together thus cooling it down.
This isn't correct. The temperature of a gas isn't related to how close together the molecules are, but their speed.
By compressing the gas, they are closer together, but the work done in compression has sped them up as well. If you wait a while and let the gas cool, they'll still be closer together, but will be moving at the original speed.
My question is given air pressure at sea level. 80 degrees at 14.70 psi, how much pressure would you have to add to cool the air to 70 degrees if it was stored in a tank for 1 cubic meter?
There is no such pressure. Applying pressure will increase the temperature in the short term, not lower it. Any temperature change by doing this is only temporary. The gas will then exchange heat with the environment and move to ambient temperature.
answered 7 hours ago
BowlOfRedBowlOfRed
20k2 gold badges35 silver badges54 bronze badges
$endgroup$
$begingroup$
Thank you for clarifying my misunderstanding, I've made a minor edit to clarify the question, If you waited until the tank, cooled back down to room temperature, How much pressure would be needed so that when releasing the gas, it came out 10 degrees cooler?
$endgroup$
– johnny 5
6 hours ago
add a comment |
$begingroup$
I will address the use of air rather than freon (or its current replacement) as a refrigerant. I believe @BowlOfRed has satisfactorily answered your other questions.
Air is not used as a refrigerant because it would not be practical. This is because a refrigerator requires the use of a working fluid that can undergo phase changes (gas to liquid in condenser, liquid to gas in evaporator) at practical operating pressures and temperatures. Air cannot undergo phase changes except at extremely low temperatures.
For example, at 1 atmosphere the boiling/condensing point of refrigerant HFC-134a (which has replaced freon for environmental reasons) is about -25 C (-13 F). This is somewhat lower than the setting of a household freezer. In contrast, the boiling/condensing point of liquid air at 1 atmosphere is -194.4 C. In order to increase the boiling/condensing temperature to that required by the refrigerator, extremely impractically high pressure would be required.
Hope this helps.
answered 6 hours ago
Bob DBob D
12k3 gold badges10 silver badges36 bronze badges
$endgroup$
$begingroup$
Thanks, I'm learning alot about thermodynamics today. I see that it would be impractical to do such a thing. But to cool the air down 10 degrees I don't think you would need to store the air in liquid form. e.g if the air at 1 atmosphere is 80 degrees, and you have air store at 2 atomsphere at 80 degrees, upon expansion what would the temperature of the air be?
$endgroup$
– johnny 5
5 hours ago
$begingroup$
e.g in this example, pass air over cool pipe, you would just release the air back into the environment at a lower temperature
$endgroup$
– johnny 5
5 hours ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f498395%2fheat-reduction-based-on-compression%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's first walk you through the usual refrigeration cycle:
We compress the freon, which heats it up. That's because we've put energy into in during the compression: It took work to push the piston.
Next we cool the gas back to room temperature. That's the big black coil on the back of the fridge.
Then we let it expand. The expansion cools the gas, because it does work (energy is removed) as it expands. Since it started at room temperature, it's cooler than room temperature after the expansion.
Finally, we let that cool gas absorb heat energy from whatever we want to keep cold. That warms the gas a bit, and gets us ready to start the cycle again.
If we didn't have step 2, the expansion in step 3 would just be the reverse of step 1 and get the freon back where it started. But since we did some of the cooling in step 2 (which we can do because it's hot from the compression), we end up with colder-than-room-temperature gas after 3.
Now, why do we do this with freon? The process I described above is for something like air which stays gaseous throughout. If you use freon (or sometimes ammonia or other chemicals) that will switch between gas and liquid during the process, it becomes more efficient. The physics is still basically the same, it's just that the boiling and condensing can transfer more heat energy without having to use really high pressures.
Back to a gas like air: If you want to calculate how much it chills as it expands, you're talking about "adiabatic expansion". The Wikipedia article is a good starting point. For that operation, there's a relation between the initial and final pressure and temperature:
$$P_i^1-gammaT_i^gamma = P_f^1-gammaT_f^gamma$$
Where $gamma = 1.6$ is a constant value for air. If we take the initial temperature to be roughly 80F or 300K, the final temp to be 70F or 295K and the final pressure to be $P_A = 10^5$ for atmospheric, then
$$ P_i = P_f (T_f/T_i)^gamma/(1-gamma) $$
$$ P_i = (10^5) (295/300)^1.6/(-0.6) = 106,000$$
That's a messy calculation, but because of the exponents ends up being that about a 6% increase in the pressure, when let out, will given you a 10F decrease in the temperature. In US units, that's from 14.7 PSIA to 15.6 PSAI; not all that much. Of course, a 10F drop isn't really enough to run a refrigerator that you'd really want...
answered 8 hours ago
Bob JacobsenBob Jacobsen
7,98213 silver badges24 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's first walk you through the usual refrigeration cycle:
We compress the freon, which heats it up. That's because we've put energy into in during the compression: It took work to push the piston.
Next we cool the gas back to room temperature. That's the big black coil on the back of the fridge.
Then we let it expand. The expansion cools the gas, because it does work (energy is removed) as it expands. Since it started at room temperature, it's cooler than room temperature after the expansion.
Finally, we let that cool gas absorb heat energy from whatever we want to keep cold. That warms the gas a bit, and gets us ready to start the cycle again.
If we didn't have step 2, the expansion in step 3 would just be the reverse of step 1 and get the freon back where it started. But since we did some of the cooling in step 2 (which we can do because it's hot from the compression), we end up with colder-than-room-temperature gas after 3.
Now, why do we do this with freon? The process I described above is for something like air which stays gaseous throughout. If you use freon (or sometimes ammonia or other chemicals) that will switch between gas and liquid during the process, it becomes more efficient. The physics is still basically the same, it's just that the boiling and condensing can transfer more heat energy without having to use really high pressures.
Back to a gas like air: If you want to calculate how much it chills as it expands, you're talking about "adiabatic expansion". The Wikipedia article is a good starting point. For that operation, there's a relation between the initial and final pressure and temperature:
$$P_i^1-gammaT_i^gamma = P_f^1-gammaT_f^gamma$$
Where $gamma = 1.6$ is a constant value for air. If we take the initial temperature to be roughly 80F or 300K, the final temp to be 70F or 295K and the final pressure to be $P_A = 10^5$ for atmospheric, then
$$ P_i = P_f (T_f/T_i)^gamma/(1-gamma) $$
$$ P_i = (10^5) (295/300)^1.6/(-0.6) = 106,000$$
That's a messy calculation, but because of the exponents ends up being that about a 6% increase in the pressure, when let out, will given you a 10F decrease in the temperature. In US units, that's from 14.7 PSIA to 15.6 PSAI; not all that much. Of course, a 10F drop isn't really enough to run a refrigerator that you'd really want...
answered 8 hours ago
Bob JacobsenBob Jacobsen
7,98213 silver badges24 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's first walk you through the usual refrigeration cycle:
We compress the freon, which heats it up. That's because we've put energy into in during the compression: It took work to push the piston.
Next we cool the gas back to room temperature. That's the big black coil on the back of the fridge.
Then we let it expand. The expansion cools the gas, because it does work (energy is removed) as it expands. Since it started at room temperature, it's cooler than room temperature after the expansion.
Finally, we let that cool gas absorb heat energy from whatever we want to keep cold. That warms the gas a bit, and gets us ready to start the cycle again.
If we didn't have step 2, the expansion in step 3 would just be the reverse of step 1 and get the freon back where it started. But since we did some of the cooling in step 2 (which we can do because it's hot from the compression), we end up with colder-than-room-temperature gas after 3.
Now, why do we do this with freon? The process I described above is for something like air which stays gaseous throughout. If you use freon (or sometimes ammonia or other chemicals) that will switch between gas and liquid during the process, it becomes more efficient. The physics is still basically the same, it's just that the boiling and condensing can transfer more heat energy without having to use really high pressures.
Back to a gas like air: If you want to calculate how much it chills as it expands, you're talking about "adiabatic expansion". The Wikipedia article is a good starting point. For that operation, there's a relation between the initial and final pressure and temperature:
$$P_i^1-gammaT_i^gamma = P_f^1-gammaT_f^gamma$$
Where $gamma = 1.6$ is a constant value for air. If we take the initial temperature to be roughly 80F or 300K, the final temp to be 70F or 295K and the final pressure to be $P_A = 10^5$ for atmospheric, then
$$ P_i = P_f (T_f/T_i)^gamma/(1-gamma) $$
$$ P_i = (10^5) (295/300)^1.6/(-0.6) = 106,000$$
That's a messy calculation, but because of the exponents ends up being that about a 6% increase in the pressure, when let out, will given you a 10F decrease in the temperature. In US units, that's from 14.7 PSIA to 15.6 PSAI; not all that much. Of course, a 10F drop isn't really enough to run a refrigerator that you'd really want...
answered 8 hours ago
Bob JacobsenBob Jacobsen
7,98213 silver badges24 bronze badges
$endgroup$
Let's first walk you through the usual refrigeration cycle:
We compress the freon, which heats it up. That's because we've put energy into in during the compression: It took work to push the piston.
Next we cool the gas back to room temperature. That's the big black coil on the back of the fridge.
Then we let it expand. The expansion cools the gas, because it does work (energy is removed) as it expands. Since it started at room temperature, it's cooler than room temperature after the expansion.
Finally, we let that cool gas absorb heat energy from whatever we want to keep cold. That warms the gas a bit, and gets us ready to start the cycle again.
If we didn't have step 2, the expansion in step 3 would just be the reverse of step 1 and get the freon back where it started. But since we did some of the cooling in step 2 (which we can do because it's hot from the compression), we end up with colder-than-room-temperature gas after 3.
Now, why do we do this with freon? The process I described above is for something like air which stays gaseous throughout. If you use freon (or sometimes ammonia or other chemicals) that will switch between gas and liquid during the process, it becomes more efficient. The physics is still basically the same, it's just that the boiling and condensing can transfer more heat energy without having to use really high pressures.
Back to a gas like air: If you want to calculate how much it chills as it expands, you're talking about "adiabatic expansion". The Wikipedia article is a good starting point. For that operation, there's a relation between the initial and final pressure and temperature:
$$P_i^1-gammaT_i^gamma = P_f^1-gammaT_f^gamma$$
Where $gamma = 1.6$ is a constant value for air. If we take the initial temperature to be roughly 80F or 300K, the final temp to be 70F or 295K and the final pressure to be $P_A = 10^5$ for atmospheric, then
$$ P_i = P_f (T_f/T_i)^gamma/(1-gamma) $$
$$ P_i = (10^5) (295/300)^1.6/(-0.6) = 106,000$$
That's a messy calculation, but because of the exponents ends up being that about a 6% increase in the pressure, when let out, will given you a 10F decrease in the temperature. In US units, that's from 14.7 PSIA to 15.6 PSAI; not all that much. Of course, a 10F drop isn't really enough to run a refrigerator that you'd really want...
answered 8 hours ago
Bob JacobsenBob Jacobsen
7,98213 silver badges24 bronze badges
edited 5 hours ago
answered 8 hours ago
Bob JacobsenBob Jacobsen
7,98213 silver badges24 bronze badges
answered 8 hours ago
Bob JacobsenBob Jacobsen
7,98213 silver badges24 bronze badges
answered 8 hours ago
Bob JacobsenBob Jacobsen
7,98213 silver badges24 bronze badges
7,98213 silver badges24 bronze badges
add a comment |
add a comment |
$begingroup$
Compression brings the molecules closer together thus cooling it down.
This isn't correct. The temperature of a gas isn't related to how close together the molecules are, but their speed.
By compressing the gas, they are closer together, but the work done in compression has sped them up as well. If you wait a while and let the gas cool, they'll still be closer together, but will be moving at the original speed.
My question is given air pressure at sea level. 80 degrees at 14.70 psi, how much pressure would you have to add to cool the air to 70 degrees if it was stored in a tank for 1 cubic meter?
There is no such pressure. Applying pressure will increase the temperature in the short term, not lower it. Any temperature change by doing this is only temporary. The gas will then exchange heat with the environment and move to ambient temperature.
answered 7 hours ago
BowlOfRedBowlOfRed
20k2 gold badges35 silver badges54 bronze badges
$endgroup$
$begingroup$
Thank you for clarifying my misunderstanding, I've made a minor edit to clarify the question, If you waited until the tank, cooled back down to room temperature, How much pressure would be needed so that when releasing the gas, it came out 10 degrees cooler?
$endgroup$
– johnny 5
6 hours ago
add a comment |
$begingroup$
Compression brings the molecules closer together thus cooling it down.
This isn't correct. The temperature of a gas isn't related to how close together the molecules are, but their speed.
By compressing the gas, they are closer together, but the work done in compression has sped them up as well. If you wait a while and let the gas cool, they'll still be closer together, but will be moving at the original speed.
My question is given air pressure at sea level. 80 degrees at 14.70 psi, how much pressure would you have to add to cool the air to 70 degrees if it was stored in a tank for 1 cubic meter?
There is no such pressure. Applying pressure will increase the temperature in the short term, not lower it. Any temperature change by doing this is only temporary. The gas will then exchange heat with the environment and move to ambient temperature.
answered 7 hours ago
BowlOfRedBowlOfRed
20k2 gold badges35 silver badges54 bronze badges
$endgroup$
$begingroup$
Thank you for clarifying my misunderstanding, I've made a minor edit to clarify the question, If you waited until the tank, cooled back down to room temperature, How much pressure would be needed so that when releasing the gas, it came out 10 degrees cooler?
$endgroup$
– johnny 5
6 hours ago
add a comment |
$begingroup$
Compression brings the molecules closer together thus cooling it down.
This isn't correct. The temperature of a gas isn't related to how close together the molecules are, but their speed.
By compressing the gas, they are closer together, but the work done in compression has sped them up as well. If you wait a while and let the gas cool, they'll still be closer together, but will be moving at the original speed.
My question is given air pressure at sea level. 80 degrees at 14.70 psi, how much pressure would you have to add to cool the air to 70 degrees if it was stored in a tank for 1 cubic meter?
There is no such pressure. Applying pressure will increase the temperature in the short term, not lower it. Any temperature change by doing this is only temporary. The gas will then exchange heat with the environment and move to ambient temperature.
answered 7 hours ago
BowlOfRedBowlOfRed
20k2 gold badges35 silver badges54 bronze badges
$endgroup$
Compression brings the molecules closer together thus cooling it down.
This isn't correct. The temperature of a gas isn't related to how close together the molecules are, but their speed.
By compressing the gas, they are closer together, but the work done in compression has sped them up as well. If you wait a while and let the gas cool, they'll still be closer together, but will be moving at the original speed.
My question is given air pressure at sea level. 80 degrees at 14.70 psi, how much pressure would you have to add to cool the air to 70 degrees if it was stored in a tank for 1 cubic meter?
There is no such pressure. Applying pressure will increase the temperature in the short term, not lower it. Any temperature change by doing this is only temporary. The gas will then exchange heat with the environment and move to ambient temperature.
answered 7 hours ago
BowlOfRedBowlOfRed
20k2 gold badges35 silver badges54 bronze badges
answered 7 hours ago
BowlOfRedBowlOfRed
20k2 gold badges35 silver badges54 bronze badges
answered 7 hours ago
BowlOfRedBowlOfRed
20k2 gold badges35 silver badges54 bronze badges
answered 7 hours ago
BowlOfRedBowlOfRed
20k2 gold badges35 silver badges54 bronze badges
20k2 gold badges35 silver badges54 bronze badges
$begingroup$
Thank you for clarifying my misunderstanding, I've made a minor edit to clarify the question, If you waited until the tank, cooled back down to room temperature, How much pressure would be needed so that when releasing the gas, it came out 10 degrees cooler?
$endgroup$
– johnny 5
6 hours ago
add a comment |
$begingroup$
Thank you for clarifying my misunderstanding, I've made a minor edit to clarify the question, If you waited until the tank, cooled back down to room temperature, How much pressure would be needed so that when releasing the gas, it came out 10 degrees cooler?
$endgroup$
– johnny 5
6 hours ago
$begingroup$
Thank you for clarifying my misunderstanding, I've made a minor edit to clarify the question, If you waited until the tank, cooled back down to room temperature, How much pressure would be needed so that when releasing the gas, it came out 10 degrees cooler?
$endgroup$
– johnny 5
6 hours ago
$begingroup$
Thank you for clarifying my misunderstanding, I've made a minor edit to clarify the question, If you waited until the tank, cooled back down to room temperature, How much pressure would be needed so that when releasing the gas, it came out 10 degrees cooler?
$endgroup$
– johnny 5
6 hours ago
add a comment |
$begingroup$
I will address the use of air rather than freon (or its current replacement) as a refrigerant. I believe @BowlOfRed has satisfactorily answered your other questions.
Air is not used as a refrigerant because it would not be practical. This is because a refrigerator requires the use of a working fluid that can undergo phase changes (gas to liquid in condenser, liquid to gas in evaporator) at practical operating pressures and temperatures. Air cannot undergo phase changes except at extremely low temperatures.
For example, at 1 atmosphere the boiling/condensing point of refrigerant HFC-134a (which has replaced freon for environmental reasons) is about -25 C (-13 F). This is somewhat lower than the setting of a household freezer. In contrast, the boiling/condensing point of liquid air at 1 atmosphere is -194.4 C. In order to increase the boiling/condensing temperature to that required by the refrigerator, extremely impractically high pressure would be required.
Hope this helps.
answered 6 hours ago
Bob DBob D
12k3 gold badges10 silver badges36 bronze badges
$endgroup$
$begingroup$
Thanks, I'm learning alot about thermodynamics today. I see that it would be impractical to do such a thing. But to cool the air down 10 degrees I don't think you would need to store the air in liquid form. e.g if the air at 1 atmosphere is 80 degrees, and you have air store at 2 atomsphere at 80 degrees, upon expansion what would the temperature of the air be?
$endgroup$
– johnny 5
5 hours ago
$begingroup$
e.g in this example, pass air over cool pipe, you would just release the air back into the environment at a lower temperature
$endgroup$
– johnny 5
5 hours ago
add a comment |
$begingroup$
I will address the use of air rather than freon (or its current replacement) as a refrigerant. I believe @BowlOfRed has satisfactorily answered your other questions.
Air is not used as a refrigerant because it would not be practical. This is because a refrigerator requires the use of a working fluid that can undergo phase changes (gas to liquid in condenser, liquid to gas in evaporator) at practical operating pressures and temperatures. Air cannot undergo phase changes except at extremely low temperatures.
For example, at 1 atmosphere the boiling/condensing point of refrigerant HFC-134a (which has replaced freon for environmental reasons) is about -25 C (-13 F). This is somewhat lower than the setting of a household freezer. In contrast, the boiling/condensing point of liquid air at 1 atmosphere is -194.4 C. In order to increase the boiling/condensing temperature to that required by the refrigerator, extremely impractically high pressure would be required.
Hope this helps.
answered 6 hours ago
Bob DBob D
12k3 gold badges10 silver badges36 bronze badges
$endgroup$
$begingroup$
Thanks, I'm learning alot about thermodynamics today. I see that it would be impractical to do such a thing. But to cool the air down 10 degrees I don't think you would need to store the air in liquid form. e.g if the air at 1 atmosphere is 80 degrees, and you have air store at 2 atomsphere at 80 degrees, upon expansion what would the temperature of the air be?
$endgroup$
– johnny 5
5 hours ago
$begingroup$
e.g in this example, pass air over cool pipe, you would just release the air back into the environment at a lower temperature
$endgroup$
– johnny 5
5 hours ago
add a comment |
$begingroup$
I will address the use of air rather than freon (or its current replacement) as a refrigerant. I believe @BowlOfRed has satisfactorily answered your other questions.
Air is not used as a refrigerant because it would not be practical. This is because a refrigerator requires the use of a working fluid that can undergo phase changes (gas to liquid in condenser, liquid to gas in evaporator) at practical operating pressures and temperatures. Air cannot undergo phase changes except at extremely low temperatures.
For example, at 1 atmosphere the boiling/condensing point of refrigerant HFC-134a (which has replaced freon for environmental reasons) is about -25 C (-13 F). This is somewhat lower than the setting of a household freezer. In contrast, the boiling/condensing point of liquid air at 1 atmosphere is -194.4 C. In order to increase the boiling/condensing temperature to that required by the refrigerator, extremely impractically high pressure would be required.
Hope this helps.
answered 6 hours ago
Bob DBob D
12k3 gold badges10 silver badges36 bronze badges
$endgroup$
I will address the use of air rather than freon (or its current replacement) as a refrigerant. I believe @BowlOfRed has satisfactorily answered your other questions.
Air is not used as a refrigerant because it would not be practical. This is because a refrigerator requires the use of a working fluid that can undergo phase changes (gas to liquid in condenser, liquid to gas in evaporator) at practical operating pressures and temperatures. Air cannot undergo phase changes except at extremely low temperatures.
For example, at 1 atmosphere the boiling/condensing point of refrigerant HFC-134a (which has replaced freon for environmental reasons) is about -25 C (-13 F). This is somewhat lower than the setting of a household freezer. In contrast, the boiling/condensing point of liquid air at 1 atmosphere is -194.4 C. In order to increase the boiling/condensing temperature to that required by the refrigerator, extremely impractically high pressure would be required.
Hope this helps.
answered 6 hours ago
Bob DBob D
12k3 gold badges10 silver badges36 bronze badges
answered 6 hours ago
Bob DBob D
12k3 gold badges10 silver badges36 bronze badges
answered 6 hours ago
Bob DBob D
12k3 gold badges10 silver badges36 bronze badges
answered 6 hours ago
Bob DBob D
12k3 gold badges10 silver badges36 bronze badges
12k3 gold badges10 silver badges36 bronze badges
$begingroup$
Thanks, I'm learning alot about thermodynamics today. I see that it would be impractical to do such a thing. But to cool the air down 10 degrees I don't think you would need to store the air in liquid form. e.g if the air at 1 atmosphere is 80 degrees, and you have air store at 2 atomsphere at 80 degrees, upon expansion what would the temperature of the air be?
$endgroup$
– johnny 5
5 hours ago
$begingroup$
e.g in this example, pass air over cool pipe, you would just release the air back into the environment at a lower temperature
$endgroup$
– johnny 5
5 hours ago
add a comment |
$begingroup$
Thanks, I'm learning alot about thermodynamics today. I see that it would be impractical to do such a thing. But to cool the air down 10 degrees I don't think you would need to store the air in liquid form. e.g if the air at 1 atmosphere is 80 degrees, and you have air store at 2 atomsphere at 80 degrees, upon expansion what would the temperature of the air be?
$endgroup$
– johnny 5
5 hours ago
$begingroup$
e.g in this example, pass air over cool pipe, you would just release the air back into the environment at a lower temperature
$endgroup$
– johnny 5
5 hours ago
$begingroup$
Thanks, I'm learning alot about thermodynamics today. I see that it would be impractical to do such a thing. But to cool the air down 10 degrees I don't think you would need to store the air in liquid form. e.g if the air at 1 atmosphere is 80 degrees, and you have air store at 2 atomsphere at 80 degrees, upon expansion what would the temperature of the air be?
$endgroup$
– johnny 5
5 hours ago
$begingroup$
Thanks, I'm learning alot about thermodynamics today. I see that it would be impractical to do such a thing. But to cool the air down 10 degrees I don't think you would need to store the air in liquid form. e.g if the air at 1 atmosphere is 80 degrees, and you have air store at 2 atomsphere at 80 degrees, upon expansion what would the temperature of the air be?
$endgroup$
– johnny 5
5 hours ago
$begingroup$
e.g in this example, pass air over cool pipe, you would just release the air back into the environment at a lower temperature
$endgroup$
– johnny 5
5 hours ago
$begingroup$
e.g in this example, pass air over cool pipe, you would just release the air back into the environment at a lower temperature
$endgroup$
– johnny 5
5 hours ago
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f498395%2fheat-reduction-based-on-compression%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
