Calculating Length Increase of Aluminium Bar Due to Temperature Change

In summary, the temperature increase of a 40mm diameter bar of aluminium results in an increase of the length of the bar in millimeters. If the temperature increase is more, the increase in length is more proportional.
  • #1
speedy46
46
0

Homework Statement



A 40mm diameter bar of aluminium is 2.5m long the bar is has its temperature raised from 20°c to 40°c. Determine the increase in length of the bar in mm. Take the Al = 24 × 10-6 °C-1
How much heat has been transferred into the bar if the SCH = 0.9kj/kg °C


Homework Equations





The Attempt at a Solution



Am stuck on where to start could anyone help please
 
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  • #2
speedy46 said:

Homework Statement



A 40mm diameter bar of aluminium is 2.5m long the bar is has its temperature raised from 20°c to 40°c. Determine the increase in length of the bar in mm. Take the Al = 24 × 10-6 °C-1
How much heat has been transferred into the bar if the SCH = 0.9kj/kg °C

Homework Equations


The Attempt at a Solution



Am stuck on where to start could anyone help please
Saying that you are stuck on where to start isn't an attempt at a solution. You much have some ideas as to where to start. Have you read your class notes or course text?
 
  • #3
increase in length = linear expansion coefficent × length of material (t2 - t1)

I do not understand where the t2 - t1 come from.

Is it the temperture increase - the temperture of the bar before it is rasied.
 
  • #4
speedy46 said:
increase in length = linear expansion coefficent × length of material (t2 - t1)

I do not understand where the t2 - t1 come from.

Is it the temperture increase - the temperture of the bar before it is rasied.

Hi speedy46! :smile:

Yes, it's the temperature increase - final temperature minus initial temperature.

Increase in length is more if the original length is more,

and it's more if the temperature increase is more.

So it's proportional to both of them (in other words, to their product). :smile:
 
  • #5
would this also work for decreassing temperture on a material?
 
  • #6
speedy46 said:
would this also work for decreassing temperture on a material?

Yes … the effect is reversible.

If a particular temperature increase makes it longer, then the same decrease will return it to its original length. :smile:

And cooling it further will make it even shorter than its normal length.

Some armies used to keep iron cannonballs inside brass "barrels". If the temperature fell below freezing point overnight, the iron and the brass would both decrease in size, but the brass would decrease faster, so the cannonballs would get closer to the top, and sometimes one of them would even be forced out of the brass barrel! :biggrin:
 
  • #7
please could you tell me if this right or not

increase in length = 24 × 10-6 × 40 mm (40 - 20) 24 ×10-6 × 40 × 20 = 0.0192
 
  • #8
speedy46 said:
please could you tell me if this right or not

increase in length = 24 × 10-6 × 40 mm (40 - 20) 24 ×10-6 × 40 × 20 = 0.0192

Looks good! :smile:
 
  • #9
thank you now i have to work out how much heat has been transferred into the bar if the SCH= 0.9KJ/ KG degrees Celsius.

What does the SHC stand for please
 
  • #11
Thank you

To work out heat transfer

Q= (t1-t2)/(l÷Ka)

so 20 - 40 = 20

20 divide by 2.5m divide by ?

I do not know what the Ka stands for
 
  • #12
Hi speedy46! :smile:
speedy46 said:
Q= (t1-t2)/(l÷Ka)

so 20 - 40 = 20

20 divide by 2.5m divide by ?

I do not know what the Ka stands for

Sorry, I've no idea what Ka is … where did you get it from? :confused:

All you need can be found in the units of the SCH in the question:
speedy46 said:
SCH = 0.9kj/kg °C

/kg °C means per kilogram, and per degree.

So you find the number of kJ by multiplying 0.9 by the weight and also by the change in temperature … and nothing else! :smile:

btw, wikipedia gives the SCH as 0.9 J /g K … can you see why that's the same?
 
  • #13
so the change in temperture is 20 degress 20-40=20

but the question does not give the weight or can i work this out.
 
  • #14
speedy46 said:
so the change in temperture is 20 degress 20-40=20

but the question does not give the weight or can i work this out.

hmm … that's very true! :rolleyes:

they only give you the volume, and you would need to know the density also. :cry:
 
  • #15
density of aluminum is 2.7g/cm³

volume of the bar is πr^2*h π * 20 2 * 2500 = 3141592.654

I changed meters into millmeters to keep them in the same units 1000*2.5= 2500mm
 
Last edited:
  • #16
so now what do I need to do knowing the volume and density
 
  • #17
speedy46 said:
so now what do I need to do knowing the volume and density

mass = volume times density,

then SCH = 0.9kj per kilogram, and per degree. :smile:
 
  • #18
2.7 3 times 33141592.654 = 61835968.21

is this correct
 
  • #19
speedy46 said:
2.7 3 times 33141592.654 = 61835968.21

is this correct

Sorry, you've lost me. :confused:

Write it out in full … and remember you need the mass in kg.

(and don't work to 10 significant figures! :rolleyes: )
 
  • #20
volume of the bar is pie times 20 squared times 2500 = 3141592
The density of aluminum is 2.7g/cm3

Mass = volume times mass

mass = 3141592 times 2.7g/cm3

mass = 61835955.34

6.1kg

is this correct am unsure
 
  • #21
speedy46 said:
volume of the bar is pie times 20 squared times 2500 = 3141592
The density of aluminum is 2.7g/cm3

Mass = volume times mass

mass = 3141592 times 2.7g/cm3

mass = 61835955.34

6.1kg

is this correct am unsure

Use the same units that you'll need at the end:

volume of the bar in cm3 is π times 2 squared times 250 = 3142
The density of aluminum is 2.7g/cm3

mass = 3142cm3 times .0027kg/cm3 = … ? :smile:
 
  • #22
tiny-tim said:
Use the same units that you'll need at the end:

volume of the bar in cm3 is π times 2 squared times 250 = 3142
The density of aluminum is 2.7g/cm3

mass = 3142cm3 times .0027kg/cm3 = … ? :smile:


3142cm3 times .0027kg/cm3 = 8.4834 kg/cm3
 
  • #23
speedy46 said:
3142cm3 times .0027kg/cm3 = 8.4834 kg/cm3

erm … you mean 8.4834 kg. :rolleyes:

So the heat transferred in kJ is … ? :smile:

(btw, it's joules with a litlle "j", but J or kJ with a big "J" … like watts and W or kW, or Newtons and N or kN :smile:)
 
  • #24
so now i place that number into the equation

0.9 times 8.4834 times 20 = 152.70 KJ
 
Last edited:
  • #25
:biggrin: Woohoo! :biggrin:

(but "kJ", not "KJ"! :smile: )
 
  • #26
152.70 j
 
  • #27
speedy46 said:
152.70 j

Nooo … 152.7 kJ ! :smile:
 
  • #28
Thank you
 

1. How do you calculate the length increase of an aluminium bar due to temperature change?

To calculate the length increase of an aluminium bar, you can use the formula: ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of thermal expansion for aluminium, L is the original length of the bar, and ΔT is the change in temperature.

2. What is the coefficient of thermal expansion for aluminium?

The coefficient of thermal expansion for aluminium varies depending on the type of aluminium and the temperature range. On average, it is around 23 x 10^-6 m/mK.

3. How does temperature affect the length of an aluminium bar?

As the temperature of an aluminium bar increases, it expands and its length increases. Conversely, as the temperature decreases, the bar contracts and its length decreases.

4. What units are used to measure the coefficient of thermal expansion?

The coefficient of thermal expansion is measured in units of inverse temperature, such as meters per meter per degree Celsius (m/m℃) or meters per meter per kelvin (m/mK).

5. Are there any other factors that can affect the length increase of an aluminium bar?

Yes, apart from temperature and the coefficient of thermal expansion, other factors such as the initial length and cross-sectional area of the bar can also affect the length increase. Additionally, the presence of any external forces or constraints on the bar can also impact its length change.

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