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Question on unit conversion with temperatures

by sgstudent
Tags: conversion, temperatures, unit
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sgstudent
#1
Apr20-13, 01:53 AM
P: 645
Hi, I was given 1Btu/(h.ft^2.F) to W/m^2.K with these conversion ratios 1m^3=1000L 1Btu=1.055*103 1m=3.2808ft and Δ1K=Δ1.8F

I can convert most of the units but I'm stumped by the temperature conversion here. Because I'm not converting a change in temperature but the discreet value of temperature here. So I'm not sure why I can just use the change in K to F for that conversion. Shouldn't we use the conversion ratio for the discreet K and F instead?

Hoping you guys can help me out with this :)
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jbriggs444
#2
Apr20-13, 05:36 AM
P: 905
The quantities measured by the units you show, what do they represent?

I suspect that they are a measure of a surface's resistance to heat flow. You get a flow of energy that is proportional to time, proportional to surface area and proportional to the difference in temperature from the hot side to the cold side.
sgstudent
#3
Apr20-13, 06:04 AM
P: 645
Quote Quote by jbriggs444 View Post
The quantities measured by the units you show, what do they represent?

I suspect that they are a measure of a surface's resistance to heat flow. You get a flow of energy that is proportional to time, proportional to surface area and proportional to the difference in temperature from the hot side to the cold side.
Sorry I'm not sure what that equation represents. I was only asked in the question to change the units. But I'm not sure why I can use the Δ1K=Δ1.8F for that though. Do you have any idea why I can do that?

jbriggs444
#4
Apr20-13, 07:15 AM
P: 905
Question on unit conversion with temperatures

Quote Quote by sgstudent View Post
Sorry I'm not sure what that equation represents. I was only asked in the question to change the units. But I'm not sure why I can use the Δ1K=Δ1.8F for that though. Do you have any idea why I can do that?
The decision on whether to use delta temperature or absolute temperature rests on what the equation represents. In this case the relevant quantity is a temperature difference.
sgstudent
#5
Apr20-13, 08:11 AM
P: 645
Quote Quote by jbriggs444 View Post
The decision on whether to use delta temperature or absolute temperature rests on what the equation represents. In this case the relevant quantity is a temperature difference.
Oh but I was told mathematically it is correct to use the change in K to F even if the units given is a discreet K or F. Is that wrong?
Redbelly98
#6
Apr20-13, 03:04 PM
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To repeat what jbriggs said, these units refer to temperature differences, not actual temperature values. "1Btu/(hft2F)" means that the density of heat flow is 1 BTU/h per square meter of material, per F of temperature difference between two sides of that material (through which heat is flowing).

It is simply not customary to include the "Δ" when expressing units like this. So we don't write it as BTU/(hft2ΔF). It's just what the convention is, and it is up to you to understand whether it is referring to a temperature difference (or change), or an actual temperature value.

Hope that helps.
sgstudent
#7
Apr20-13, 08:13 PM
P: 645
Quote Quote by Redbelly98 View Post
To repeat what jbriggs said, these units refer to temperature differences, not actual temperature values. "1Btu/(hft2F)" means that the density of heat flow is 1 BTU/h per square meter of material, per F of temperature difference between two sides of that material (through which heat is flowing).

It is simply not customary to include the "Δ" when expressing units like this. So we don't write it as BTU/(hft2ΔF). It's just what the convention is, and it is up to you to understand whether it is referring to a temperature difference (or change), or an actual temperature value.

Hope that helps.
Hi thanks so much. This helps immensely :)

So if we were to know that the K is actually the discreet value and not the temperature difference, then we would have to use the other conversion ratio to solve it? So actually if I weren't taught that formula I wouldn't be able to solve it?

Also, why does temperature work in a different way as Litres. For example if we have a change in 50L and we want to change it to , we can just change it directly to cm3 we can just use the unit conversion of 1L=1000cm3 but for temperature we cannot do that?

And thanks so much. This has been bugging me ever since my tutorial.
Redbelly98
#8
Apr20-13, 09:09 PM
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Quote Quote by sgstudent View Post
Hi thanks so much. This helps immensely :)

So if we were to know that the K is actually the discreet value and not the temperature difference, then we would have to use the other conversion ratio to solve it?
For discrete temperatures, it is not simply a conversion ratio. There is a formula, e.g. subtracting 273 from the Kelvin value to get Celsius, and then apply the usual conversion from C to F. But surely you have been taught this?

So actually if I weren't taught that formula I wouldn't be able to solve it?
Not sure if I understand your question. Yes -- if you were not taught how to convert between different temperature scales, you would be unable to do so until you learned how. But that is an obvious thing to say, so I am wondering if you really meant that or something else?

Also, why does temperature work in a different way as Litres. For example if we have a change in 50L and we want to change it to , we can just change it directly to cm3 we can just use the unit conversion of 1L=1000cm3 but for temperature we cannot do that?
Because 0 L is the same as 0 cm3, so converting the quantity works by using a conversion factor, just as converting a change in the quantity works.

But, for example, 0 C is not 0 F. So converting between C and F is more complicated than doing a simple multiplication.
sgstudent
#9
Apr22-13, 07:53 AM
P: 645
Quote Quote by Redbelly98 View Post
For discrete temperatures, it is not simply a conversion ratio. There is a formula, e.g. subtracting 273 from the Kelvin value to get Celsius, and then apply the usual conversion from C to F. But surely you have been taught this?


Not sure if I understand your question. Yes -- if you were not taught how to convert between different temperature scales, you would be unable to do so until you learned how. But that is an obvious thing to say, so I am wondering if you really meant that or something else?


Because 0 L is the same as 0 cm3, so converting the quantity works by using a conversion factor, just as converting a change in the quantity works.

But, for example, 0 C is not 0 F. So converting between C and F is more complicated than doing a simple multiplication.
Thanks so much for the detailed explanation. What I meant was if I had a formula like PV=nRT and so if I were to convert the units of R from K to F I would have to use the conversion of the discreet temperature values rather than the change in variables?
MrAnchovy
#10
Apr22-13, 08:05 AM
P: 418
Quote Quote by sgstudent View Post
If I had a formula like PV=nRT and so if I were to convert the units of R from K to F I would have to use the conversion of the discreet temperature values rather than the change in variables?
Temperatures measured in Farenheit cannot be used in the equation PV=nRT (what would this imply at 0F?)
jbriggs444
#11
Apr22-13, 09:29 AM
P: 905
Quote Quote by MrAnchovy View Post
Temperatures measured in Farenheit cannot be used in the equation PV=nRT (what would this imply at 0F?)
Let me add a little detail to this.

Suppose that you had PV=nRT where T was expressed as a reading using the Kelvin scale and R was expressed in units of [mumble] per degree Kelvin. You could change from the Kelvin scale to the Rankine scale (the Rankine scale is Fahrenheit degrees above absolute zero). Then the units on R would change from degrees Kelvin to degrees Fahrenheit.

But if you wanted to use the Fahrenheit scale, that's not just a "linear" transformation. That's an "affine" transformation. The units on R would be the same, but you would need to change the form of the equation. In painful detail, that's...

PV = nRkelvinTKelvin
becomes... PV = nRRankineTRankine
becomes... PV = nRRankine(TFahrenheit+459.67)
becomes... PV = nRFahrenheit(TFahrenheit+459.67)
sgstudent
#12
Apr22-13, 06:38 PM
P: 645
Quote Quote by jbriggs444 View Post
Let me add a little detail to this.

Suppose that you had PV=nRT where T was expressed as a reading using the Kelvin scale and R was expressed in units of [mumble] per degree Kelvin. You could change from the Kelvin scale to the Rankine scale (the Rankine scale is Fahrenheit degrees above absolute zero). Then the units on R would change from degrees Kelvin to degrees Fahrenheit.

But if you wanted to use the Fahrenheit scale, that's not just a "linear" transformation. That's an "affine" transformation. The units on R would be the same, but you would need to change the form of the equation. In painful detail, that's...

PV = nRkelvinTKelvin
becomes... PV = nRRankineTRankine
becomes... PV = nRRankine(TFahrenheit+459.67)
becomes... PV = nRFahrenheit(TFahrenheit+459.67)
Hi I'm not too sure what that all means. Why must I convert them to rankine first then Fahrenheit?

The question for the Units conversion for R was this 82.06(atm.cm3)/gmole.K to (psia.ft3)/lbmole.F

So what I used was the conversion ratio of Δ1K=Δ1.8F for this question. But I don't think I can do that because those temperature units doesn't represent the change in temperature right? So must I just substitute 1K for -457.87F instead?

But the answer for that conversion for the R in the gas law was 10.73(psia.ft3)/gmole.K

Sorry for not understanding the explanation and thanks for helping me out.
MrAnchovy
#13
Apr23-13, 04:15 AM
P: 418
Quote Quote by sgstudent View Post
I'm not converting a change in temperature but the discreet value of temperature here.
This is incorrect, and is at the root of your confusion.

A BTU is the amount of energy required to raise the temperature of a cubic foot of water by one degree Fahrenheit, so it certainly does relate to a change in temperature.

The gas equation relates to absolute temperature and so can only be used with a scale that is zero at absolute zero. If you want to express the gas constant in a non-zero based scale such as Fahrenheit you need to apply the absolute zero correction to T to get PV=nRarbitrary(T-T0).
sgstudent
#14
Apr23-13, 06:04 AM
P: 645
Quote Quote by MrAnchovy View Post
This is incorrect, and is at the root of your confusion.

A BTU is the amount of energy required to raise the temperature of a cubic foot of water by one degree Fahrenheit, so it certainly does relate to a change in temperature.

The gas equation relates to absolute temperature and so can only be used with a scale that is zero at absolute zero. If you want to express the gas constant in a non-zero based scale such as Fahrenheit you need to apply the absolute zero correction to T to get PV=nRarbitrary(T-T0).
Oh! But if i use the formula PV=nRΔT won't it be wrong? Because V is directly proportional to temperature and not change in temperature right?

Or if we set the initial temperature to be 0 degrees then we can use it?

But actually if something is non zero why can't we use it?
jbriggs444
#15
Apr23-13, 06:48 AM
P: 905
Quote Quote by sgstudent View Post
Oh! But if i use the formula PV=nRΔT won't it be wrong? Because V is directly proportional to temperature and not change in temperature right?
Right. Because that formula expects temperatures to be expressed on an absolute scale.

For purposes of assigning units of measurement to conversion factors, that does not matter. The conversion factor for changing from one absolute scale to another is the same as the conversion factor for changing from one temperature difference measure to another.

And that makes sense. After all, an absolute temperature scale just expresses the difference between the measured temperature and absolute zero.
MrAnchovy
#16
Apr23-13, 10:07 AM
P: 418
Quote Quote by sgstudent View Post
Or if we set the initial temperature to be 0 degrees then we can use it?
There is no "initial temperature" in the equation PV=nRT: perhaps I confused you by using the symbol T0 to mean the value of absolute zero.

In the 18th century the work of Boyle, Charles and others explored the relationship between changes in pressure, temperature and volume of a gas and worked in Farenheit or Celcius/Centigrade; they observed the proportionality of changes but did not have an adequate model of the underlying physical processes.

In the 19th century the work of Joule and Thomson (who later adopted the title Lord Kelvin) revealed a new understanding of thermodynamics implying more fundamental scales of measurement which recognised the concept of absolute zero and the covertability of heat and mechanical work. It is no accident that we still use their names today when measuring these quantities in science, rather than units such as Farenheit and Horse Power.
sgstudent
#17
Apr26-13, 02:00 AM
P: 645
Quote Quote by MrAnchovy View Post
There is no "initial temperature" in the equation PV=nRT: perhaps I confused you by using the symbol T0 to mean the value of absolute zero.

In the 18th century the work of Boyle, Charles and others explored the relationship between changes in pressure, temperature and volume of a gas and worked in Farenheit or Celcius/Centigrade; they observed the proportionality of changes but did not have an adequate model of the underlying physical processes.

In the 19th century the work of Joule and Thomson (who later adopted the title Lord Kelvin) revealed a new understanding of thermodynamics implying more fundamental scales of measurement which recognised the concept of absolute zero and the covertability of heat and mechanical work. It is no accident that we still use their names today when measuring these quantities in science, rather than units such as Farenheit and Horse Power.

Hi sorry for bringing this up again, but when my teacher explained this question
(d) 82.06 (atm∙cm3)/(gmole∙K) to (psia∙ft 3)/(lbmole∙F)
Given:
1m3 = 1000 L
1 kg = 2.2046 lb
1 in = 2.54 cm
1 m = 3.2808 ft
1 W = 1 J/s
1 Btu = 1.055 x 103 J
Change in 1 K = Change in 1.8F
1 atm = 14.7 psia

To me, he explained that the fact that its a per kelvin as indicated by the divide line, it would instantly mean that the temperature units conversion are of "change in".

However, is this true? Because as MrAnchovy and the rest of you explained I cannot use Fahrenheit or degrees Celsius in the absolute form as that would mean that volume is is not proportional to temperature and the other laws where temperature is a variable.

So I was thinking that the better explanation was what MrAnchovy pointed one that in that question the temperature refers to T-T0. So in this case, it would make sense to use the change in conversion ratio.

Do you guys have the same idea here?
MrAnchovy
#18
Apr26-13, 07:48 AM
P: 418
This is ridiculous.
  1. This is nothing to do with Mathematics
  2. This is homework and should be in the appropriate section
  3. The question you have been set is pointless - nobody uses units like those any more. There is some sense in working with conversions of things like BTUs, lb.ft, atm etc. as these are used for historical reasons in some branches of engineering. But lb-mole?
  4. You simply cannot express the gas constant in Farenheit because the product of temperature and volume is not proportional to the temperature in Farenheit. If you want to express it on this scale you MUST use the Rankine scale. Whoever set this question is a fool.
  5. 'He explained that the fact that its a per kelvin as indicated by the divide line, it would instantly mean that the temperature units conversion are of "change in"'. This explanation is rubbish. Does the "divide line" mean that there is a "change in" the number of molecules too?


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