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 Ioan #1 Posted : 13 September 2012 23:23:02(UTC) Rank: MemberGroups: Registered Joined: 17/06/2012(UTC)Posts: 26Was thanked: 1 time(s) in 1 post(s) Hi all,It seems to me that the temperature units handling is strange (?)I attached an example. In engineering it is often used °C unit for the temperature.Maybe it would useful to use different units for the absolute temperature and for the temperature difference ?Regards,IoanFile Attachment(s): temperature.sm (7kb) downloaded 171 time(s).
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 Peter #2 Posted : 14 September 2012 07:24:15(UTC) Rank: MemberGroups: Registered Joined: 04/10/2010(UTC)Posts: 16Location: Hörselgau, GermanyWas thanked: 7 time(s) in 6 post(s) Hello Ioan,here you must use the unit K for temperatur difference, becausealfa is defined as: "change of length [micrometer] per length [m] and per temperstur difference [K]"Regards,Peter
 Ioan #3 Posted : 14 September 2012 10:29:07(UTC) Rank: MemberGroups: Registered Joined: 17/06/2012(UTC)Posts: 26Was thanked: 1 time(s) in 1 post(s) Hello Peter,Thanks for your answer.I note that in Mathcad, all the situations in my example are accepted because they reserved different unit notation for the absolute temperature and for the temperature change. Why not to do the same for the engineering practical viewpoint?The temperature change values are the same in °C and K.Of course the basic SI unit for temperature is the Kelvin and the temptation is to reduce all to K (!?).Regards,Ioan
 Peter #4 Posted : 14 September 2012 10:57:20(UTC) Rank: MemberGroups: Registered Joined: 04/10/2010(UTC)Posts: 16Location: Hörselgau, GermanyWas thanked: 7 time(s) in 6 post(s) Hello Ioan,if you use °C as unit for the temperature difference, than SMath calculates intenal with unit K.This is correct.Because of the definition: 0°C=273.15K a DeltaT of 5°C would be a diff. of 278,15K !If MathCAD accepts temp. diffs. with unit °C, than that is im my opinion an error (at least not correct)Regards,Peter
 Davide Carpi #5 Posted : 14 September 2012 12:10:38(UTC) Rank: Advanced MemberGroups: Registered, Advanced MemberJoined: 13/01/2012(UTC)Posts: 2,279Location: ItalyWas thanked: 1157 time(s) in 760 post(s) Originally Posted by: Peter Hello Ioan,if you use °C as unit for the temperature difference, than SMath calculates intenal with unit K.This is correct.Because of the definition: 0°C=273.15K a DeltaT of 5°C would be a diff. of 278,15K !If MathCAD accepts temp. diffs. with unit °C, than that is im my opinion an error (at least not correct)Regards,PeterI agree with you Note that you can simply use the °C temperature giving as input a temperature difference (as required by the definition of the equation)regards,w3b5urf3rEdited by user 14 September 2012 12:45:02(UTC)  | Reason: Not specifiedDavide Carpi attached the following image(s): If you like my plugins please consider a donation to SMath Studio; for personal contributions to me: paypal.me/dcprojects 1 user thanked Davide Carpi for this useful post. on 14/09/2012(UTC)
 ElSid #6 Posted : 26 September 2012 01:46:04(UTC) Rank: Advanced MemberGroups: Registered Joined: 05/03/2009(UTC)Posts: 302Location: USAWas thanked: 10 time(s) in 9 post(s) This is an issue for me as well. MathCAD has a function called delta T under their units drop down (version 14 at least). Trying to figure out how to write a function that would fix this issue in SMath. I know it took MathCAD a long time to address this issue For me its as simple as 100C - 80C needs to equal 20C and not -253.15C. It appears that they defined DC=K (kelvin) and DF=(5/9)KEdit 1:I can confirm. MathCAD does define the deltas as mentioned above. I whole heartedly disagree with Peter and w3b5urf3r. I use the "unit"/scale value of DT ALL the time. An example would be Heat Transfer Coefficients. Take a look at and units on any chart. If you do the math, the "temperature" in the coefficient's units is based on a change (DT). I have gotten into a habit of re-writing these coefficients with a DT such that my calculation output will be a direct unit (be it C or F).Edit 2: Post from PTChttp://blogs.ptc.com/201...ss-of-temperature-units/Edited by user 22 December 2012 02:17:37(UTC)  | Reason: Confirmed scale
 mkraska #7 Posted : 28 January 2013 02:41:25(UTC) Rank: Advanced MemberGroups: Registered Joined: 15/04/2012(UTC)Posts: 1,608Was thanked: 885 time(s) in 561 post(s) One more example from the chapter "What you see is not always what you get"I just played around with temperature units and found an issue that might be a pitfall for the unaware user. Type the expression -10°C and try to display the value in units °C. You get-10°C=-556,3°COn the LHS, times (10 by °C) has higher priority than minus (0 - 10). Thus 10°C is evaluated to 283,15K and then subtracted from zero, giving -283,15K. This is then displayed in °C by subtracting another 273,15 from the value. In the result, everything to the left of the unit is treated as if it was in braces. The corresponding survival rule is: "Negative temperature values in non-absolute units (°C, °F, °Re) need to be bracketed"The other rule is already given in the previous discussion:"Always display temperature differences in absolute units (K, °Ra)"Edited by user 28 January 2013 02:43:00(UTC)  | Reason: Not specifiedmkraska attached the following image(s): Martin Kraska Pre-configured portable distribution of SMath Studio: https://en.smath.info/wi...th%20with%20Plugins.ashx
 mikekaganski #8 Posted : 28 January 2013 03:15:57(UTC) Rank: Advanced MemberGroups: Registered Joined: 17/01/2013(UTC)Posts: 296Location: Khabarovsk, RussiaWas thanked: 151 time(s) in 107 post(s) http://ru.smath.info/for...ta-unarnogho-minusa.aspxThis is not a unit issue, it is a problem with operators priority Best regards, Mike Kaganski
 Frame #9 Posted : 19 October 2016 16:33:31(UTC) Rank: NewbieGroups: Registered Joined: 07/04/2016(UTC)Posts: 3Location: Indiana Sorry to bring up a seemingly old topic with the temperature units but after using the forums Search tool I could not find a satisfying answer.I do not understand how SMath arrives at the answer shown. I assumed that SMath would have converted to kelvin for the calculations and then converted back to Fahrenheit to display the solution. It does not seem to do that?Anyways, thanks for entertaining my question. Temperature Units.sm (3kb) downloaded 24 time(s).
 mikekaganski #10 Posted : 19 October 2016 16:47:12(UTC) Rank: Advanced MemberGroups: Registered Joined: 17/01/2013(UTC)Posts: 296Location: Khabarovsk, RussiaWas thanked: 151 time(s) in 107 post(s) It does convert to Kelvins, then subtracts.But then you take the temperature difference, and try to show it as temperature. SMath then shows you how much the resulting temperature is higher than zero Farenhait - i.e. how much the 1388.705 K is greater than 255.3722 K.You should have used units for temperature difference, i.e. 'Δ°F Best regards, Mike Kaganski
 Frame #11 Posted : 19 October 2016 17:27:47(UTC) Rank: NewbieGroups: Registered Joined: 07/04/2016(UTC)Posts: 3Location: Indiana mikekaganski,Thank you for the quick reply. Please forgive me for my elementary understanding of temperature units here. It does not seem logical to me that the units would change during a simple subtraction operation. You explain that I have taken the temperature difference and then ask SMath to show it as temperature. Why would we differentiate the two as different units? Is this a scalar vs vector unit issue?I understand that the temperature units I am using as inputs are not absolute but it seems like by converting to an absolute scale before the operations and converting back after the operations would result in a correct answer every time. However, I also understand and appreciate that there could be much more complicated situations the software would have to be prepared for but using a special unit of temperature difference seems silly to me.Also, please do not take my questions as demeaning. I am not interested in being negative about this wonderful free software. I am however very interested in learning from this community!Thanks!
 mikekaganski #12 Posted : 19 October 2016 17:36:50(UTC) Rank: Advanced MemberGroups: Registered Joined: 17/01/2013(UTC)Posts: 296Location: Khabarovsk, RussiaWas thanked: 151 time(s) in 107 post(s) 1. T_F := 2100°F = 255.3722 K + (2100*5/9) K = 1422.0389 K2. T_I := 60°F = 255.3722 K + (60*5/9) K = 288.7056 K3. T_F - T_I = 1422.0389 K - 288.7056 K = 1133.3333 K4. 1133.3333 K = (1133.3333 K - 255.3722 K)*9/5 = 1580.33°FPlease note the subtraction in last (fourth) position.In third position, the two additions are compensated; while the last one has no counterpart.Temperature in Farenhait is measure how many Farenhait degrees are above zero Farenhait. That's the culpit.Think about this.You write: T_I := 60°F + 5°F.What do you mean? Do you mean that some body has that temperature of 60°F?And is 5°F the number of degrees that the body got from heating (temperature difference)?I (human) see your intention. Does the program?SMath uses base SI units to keep the values. So it will make the following:60°F + 5°F = 255.3722 K + (60*5/9) K + 255.3722 K + (5*5/9) K = 288.7056 K + 258.15 K = 546.8556 K = 524.67 °FAddition will make this apparent.So, you need to use 60°F + 5Δ°F to let program know. And at the same time, to teach yourself discipline in thinking about non-standard temperature scales that have that shifted zero point.Edited by user 19 October 2016 17:53:57(UTC)  | Reason: Not specified Best regards, Mike Kaganski
 Frame #13 Posted : 19 October 2016 18:10:55(UTC) Rank: NewbieGroups: Registered Joined: 07/04/2016(UTC)Posts: 3Location: Indiana I now understand.2100-60=20401580+460=2040 -->being that Absolute zero is -460 F. and as you said above zero Fahrenheit, Therefore: 0-(-460)=460)This was not immediately obvious to me.Thank you for taking the time to explain it to me.
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