Hi Art:

 

I composed the letter below and the attachment to respond to you questions and those from others.  My emails to LML keep bouncing.  Marv and I are grappling with the problem, but to help with the solution, please forward to LML from your email, and let’s see if that works.   Let me know if I answered your questions.

 

Cheers,

 

Fred

____________________________________________________

 

I have received questions both on the forum and off concerning my earlier comments about cooling drag.  They sent me scurrying back to my books and calculator. 

 

I spent some time generating a self-consistent set of numbers for cooling drag, pressures, temperatures, etc. with a uniform set of assumptions.  Some guesstimates are required, particularly concerning losses from friction in various locations, so I have listed everything assumed and calculated, and invite investigation and questions.  The results are summarized in the ugly sketch attached.  It shows the flow conditions at various points along the cooling air pathway assuming you have the ultimate set up, no leaks, cowl flaps perfectly adjusted to get the right exit area, etc.   

 

Air enters from the left, is partially compressed in front of the inlet by the slowing of the air flow, goes through the inlet, then through a short diffuser where it is slowed more, and then dumps into a large plenum over the engine where the flow is a fairly slow moving mess.  Then the air flows through cylinders and oil cooler (which can take a lot of flow) into the space below the engine (where the flow is again a mess, but a slow moving mess), and then accelerates through the outlets, the area of which is controlled by cowl flaps which are optimally adjusted.   I have assumed 9000 feet, 200 knots IAS, 230 knots TAS as an interesting speed for us aspirated engine guys to investigate.  I have used pressures in inches of water since the engine pressure drop charts use these units of measure.  The inlet velocity ratio is assumed to be 0.4 (velocity through cowl inlets divided by TAS) which seems to be roughly what the latest and fastest designs are using in the fastest production aircraft and speed mods offered by LoPresti.

 

Comments:

1)     Flow is controlled at the outlet with cowl flaps, so we can get a lot of compression in front of the inlets where the compression process is frictionless.  With an inlet velocity of 0.4, we get about 86% of the total dynamic pressure in front of the inlet. 

2)     We are forced to accept a short diffuser behind the inlets which further slows the flow, but friction losses eat up an assumed 2 inches of water pressure.

3)     The velocity in the plenum is anybody’s guess as the flow is a turbulent mess, but the velocity is low, so losses are low.  But there are losses in the plenum, particularly with the 550 Continental engines because of the intake manifold and other clap trap above the cylinders.

4)     Note that the compression raises the temperature about 12F so the assumed temperature above the fins is about 40F.  We can use this when we go to the engine cooling charts to make an estimate of the air flow required.

5)     Engine cooling charts are published assuming full power and 475F CHT (red line) and cruise with 435F CHT.  The charts suggest that the engine pressure drop will be about 5 inches of water at 435f CHT.  But we like 350F or thereabouts for our CHT and that requires more airflow for more cooling.  A real rough guess is about 20-25% more air flow, and this requires about 8 inches of pressure drop across the engine to force the increased air flow through the fins.   

6)     The flow below the engine is chaotic, but slow so losses are low.  It is also hot.  With 350F CHT, it is probably 100-120F higher than ambient, or perhaps 150F.  (There is a greater temperature rise across cylinder heads, and much less across the oil cooler, and it all mixes together.)  Because there is a lot of clap trap below the engine, there are more losses which I guesstimate at around 1.0 inch of water.

7)     As you approach the exit (“discharge nozzle”) the flow is accelerated, hopefully smoothly, efficiently, and directly aft to regain as much momentum as possible.  However, our exit nozzle is filled with exhaust pipes, pipe brackets, and other clap trap in the higher velocity area as well as friction losses along the walls.  My guess is we lose another 2 inches of water pressure.  This leaves about 12 inches of static pressure converted into dynamic pressure in the exiting flow.  With the warm, lower density air I get about 170 knots discharge velocity directly aft for our perfect cooling system.   The loss in velocity for the cooling flow is then 230 KTAS minus 170 KTAS outlet, or 60 knots. 

 

Cooling drag is equal to the momentum loss of the cooling air.  It is cooling air mass flow (pounds per second) times the velocity lost during passage through the cowl and engine.  If we had lower pressure losses, we would have more pressure at the exit to accelerate the flow aft faster. 

 

If we could get the exit velocity = free stream velocity (the true air speed), cooling drag would be zero. 

 

If exit flow was faster than the free stream velocity we would get thrust. 

 

If we are moving really fast with lots of ram pressure and low losses and we raised the temperature of the air stream up very high to lower the air density a lot and allow it to be accelerated aft really fast, we would have a ram jet.

 

But for our air cooled engine with lots of friction inside the cowl and engine and only modest temperature rises, we can not get thrust, even for our prefect case above.

 

Now for a cross check.  Let’s assume our IO-550 is pumping out about 225 HP (about 73%).  If you go to a Lycoming power chart and assume it is the same for the Continental (pretty good assumption), for 40F above the engine and 435F CHT the chart shows you need 3 lbs/sec. of cooling air and 5 inches of water pressure drop across the engine to get this flow. 

 

But we want 350F CHT to let our jugs live longer, so the required air flow is (rough guesstimate) 3.7 lbs/sec. and then the required pressure drop to get this flow is about 8 inches of water, the figure I used above.   And if you calculate the cowl inlet area required to admit this flow rate at 40% of the free stream velocity (TAS), you get two 6 inch diameter inlets.

 

Moving right along, if we assume that the prop efficiency at 230 KTAS and 225 HP is 85%, then the thrust = drag = 272 pounds.  Now we compute the cooling drag which is the momentum loss which is the cooling air flow rate times the velocity loss of 60 knots times some constants to make the units work and we get 12 pounds for the cooling drag or about 4% of total drag for our perfect zero leakage cooling system. 

 

In my earlier message I noted that your run-of-the-mill spam can had cooling drag of about 6-7%.  Our hot rods have about half the drag as a typical Bonanza or C210, so cooling drag becomes maybe 14%.  If we work really hard maybe we can get it down to 4%.  You can now see what it takes.

 

Your turbo guys at 25,000 feet have a much more serious problem because there is little to work with.  Assuming 280 knots TAS, the IAS = 190 knots and you thus have about 10% less ram pressure available to start with, maybe 23 inches of water.  The air is colder, but it is also a lot less dense so if I really extrapolate off my power chart, and assume 262 HP (75% for a TSIO 550) and 350 CHT, the pressure drop required across the engine is 12-14 inches of water, call it 5 inches more than at low altitude.  So the high altitude means that you lose 3+5= 8 inches of water pressure compared to the low altitude case above as the flow approaches the exit.  Instead of 12 inches of water to accelerate the flow aft, you have about 4 inches, so the exit velocity is much lower requiring a larger exit area,  and the momentum loss much higher which means cooling drag is much higher.  And that is the ideal case.  Most LIV cooling systems are far from ideal, so my guess is that for the stockers (no mods) cooling drag is somewhere around 20-25% of total drag.

 

You can see for the high altitude case cooling drag will always be a bigger piece of the drag pie even if we assume a perfect, leak free cooling system.  However, if you run this configuration (big exit area) at lower altitude, you will have a lot of excess cooling drag you could shed – if you could close cowl flaps, shrink the exit area and thereby accelerate the flow aft to a higher velocity (lower momentum loss).

 

Closing comments: the above discussion and figures assume a perfect leak-free system with optimal inlet and exit areas.  Optimal exit area for cruise flight is too small for slow, high power climb on a hot day (you will cook the heads), so it requires adjustable exit area meaning cowl flaps.  And low leakage means a plenum, careful attention to plugging leaks, and also plugging leaks from inside the cowl to the outside. 

 

Bottom line: the best you can get over stock is probably 10-12 knots.  If you just minimize leakage with a plenum and closing off leaks, you might get 5 knots which has been reported by Chris Z.  If you don’t want cowl flaps, you can close down the inlets (velocity ratio much higher than 0.4) to minimize flow rate and get another 2-3 knots, but you will have to climb at high IAS to stay cool. 

 

If you want all you can get (10-12 knots) then you will have to do it all, and do it to perfection.   

 

Cautions: I am getting old and forgetful and probably forgot something.  Your mileage may vary.

 

Questions and comments welcome.

 

Fred Moreno