If we have a gas (like air) moving through the narrow throat of a Venturi Tube, do we get a lower temperature to go with the lower pressure at that throat?
Here's what I've worked out:
For Venturi tube problems with use the equation of continuity,
$A_1 v_1=A_2 v_2$
and the Bernoulli equation
$\frac{1}{2} \rho v_1^2+P_1=\frac{1}{2} \rho v_2^2+P_2$
With a smaller area, we get a larger velocity, with a larger velocity we get a lower pressure.
What I'd like to do next is to bring in the ideal gas equation law.
$PV=nRT$
If the density of air is constant throughout, then we would get
$\frac {P_1}{T_1}=\frac {P_2}{T_2}$
Which implies that a lower pressure gives a lower temperature.
Something about this troubles me, though. Like there's a hidden assumption that I'm getting wrong.
Does this also imply that some of the kinetic energy of random motion of the gas in the case of the slower region is being converted to kinetic energy of the uniform fluid motion in the throat region?
[![airflow moving through a periodic chamber][1]][1]
Best Answer
All of your equations and results are correct. The throat temperature, $T_2$, is in fact lower than the upstream temperature, $T_1$, for a Venturi tube. An alternative way of looking at this problem is by using an energy balance of the system. For instance, disregarding any heat or work added to the system, and assuming steady-level flow, we have,
$$ h_1 + \frac{1}{2}V_1^2 = h_2 + \frac{1}{2}V_2^2 $$
where $h$ is the enthalpy of the flow, which for a calorically perfect gas can be expressed as,
$$ h = c_p T $$
Thus,
$$ c_p T_1 + \frac{1}{2}V_1^2 = c_p T_2 + \frac{1}{2}V_2^2 $$
Solving for $T_2$,
$$ T_2 = T_1 + \frac{1}{2 c_p} \left(V_1^2 - V_2^2\right) $$
Which from conservation of mass we know $V_1^2 - V_2^2 < 0$, and hence $T_2 < T_1$.