(49g) (50g) Water boiling temperature = f(altitude/pressure + temp); Water density

[Gil]

Completely new versions, following questions of Albert CHAN.

My thanks to him.

The 3 old ones, with inconsistent results between themselves, are named now with ...BOIL.OLD (with capital letters & OLD-suffix); they are to be found at the end of the directory WATER; the best would be to delete them.

The 1st new program 'MT—>Boil.H²O' means:
- Give in stack level 1 the altitude (M stands for meters);
- Or give in stack level1, in {}, the altitude & the local air temperature (T stands fort Temperature, that is to be entered in degrees Celsius), to get the corresponding boiling temperature of water.

Its code is:
\<< "1 stack Arg:
. 
Alt

. or 
{Alt T.Air}

with 
{}


Alt in [m]
T.air in [\^oC]

" DROP STD DUP TYPE 5 ==
IF
THEN OBJ\-> DROP NEG 15 + OVER /
ELSE .0065
END .0289644 9.80665 8.31446261815 \-> a M g R
\<< 'Alt.m' STO "At " Alt.m + "m:" + 15 a Alt.m * - 1 RND "Air [\^oC]" \->TAG DUP 'T.Air' STO '101325*(1-a*Alt.m/(15+273.15))^(M*g/(R*a))' \->NUM P\->Boil.H\178O
\>>
\>>

The 2nd program 'P—>Boil.H²O'
is to be used directly (without using 'MT—>Boil.H²O')
when already knowing the local pressure (independently of altitude and temperature).
- Give the pressure in Pascals, to get the corresponding boiling temperature of water.

Its code is:
\<< "1 Arg:
. 
Pressure
 in [Pa]

" DROP STD "P [Pa]" \->TAG DUP 'P' STO "
GOFF.GRATCH 1984
\->" 'LOG(P/100)=-(7.90298*(373.15/(T1+273.15)-1))+5.02808*LOG(373.15/(T1+273.15))-.00000013816*(10^(11.344*(1-(T1+273.15)/373.15))-1)+.0081328*(10^(-3.49149*(373.15/(T1+273.15)-1))-1)+LOG(1013.246)' 'T1' 100 ROOT "Boil.\^oC H\178O" \->TAG "
ARDEN
 
BUCK 1996
\->" 'P=611.21*e^((18.678-T2/234.5)*(T2/(257.14+T2)))' 'T2' 100 ROOT "Boil.\^oC H\178O" \->TAG 3 FIX
\>>

The references
are under the program REFER.

The two used formulae used here for boiling temperature of water =f(pressure) are:
- Arden Buck equation, 1996;
- Goff-Gratch equation, 1984.

Both formulae give quite similar boiling temperatures to 0.03 degrees Celsius.

Example Everest, altitude 8848 meters.
1)Type 8848
2)ENTER
3)Press MT—>Boil.H²O

Result
"At 8848m:"
:Air [°C]: -42.500
[Pa]: 31444.626
"
GOFF.GRATCH 1984—>"
:Boil.°C H²O: 70.205
"
ARDEN
 
BUCK 1996—>"
:Boil.°C H²O: 70.180

Suppose you will travel in summer to Everest.
You could check that average temperature is then equal to -19 degrees Celsius, and not -42.5.
Then
1) Write {8848 -19} (don't forget the {})
2)ENTER
3)Press MT—>Boil.H²O

The new results are:
"At 8848m:"
:Air [°C]: -19.000
[Pa]: 33184.930
"
GOFF.GRATCH 1984—>"
:Boil.°C H²O: 71.457
"
ARDEN
 
BUCK 1996—>"
:Boil.°C H²O: 71.432

Of course, in both cases the calculated pressure is an approximation.
It seems that, at Everest, the mean pressure in July is about 253 torrs.
101325 PA —> 760 torrs.
253 torrs = 101325/760 torrs×253=33731.
In that case we should use
directly the program Press P—>Boil.H²O.
1)Write 33731
2)ENTER
3)Press P—>Boil.H²O

The corresponding results are then:
[Pa]: 33731
"
GOFF.GRATCH 1984—>"
:Boil.°C H²O: 71.839
"
ARDEN
 
BUCK 1996—>"
:Boil.°C H²O: 71.814

For the time being, I let g as a constant, but it should diminish with the altitude: g=go×(Re/Re+h)².

Regards and enjoy!
Gil