CoolProp units

CoolProp units - Getting 'strange' units and can't override them - Сообщения

#1 Опубликовано: 01.08.2017 07:43:27
Peter J Francis

Peter J Francis

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Hi all ,
I'm putting together a sheet using the CoolProp wrapper
It generally works really well except for the following

Where I'm getting strange units for the Specific Heat capacity of Air
I've tried overriding the units but still don't get what I want (J/kg/K)

I've attached an extract from the sheet --> UnitsProblem.sm (7 КиБ) скачан 94 раз(а).

Probably a really simple answer but I can't seem to find it !!

Units.png
Peter J Francis Twitter: @PeterJFrancis Thermal Engineer , Maker , Pilot
#2 Опубликовано: 01.08.2017 09:37:11
Martin Kraska

Martin Kraska

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Can't reproduce. J/(kg*K) works. Paste 'J/{'kg*'K} to the square placeholder to the right of the unit when the region has focus.

2017-08-01 14_35_35-SMath Studio 0.98.6398 - [UnitsProblem.sm_].png
Martin Kraska Pre-configured portable distribution of SMath Studio: https://en.smath.info/wiki/SMath%20with%20Plugins.ashx
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Peter J Francis 02.08.2017 05:18:00
#3 Опубликовано: 01.08.2017 17:05:37
Jean Giraud

Jean Giraud

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Can't reproduce either "Units don't match",
because there is NO heat capacity unit in Smath.
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Peter J Francis 02.08.2017 05:18:00
#4 Опубликовано: 01.08.2017 18:52:12
CBG

CBG

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You just have to change the units.

Best Regards

Carlos

Units_Change_1.png
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Peter J Francis 02.08.2017 05:18:00
#5 Опубликовано: 02.08.2017 05:28:04
Peter J Francis

Peter J Francis

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Thanks All,
You are all correct (I knew the answer was going to be obvious ) , pasting in the units J/kg K works fine

Its just that Gy/K seems an odd default unit

Thanks for your help

Peter
Peter J Francis Twitter: @PeterJFrancis Thermal Engineer , Maker , Pilot
#6 Опубликовано: 04.12.2017 16:09:32
usamakhan

usamakhan

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hey fellows, how do I calculate WET steam properties through this extension,lets suppose at 16Bar Pressure & 0.7 quality 7 which IFP standard is currently being used?
#7 Опубликовано: 05.12.2017 10:27:10
Jean Giraud

Jean Giraud

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Wrote

how do I calculate WET steam properties through this extension,lets suppose at [b]16Bar Pressure



Good question. The vapor phase is dry and properties don't change.
That portion of water that drags at the bottom of piping must be
taken into account for massflow calculation.
At the speed steam flows, better be dry. If not => will destroy
my control valves. Puzzling question ?
#8 Опубликовано: 05.12.2017 11:17:30
usamakhan

usamakhan

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Wrote

Wrote

how do I calculate WET steam properties through this extension,lets suppose at [b]16Bar Pressure



Good question. The vapor phase is dry and properties don't change.
That portion of water that drags at the bottom of piping must be
taken into account for massflow calculation.
At the speed steam flows, better be dry. If not => will destroy
my control valves. Puzzling question ?



The property(s) that does change when you add water to steam is NOT its pressure or temperature but its enthalpy ...the most major parameter which defines the amount of work that can be extracted

Yes wet steam at very high velocities can be micro bullets for control valves & turbine blades causing erosion
#9 Опубликовано: 05.12.2017 12:00:21
Jean Giraud

Jean Giraud

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Wrote

The property(s) that does change when you add water to steam is NOT its pressure or temperature but its enthalpy ...the most major parameter which defines the amount of work that can be extracted

Yes wet steam at very high velocities can be micro bullets for control valves



Worst than erosion => cavitation.
Can't help more. Steam out of the balloon is dry, isn't ?
If you have wet steam out of the balloon, you have a serious level control problem.

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