
#1
Apr2013, 01:53 AM

P: 636

Hi, I was given 1Btu/(h.ft^2.°F) to W/m^2.K with these conversion ratios 1m^3=1000L 1Btu=1.055*10^{3} 1m=3.2808ft and Δ1K=Δ1.8°F
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 :) 



#2
Apr2013, 05:36 AM

P: 748

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. 



#3
Apr2013, 06:04 AM

P: 636





#4
Apr2013, 07:15 AM

P: 748

Question on unit conversion with temperatures 



#5
Apr2013, 08:11 AM

P: 636





#6
Apr2013, 03:04 PM

Mentor
P: 11,988

To repeat what jbriggs said, these units refer to temperature differences, not actual temperature values. "1Btu/(h·ft^{2}·°F)" 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/(h·ft^{2}·Δ°F). 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. 



#7
Apr2013, 08:13 PM

P: 636

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 cm_{3} we can just use the unit conversion of 1L=1000cm^{3} but for temperature we cannot do that? And thanks so much. This has been bugging me ever since my tutorial. 



#8
Apr2013, 09:09 PM

Mentor
P: 11,988

But, for example, 0 °C is not 0 °F. So converting between °C and °F is more complicated than doing a simple multiplication. 



#9
Apr2213, 07:53 AM

P: 636





#10
Apr2213, 08:05 AM

P: 388





#11
Apr2213, 09:29 AM

P: 748

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 = nR_{kelvin}T_{Kelvin} becomes... PV = nR_{Rankine}T_{Rankine} becomes... PV = nR_{Rankine}(T_{Fahrenheit}+459.67) becomes... PV = nR_{Fahrenheit}(T_{Fahrenheit}+459.67) 



#12
Apr2213, 06:38 PM

P: 636

The question for the Units conversion for R was this 82.06(atm.cm^{3})/gmole.K to (psia.ft^{3})/lbmole.°F So what I used was the conversion ratio of Δ1K=Δ1.8°F 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 457.87°F 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. 



#13
Apr2313, 04:15 AM

P: 388

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 nonzero based scale such as Fahrenheit you need to apply the absolute zero correction to T to get PV=nR_{arbitrary}(TT_{0}). 



#14
Apr2313, 06:04 AM

P: 636

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? 



#15
Apr2313, 06:48 AM

P: 748

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. 



#16
Apr2313, 10:07 AM

P: 388

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. 



#17
Apr2613, 02:00 AM

P: 636

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 1.8°F 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 TT0. So in this case, it would make sense to use the change in conversion ratio. Do you guys have the same idea here? 



#18
Apr2613, 07:48 AM

P: 388

This is ridiculous.



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