Actually, the “H” energy head factor in the equation  H = Pi*n /(pgQ) is inversely related to restriction.  The higher “H” the more restriction.

 

If we rearrange the formula and solve for flow “Q”, we have   flow Q = Pi*n/(pgH) which says (leaving other factors along) that if we reduce the factor “H” (head factor – which in essence means reducing restriction so head “H” is lowered) then Q will increase.  Ergo, to increase Q reduce H meaning reduce your restriction.  Well, I think that’s what it says {:>)

 

However, Jeff,  I must admit my only real experience with “restrictor” plates and the rotary were the two occasions I attempted to insert the coolant thermostat back into the engine – in both instances I almost fried my engine.  The engine would get overly hot (like Pinging hot)  and the thermostat never seemed to open or at least the system did not cool adequately for sure.    I even drilled 1/8” holes around the rim of  one to ensure no air could be trapped based on recommendation from someone on the list.  I tried a couple of thermostats in case I had a “bad” one.  No go, and I still fly without a thermostat {:>).  I know other’s fly with a thermostat – but, just didn’t work out for me.

 

So I decided “Free Flow” - to heck with restrictions – and have cooled every since.  Like I mentioned I can see situations were a restriction might be called for (Lynn has mentioned it as well), but for my system I want it free flowing (well, as much as possible).

 

 

 

 

But Ed, does it mathematically show that you now want to install a restrictor plate?

Or are you sticking to this original statement?: The entire ideal is to transfer heat from engine to air.  Holding the coolant longer in the block (via a flow restrictor) will indeed increase the temperature and therefore the heat content of the coolant.  However, my view is the ideal is NOT to hold the heat in the block but to promote its rapid exit from the block to the radiator. 

Jeff

 

 

Here is a formula for a centrifugal pump that clearly? Shows that Tracy and Lynn are correct

 

Energy Usage

The energy usage in a pumping installation is determined by the flow required, the height lifted and the length and characteristics of the pipeline. The power required to drive a pump (P_i), is defined simply using SI units by: by:

where:

P_i is the input power required (W)

ρ is the fluid density (kg/m³)

g is the gravitational constant (9.81 m/s²)

H is the energy Head added to the flow (m)

Q is the flow rate (m³/s)

η is the efficiency of the pump plant as a decimal

 

One can see that if Q the flow rate becomes zero (by blocking the exit) then the power required to drive the pump Pi also becomes zero.  So block the pump and lower the flow and the power required drops – or with the same power, the pump can spin faster.  There is always some flow around the vanes of a centrifugal pump in reality, so the power does not cause the pump to spin to infinity rpm but it equalizes at a higher rpm than when considerable (unblocked) flow is the condition.

 

Is this fun or what?

 

NO!  I meant exactly what I wrote.  It is admittedly counter-intuitive but true none the less.  Did you attempt to prove it to yourself with the suggested test?   Only takes a few seconds :>)

Tracy

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