Help with boost formula conversion - Fahrenheit to Celcius

[TERRA Operative]

Ok gurus, I'm working out some stuff for superchargers etc. And I'm trying to convert a formula that calculates temperature gain per unit of boost from Fahrenheit to Celsius.

The formula is a s follows:

Temperature gain = [(14.7psi + boost psi/14.7psi)0.28-1]x(460oF+90oF)

- the 14.7psi is atmospheric pressure
- the boost psi is the amount of boost being produced
- the 460oF is absolute zero
- the 90oF is atmospheric temperature

We then divide the result by the compressor efficiency (say, 0.55) to get the final temp rise due to compression.

The exponent 0.28 in the numerator is determined by the gas constant, it indicates the extent to which a gas heats up when compressed.
It multiplies the absolute ambient temperature (550oF) by a pressure ratio factor (the bit in the curved brackets) to find the temperature the charge will be raised.

The specific gas constant of dry air is 287.06 J kg−1 K−1, based on a mean molar mass for dry air of 28.9645 g/mol.
Whatever that means.....

I just want to avoid too many conversions, as I'm writing an Excel spreadsheet to calculate all the variables to size up a supercharger to an engine to an intercooler etc etc. It just ends up too messy otherwise...

Can someone help convert this to Celcius? I have tried just plugging in the values at Celcius (including the absolute zero values) but my results don't match....


Help!

[Billzilla]

o.oooooooooooooooooooooooo1 seconds of searching finds this ->
http://www.wbuf.noaa.gov/tempfc.htm

That silly foreignheight stuff should be banned, only one country uses it anymore these days anyway.

[oldcorollas]

Ok gurus, I'm working out some stuff for superchargers etc. And I'm trying to convert a formula that calculates temperature gain per unit of boost from Fahrenheit to Celsius.
The formula is a s follows:
Temperature gain = [(14.7psi + boost psi/14.7psi)0.28-1]x(460oF+90oF)
- the 14.7psi is atmospheric pressure
- the boost psi is the amount of boost being produced
- the 460oF is absolute zero
- the 90oF is atmospheric temperature
We then divide the result by the compressor efficiency (say, 0.55) to get the final temp rise due to compression.
The exponent 0.28 in the numerator is determined by the gas constant, it indicates the extent to which a gas heats up when compressed.
It multiplies the absolute ambient temperature (550oF) by a pressure ratio factor (the bit in the curved brackets) to find the temperature the charge will be raised.
The specific gas constant of dry air is 287.06 J kg−1 K−1, based on a mean molar mass for dry air of 28.9645 g/mol.
Whatever that means.....
I just want to avoid too many conversions, as I'm writing an Excel spreadsheet to calculate all the variables to size up a supercharger to an engine to an intercooler etc etc. It just ends up too messy otherwise...
Can someone help convert this to Celcius? I have tried just plugging in the values at Celcius (including the absolute zero values) but my results don't match....
Help!

i think the 0.28 is wrong (for celcius)

(14.7psi + boost psi/14.7psi)
do you mean
(14.7psi + boost psi)/14.7psi??
in which case it is just absolute pressure in atmospheres.. (edit, i guess you can call it pressure ratio )

460oF+90oF is just absolute temperature.. so if anything you just need to convert to K..

but really, you will need to see how the 0.28 was derived, because you need to change F to K, and psi to kPa or similar... and that will change the 0.28 (i think)


maybe this will help..
http://www.engineeringtoolbox.com/in...ant-d_588.html

eg
The Universal Gas Constant - Ru - in alternative Units
J/(mol.K) : 8.314472
lbf.ft/(lbmol.oR) : 1545.349

or
"the imperial system the units for the individual gas constant are ft lb/slug oR. In the SI system the units are J/kg K"

good luck

[TERRA Operative]

o.oooooooooooooooooooooooo1 seconds of searching finds this ->
http://www.wbuf.noaa.gov/tempfc.htm

That silly foreignheight stuff should be banned, only one country uses it anymore these days anyway.

It's going in an excel spreadsheet, so conversions are going to be a little messy and introduce inaccuracies...

i think the 0.28 is wrong (for celcius)

(14.7psi + boost psi/14.7psi)
do you mean
(14.7psi + boost psi)/14.7psi??
in which case it is just absolute pressure in atmospheres.. (edit, i guess you can call it pressure ratio )

Yea, pressure ratio, it's expanded out to allow the insertion of whatever boost is required. I forgot the brackets..

I think the coefficient may be for Fahrenheit too. I just gotta find the right one.

460oF+90oF is just absolute temperature.. so if anything you just need to convert to K..

Yep, got that.

but really, you will need to see how the 0.28 was derived, because you need to change F to K, and psi to kPa or similar... and that will change the 0.28 (i think)

maybe this will help..
http://www.engineeringtoolbox.com/in...ant-d_588.html

eg
The Universal Gas Constant - Ru - in alternative Units
J/(mol.K) : 8.314472
lbf.ft/(lbmol.oR) : 1545.349

or
"the imperial system the units for the individual gas constant are ft lb/slug oR. In the SI system the units are J/kg K"

good luck

Hmmm, unfortunately, I missed my fluid dynamics classes in primary school... I'll do some more research with what you've suggested.

[big_zop]

- the 460oF is absolute zero

That should be negative 460degF (absolute zero is -273.15degC).

You cannot have an absolute temperature in Fahrenheit - absolute is in Kelvin. Where did you get this formula and is it in SI or imperial?

The specific gas constant of dry air is 287.06 J kg−1 K−1, based on a mean molar mass for dry air of 28.9645 g/mol.
Whatever that means.....

I believe it means that it takes 286.06 Joules to heat up 1kg of whatever the item is by 1 degree Celcius.

I just want to avoid too many conversions, as I'm writing an Excel spreadsheet to calculate all the variables to size up a supercharger to an engine to an intercooler etc etc. It just ends up too messy otherwise...

Excel is easy once you get the hang of it. Depending on how much of the spreadsheet you want to be able to understand later on (ii: you dont want to look at a formula and go "what the hell is that...").

Try separating all of your variables (including conversions) till you know the final answer is correct. Then you can start to combine it till it is as neat as you desire. Also, if oyu would prefer to break it up into stages, then colour code each thing and make sure you work out your units and put them on too.

Here is something i created for an assignment (it has been modified since):



blue - inputs (but in the assignment they were fixed)
light yellow - variable inputs
green - existing formulas or calculations
dark yellow - outputs

[Draven]

to save the all those boost numbers, you could just have "boost (in Bar) +1", which is just absolute pressure in bar - it's the same number without all the dividing etc

[The Witzl]

farenheit to celcius = ([*F] - 32) / 1.8

Celcius to Farenheit = [*C] x 1.8 + 32.


i remembers my physics formulas still .
NINJA EDIT..... i did a booboo

[Hiro]

That should be negative 460degF (absolute zero is -273.15degC).

You cannot have an absolute temperature in Fahrenheit - absolute is in Kelvin. Where did you get this formula and is it in SI or imperial?

You can change Fahrenheit to Rankine for absolute temperatures. Same deal as with Kelvin, -460degF = 0 Ra.

[oldcorollas]

OMG i forgot abotu Rankine
so that would be the temp unit in this then? lbf.ft/(lbmol.oR)

LOL pound force x feet / pound x moles x Rankine....

no wonder they crashed the mars lander


since
J/(mol.K) : 8.314472
lbf.ft/(lbmol.oR) : 1545.349

mol is common so you can ignore that
J/K vs lbf.ft/lb .R

R = K x 9 / 5
foot pound-force (ft.lbf) to J: 1.35582 (ie, 1J = 0.737561033175495 ft.lbf)

i'm not sure where that other lb comes from.. unless it is lb.mol instead of mol?


also
gas constant = 6.132440 lbf·ft·K-1·g-mol-1

but you still need to work out how that exponent is derived. is it empirical?

can you not just plot using psi and F, and then convert those numbers and find new exponent?
6.132440 lbf·ft·K-1·g-mol-1

[oldcorollas]

is that 0.28 an exponent (as in.. to the power of)? or just a fudge factor?

if a fudge factor, your equation is wrong
if exponent, the numbers seem to work out.

easier to think of equation like this
dT = [(boost ratio)^0.28-1]x(oR)

then
dT/oR = (boost ratio)^0.28-1

where T is in F and oR is in rankine so you can change them to C and K.. since it is just a ratio of
temp rise/ambient temp

ie converting..
F/R = ((9C/5)+32) / (9K/5)
= (9C+160)/9K (maybe)

edit: for equation, don7t need the +32
so F/R = ((9C/5)+32) / (9K/5)
= (9C+160)/9K (maybe)
is actualyl just 9C/9K ... so F/R = C/K????　hmmmmmmmmmm

so
dT/oR = (boost ratio)^0.28-1
becomes
dC/K = (boost ratio)^0.28-1

1+dC/K = (boost ratio)^0.28

take log of both sides

log[1+(dC/K)] = 0.28 log(boost ratio)
which is same(i hope) as log[1+(dT/R)] = 0.28 log(boost ratio)

and theeennn

you can graph to find the exponent...

[oldcorollas]

well, in the end... (after playing with excel and equations)

dT (in celcius) = K x [ (boost ratio)^0.28 -1 ]

gives the same result as original equation.....

[TERRA Operative]

o.oooooooooooooooooooooooo1 seconds of searching finds this ->
http://www.wbuf.noaa.gov/tempfc.htm

That silly foreignheight stuff should be banned, only one country uses it anymore these days anyway.

It's going in an excel spreadsheet, so conversions are going to be a little messy and introduce inaccuracies...

i think the 0.28 is wrong (for celcius)

(14.7psi + boost psi/14.7psi)
do you mean
(14.7psi + boost psi)/14.7psi??
in which case it is just absolute pressure in atmospheres.. (edit, i guess you can call it pressure ratio )

Yea, pressure ratio, it's expanded out to allow the insertion of whatever boost is required. I forgot the brackets..

I think the coefficient may be for Fahrenheit too. I just gotta find the right one.

460oF+90oF is just absolute temperature.. so if anything you just need to convert to K..

Yep, got that.

but really, you will need to see how the 0.28 was derived, because you need to change F to K, and psi to kPa or similar... and that will change the 0.28 (i think)

maybe this will help..
http://www.engineeringtoolbox.com/in...ant-d_588.html

eg
The Universal Gas Constant - Ru - in alternative Units
J/(mol.K) : 8.314472
lbf.ft/(lbmol.oR) : 1545.349

or
"the imperial system the units for the individual gas constant are ft lb/slug oR. In the SI system the units are J/kg K"

good luck

Hmmm, unfortunately, I missed my fluid dynamics classes in primary school... I'll do some more research with what you've suggested.

[oldcorollas]

[QUOTE=oldcorollas]well, in the end... (after playing with excel and equations)

dT (in celcius) = K x [ (boost ratio)^0.28 -1 ]

gives the same result as original equation.....[/QUOTE]
..........

calculate some temp differences for different boost pressure and different inlet temps.

then convert the temps seperately to K or delta C, and then convert boost to boost ratio, and plug into above equation and check the numbers...

edit: to make the thinking easier.. think of it as dR/R = dK/K
ie, from absolute zero, the ratio of increase in temp is same, regardless of the units...
ie 2 x A = 2A
2 x B = 2B

[TERRA Operative]

Woah, I was missing posts in this thread that just appeared then....


I got it working as per below:

=((F12 + 273) + ((((((J12 + 14.696) / 14.696)^0.28) - 1) * (F12 + 273)) / 0.55)) - 273

F12 = atmospheric temp
J12 = boost in psi

I compared it to a dyno printout, and it's pretty accurate.

[oldcorollas]

Woah, I was missing posts in this thread that just appeared then....


I got it working as per below:

=((F12 + 273) + ((((((J12 + 14.696) / 14.696)^0.28) - 1) * (F12 + 273)) / 0.55)) - 273

F12 = atmospheric temp
J12 = boost in psi

I compared it to a dyno printout, and it's pretty accurate.

is that different to
dT (in celcius) = K x [ (boost ratio)^0.28 -1 ]
??
do your 273's cancel out?

seems easier to just convert to K and boost ratio......