This has gotten really difficult really quick. (for me anyways, but that's what makes it fun)
>>The main culprit we are talking about here is FRICTION.
agree
>>The formula for calculating the pressure drop due to wall friction is:
>>deltap=(4*f*L/D)*(1/2)*rho*u^2
>>where:
>>dletap = pressure drop between two points in a pipe
>>f = friction factor
>>L = pipe length between the two points
>>D = pipe diameter
>>rho = fluid density
>>u = fluid mean velocity
>>Please forgive the crappy naming convention, but we normally work in greek letters.
Nonsense, your doing us a favor by translating.
>>So if the length of pipe from the radiator to the engine is fixed, what is the difference <<in pressure drop due to the change in pipe diameter?
So you haven't really used your full deltap formula to find the pressure drop, your just comparing the 2 diameters to find what the difference in the drop would be?
>>f, the friction factor is basically constant over the range of velocities we're looking at. >>L is constant, as is rho. So all that is changing is the pipe diameter and the >>velocity. To find the velocity we must find the change in cross-sectional area:
>>A = pi*(D^2)/4
so * means multiply? and ^2 means squared? (sorry I've never seen math on a basic keyboard)
>>where:
>>A = cross-sectional area
>>pi = 3.14....
>>so the change in crossectional area is
>>A2/A1 = (D2^2)/(D1^2)
>>For us that is (1.5^2)/(1.75^2) = .735.
so the cross section of the 1.5 D pipe is .735 or about 3/4 the area of the 1.75 D pipe.
I don't understand how you went from the circle area formula to only comparing the diameter squared? Why are you comparing the diameter squared and not the area?
>>So what does that do to velocity?
>>u = Q/A
>>where:
>>Q = volumetric flow rate
>>So if we say that we want the same volumetric flow rate.
>>u2/u1 = A1/A2 = 1.361
>>so D goes down, u goes up and everything else is fixed.
so pipe diameter goes down, velocity goes up IF volume/flow remains the same
>>What does this do to our pressure drop through the tube?
>>deltap2/deltap1 = (u2^2*D1)/(u1^2*D2) = (1.361^2)*(1.75/1.5) = 2.161.
>>So if you can maintian the flow rate, the pressure drop, and therefore power >>necessary to create it, is approximately double for the 1.5" tube.
The math is lost on me at this point but I can understand your end result.
To achieve the same volume/flow going from a 1.75 tube to a 1.5 tube takes twice as much HP at the pump because friction increases with the increased velocity. That's assuming that the pump is capable. This also brings into question how much HP does a water pump use.
from the Stewart site
"Stewart high-flow water pumps deliver up to 180 GPM (gallons per minute) of coolant flow (at 8,000 RPM), yet consume just 2.26 horsepower (at 4,000 RPM)!"
so a decrease in tube size would bring waterpump HP drain to 4.5HP. Maybe that's why the original 40's used 1.75 because they were chasing a couple HP, like all racers do, and the person saying 1.5 works fine is correct as far as he's concerned because the system cools just as well and the HP loss is not noticeable. OR what if the rad already has more resistance than the 1.5 tube, would it matter then?
>>On the other hand, if you can only maintain a certain pressure drop, the flow velocity >> is going to decrease to:
>>u2/u1 = 0.92
>>and so the volumetric flow rate will decrease to:
>>(u2/u1)*(A2/A1) = 0.92*0.735 = .680
>>So your flow rate would be approximately 68% that of the 1 3/4 pipe.
Is this figure dependent on a waterpump not providing ANY increase in flow/volume? This figure compares closely with the direct size comparison of .735.
>>The truth is something between these two cases. As you increase the restriction to the >> pump, the pressure will increase some, and the flow rate will go down some, but at >>least this gives you some idea.
>>Also remember that we are talking about the the pipes alone. Anyone with an >>electrical background can think of the radiator and pipes as resistors in series. If the >>resistance of the radiator is >> than that of the pipes, then the pipe diameter does not >>matter.
Until that pipe gets so small that it has more resistance than the radiator? Now this is how we're going to find out how small of a pipe is possible.
So your saying that if the rad resistance is greater than the 1.5" pipe then going to a 1.75" pipe isn't going to decrease the waterpump HP drain? I'm kind of lost on this one too, it doesn't make sense to me. Are you saying that if the loss at point B is greater than the loss at point A the the loss at point A does not exist?
>>So with the double pass radiator you have effectively reduced your cross-sectional >>area by half, and increased your length by double. The same calculations above can >>be used to find the difference in pressure and flow.
>>Maybe Kalun can run through those for us!
ha ha ha, (don't laugh at me, laugh with me) (they didn't teach this level of math at the school of hard knocks that I went to)
>>If we found out how many, what size, and length the radiator tubes are, then we >>could find the relative resistance of the radiator to the coolant pipes.
>>Adam
OK here are the sizes, carefully measured with wire gauges made up of bailing wire, wire feed wire, electric fence wire (reminding me of a shocked cat), solder wire, and a couple other wire sizes that were laying around. Inserted down in one of the rad tubes via the inlet neck of the radiator. The bailing wire came the closest at .048.
The outer dims of the rectangular tubes are 1.00 x .080. So with the .048 wire fit, that means a wall thickness of .016 which sounds about right. So the inner dim is .968 x .048.
The ERA spec rad has 2 rows of tubes with 64 tubes per row so total 128 tubes @ .968 x .048 x 12" long. Or 64 tubes 24" long with a double bypass. At first I thought why not just add up the cross sectional area, which totals 5.95 sq. in. which is way more than the 2.4 sq. in. of the 1.75 tube. But I'm guessing you can't do that because of that FRICTION culprit, which exists at the walls of the tubing. So somehow you have to bring the wall area into the equation.
I think you need to compare the ratio of wall surface to cross section area of the total rad tubes to the same ratio of the transfer tubes.
I'll compare total cross section and total wall area. I'm assuming it's correct to only compare the transfer tubes going one way to the radiator because of the series/resistance thing. (Maybe that's why on some production cars the return line is smaller than the supply line to the radiator, hey why can't we save some size/room/weight here too?)
OK, a 1.5D tube has a cross section area (A = pi*(D^2)/4) or (pi x radius^2) of 1.77 sq. in.
the wall surface of that same tube with a circumference (pi x D) of 4.71" x 108" = 508 sq. in.
So the ratio of cross section to wall surface of the 9' x 1.5" tube is 1.77/508 or about .0035.
The radiator has 128 tubes with a total cross section area, wait a minute, let's go with a double bypass. So the radiator has 64 tubes .968 x .048 so total cross section of 64 x .968 x .048 = 2.97 sq. in.
The total wall surface of the rad tubes .968 x 2 + .048 x 2 x 24" x 64 = 3,121 sq. in.
So the ratio of cross section to wall surface in the dbl bypass rad is 2.97/3,121 or .00095.
so comparing the two ratios .0035/.00095 = 3.68. So this radiator has 3.68 times the resistance/FRICTION of the 1.5"x 108" tube in question.
This is shade tree engineering math only, it may be TOTALLY incorrect, please grind the numbers Adam.
Ron Davis ERA spec dbl. pass aluminum rad.
64 12" tubes going up and 64 12" tubes going down so in effect
64 tubes 24" x .968 x .048 inside dimension compared to 1.5" and 1.75" x 108" tubing.
If you have time compare this to a standard rad with
128 tubes 12" x .968 x .048 so we can debunk the Stewart claim that a dbl bypass has 16 times the resistance.
and analyze/compare the possibility of having an even smaller return transfer tube. (time dependent also)
++++++++++++++++++
I think this information could be valuable to manufactures and scratch builders.
This car is very crowded/ compact to build. One of the most crowded areas is in the firewall to engine area. The rad transfer tubes both go through this area, if they could be reduced in size it would only help matters.
Another thing I don't think has been considered is the advances in radiator technology. The move to aluminum and the change in core tubing shape means a more efficient radiator than the brass unit in the original 40's. Radiators have changed but the transfer tubing size has stayed the same. If the radiator takes less flow to achieve the same cooling then why can't the transfer tubes flow less and thereby be smaller?
This gains advantages in weight and cost by reduction in tube size and coolant volume and the afore mentioned space considerations.
And don't think that this amount of weight reduction doesn't matter. A race cars weight is reduced in this very fashion. After the change from iron to aluminum heads, or cost prohibitive changes like a carbon fibre body, other weight reductions are on a much smaller scale and a large reduction is achieved by many smaller ones.