This is great! I was just wondering, wouldn't you use the area of the pipe and the entrance at with pi(r^2) and 2alpha? I know it says for linear expansion but I figured if we're given diameter and not length, or both, the it's Area expansion instead of linear or volume. Please cure my overthinkingness lol
If both the rod and the plate are going to be heated, and the rod doesn't fit at 20 degrees C, then it will not fit at any temperature if both the rod and the plate are made of the same material.
In interference fits, it is common to cool parts & heat the outer body. This method will shrink the shaft ( maybe using liquid nitrogen) & heating the housing that it will be inserted into. You can then make the interference greater, which increases the holding capacity.
+Archer June Diaz Since Al has to expand more than the steel for the sizes to become the same The change in Al will be 0.01 cm more than the change in the steel
Another explanation: We want a temperature increase such that L Al + delta Al = L S + delta S. Plugging in the lengths, we get 3.99 cm + delta Al = 4 cm + delta S and subtracting 3.99 cm from both sides, we get delta Al = 0.01 cm + delta S.
great video thank you!!! I watched it because I fix appliances for a living and I was having a hard time to remove an aluminum pulley out of a steel shaft from a washer's motor... the shaft was a hexagon about one centimeter... I tried with a torch but didn't work so I left the motor in the freezer of my refrigerator overnight and... yay!!!... expansion didn't work for me it's time but it's there any video about contraction calculation???
Michael, The equation is correct. Since the coefficient of expansion for Aluminum is larger, it stands to reason that the change in length for Aluminum is greater than the change in length for steal.
I don't like to make such general statements. In every situation, you need to find a way to equate what is physically happening. In this case we could recognize that Al expands more than steel thus the expansion of Al will be 0.01 cm more than the expansion of steel when you can slip the pin in the hole. It is better to think about each problem in a similar fashion.
Dude!!! you have the same question that I have.... this is what I post after watching that video: great video thank you!!! I watched it because I fix appliances for a living and I was having a hard time to remove an aluminum pulley out of a steel shaft from a washer's motor... the shaft was a hexagon about one centimeter... I tried with a torch but didn't work so I left the motor in the freezer of my refrigerator overnight and... yay!!!... expansion didn't work for me it's time but it's there any video about contraction calculation???
is it because of the coefficient of the expansion the diameter of the al will be larger? because if we base on the chnge of length alone it does not make sense if you add 0.01 to the steel when its length is already larger than al, right?
Sir, during thermal expansion of diametrical components, whether it expands circumferencially or diametrically? Can you please clear this through comment section.
Can you please explain why the hole in the plate does not get small when we heat it because the metal expand in all directions so why does it not expand on the inner side of hole???
One way to think about is like this: 1. Take a flat piece of metal (without a hole) and draw a circle on it, the size of the hole. Now heat the metal. What will happen to the circle? If you let it cool and cut out the circle. And then just heat the circle, what will happen to the circular piece when you heat it?
@@MichelvanBiezen The circular piece will expand. But can you please explain this on a molecular level that what exactly resist the molecules to move inside the whole because they also have the space there?.......
Think of a Roman arch. What happens to the arch when you heat the stones in the arch? They expand, which means they need more room, so the length of the arch needs to increase to make room for the larger stones.
I cant agree with math here aluminium change size 24um per degree steel 11 um per degree so if i simple heat 100 degree rod it wilk give me total size 4.011cm for aluminium if i heat for 100 degre it will grow to 4.014cm so its clear will fit i made misteake somewhere??
Without knowing the initial temperature, I don't believe you can solve a problem. (unless other information is given from which we can infer the temperature).
Is it correct to set both final lengths equal to each other and then solve for the change in temperature, like this? L0(st)*K(st)*T = L0(al)*K(a)*T ...where K is the expansion coefficient.
M. Taiseer I think I meant the initial lengths. It was a long time since I last worked on this stuff, but my equation seemed to give the right answers when I used it.
M. Taiseer I also think the left and right side of the equation should give the difference between the final temperature and initial temperature for both respective materials.
M. Taiseer I don't clearly remember what I did in that equation, but I think these were the left and right sides: T-T0(st) T-T0(al) where T is the final temperature and T0 the initial temperature for each material. So I guess my intended equation should look like this: L0(st)*K(st)*(T-T0(st)) = L0(al)*K(al)*(T-T0(al))
if i only add heat in the aluminum plate, and take this expresion 4cm (Lo) = 3.99cm(Lo) + 24.10^-6 *(Tm-20)*3.99(Lo) if i clear Tm, and that gave me 124 C degrees... That could be a anwers if we want spare energy and dont want heat the bar steel?
is it because of the coefficient of the expansion the diameter of the al will be larger? because if we base on the chnge of length alone it does not make sense if you add 0.01 to the steel when its length is already larger than al, right?
From a production point of view, it is easier to heat objects than it is to cool them. Also the process works better if they both are at the same temperature, so there is no heat transfer during the process that could cause problems.
I've actually made the mistake in cooling than heating. Part was stuck on a mandrel, using dry ice, i froze the mandrel, shrinking it and therefore the part did come off. However, 4 days later that part turned to rust. Never again
what I like about your videos is that you make physics simple
Thank you for this, you are so much better at explaining this than my professor.
You are so great !!!!!! i have thermodynamics test coming up and i feel so much more confident thanks to you
Great! Glad you found our videos! 🙂
This is great! I was just wondering, wouldn't you use the area of the pipe and the entrance at with pi(r^2) and 2alpha? I know it says for linear expansion but I figured if we're given diameter and not length, or both, the it's Area expansion instead of linear or volume. Please cure my overthinkingness lol
It is just easier to use the linear expansion of both the hole (diameter) and the cylinder.
hi Michel. am from mauritius..and how you put the formula in the calculated ...thank ..
That is hard to explain like this espcially since the method of entry varies a lot depending on what type of calculator you have.
Ok I understand now..thank you
That is the starting difference between the diameter of the hole and the diameter of the shaft. The hole starts out 0.01 cm smaller than the shaft.
Would the temperature difference be large if the two objects were the same type of metal
If both the rod and the plate are going to be heated, and the rod doesn't fit at 20 degrees C, then it will not fit at any temperature if both the rod and the plate are made of the same material.
In interference fits, it is common to cool parts & heat the outer body.
This method will shrink the shaft ( maybe using liquid nitrogen) & heating the housing that it will be inserted into. You can then make the interference greater, which increases the holding capacity.
Initially the diameters differ by 0.01
Been looking for this!
Glad you found it. 🙂
i have a question why you did not use the area expansion why you use the linear expantion
You could, but it is easier to solve using the diameter.
Why is it you add the steel by 0.01cm rather than aluminium ?
+Archer June Diaz
Since Al has to expand more than the steel for the sizes to become the same
The change in Al will be 0.01 cm more than the change in the steel
Thank you sir!
Another explanation: We want a temperature increase such that L Al + delta Al = L S + delta S. Plugging in the lengths, we get 3.99 cm + delta Al = 4 cm + delta S and subtracting 3.99 cm from both sides, we get delta Al = 0.01 cm + delta S.
is it possible to have a negative delta T?
It depends on how you define it. When you define it as: Tf - To and To is higher than Tf then you will get a negative delta T.
great video thank you!!! I watched it because I fix appliances for a living and I was having a hard time to remove an aluminum pulley out of a steel shaft from a washer's motor... the shaft was a hexagon about one centimeter... I tried with a torch but didn't work so I left the motor in the freezer of my refrigerator overnight and... yay!!!... expansion didn't work for me it's time but it's there any video about contraction calculation???
That's great. Contraction is essentially the same as expansion with a negative.
sir, wouldnt it be the other way round for the addition of .01. it should be with the change in al + .01 equals the change in steel??
Michael,
The equation is correct. Since the coefficient of expansion for Aluminum is larger, it stands to reason that the change in length for Aluminum is greater than the change in length for steal.
wonderful explanation sir.....respect from Nepal (a beautiful country)
It certainly is. Welcome to the channel!
thank you sir
Sir your videos really help.....thank you so much!
Sir may I know whether this equation can be used only when two objects are being heated/cooled
I don't like to make such general statements. In every situation, you need to find a way to equate what is physically happening. In this case we could recognize that Al expands more than steel thus the expansion of Al will be 0.01 cm more than the expansion of steel when you can slip the pin in the hole. It is better to think about each problem in a similar fashion.
Dude!!! you have the same question that I have.... this is what I post after watching that video:
great video thank you!!! I watched it because I fix appliances for a living and I was having a hard time to remove an aluminum pulley out of a steel shaft from a washer's motor... the shaft was a hexagon about one centimeter... I tried with a torch but didn't work so I left the motor in the freezer of my refrigerator overnight and... yay!!!... expansion didn't work for me it's time but it's there any video about contraction calculation???
Thank you for lecture. from South Korea
Welcome to the channel
is it because of the coefficient of the expansion the diameter of the al will be larger? because if we base on the chnge of length alone it does not make sense if you add 0.01 to the steel when its length is already larger than al, right?
That is correct.
Wish i had your knowledge of physics..great vid!
This a very useful lesson, thank you
You are welcome!
This information is gold
Glad you find our videos helpful.
Sir, during thermal expansion of diametrical components, whether it expands circumferencially or diametrically? Can you please clear this through comment section.
Imagine a plate without a hole. Then draw a circle on the plate. Now heat the plate. What will happen to the drawn circle?
May I know why the Diameter will increase?
Make the plate solid (remove the hole) and draw a line where the hole is. Will the line move outward when you heat the plate?
Can you please explain why the hole in the plate does not get small when we heat it because the metal expand in all directions so why does it not expand on the inner side of hole???
One way to think about is like this: 1. Take a flat piece of metal (without a hole) and draw a circle on it, the size of the hole. Now heat the metal. What will happen to the circle? If you let it cool and cut out the circle. And then just heat the circle, what will happen to the circular piece when you heat it?
@@MichelvanBiezen The circular piece will expand. But can you please explain this on a molecular level that what exactly resist the molecules to move inside the whole because they also have the space there?.......
Think of a Roman arch. What happens to the arch when you heat the stones in the arch? They expand, which means they need more room, so the length of the arch needs to increase to make room for the larger stones.
I cant agree with math here aluminium change size 24um per degree steel 11 um per degree so if i simple heat 100 degree rod it wilk give me total size 4.011cm for aluminium if i heat for 100 degre it will grow to 4.014cm so its clear will fit i made misteake somewhere??
The answer in the video is correct. Note that they don't start out at the same size.
@@MichelvanBiezen thx
why do you mix centigrade dergees with degrees centigrade?
degrees centigrade is the actual temperature. Centigrade degrees is the temperature difference.
there is a hole in middle of copper plate if we heat the plate what will happen with diameter of hole???
The hole will expand (will get bigger)
@@MichelvanBiezen Thanks
why do we have to heat both of them instead of heating only aluminium
It makes it easier to slip the ring over the rod if they are both heated.
If Al expands more than the steel, why do we need to add the 0.01?
does the height/length of the cylinder change as well?
Yes, it does.
Sir what will we do if we don't have any temperature of any metal
Without knowing the initial temperature, I don't believe you can solve a problem. (unless other information is given from which we can infer the temperature).
But if you heat something shaped circle wouldn't the circle expand in all directions ?
+The Man Upon heating, both the diameter of the rod and the diameter of the hole will increase.
+Michel van Biezen Oh, yes! For sure!
Very interresting. I thought actually the whole in the aluminium would get smaller. Good to know that I was wrong.
👍
Hi Michel,
Can you show how you Calculate that with the Calculator?
That would be great.
Greetings
What if i was given Fdeg instead of Cdeg with lengths in meters unit?
Simply convert from F deg to C deg if the coefficient of expansion is given to you in C-deg
+Michel van Biezen helped me lots, thank you! :)
Thank you so much! Very helpful!
You are welcome.
Fantastic explanation, thanks so much!
Is it correct to set both final lengths equal to each other and then solve for the change in temperature, like this?
L0(st)*K(st)*T = L0(al)*K(a)*T
...where K is the expansion coefficient.
Laurelindo,
Try it and see if you get the same answer.
That is part of the learning process.
M. Taiseer I think I meant the initial lengths.
It was a long time since I last worked on this stuff, but my equation seemed to give the right answers when I used it.
M. Taiseer I also think the left and right side of the equation should give the difference between the final temperature and initial temperature for both respective materials.
M. Taiseer I don't clearly remember what I did in that equation, but I think these were the left and right sides:
T-T0(st)
T-T0(al)
where T is the final temperature and T0 the initial temperature for each material.
So I guess my intended equation should look like this:
L0(st)*K(st)*(T-T0(st)) = L0(al)*K(al)*(T-T0(al))
Isn't it circle soo we use pie/4 x d^2 ... It's not rectangular
The length of a rod and the diameter of a circle will change by the same amount.
I hope that you can make lecture about material and enegy balance related to chemical eng
if i only add heat in the aluminum plate, and take this expresion
4cm (Lo) = 3.99cm(Lo) + 24.10^-6 *(Tm-20)*3.99(Lo)
if i clear Tm, and that gave me 124 C degrees... That could be a anwers if we want spare energy and dont want heat the bar steel?
thanks a lot about your lecture
Sir Michel, to make it easier, just make the final lengths of steel and aluminum equal.
I got the same answer using this idea.
Very nice.
Thank you. Glad you liked it.
how to find final temperature??
Thanks 🤗
Welcome 😊
Sir, please make videos on thermodynamic potentials 😀
it is Ds-Dl = 0.01cm . Diameter of aluminum will be larger when it changed than diameter of steel. That's why he added 0.01 to Ds .
is it because of the coefficient of the expansion the diameter of the al will be larger? because if we base on the chnge of length alone it does not make sense if you add 0.01 to the steel when its length is already larger than al, right?
Sir please upload video on thermodynamic potential
you should do a question or a continuation of this question where 2 materials act upon each other so you can't apply the equation directly
Is T1 always 20
You can let T1 be whatever temperature you like. We just used that here in this example.
Thank you
why do we have to heat aluminium 0.01 greater than steel
Thank you sir
You're Welcome!
Thanx Sir
thank you so much
Arigato, sir.
Love the bow tie
thank you
why cant you just cool the steel and fit in
From a production point of view, it is easier to heat objects than it is to cool them. Also the process works better if they both are at the same temperature, so there is no heat transfer during the process that could cause problems.
Ok thanks helped a lot
I've actually made the mistake in cooling than heating. Part was stuck on a mandrel, using dry ice, i froze the mandrel, shrinking it and therefore the part did come off. However, 4 days later that part turned to rust. Never again
Thank you very much! :)) Also, awesome bowtie! ;)
Or, Interference fit.
he sounds a bit like gru from despicable me!
Where did .01 came from?
WHERE FROM 0.01CM
Thanks for this lecture
You are welcome.