Explanation Needed: Thermodynamics of a hot/cold water jet machine

I didn't know where to begin with this problem. I eventually found a solution online, which is why I'm reposting this question with an answer. I was wondering if anyone can explain the one question I have.

A self-contained machine only inputs two equal steady streams of hot and cold water at temperatures $T_1$ and $T_2$. Its only output is a single high-speed jet of water. The heat capacity per unit mass of water, $C$, may be assumed to be independent of temperature. The machine is in a steady state, and the energy in the incoming jets is negligible.

$(a)$ What is the speed of the outgoing jet in terms of $T_1$, $T_2$ and $T$, where $T$ is the temperature of the outgoing jet?

$(b)$ What is the maximum possible speed of the jet?

$(a)$ The heat intake per unit mass is $\Delta Q = (C(T_1 - T) - C(T - T_2))/2$

Since the system is in a steady state, $\frac{v^2}{2}=\Delta Q$, therefore $v=\sqrt{C(T_1 + T_2 - 2T)}$

why is that the case? Why does all the heat go towards its kinetic energy per unit mass? That is my one question

$(b)$ $\Delta S = \frac{C}{2}[ln(\frac{T}{T_1})+ln(\frac{T}{T_2})] \ge 0$

So $T\ge \sqrt{T_1 T_2}$

and $v \le v_{max} = \sqrt{C(T_1 + T_2 - 2\sqrt{T_1 T_2})}$

What an insidious problem, no?

why is that the case? Why does all the heat go towards its kinetic energy per unit mass? That is my one question

Basically there are two places for the energy to go in the output stream:

• Thermal energy
• Kinetic energy

The problem through (a) is a direct energy balance problem. In this part, we're treating $T$ as independent. If $T$ is the average of the inlet temperatures, then the velocity will be zero because all energy is accounted for, and $\Delta Q$ will be zero.

Mentally, I see a thermal cycle with the $T_1$ feeds the boiler and $T_2$ feeds the condenser. That cycle outputs useful work. The more useful work it produces, the more the average temperature of the streams is lowered. This is similar to a thermal power plant. The boiler produces more heat than the condenser rejects. Typical efficiency is 33%, so the condenser removes only 2/3rd of the boiler's heat. The rest of the heat went to turning the turbine because heat is a form of energy and energy is the ability to do work.

In your case, the stream might have been accelerated by a pump that feeds into a cavity that has a nozzle where the stream is accelerated. The more work the pump does, the more the temperature of the water has to decrease because this is a closed system.

Practically, either the temperature change would be miniscule or the velocity would be gigantic. The reason is that thermal vibrations are fast compared to speeds we're used to. You can look at it this way in terms of your problem too! The average kinetic energy of the molecules is the same going in and going out - it's just that some of that kinetic energy going out is from the bulk motion of the stream so we don't count it as temperature. That temperature would have to be measured by a thermometer moving along with the stream.

• Thanks for your answer. When you say "the average temperature of the streams is lowered" you are referring to the temperature of the output stream yes? Efficiency is the measure of heat energy used for work, so in this case work refers to the increase in speed of the water jet. When you talk about thermal vibrations, are those vibrations responsible for the increase in speed? Or are you describing the speed at which the temperature changes? I see your point, I think. That the "leftover" thermal energy must be now bulk kinetic energy, otherwise it would be warmer, and T is independent here. – walczyk Aug 20 '13 at 21:11
• @walczyk I reference the average temperature because that's a proxy for thermal energy per unit mass that can be compared for the inlets ($0.5 (T_1+T_2)$) and the outlet ($T$). Hope that's more clear now. Oh, and yes, work is the velocity increase in water. That energy is low entropy (effectively zero), so it's treated separately from thermal. But work can travel through other intermediaries. It can be manifested in torque of the shaft of your pumping device, for instance. That's just transmitting the work, and it winds up as velocity. – Alan Rominger Aug 20 '13 at 21:36