Are there thermodynamic obstacles to thermoelectric generation on the surface of Venus?

Question

There has recently been a NASA design challenge for a clockwork Venusian terrain navigation system which uses no sensitive electronics or hydraulics. I was thinking that rovers and probes on Venus could likewise benefit from a clockwork power plant to recharge themselves and operate indefinitely. Intuitively, the idea of a thermoelectric plant drawing on atmospheric pressure and heat "seems" like a perpetual motion machine, but I can't think of a real reason it would not work.

Assume a configuration similar to a nuclear fission power plant: A high pressure primary coolant system carries hot water into coils in a steam generator, which saturates steam in a secondary system. This in turn conveys the steam's enthalpy to turbine stages which convert the heat energy into mechanical motion. Low enthalpy steam is condensed via some condenser (which is not entirely necessary, but is normally included as a matter of efficiency). Condensate then re-enters the steam generator to pick up heat from the primary coils, then the process repeats.

It seems that any arbitrary size thermoelectric facility could operate indefinitely in this way, barring mechanical breakdown, but again it has been argued by some that this is suggesting a perpetual motion machine. Mostly because the electricity generated would likely be driving the condenser, seeming like we are using a quantity of heat as an energy source to cool itself.

Many comments suggest that a "heat sink" or "cold reservoir" in the form of lower heat applied to the condensate is required from outside the system to make this work. The condenser only increases efficiency, reduces pollution, reduces noise, and saves fuel. None of those things matter at all on Venus. A condenser is not NECESSARY to operate a steam power plant. But, if this is your belief, I would like to know a way to calculate how large a sink is needed to produce a given power at a given efficiency and input enthalpy. As an example Let's assume a 50MW plant with primary coolant pipes submerged in a pond of lead as a heat source. The pond may have CRES steel radiator fins extending into the atmosphere to increase the surface area drawing heat from the continuous wind. For discussion, I have designed the steam system model here. This system draws 1198 BTU per lb of water from the environment, with a mass flow of about 12,000 lb steam per hour. The turbine has been designed with 97% blade efficiency (not CHT efficiency as this turbine consumes no fuel), 81.2% isentropic efficiency, and reduce the steam from $620^\circ$C @ 1524 PSIA to $570^\circ$C @ 1211.46 PSIA (ambient temp and pressure is $467^\circ$C @ 1366 PSI. So the pressure and temperature differential on the system is very small. At the final stage, the 1211.46 PSI steam will flash to water at $569.993^\circ$C, so condensation requires a very small heat sink. This will be provided by an electric refrigeration unit, removing 8,214 BTU per hour from the output steam to condense it. Cooled condensate is then pumped back into the steam generator with a 320 PSI water pump.

Given such a 50MW system, if a "heat sink" is needed, how can we calculate how much heat in Watts this sink must be able to dissipate to prevent the system from stopping?

I've considered that elemental mercury vapor may be a better choice than water steam in this application, as it has a higher boiling point closer to ambient conditions, can not disassociate (needs no blowdown), and automatically lubricates the whole system. But this and other design considerations are beyond the scope of the thermodynamic question. I include this information for reference only.

I assume starting such a facility would be the opposite of a terrestrial steam plant, wherein the boiler pressurizes and heats the steam to an operating enthalpy before opening the valve to the turbines. In this case, the condensate stage must be cooled until liquid water forms, after which the valve between the turbines and condenser is opened to draw steam through the turbines by a negative pressure.

I argue that as long as the plant maintains a sufficient power load, turbine stages will reduce steam enthalpy until it condenses, and sustaining angular force on the turbine blades can be maintained indefinitely. The plant could, for example, be processing bulk CO2 into oxygen and carbon, with little real concern for efficiency due to the abundance of heat energy available.

Other than environmental challenges of construction, is this a sound thermodynamic process?

Answer 1

Don't think of internal machinery, just pack your system into a box ,and you have a simple heat engine .

The second law of thermodynamics puts a fundamental limit on the thermal efficiency of all heat engines. Even an ideal, frictionless engine can't convert anywhere near 100% of its input heat into work. There is always some heat that has to be discarded to surrounding and your system doesn't have a sink.

A significant amount of energy is needed for cooling and the energy obtained from turbines will likely be lesser than the amount of energy used for condensing steam.

Your system won't produce an ounce (correction: joule) of energy rather it would only dump energy to surroundings.

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Relevant data on Venus

Room temp: 471°C Pressure. :40 atm BP of water:$\approx 250$°c

Answer 2

This will be provided by an electric refrigeration unit, removing 8,214 BTU per hour from the output steam to condense it.

No, there is no obstacle to thermoelectric power generation. There is an obstacle to net thermoelectric power generation!

Invoking the need for a refrigerator smacks of 'turtles all the way down'.

Answer 3

As Nick said, there is no obstacle to doing this on Venus, but net positive work will not be possible without some temperature differential provided by the environment. While it is convenient to look at a thermoelectric plant on Venus like nothing more than a modern nuclear propulsion plant as found on a submarine with a planet-sized reaction chamber, the problem is that in this case, the entire plant is now operating inside that reaction chamber. A submarine adds heat of fission to make a hot temperature, but also has its condensate venting into ambient temperature equipment. So a temperature differential exists within the system. This is not explicity stated in the laws of thermodynamics but it is logically implied. Here is why that is necessary:

Most discussions about the second thermodynamic law focuses on adding heat to some ambient system. Your thought is that if you start with a very large heat source, you don't need to add anything until "Venus is a cold, dead planet" as commented. And everything the question posits is true, however it is omitting any process for waste heat disposal, without which the process can not run on indefinitely. The math for this has been done by a 19th century French army mechanical engineer named Nicolas Sadi Carnot, who through meticulous calculations derived an ideal heat engine, with which we can calculate the maximum theoretical attainable efficiency of any heat engine. That final calculation reduced the entire efficiency problem down to nothing more than two variables: the hot and cold temperatures of the system: $$\eta_{max} = 1 - \frac{T_L}{T_H}$$

As you can see, a nuclear propulsion plant's condensate environment both within the hull of the ship and through the sea chest provide a relatively low absolute T$_L$, which divided by the fission reactor's very large heat source (in Kelvin or Rankine absolute measurements), will yield a small fraction. That plant's maximum theoretical - or Carnot efficiency will be 100% less the ratio of those two temperatures, which will be a net positive number.

Now, if you place that entire plant inside the reactor vessel (by building it on Venus), irreversible heat flow will at some point occur between the components no matter how well they are insulated, and friction waste heat will be generated with no path to flow out, and the condensate will be in a room which is the same temperature as the heat source. As you can see, once the T$_L$ becomes equal to T$_H$, your Carnot efficiency becomes 1-1, or zero. Thus, net output will be your input time 0% efficiency in the best case scenario.

So to use Venus as your entire heat source, this would work if you could somehow provide a path for the heat to move off the planet. Obviously, this is hardly practical, so the plant will need to exploit a largeish natural heat differential on the planet surface, or possibly in the atmosphere.

What is interesting about this scenario is that although it looks like there is a greater opportunity provided by larger temperatures, you will need an even greater temperature differential on Venus than on earth for the same theoretical efficiency. Consider an identical plant operating in 300$^\circ$C temperature differential: The terrestrial plant operates in room temperature with T$_L$ = 298$^\circ$K and adds heat to make T$_H$ = 598$^\circ$K. The same plant on Venus operates by cooling condensate (possibly by ammonia evaporation or another endothermic process), from an ambient T$_H$ = 740$^\circ$K down to T$_L$ = 440^$\circ$K. The Carnot efficiency of the two looks like this:

  Terrestrial Plant    Venusian Plant    
$$\eta_{max} = 1 - \frac{298^\circ}{598^\circ}$$$$\eta_{max} = 1 - \frac{440^\circ}{740^\circ}$$
     = 50.16%             = 40.54%            

As you can see, all other things being equal, your maximum theoretical output from the Venusian plant with cooling will only be 75% of what the same plant on earth can do with heating. And if you follow this math, heating the Venusian plant 300$^\circ$ instead of cooling the condensate will make matters worse. In general, Carnot efficiency of any one heat engine will require a greater heat differential input to the system as the mean of the high and low temperatures increases.

To answer the specific question then, assuming you found a heat sink to reduce your condensate from 620 (600$^\circ$K) to a flashpoint of 570 (572$^\circ$K) to match your outlet steam temperature. Your 50MW plant will loose 3% to turbine efficiency (1.5MW); 10% (5MW) to pump condensate; and 2.5kW to condensation. Your theoretical max output would be:
50MW x $\eta_{max}$
= 50MW x (1-0.953)
= 2.3MW.

Therefore, if your external power source (such as a windmill) added 4.3MW, and removed all waste heat by some external heat sink, your power plant would run at a net zero output.

Now, to even reach a theoretical 50% output (25MW), you need to find a way to cool condensate down to 233$^\circ$F when it returns to the steam generator. This means reducing your 282.5klp/hr, 570$^\circ$F outlet steam by 337$^\circ$F; or providing 95.2 x10$^6$ BTU per hour of condensate cooling. That is 27.9MW of cooling.

As far as using other fluids such as mercury for steam, Carnot also talked about this:

The motive power of heat is independent of the agents employed to realize it; its quantity is fixed solely by the temperatures of the bodies between which is effected, finally, the transfer of caloric.

— Carnot 1890, p. 68

So to the other point, there is nothing specifically beneficial about using water steam for a heat engine. Possibly an elemental steam which can not dissociate will provide a small efficiency advantage, but this may be counteracted by the wear and tear on equipment needed to pump the enormous density of liquid mercury.