I've had my scoob for a week and I notice the intercooler(?) that sits under the air scoop has already got 2 dents in it from where stone chips have gone in.

what do I do?

 

More likely to be bees/wasps etc than stones getting into the intercooler.

The fins are easily bent, and can be straightened out again by using a cocktail stick or other such implement.

I understand several Scooby related tuning firms sell modified cocktail sticks in a range of materials (from carbon fibre to titanium) and different colours for this specific purpose

 

My fins were fine for 30,000 miles - two dents. Now it looks like if someone has bounced a squash ball a few times down the lefthand side. Must get one of those titanium (for lightness and strength) cocktail sticks to sort it

 

Get an intercooler cover. Roger Clark, GGR, PE and others sell them, seemed a bit gimmicky but I've just had one fitted
About £35-£50, there seem to be two types, one that rivets onto the intercooler (PE) and one the goes inside the bonnet skin (RC/GGR).
The PE looks better: and I would imagine is more substantial but I went for the other. With the PE it would protect it from straying spanners when the bonnets up, but as I hardly ever go under there anyway....

 

How does fitting one of those affect my 2yrs and 51 week warranty?

 

Surely a careful hand and a little screw driver (like the little plug ones)would nail it...save a few quid??

..Ian

 

This may seem a silly question, but...

If the intercooler fins get bent then cooling is reduced as airflow is reduced, but surely if you put another grill (ie a protector) over the intercooler, won't this also reduce airflow and cooling??

Martin

 

I tend to agree with Dr on this. Looking at the photos, if the ratio of the thickness of the wire mesh to the void area in between is say 1:5, then the area blocked off will be in the ratio of 1x1:5x5, i.e. 1:25, or a 4% reduction in direct air flow. This takes quite a few bent fins to achieve. I may be wrong, if on the other hand, the volume of air entering the air scoop at a given speed should remain the same and the reduction in exposed surface area is compensated by a corresponding increase in air flow speed through the fins, without affecting cooling.

 

Surely that air reduction should read 4.12356% !!!!!!

 

If the fins are bent less air will enter at any given speed. The same applies if a mesh guard of low free area is used. This is because a restriction of any kind will create an additional resistance, and air will tend to be directed over the opening so less will be scooped up.

Lokokee, you are correct in suggesting that the velocity through the remaining undamaged finns will increase, but the total airflow will still be less.

I would suggest a guard with high free area, alternatively you can buy combs for straightening damaged finns from any refrigeration or air conditioning wholesaler.

I wish I knew as much about Scoobies as I know about air conditioning, life would be so much more interesting.

 

The earlier posting was done at 6 o'clock in the morning after an all-night poker session and second-hand smoke from three sides, so the brain must be running on near empty. The correct calculation on the airflow reduction should be as follows: Assuming the thickness of the mesh/grid material is 1 unit and the void is 5 units square, then each cell is 6 units square, i.e. the void plus half the thickness (the mid-point) of the mesh at the sides surrounding it. The nett usable area is then 25 (5x5) out of a total area of 36 (6x6), or 69.44%, a reduction of 30.56 %, which is much more substantial.

Nutty, thanks for the correction. Resistance to airflow at the intercooler creates a back pressure at the air scoop, reducing total air intake. But your comb would not work because the fins are in a spiral, not amenable to combing as in airconditioners.

[This message has been edited by lokokkee (edited 27-06-2000).]

 

actually each cell would be 5 and 1/2 units (the remaining 1/2 unit would belong to the cell next to it)

JD(pedant mode)

[This message has been edited by JayDee (edited 27-06-2000).]

 

JayDee, count it this way: Left side (half unit)+ void (5 units) + right side (half unit) = 6 units. Top side (half unit) + void + bottom side (half unit) = 6 units. Therefore, each intermediate cell is 6 x 6 square inclusive of the sides. Your calculation applies to those cells along the sides.

 

Daft question - Are they any dangers in hoovering out the remains of all the dead flies from the intercooler. I started picking them out last night but then had this brainwave...