- #36
Anachronist
Gold Member
- 119
- 58
How embarrassing. I was focusing on the end result in Example 2 and trying to back out temperature from entropy, and somehow forgot about equation 5.Chestermiller said:You use Eqn. 5 of Example 2.
Using γ=1.4, P0=5 bars, PF=1 bar, T0=20°C=293K:
Adiabatic expansion: ##T_F= T_0 \left ( \frac{P_F}{P_0} \right ) ^\frac{\gamma - 1}{\gamma} = 185\text{K} = -88°\text{C}##
Eq 5 from example 2: ##T_F=T_0 \left[1-\frac{(\gamma -1)}{\gamma}\frac{(P_0-P_F)}{P_0}\right] = 226\text{K} = -47°\text{C}##
The equation 5 result looks more reasonable but still seems colder than what would be observed in the real world. I would guess that's likely due to two things not accounted for here: (1) the air is being forced through a constriction (does that matter?), and (2) water vapor in the air effectively chokes down the reduction in temperature, so it doesn't go as low.