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Old Apr 5, 2018 | 05:44 PM
  #61  
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Originally Posted by iwasmenowhesgone
In a very related question......does this spring go the entire distance from the outlet of the radiator to the inlet of the water pump? I believe it is shorter than that, based on pictures of the spring I have seen, sooooo if it does not fill the hose the entire way, why does the hose not collapse in the area without the spring?
The hose on each end is supported by the radiator opening and the waterpump inlet...

Take a 2 x 4 - span it across a couple of concrete blocks. In the middle of the span - it flexes the most...same with the radiator hose
Old Apr 5, 2018 | 05:58 PM
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Originally Posted by Richard454
Beating a dead horse is not....
It's not very nice to call "resdoggie" a dead horse....
Old Apr 5, 2018 | 07:08 PM
  #63  
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Originally Posted by iwasmenowhesgone
In a very related question......does this spring go the entire distance from the outlet of the radiator to the inlet of the water pump? I believe it is shorter than that, based on pictures of the spring I have seen, sooooo if it does not fill the hose the entire way, why does the hose not collapse in the area without the spring?
You brought up a great question that is pertinent to this thread, thank you. (You jogged an old memory cell of mine).
It is IMPORTANT to center the spring in the hose, and favor the radiator side.
Back in '73 a guy came in our service station with a Vette that was SCREAMING like no tomorrow. The car, not the guy. He had just changed the hoses.
The spring was too close to the water pump inlet, moved enough to grab the impeller, and proceeded to wrap itself around the impeller, causing a shrieking sound like you never heard.
Needless to say that was what I spent that evening repairing.
Old Apr 6, 2018 | 02:01 PM
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Originally Posted by lionelhutz
It's not very nice to call "resdoggie" a dead horse....
That's OK, call me any names you like - just don't call me late for dinner!!!

"park" is dwell time. An important consideration in designing a heat exchanger.
Old Apr 6, 2018 | 02:42 PM
  #65  
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Originally Posted by resdoggie
"park" is dwell time. An important consideration in designing a heat exchanger.
LOL, not really. That makes it clear why you don't understand that the author is writing wives tales. You also believe them.

Last edited by lionelhutz; Apr 6, 2018 at 02:43 PM.
Old Apr 6, 2018 | 03:57 PM
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LOL, no, the author is correct and here's why.

If coolant at 180*F enters a 10' foot long tube with air blowing across at 75*F and exits at 170*F, will the same coolant entering a tube 100' long with the same velocity exit at the same temperature? This is a simple yes or no. If the answer is no, will the coolant temperature be lower exiting the 100' tube? Yes or no?

Now increase the speed by 10 fold. Will the coolant exit the 10' tube at 170*F or at a higher or lower temperature. What about the 100' tube?

By extending the tube you increase dwell time for the coolant to lose more heat through the tube to the air passing across. But if you increase the velocity of the coolant, dwell time is decreased and the coolant will not lose the same amount of heat when it flowed slower.

At some point, if the coolant travels too fast through a tube, there isn't enough TIME to lose enough heat i.e. cool. Hence, speed of the coolant and dwell time effect the cooling of a fluid and of course material selected and ambient conditions.

And that is how engines will eventually overheat without a TS. Coolant moves through the radiator too quickly because the TS has been removed. But not when its -20*F outside. In this case, the engine will never reach optimum engine temp which is 180*F or so.

Btw, have you removed your TS and run straight water?
Old Apr 6, 2018 | 06:30 PM
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Wow. Yet another believer who completely forgot moving the coolant slower through the radiator ALSO results in the coolant moving slower through the engine. a believer who completely forgets it's a closed system with a fixed amount of coolant.

If coolant comes from the radiator at 170* and stay in in the engine for 1 minute before exiting at 190*, will it still exit the engine at 190* if you kept it in the engine for 10 minutes?

Here's a simple explanation why he's writing wives tale. To keep it as stupidly simple as I can, I will assume BOTH the air flow through the radiator and the coolant flow through the engine stay constant ALL the time. I will also assume the temperature difference between the coolant and air at the radiator makes no difference in it's efficiency.

1st case - The engine idles and produces about 2000 W of heat which raises the engine coolant temperature by about 10*F. The coolant goes through the radiator and loses that 2000 W of heat which drops it by 10*F again before it enters the block again. This means the engine is running at about 180*F where the coolant enters the block and 190*F where the coolant exits the heads. Great, all balanced and the engine temperature is nice and stable throughout.

2nd case - I throw the car in drive and start climbing a hill. The engine starts producing 20000 W of heat. This means the coolant temperature in the engine is raised by 100*F as it exits. The coolant then goes through the radiator and sheds the same 200 W of heat as before. This means the first time the coolant circulates it will still start at 180*F as it enters the block but it is now 280*F as it exits the heads. Then, it goes through the radiator and cools to 270*F. So, on the second pass the coolant is 270*F as it enters the block and 370* as it exits the heads. This doesn't make for a very happy engine.

For the second case, The first thing to do is increase the coolant flow by about 10X to bring the temperature differential across the engine back to 10*F. The next thing to do is get the radiator to do 20000W of cooling instead of 2000W. I don't know how much more airflow, but I'd bet any money you want that a lot more airflow will be the right answer in accomplishing that goal.
Old Apr 6, 2018 | 07:50 PM
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Originally Posted by lionelhutz
Wow. Yet another believer who completely forgot moving the coolant slower through the radiator ALSO results in the coolant moving slower through the engine. a believer who completely forgets it's a closed system with a fixed amount of coolant.

If coolant comes from the radiator at 170* and stay in in the engine for 1 minute before exiting at 190*, will it still exit the engine at 190* if you kept it in the engine for 10 minutes?

Here's a simple explanation why he's writing wives tale. To keep it as stupidly simple as I can, I will assume BOTH the air flow through the radiator and the coolant flow through the engine stay constant ALL the time. I will also assume the temperature difference between the coolant and air at the radiator makes no difference in it's efficiency.

1st case - The engine idles and produces about 2000 W of heat which raises the engine coolant temperature by about 10*F. The coolant goes through the radiator and loses that 2000 W of heat which drops it by 10*F again before it enters the block again. This means the engine is running at about 180*F where the coolant enters the block and 190*F where the coolant exits the heads. Great, all balanced and the engine temperature is nice and stable throughout.

2nd case - I throw the car in drive and start climbing a hill. The engine starts producing 20000 W of heat. This means the coolant temperature in the engine is raised by 100*F as it exits. The coolant then goes through the radiator and sheds the same 200 W of heat as before. This means the first time the coolant circulates it will still start at 180*F as it enters the block but it is now 280*F as it exits the heads. Then, it goes through the radiator and cools to 270*F. So, on the second pass the coolant is 270*F as it enters the block and 370* as it exits the heads. This doesn't make for a very happy engine.

For the second case, The first thing to do is increase the coolant flow by about 10X to bring the temperature differential across the engine back to 10*F. The next thing to do is get the radiator to do 20000W of cooling instead of 2000W. I don't know how much more airflow, but I'd bet any money you want that a lot more airflow will be the right answer in accomplishing that goal.
"This means the engine is running at about 180*F where the coolant enters the block and 190*F where the coolant exits the heads."

Wrong logic because it doesn't happen this way. Does your temp gauge fluctuate between 180-190-180 and so on? If it does you have cooling problems. Proper functioning cooling systems will remain steady under most changing engine loads. Lots of guys in here will attest to this and some have really BIG hp numbers. My wife's CTS-V stays pegged at 185 and has never moved once it reaches that normalized temperature. My car has the temp sending unit at the TS housing. It doesn't do this either. I should stop here but I'll keep going to try to get you out of the myth, lol.

You failed to mention anything about the TS and its vital function! If the coolant comes in at 170 and the TS is rated at 180*, the coolant may already be flowing throughout the system but not at full flow. The engine will normalize at 180 - steady state, steady flow. Now, as engine load increases, using your example of going up a hill, the engine coolant will get hotter in the block. Time in the block is meaningless because the TS doesn't run on a timer. But time in the rad does and that is dwell time which is controlled by the TS's valve modulating up and down to control the volume of coolant required to enter the block. As coolant enters the block, the hotter coolant is pushed out through the TS. The TS will sense the cooler coolant and begin closing again until set point is reached, 180 and back to steady state, steady flow until the engine load changes again or the coolant entering the engine changes due to ambient air temp change, air flow through rad changes by example. The rad's capacity is designed to meet the cooling demand for whatever the designed maximum load of the engine will be. It is not sized to equal volume of coolant in the block to equal volume in the rad, not that you said that. So equal time in the block to equal time in the rad is utter nonsense. Its the heat contained in the coolant that needs to be rejected in the rad is what's important and not the time it takes to heat it up.

Consider this. Does a non-A/C car have the same size rad as an A/C car for a SB? Nope. But the volume of coolant in the block doesn't change but in the rad it does. Why a bigger rad? More dwell time required to cool the coolant for the added engine load while the A/C is operating because we can't increase the size of the block's coolant capacity. So, that dwell is increased by the addition of cooling tubes. You know, like lengthening the tube in my previous post to reject more heat?

Are you starting to get it, Lionel?
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Old Apr 7, 2018 | 10:16 AM
  #69  
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Originally Posted by resdoggie
Does your temp gauge fluctuate between 180-190-180 and so on? If it does you have cooling problems.

You failed to mention anything about the TS and its vital function!

Time in the block is meaningless

So equal time in the block to equal time in the rad is utter nonsense
Statement #1 - The temperature gauge sensor doesn't move between the block inlet and outlet. constant flowing coolant which enters the block at 180* and exits the block at 190* will be at a fixed temperature at the sensor.

Statement #2. - Last paragraph. Must increase flow to keep engine cool. You apparently can't read.

Statement #3. - The engine produces X number of BTU/m. Less BTU/m when doing nothing and more BTU/m when working hard. Each BTU/m will raise the temperature of 1lb of water by 1*F in a minute. If you don't keep the water flowing fast enough it will be too hot by the time it exits the engine. The thermostat lets the water flow fast enough that the water AT the location where it is exiting the engine is not too hot.

Statement #4 - There is a fixed volume of coolant in the loop. Each "unit" of coolant MUST spend the same amount of time flowing through the radiator as it does flowing through the engine block only when it is making that loop. But, it will actually spend more time in the engine side when it is bypassed from the thermostat back to the inlet of the water pump. Time to get hot enough it needs to be let to the radiator. You know, times when the thermostat is mostly closed.


Your last 2 statements show a complete lack of any actual knowledge on how the system really works.

Last edited by lionelhutz; Apr 7, 2018 at 10:20 AM.
Old Apr 7, 2018 | 10:32 AM
  #70  
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high volume/flow water pumps
Old Apr 7, 2018 | 12:39 PM
  #71  
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Originally Posted by lionelhutz
Statement #1 - The temperature gauge sensor doesn't move between the block inlet and outlet. constant flowing coolant which enters the block at 180* and exits the block at 190* will be at a fixed temperature at the sensor.

??????????? The coolant doesn't enter the block at 180. It's a lot cooler than that having lost some of its heat through the radiator before re-entering the block. It's obvious by your statements that you don't have a grasp on how the TS interacts with the cooling system.


Statement #2. - Last paragraph. Must increase flow to keep engine cool. You apparently can't read.

???????

Statement #3. - The engine produces X number of BTU/m. Less BTU/m when doing nothing and more BTU/m when working hard. Each BTU/m will raise the temperature of 1lb of water by 1*F in a minute. If you don't keep the water flowing fast enough it will be too hot by the time it exits the engine. The thermostat lets the water flow fast enough that the water AT the location where it is exiting the engine is not too hot.

When the cooled coolant flow is increased to handle the increased engine load, the TS will will close slightly to maintain 180 for the new engine load or if the load continues to increase, still climbing a hill, the TS will need to open even further to allow more cooled coolant into the block to maintain 180. If you keep going up the hill and its 100F outside temperature, at some point the TS will be wide open permitting full flow of cooled coolant into the block trying to maintain 180. The cooling temp may rise to 200 and level off if NO additional loads are put on the engine. You now have surpassed the rad's capacity to maintain 180. But, if the load is increased again or the outside temp climbs to 110, your out of cooling capacity and the engine will eventually over heat. Now when you come down the hill, engine load is removed and less heat transferred to the coolant so the coolant temps will drop and the TS will eventually begin to reduce the flow when the coolant temp returns to 180 - steady state, steady flow until the next load change.


Statement #4 - There is a fixed volume of coolant in the loop. Each "unit" of coolant MUST spend the same amount of time flowing through the radiator as it does flowing through the engine block only when it is making that loop. But, it will actually spend more time in the engine side when it is bypassed from the thermostat back to the inlet of the water pump. Time to get hot enough it needs to be let to the radiator. You know, times when the thermostat is mostly closed.

Again, ?????????! Get off this "time in rad and block" argument. Its meaningless. Its the heat transferred to the coolant we're interested in and how much rad capacity is required to reject that heat in the coolant regardless of how much time it takes to transfer the heat from the cylinder to to the coolant. The only time factor is the dwell time in the rad to give the rad sufficient time to cool the coolant. A rad can never be too big but it sure as hell can be too small for the expected cooling requirements. If the water pump is pumping 1 gpm through the TS and the rad holds 1 gal, for example, it will take 1 min for the pump to pump the entire rad back into the block to replace the hot coolant leaving the block. Dwell time = 1 min for the rad to cool the coolant enough to maintain 180 in the block. Now, I put A/C in my car or I add 200 HP. The existing rad is not designed for this additional engine power requirements. I now put in a rad with a 2 gal capacity to hold coolant. Water pump is still the same. Now, how long does it take to change out 2 gal? Yep, 2 min. So I have now increased my dwell time by an extra minute which means I have increased my rads coolant capacity by doubling the dwell time. But that will only be required when I've fully loaded the engine and have the A/C turned on to maintain 180 coolant temp. Now, if I put the 1 gal radiator back in for the added hp and A/C, whats gonna happen under full load? Most folks following along would say the car will not maintain 180 but will be above that temp and eventually under the right conditions, e.g. 110F outside temp, will over heat. Why? Well, the smaller rad only has 1/2 the dwell time to reject the heat but it can't reject enough in 1 min to cool the engine with under full load. But the larger rad has twice the dwell time to reject the heat and maintain 180. So, this is how it works. More engine mods, more accessories, more HP requires a bigger rad for the same water pump rating.

Your last 2 statements show a complete lack of any actual knowledge on how the system really works.
Back in college, many decades ago studying marine systems engineering design, I had to design an oil/water heat exchanger from scratch, as part of my final year project - design the propulsion plant with auxiliaries for a 600' loa oil cargo carrier. So, how many heat exchangers have you designed from scratch?

I'm ending it here, for you, because you just want to argue for the sake of arguement and have offered nothing sensible to support the myths as being true - because they are just that, myths.
Old Apr 7, 2018 | 06:20 PM
  #72  
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Originally Posted by resdoggie
If you keep going up the hill and its 100F outside temperature, at some point the TS will be wide open permitting full flow of cooled coolant into the block trying to maintain 180.
Thanks for agreeing with me. The thermostat does open more under a higher load to allow more flow. The radiator in an existing system holds a CONSTANT FIXED amount of coolant, such as 2 gallons. So, more gallons per minute of flow means that that each 2 gallons of coolant will remain in the radiator for less time. Completely the opposite of any theory including a slower flow or more dwell time to get more cooling.

Originally Posted by resdoggie
The only time factor is the dwell time in the rad to give the rad sufficient time to cool the coolant.
WRONG. The the GPH of flow into and out of the engine is the ONLY important part. You MUST flow the coolant into and out of the engine at the correct GPM to keep the engine at a constant temperature. The correct flow rate will vary as the engine load changes. The harder the engine is working, the more GPM of coolant flow is required into and out of the engine.

At no point does the cooling system flow the coolant at a certain flow rate to make the radiator happy. The flow rate IS whatever flow is required to make the engine happy.

I stressed into and out of the engine, not through the engine for a reason. There is a difference. Flow into and out of the engine is in via the lower radiator hose and out via the upper radiator hose. Flow through the engine includes both recirculating flow and flow into and out of it via the radiator hoses. Coolant DOES NOT return from the radiator at the correct temperature for the engine. It is mixed with recirculating coolant.

As for the rest, you're simply introducing a straw man argument in an feeble attempt to support your previously wrong claims. The radiator volume does not change for a existing system which had it's thermostat removed. So, any claims which include the volume of the radiator changing are null and void.

Last edited by lionelhutz; Apr 7, 2018 at 09:41 PM.
Old Apr 8, 2018 | 06:29 AM
  #73  
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As the saying goes:
An ounce of prevention is worth a pound of cure.
Cheap insurance.
Old Apr 8, 2018 | 01:26 PM
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I'm wondering why lionelhutz chooses to discuss/argue this with all of us "dummies", when he obviously has thermodynamic skills way beyond our feeble capabilities....

P.S. "Book learning" is only the beginning of understanding....
Old Apr 8, 2018 | 02:11 PM
  #75  
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My father use to say- "If you are the smartest person of the group- you need to find a new group!"
Old Apr 8, 2018 | 06:04 PM
  #76  
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This same argument comes up once a year, goes 100 pages, gets locked and unlocked, then dies, with sometimes a conclusion.
Want a headache?
Model T Ford has NO water pump and NO thermostat, and NO pressured cap and never gets hotter than 212*. (Unless it boils dry).
Old Apr 8, 2018 | 11:11 PM
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Just pointing out how pathetic the article you linked was Richard454. I would have though you'd have known more and not posted a link to such pathetically written nonsense, with the "myth debunking" including 2 myths......

But then it's also sad to be posting links to such nonsense and then making fun of a person who points out that it was wrong.

Last edited by lionelhutz; Apr 9, 2018 at 01:03 AM.

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Old Apr 9, 2018 | 12:04 AM
  #78  
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Not a "team" player....

It's good to have 'dialogue' on an open Forum. But character attacks, belittlement, and mockery are not part of the deal. Although you have every right to participate on this site...we would appreciate it if you found another site to berate.
Old Apr 9, 2018 | 12:07 AM
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Originally Posted by 7T1vette
Not a "team" player....

It's good to have 'dialogue' on an open Forum. But character attacks, belittlement, and mockery are not part of the deal. Although you have every right to participate on this site...we would appreciate it if you found another site to berate.
who is we?
Old Apr 9, 2018 | 12:21 AM
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Originally Posted by 7T1vette
Not a "team" player....

It's good to have 'dialogue' on an open Forum. But character attacks, belittlement, and mockery are not part of the deal. Although you have every right to participate on this site...we would appreciate it if you found another site to berate.
But it's OK when you do it, or maybe you're admonishing yourself???



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