Website calculator, safety margins

[engunear]

I've been messing around with a Leibig condenser and comparing its performance to the web calculator on the parent page and have found some interesting things. This introduction is written in the simplest non-maths language I can find. There is a more technical version that will be attached as a pdf. If this description is a bit too didactic, my apologies. Its hard to pitch the writing for all the audience for this site.

First, consider the thermal conductivity equation. This is the basis of the “heat transfer coefficient” (HTC) used by the parent website calculator. The parent gives some textbook values but I'm about to calculate what it should be for the condensers we use. The thermal conductivity tells you how much heat is transferred through a solid, and is specified for cube 1m on each side, in watts per degree C. It captures how good that material is compared to others e.g. copper has high conductivity, wood is low.

Then this can be scaled linearly going up with area and down with length the heat has to travel through and the area it gets to use; if there is half the area, half the heat gets conducted. If it is half as thick, it takes half the temperature difference for the same power. If we keep everything in MKS units then the numbers are easier to work out.
So for a piece of material which is not a cube, where A is the area and L the length the heat has to travel through:

Power = dT (CA/L)

The CA/L is grouped as it is fixed for a piece of material.

Instead of thinking about conduction, we can think about its reciprocal resistance:

Power = dT/R

The thermal resistance is the reciprocal of the conductivity, and is

R = L/(CA)

The thermal resistance of the condenser can be calculated by adding the resistance of the various components in the path of the heat. The website calculator gives 850 W/sq M/C – the total thermal resistance of 1 square meter of whatever makes up the walls of the Leibig.

Assuming the condenser has a copper pipe, its thickness is 1mm, its thermal conductivity is 385W/mK and its resistance contribution is:

Rc = 385/0.001=385,000 W/sq m.

So the copper makes a small contribution. It might be the boundary layer of water between the pipe and the (assumed to be turbulent) liquid. This boundary layer seems to be around 25 microns thick. So its contribution is

Rw = 0.6/25E-6 =24,000 W/sq m.

It is not the water. Looking at Leibig condensers on the web from the period of the textbooks on condenser designs, they are all made of glass. Assume the glass is 1mm thick:
Rw = 0.6/0.001 = 600 W/sq m. At around 0.75mm we get the 850 W/m/C. Bingo! So perhaps the 850W number is for a glass condenser. If so, then the website calculator default value is encouraging us to build condensers that are too long, especially after a safety margin is added.

You can measure C for a Leibig condenser by turning the flow up as high as possible, and poking a thermocouple into it to measure the length of the region where condensation is occurring. The condensation area seems to be quite compact. From the length calculate the area, the temperature difference is known (if the flow is high the output temperature is equal to the input temperature) so the dT is just the difference between the water and the boiling point. Because the inlet temperature equals the outlet temperature the complicated effects considered by the website calculator are not relevant.

I've measured this once and got about 3000 W/degree per C meter. The condenser has a 900mm cooled region and an active region that is closer to 300mm in some cases. I will measure it carefully when time permits, and have been thinking about safety margins too ...

[engunear]

See Graph 1:

There is talk in various forum threads about turbulators increasing the efficiency of a condenser. At first I puzzled over this, thinking that a condenser must obey the well-known law of physics “The Conservation of Energy”, so the heat out is equal to the heat in, regardless of how much turbulence there is in the system.

There is a simple equation that relates the boiler input power, the flow rate f, the difference between the input and output temperatures of the cooling water and the specific heat of water s.

P=fs(Tout-Ti)

Just to be sure, I measured the flow, inlet temperature and outlet temperature with two heads at various flow rates. Graph 1 compares theory and measured values of temperature and flow rate for the two condensers. It is difficult to get this right, the boiler has to be insulated to reduce heat loss, pipes have to be short as they lose heat, the boiler can lose a lot of heat if placed in a concrete floor etc. The distillate temperature also carries away about 10% of the total power, so a Nixon offset design, which produces hot distillate produces a slightly different value to a Leibig that produces cooler distillate. This is mentioned on the parent page. The Nixon head had an additional cooler so it was possible to measure with and without this.
The flow equation also says that for a given power, there is a minimum flow below which the condensation is incomplete, and a flammable liquid can be vented into the still area no matter how good or how long the condenser is..

The minimum flow is given by (Tb=boiling point of distillate, Ti = inlet water temperature)

f=P/s/(Tb-Ti)

For a 2KW boiler (P=2000), with water (Tb=100), with an intlet temperature of 10C (Ti=10), and with the specific heat of water being 4184 J/K, this has a value of 319ml/min and for ethanol this is 423 ml/min.

The effect of this can be seen in the diagram, at around 500 ml/min where the outlet temperature shoots up towards 100C. For ethanol the value is higher because Tb is 78C.

Conclusions:
1. Classical physics works. A trivial result I know.
2. There is an equation that tells you when your still will start venting flammable gas and turn dangerous.
3. Thought for the day: a steam engine is a still for doing work.

[engunear]

It is important to operate a still with a safety margin in the condenser so that if, for example, the coolant pressure drops, then you don't have flammable vapour emitted. Water pressure fluctuates throughout the day – where I live it drops in the late afternoon. Flow has to be higher than the critical flow, but by how much?

Length does not vary through the day, so the situation is simpler. A condenser also has to be longer than the minimum length, but by how much? Making too large a condenser makes it heavy so it is hard to support and mechanically stresses the boiler connection (interestingly this gets much worse as the condenser gets longer because it acts like a lever and the torque gets larger as the weight is further from the mounting)

So taking an example: consider condenser for alcohol with a 2KW boiler, where the inlet water can be as high as 20C, and made with 18mm (ID) copper pipe.

Graph 4 shows how the active length varies with outlet temperature. The reason for this is that the outlet temperature is easier to measure than flow. In choosing the length and adjusting the flow, we are trying to keep the active length shorter than the total length.

As part of the commissioning of a new condenser, it is probably worth sticking a thermometer into it, to measure the active length and check that it is working. If for some reason the design had a heat transfer coefficient is different to that used in the calculations, then the results would be wrong.

Graph 5 shows a version of Graph 4 in which the flow is divided by the minimum flow and the length by the minimum length. I think can be used for any power and any material but this has not been checked.

To produce a small, functional design, following this method:
1. From the power, and the planned material, calculate the minimum length using the equation: X=P/(C.diameter.pi.(Tb-Ti)). Pi=3.14159... where C is the heat transfer coefficient per unit length of material (say 3000W/K/sq m for copper/water).
2. Choose the safety margin for the length (say 2X).
3. Find the minimum flow for the chosen length from F=P/(s.(Tb-Ti)) with s=4180, P=2kW, Tb=78, Ti=20 this is 0.0059 liters per second or 358ml per minute.
4. Choose the flow safety margin (say 3X).
5. From the Graph 5, find the outlet temperature at this flow
6. Mark a redline on the output temperature monitor at the value corresponding to the minimum flow for that length.
7. Set the flow and relax.

Example: Power =2kW, diameter=18mm, inlet temperature=20C (worst case), Tb=78C (ethanol).
1. Minimum length X=2000.0/(3000*0.018*3.14159*(78-20))=0.203
2. Planned Length=2X*0.203=0.406
3. Minimum flow = F = 2000.0/(4180*(100-20)) = 0.0059 liters/sec = 0.358l/min.
4. Flow with safety margin = 1.076 liters per minute.
5. Looking for 3X on the green line, and reading horizontally to temperature we see we need a 38C outlet temperature. Reading vertically the active length of the condenser is about 1.3 times the minimum length, so the 2X safety factor on length appears OK.
6. Monitor output temp regularly.

[engunear]

Finally, I'm attaching a longer document that has even more (!) detail. Then off to cut my 900mm Leibig down to size and try it out (with thermocouples of course).