Thermosyphon warming process

[Bob Culver]

In the 1960s I used to regularly drive my RP 60 miles to where I was staying.Flat out most of the way and the last two miles flat out full revs in 3rd and 2nd. Wisps of steam from the radiator cap and the time it was garaged the whole engine was so hot could not touch even the distant end of dynamo.

[Duncan Logan]

I did something similar a few years ago. I found that the main heat flow in the radiator was across the top and diagonally through to the bottom hose. The lower left side portion of the radiator didn't get anywhere near the temperature change. My ambient at the time was 34C.

If I find the images I will put them up.

[Graham Barker]

Duncan,  do you think that part of the lower core was blocked?  Or was the flow just bypassing the one bottom corner to go directly to the outlet?
Regards  
Graham

[Dave Mann]

The photos are interesting especially the cold carburetor, it's a pity you couldn't take one on power. Going along the A50 at initially 45 mph yesterday after a few miles the speed crept up to 53 mph. Was that as the induction system warmed up providing better fuel vaporization? I've noticed this characteristic on hot days.

[Andy Bennett]

  P7030_sn1549(black)_56.jpg (Size: 10.32 KB / Downloads: 156)

Dave
This one was taken of a Ruby having just arrived at my house and not switched off so as close as can be to running on the road. 
apologies only black and white but shows even more clearly just how cold the carb is.

Andy

[Colin Morgan]

Looked a bit further into thermosyphon cooling to think about what might be going on inside. (If anyone spots errors or numerical mistakes, I will be happy to make corrections.)

The water circuit comprises the block + head, top hose, radiator header tank, radiator tubing and bottom tank/bottom hose back up to the block.

Assuming the cooling system contains just water (no antifreeze) and the engine is putting out a useful 7 BHP – (the car is doing a steady 35mph or so?) then, if the engine is 40% efficient, the power output of 7 BHP (= 5.1KW), gives the radiator 7.6KW of waste heat to dissipate at equilibrium, ignoring other heat losses. (Correction - this amount of heat being lost by the radiator would be from about 1/3 of the total output of the engine - with in practice about 1/3 of the energy from combustion being useful work, 1/3 being taken away by the exhaust gases, and 1/3 lost through the radiator. So the 7.6KW of heat through the radiator would occur when the engine was putting out 7.6KW of useful work, so about 10 BHP - which would correspond to driving on the level at a speed of about 40 mph.)

If the temperature is measured at the top of the radiator and found to be 80 C, and at the bottom 40 C, then the amount of water in circulation can be calculated. The water flow rate is the KW of heat removed divided by the heat capacity of water x temperature difference top to bottom in the radiator. So, this is 7.6KW divided by (4200J/Kg C x 40 C) which is 0.045 Kg per second (or litres per second), so about 45 g/second or cc/second (or 1.8 fl oz, or about one egg’s volume a second). This isn’t much and shows how good a coolant water is, with its very high heat capacity?

Given a total cooling water capacity of about 11.5 pints, which is 5.4 litres, this flow rate means that the water takes an average of 119 seconds (5.4 litres/0.045 litres/sec), or almost 2 minutes, to go around the whole system.

The top hose is about 40mm diameter. At this flow rate the water is travelling on average at 0.04m/s or about 40mm or 1.5” per second. The radiator has about 45 crimped tubes about 50mm x 2mm in section, which gives an average speed of flow of about 10mm/s down though the radiator.

From the estimated water speeds, time taken by the water to past through various parts of the circuit can be calculated: Top hose about 9 seconds, radiator top tank (if water is 3/4” deep) 18 seconds, rest of radiator about 45 seconds, bottom hose 9 seconds, inside block and head (assuming flow area is 40% of 11” x 5” and it is 5” high) 40 seconds. (This adds up to 121 seconds – which is similar to the estimated time from the heat balance of 119 seconds.)

[Mark Dymond]

Interesting 2 minutes is slower than I would have guessed but your logic is sound. Did you get the 40 degree temperature drop from anywhere or is it an estimate.
I was once told when looking a stationary diesel engine that  one third of the heat went into work, one third out the exhaust and one third out the radiator.

[Colin Morgan]

You are right, I have forgotten about the heat lost in the exhaust. It seems a better estimate of where the energy from combustion goes would be, as you say, 1/3 mechanical work, 1/3 heat lost in the exhaust gases, and 1/3 removed by the cooling system.

I have measured the top temperature in the radiator in the past, and the temperature of the bottom hose was estimated - it was about as warm as bath water after a run. So the temperatures are about right? (Ambient temperature was about 15 degrees.) Therefore, taking the above calculation where 7.6KW is lost through the radiator, and the 1/3, 1/3, 1/3 split this would correspond to a useful work output of 7.6 KW which is 10.2 bhp - so the engine would need to be working harder to produce that heat load on the radiator and the flows given above - perhaps a speed of more like 40 mph?

[Dave Mann]

I've never experienced this performance increase in the depths of winter. Also when we were testing 1500 bhp gas engines I was quite surprised at how fast the engine cooled down on reducing from full power to idle.

[John Cornforth]

Hi

I have noticed the gradual performance increase with time on the road, but always put it down to the slow warming and thinning of the gearbox and back axle oil, thereby reducing drag.