Display Moreas long as we are nitpicking
Adiabatic temperature increaseThe outlet pressure is 300 pascals more than the inlet pressure..from fanspec.
Pin 101300, Pout 101600 pascals
To/Tin =(Po/Pin) exp(0.286) … the exponent is related to Cp, Cv via theory.
For Tin = 25C, 288K
To = 288x(101600/101300) exp(0.286)=288.2432 K…
25.2432C
Adiabatic temperature rise= 0.243 Cwhich is a large proportion of 0.35 C.(actually measured)
I am struggling with this. Maybe I am "missing something" too.
How is this in any way relevant to anything?
The fan is there to cause movement of air through the calorimeter. It does this by behaving like a compressor, causing a pressure gradient which generates the flow.
Whatever heat it inputs to the system is needed due to the need for air flow. This will come out in the calorimetry. I can't be bothered to look for it, but presumably TM has allowed for this. Even if he hasn't, it doen't make any difference.
I like rough approximations to start with, to make sure one is not calculating to 5 decimal places whilst being an order of magnitude out of ball park.
I have done a rough approximation based on the consistency of the calorimeter with R19 and R20. In one paper, reporting on R19, the cal run is 100W in, and it results in delta T of ~5C. With the excess heat run, the cop is 1.8 and the delta air T is 9C. So, 100/5 = 20W/C. If we then apply this to the excess heat run, we have 80W of excess, which calcs as 80W/(20W/C) = 4C. Then, 5C + 4C = 9C, which is the reported delta T. With the high cop R20 run, it is a bit more difficult because the 50W input for cal is difficult to discern on the scale, but it looks like about 60% towards half way, which would mean 2.6C. Thus, 2.6 * 20 is 52C, so within the likely variations, this is consistent enough, for me with this type of analysis.
So, the 0.2432C rise that you have calculated as adiabatic increase would give us .024(32) * 20 = 4.8 W. This, as expected, is the power of the compressor. It is a) allowed for, and b) totally to be expected.
As an aside, 25C ambient is actually 298.15 K. I have also calculated the more accurate answer as 298.15 *(101600/101300) ^0.286 = 298.402. I took the liberty of rounding that to + 0.25C. Which gives us 0.25*20 = 5W ie, no difference in the context.
So what is the 0.35C measured that you refer to as this being a large proportion of?
Also, what is the point please? The data in the reports is such that it renders the seemingly never ending arguments over minutiae completley irrelevant when considered in the context.