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1-Newbie

## Samme numbers, different results???

Can anyone please tell explain to me why these two calculations give two different results???

1 ACCEPTED SOLUTION

Accepted Solutions

Search the forum for "temperature" and you'll get a lot of info.

Mathcad's base unit for temperature is Kelvin.  Since 0°K does not equal 0°C, you end up with issues that need addressed.

If you clarify the meaning behind each of the temperatures in your equation, we'd be happy to help you sort it out.  If you replace all the units with Δ°C, then you get the result you're expecting.  However, that might not be the best way to represent whatever you're trying to calculate.

17 REPLIES 17
24-Ruby IV
(To:bbomst�rk)

300C=(300+273.15)K etc

Could you elaborate on that? Can't see what Kelvin has to do with this!?

19-Tanzanite
(To:bbomst�rk)

Kelvin has everything to do with it. Most units convert with just a scaling factor, but temperature units do not. There's also an additive component. If you write 300C+1C what you usually mean is increase the temperature of 300C by 1C. That is, 300C+1deltaC, where 1deltaC=1K, as opposed to 1C=274.15K. There is no other way to handle temperature units in unit aware software, because the software cannot know what you mean by 300C+1C. Add 1K, or add 274.15K? The user has to distinguish between the two possibilities, because only the user knows what they want.

23-Emerald I
(To:RichardJ)

Welcome to the wonderful world of empirical formulas.

24-Ruby IV
(To:Fred_Kohlhepp)
 Fred Kohlhepp написал(а): Welcome to the wonderful world of empirical formulas.

I see in reference books more pseudoempirical than empirical formulas!

Search the forum for "temperature" and you'll get a lot of info.

Mathcad's base unit for temperature is Kelvin.  Since 0°K does not equal 0°C, you end up with issues that need addressed.

If you clarify the meaning behind each of the temperatures in your equation, we'd be happy to help you sort it out.  If you replace all the units with Δ°C, then you get the result you're expecting.  However, that might not be the best way to represent whatever you're trying to calculate.

1-Newbie
(To:MJG)

It's a part of this formula where the fraction above is the one in the ln(...). So that specific fraction should just come out as a number without any unit. How can I do that without deleting the temperature units

1-Newbie
(To:bbomst�rk)

You're right,  Δ°C does work and though it's not the best solution it might be the only one

Without being familiar with this equation, it's hard to give much advice.

Firstly, are you certain that the temperatures should be in °C?  You'll get a different result if you use another temperature unit (°K, °F, °R).

Secondly, what do the variables Θf, Θi, and β represent?

1-Newbie
(To:MJG)

β reciprocal conductor resistance temperature coefficient at 0 °C. β in °C.

Θi onset temperature in °C. Values can be taken from IEC 60287-3-1. If there is no defined values in the national tables should ambient temperatures of 20 °C at a depth of 1 m selected.

Θf final temperature in °C

As expected, this is over my head.

If you don't like using Δ°C for all the units, you could consider omitting the units for these variables.  For clarification, you could put °C in a text box beside the input variable.

1-Newbie
(To:MJG)

This could also be a solution. Thank you!

19-Tanzanite
(To:bbomst�rk)

Someone must be doing something weird with the units to get that equation, because the reciprocal of the resistance temperature coefficient would normally have units of °C/Ohm

24-Ruby IV
(To:bbomst�rk)

More interesting

24-Ruby IV
(To:ValeryOchkov)

19-Tanzanite
(To:bbomst�rk)

You can't add temperatures in C (or F) like that.

300C=573.15K and 234.5C=507.65K. Therefore 300C+234.5C=1080.8K=807.65C.

When you add them mentally and get the result 534.5C what you are actually doing is removing the units to get just the numeric values, adding the numeric values, and then putting the units back on. Look up temperature units in the help.

24-Ruby IV
(To:bbomst�rk)
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