Thanks for doing this, but this is a very frustrating result. Hard to be confident of anything based on it.
I don’t think treating the ‘control’ result as a baseline is reasonable. My best-guess analysis is as follows:
Assume that
dTin/dt = r ((Tout—C) - Tin)
where
Tin is average indoor temperature
t is time
r is some constant
Tout is outdoor temperature
C is the ‘cooling power’ of the current AC configuration. For the ‘off’ configuration we can assume this is zero.
r obviously will vary between configurations, but I have no better idea than pretending it doesn’t so that we can solve for it in the control condition and then calculate C for the one-hose and two-hose conditions.
Results?
Using the average temperature difference to approximate dTin/dt as constant, we get:
In the ‘off’ configuration:
0.5 hours * dTin/dt = 0.5 hours * r * (14 degrees) = 0.889 degrees
Giving r = 0.127 (degrees per degree-hour)
In one-hose:
1 hour * dTin/dt = 1 hour * r * (19.1111 - C) = 0.3333 degrees
Giving C = 16.486 degrees
In two-hose:
0.5 hours * dTin/dt = 0.5 hours * r * ( 22.944 - C) = −0.555
Giving C = 31.693 degrees
Also finding that the two-hose version has roughly double the cooling power!
Thanks for doing this, but this is a very frustrating result. Hard to be confident of anything based on it.
I don’t think treating the ‘control’ result as a baseline is reasonable. My best-guess analysis is as follows:
Assume that dTin/dt = r ((Tout—C) - Tin)
where
Tin is average indoor temperature
t is time
r is some constant
Tout is outdoor temperature
C is the ‘cooling power’ of the current AC configuration. For the ‘off’ configuration we can assume this is zero.
r obviously will vary between configurations, but I have no better idea than pretending it doesn’t so that we can solve for it in the control condition and then calculate C for the one-hose and two-hose conditions.
Results?
Using the average temperature difference to approximate dTin/dt as constant, we get:
In the ‘off’ configuration: 0.5 hours * dTin/dt = 0.5 hours * r * (14 degrees) = 0.889 degrees
Giving r = 0.127 (degrees per degree-hour)
In one-hose: 1 hour * dTin/dt = 1 hour * r * (19.1111 - C) = 0.3333 degrees
Giving C = 16.486 degrees
In two-hose: 0.5 hours * dTin/dt = 0.5 hours * r * ( 22.944 - C) = −0.555
Giving C = 31.693 degrees
Also finding that the two-hose version has roughly double the cooling power!