On Sun, 23 Apr 2017 21:42:59 +0100, Vir Campestris wrote:

On 22/04/2017 22:43, Johnny B Good wrote:
On Sat, 22 Apr 2017 12:13:31 +0100, charles wrote:

In article , Tim Streater
wrote:

One of the laws of thermodynamics IIRC - efficiency is related to the
difference between input and output temps - in degrees K, not C.

my elementary physics suggests that if you are only refering to a
temperature difference the answer is the same in °K and in °C.°

The main benefit of using deg K instead of Deg C is the complete and
utter absence of negative numbers by which to confuse the mathematics.
:-)

It's true enough that as far as steam engines are concerned, this is
unlikely to effect calculations involving deg C, but there are other
heat engines designs based on fluids with much lower freezing and
boiling points than zero deg C (the triple point of water to within an
accuracy of one decimal place).

FFS Kelvins aren't degrees.

Well, assuming by "Kelvins" you meant Degrees Kelvin and by "degrees"
you meant degrees Celcius, they are! If the standard atmospheric pressure
had been a higher value than it's currently assumed, those degrees Celcius
would simply have been a larger sized increment across the arbitrary 100
degrees Mr Celcius had chosen to span the freezing and boiling points of
pure water and a smaller starting volume of gas than the 273cc's worth at
zero degrees Celcius would have been determined as the basis for
establishing where absolute zero would be in the (now) larger scale units
of degrees Celcius.

Anyhow, putting that aside, degrees Kelvin are directly based on degrees
Celcius no matter that there is an element of arbitrariness in the Celcius
scale. The difference is merely in where the zero point is set on the
scales. In the case of degrees Kelvin, this was based on an Absolute Zero
derived from the behaviour of gases which all exhibited the same
expansion coefficient throughout their gas phase, allowing the Absolute
Zero point to be extrapolated beyond the limits set by each particular
gas's liquifaction points at NTP.


And it must be absolute temperature for the equations to work.

My bad, I just assumed that mere temperature differences would
suffice. :-(


Just consider what the results would look like for Tin-Tout/Tout if Tout
was negative.

Obviously, I *didn't* consider negative temperatures, just temperature
differences. :-(

--
Johnny B Good