run straight coolant - NO water

[Gokhan]

Originally Posted By: Al
Heat transfer wise it doesn't matter whether you use 100% water or 100% coolant. I know it seems counter intuitive..the only thing water does is hold more btu's. This means the temp will rise slower. The coolant to air transfer surface in the radiator is the main facter affecting engine steady state temperature.

We wrote a Fortran IV program on that subject when I was in engineering school almost 50 years ago.

That's not entirely true and you don't need a computer program.

Without writing the equations, the rise of the coolant temperature between water inlet and outlet of the engine is proportional to the heat generated by the engine and inversely proportional to the heat capacity of the coolant. So, if the heat capacity of the coolant was infinite, the coolant temperature would be the same at the inlet and outlet. Since this is not the case and water has about 80% more heat capacity than antifreeze [1.00 kcal/(kg.C) vs. ~ 0.56 kcal/(kg.C)], that means the inlet and outlet temperature difference will be about 80% higher -- almost twice!

However, you are right that the average coolant temperature (between inlet and outlet) will be the same. However, do you really want to raise the temperature gradient in your engine and radiator by a factor of two? This could strain both the engine and radiator, as well as your hoses.

 

[edhackett]

Originally Posted By: A_Harman
Originally Posted By: turtlevette
If you use straight antifreeze, use propylene glycol ie Sierra. Conventional antifreeze, ie ethylene glycol, does not protect for very cold temps at 100 percent. Freeze protection for EG peaks at 70% concentration then lessens as you further concentrate towards 100%.

Nice graph. At 70% EG/30% H20, the freezing point is -90F. I think I would opt to stay home at a balmy -40F.


That graph is not correct, it has the freezing point of 100% EG about 10C degrees too low. There are a bunch of charts out there that are off to a lesser or greater degree. This one is is as close to correct as I could find. Actual freeze point for 100% EG is about +10F, -12C.

Ed

 

[Al]

Originally Posted By: Gokhan

That's not entirely true and you don't need a computer program.

Without writing the equations, the rise of the coolant temperature between water inlet and outlet of the engine is proportional to the heat generated by the engine and inversely proportional to the heat capacity of the coolant. So, if the heat capacity of the coolant was infinite, the coolant temperature would be the same at the inlet and outlet. Since this is not the case and water has about 80% more heat capacity than antifreeze [1.00 kcal/(kg.C) vs. ~ 0.56 kcal/(kg.C)], that means the inlet and outlet temperature difference will be about 80% higher -- almost twice!

You do need a computer program bc we are taking non steady state when heating up and steady state when the temp equalizes. An infinite capacity is not viable bc then the temp would not go up in the engine nor would it go down in the radiator..remember flow is assumed.

Ultimately btu's picked up in the engine are released via the radiator. It turns out changing the flow rate of the pump doesn't change things a whole lot unless the pump is not adequate to carry initial BTU buildup away from the engine.

Again Radiator convection area ultimately controlls the steady state temp of the engine.

Also remember if temp doesn't go up much in the engine much bc of high heat capacity it doesn't go down much in the radiator (bc of high heat capacity). Like I said..counter intuitive unless you have taken a heat transfer course or two or three. If you think I'm wrong..that's OK. I can accept that.

 

[Gokhan]

Originally Posted By: Al
Originally Posted By: Gokhan

That's not entirely true and you don't need a computer program.

Without writing the equations, the rise of the coolant temperature between water inlet and outlet of the engine is proportional to the heat generated by the engine and inversely proportional to the heat capacity of the coolant. So, if the heat capacity of the coolant was infinite, the coolant temperature would be the same at the inlet and outlet. Since this is not the case and water has about 80% more heat capacity than antifreeze [1.00 kcal/(kg.C) vs. ~ 0.56 kcal/(kg.C)], that means the inlet and outlet temperature difference will be about 80% higher -- almost twice!

You do need a computer program bc we are taking non steady state when heating up and steady state when the temp equalizes. An infinite capacity is not viable bc then the temp would not go up in the engine nor would it go down in the radiator..remember flow is assumed.

Ultimately btu's picked up in the engine are released via the radiator. It turns out changing the flow rate of the pump doesn't change things a whole lot unless the pump is not adequate to carry initial BTU buildup away from the engine.

Again Radiator convection area ultimately controlls the steady state temp of the engine.

Also remember if temp doesn't go up much in the engine much bc of high heat capacity it doesn't go down much in the radiator (bc of high heat capacity). Like I said..counter intuitive unless you have taken a heat transfer course or two or three. If you think I'm wrong..that's OK. I can accept that.

I'm talking about the steady state:

heat generated by engine * time interval = coolant flow rate * specific heat capacity of the coolant * temperature rise of the coolant between inlet and outlet * time interval

Time interval cancels and you can see that temperature rise of the coolant between inlet and outlet is inversely proportional to the specific heat capacity of the coolant for a given amount of heat generated by engine in a time interval.

I concur that the average temperature of the coolant (between inlet and outlet) wouldn't be affected much by the specific heat if at all, as the outlet temperature rises and radiator must dissipate the same heat, the inlet temperature must fall. That's because the amount of heat dissipated by the radiator is proportional to the average coolant temperature. However, as I am emphasizing, lower the specific heat, higher the outlet temperature and lower the inlet temperature -- or higher the temperature difference between inlet and outlet -- will be. This is not very desirable.

 

[A_Harman]

Originally Posted By: edhackett
Originally Posted By: A_Harman
Originally Posted By: turtlevette
If you use straight antifreeze, use propylene glycol ie Sierra. Conventional antifreeze, ie ethylene glycol, does not protect for very cold temps at 100 percent. Freeze protection for EG peaks at 70% concentration then lessens as you further concentrate towards 100%.

Nice graph. At 70% EG/30% H20, the freezing point is -90F. I think I would opt to stay home at a balmy -40F.


That graph is not correct, it has the freezing point of 100% EG about 10C degrees too low. There are a bunch of charts out there that are off to a lesser or greater degree. This one is is as close to correct as I could find. Actual freeze point for 100% EG is about +10F, -12C.

Ed


Why the differences between different graphs of EG? Are there additive differences that cause different freezing points?

 

[Al]

Originally Posted By: Gokhan

heat generated by engine * time interval = coolant flow rate * specific heat capacity of the coolant * temperature rise of the coolant between inlet and outlet * time interval

Time interval cancels and you can see that temperature rise of the coolant between inlet and outlet is inversely proportional to the specific heat capacity of the coolant for a given amount of heat generated by engine in a time interval.

Although your equatiion is corrrect it doesn't address the mechanism of the heat getting into the coolant it just shows a heat balance.

Heat enters the coolant bc of difference in fluid temp, configuration of the mechanism (including surface area), and velocity of the fuid. Since temperature is controlled by the T-stat, configuration doesn't change, and neither does flowrate. So the temp entering the fluid is strictly a function of Temperature difference iof the coolant and engine surface.

The same equation is applied to the Radiator which will again be affected by configuration of the radiator (including area), velocity of the fluid and fluid temperature.. In this case it is a bit more comlicated bc you have convection of the coolant into the radiator core and then convectiion of the heated core to the air. Again none of this is changed by the type of fluid....the air flow around the radiator is again not affected by the fluid.

So the correct applicable equation for steady state is: Heat picked up by the engine = heat rejected by the radiator. And steady state conditions make it prety simple.

During heatup you find that the fluid is slower to rise bc heat transfer in to the lower temp coolant is more effective. But at the same time you notice that this lower temp fluid finds it more difficult to reject its heat bc the temp of the fluid is COOLER.

Readers digest version: At steady state if the T-Stat is controlling temp and the capacity of the radiator is adequate the fluid temperature wil br controlled by T-Stat seting. If the capacity of the radiator is insufficient and the T-Stat is wide open..the new steady state temp will rise until heat rejected at the radiator = heat transferred to the fluid by the engine.

The only solution to keeping an engine cooler when the T-Stat is wide open at steady state conditions is to get a larger radiator.

Hopefully someone who is well versed in heat transfer (and wasn't watching women in class) will chime in.

 

[Garak]

Originally Posted By: A_Harman
Why the differences between different graphs of EG? Are there additive differences that cause different freezing points?

That's a good question. I should see if I can find my lab book from the experiment I did in undergrad chemistry many years ago and see how that turned out.

 

[440Magnum]

Originally Posted By: Al
Originally Posted By: 901Memphis
You will over heat if you run straight coolant unless you're system is older and oversized for the job.

You need water at least 30% for proper heat transfer.

Heat transfer wise it doesn't matter whether you use 100% water or 100% coolant. I know it seems counter intuitive..the only thing water does is hold more btu's. This means the temp will rise slower.


There's a big "oops" in your logic. Its a circulating system- X number of milliliters of water (or antifreeze mix) are circulated per second through the hot engine, and that number doesn't change when you switch from pure water to pure antifreeze (or a mixture). If each milliliter of water holds 30% more heat energy for the same delta-T than each gram of coolant does, then the total heat rejection capacity of the system increases by the same 30%.

Water/coolant mix DOES result in a higher cooling system heat transfer capacity than straight coolant. End of story.

 

[Gokhan]

Originally Posted By: Al
[Although your equatiion is corrrect it doesn't address the mechanism of the heat getting into the coolant it just shows a heat balance.

...

So the correct applicable equation for steady state is: Heat picked up by the engine = heat rejected by the radiator. And steady state conditions make it prety simple.

...

I can use the right side (radiator side) of the equality (which I did when I was doing my calculations) to show that (T_outlet + T_inlet)/2 will not depend on the specific heat of the coolant, albeit with certain assumptions. This is what I said before. Nevertheless, as I have been trying to explain to you, for a given engine heat power generation and coolant flow rate, T_outlet - T_inlet is inversely proportional to the specific heat of the coolant. This is only determined by the left side of the equality (engine side) and the radiator doesn't enter into the picture in determining that.

Having T_outlet - T_inlet larger is not desirable. Therefore, it's better to have a coolant with more specific heat (using more water in the mix). 50% concentration is ideal for automotive coolants in many ways.

 

[Gokhan]

Originally Posted By: Garak
Originally Posted By: A_Harman
Why the differences between different graphs of EG? Are there additive differences that cause different freezing points?

That's a good question. I should see if I can find my lab book from the experiment I did in undergrad chemistry many years ago and see how that turned out.

Additives play little or no effect in the melting and boiling points.

The answer is simple. Ethylene glycol has a melting point of 12.9 °C (8.8 °F), whereas propylene glycol has a melting point of 59 °C (74 °F) -- a huge difference.

 

[Al]

Originally Posted By: 440Magnum


There's a big "oops" in your logic. Its a circulating system- X number of milliliters of water (or antifreeze mix) are circulated per second through the hot engine, and that number doesn't change when you switch from pure water to pure antifreeze (or a mixture). If each milliliter of water holds 30% more heat energy for the same delta-T than each gram of coolant does, then the total heat rejection capacity of the system increases by the same 30%.

Water/coolant mix DOES result in a higher cooling system heat transfer capacity than straight coolant. End of story.


Well higher heat of Enthalpy (as I have said) has no effect on STEADY STATE OPERATION). It only comes into play as the coolant warms up and the btu's are absorbed with a slower temperature increase.

Once steady state is reached (as defined by the T-Stat setting)..the heat balance is very simple: heat absorbed by the engine = heat rejected by the radiator. And as mentioned mass flow is higher with water than glycol but volume flow rate which means the fluid will flow at the same rate by engine surface and radiator core surface means transfer coefficient.

The only variatiion is that on its path through the engine the fluid will not get as hot which at first glance seems to imply more heat will be picked up....BUT and this is a big BUT. The reverse occurrs in the radiator..this cooler temperature as it hits the radiator means that less heat will be rejected by the radiator and hence the fluid will enter the engine on the next cycle at a higher temp. All this occurrs in an intergral equation and thats why a computer probram is mentioned. I "cheated" a bit on the explanation.

This is why Gokhan's argument fails. He is also fixated on Heat Capacity (More correctly called Heat of Enthalpy). I'm done. I'll let another Mechanical Engineer try to explain it better than I. Although it is almost impossible without a computer simulation.

 

[Gokhan]

Originally Posted By: Al
Originally Posted By: 440Magnum


There's a big "oops" in your logic. Its a circulating system- X number of milliliters of water (or antifreeze mix) are circulated per second through the hot engine, and that number doesn't change when you switch from pure water to pure antifreeze (or a mixture). If each milliliter of water holds 30% more heat energy for the same delta-T than each gram of coolant does, then the total heat rejection capacity of the system increases by the same 30%.

Water/coolant mix DOES result in a higher cooling system heat transfer capacity than straight coolant. End of story.


Well higher heat of Enthalpy (as I have said) has no effect on STEADY STATE OPERATION). It only comes into play as the coolant warms up and the btu's are absorbed with a slower temperature increase.

Once steady state is reached (as defined by the T-Stat setting)..the heat balance is very simple: heat absorbed by the engine = heat rejected by the radiator. And as mentioned mass flow is higher with water than glycol but volume flow rate which means the fluid will flow at the same rate by engine surface and radiator core surface means transfer coefficient.

The only variatiion is that on its path through the engine the fluid will not get as hot which at first glance seems to imply more heat will be picked up....BUT and this is a big BUT. The reverse occurrs in the radiator..this cooler temperature as it hits the radiator means that less heat will be rejected by the radiator and hence the fluid will enter the engine on the next cycle at a higher temp. All this occurrs in an intergral equation and thats why a computer probram is mentioned. I "cheated" a bit on the explanation.

This is why Gokhan's argument fails. He is also fixated on Heat Capacity (More correctly called Heat of Enthalpy). I'm done. I'll let another Mechanical Engineer try to explain it better than I. Although it is almost impossible without a computer simulation.

None of my arguments fail. In fact, you have just corrected your claims to what I've been saying all along after finally reading and partially understanding my posts regarding the inlet and outlet temperatures.

 

[Garak]

Originally Posted By: Gokhan
The answer is simple. Ethylene glycol has a melting point of 12.9 °C (8.8 °F), whereas propylene glycol has a melting point of 59 °C (74 °F) -- a huge difference.

My lab in the day was to determine melting points of various ethylene glycol and water solutions. I meant I'd like to compare those results to the graphs we've been shown in the thread.

And you better check your above conversions. They are totally out of whack.

 

[Gokhan]

Originally Posted By: Garak
Originally Posted By: Gokhan
The answer is simple. Ethylene glycol has a melting point of 12.9 °C (8.8 °F), whereas propylene glycol has a melting point of 59 °C (74 °F) -- a huge difference.

My lab in the day was to determine melting points of various ethylene glycol and water solutions. I meant I'd like to compare those results to the graphs we've been shown in the thread.

And you better check your above conversions. They are totally out of whack.

Oops, the minus signs got deleted.

It should be:

Ethylene glycol has a melting point of -12.9 °C (8.8 °F), whereas propylene glycol has a melting point of -59 °C (-74 °F) -- a huge difference.

 

[turtlevette]

I think both of you have some things right and some things wrong.

The specific heat is less important than the heat conductivity which I mentioned in a thread below.

It's all about how fast you can move the heat out of the block. delta T

for an example check out the heat conductivity of Cu vs Al. Then check out the specific heat of Cu vs Al.

Although Al can "hold" more heat, Cu is a better heat conductor.

In this case, the water antifreeze mixture has both better heat conductivity and higher specific heat, so Gokhan's math still works.

 

[Al]

Originally Posted By: turtlevette
so Gokhan's math still works.



Of course the equation is correct..its a heat balance. Just doesn't get to the root of the problem.

And condctivity doesn't have a lot to do with it either. Carrying heat away from the engine (holding heat) is fine for the engine but it also means it will "hold" more heat coming out of the radiator. The engine AND the radiator is the closed system under discussion. You are thinking of just an OPEN system consisting of the engine only.

And even if the condictivity helps get more heat into the cooling fluid (which I doubt is very significant..I just don't know...I'll do a bit of checking). even so it will affect the convection at the RADIATOR the same way and just cancel out.

Thats where all of you guys are making your mistake. What ever happens in the engine also happens in the radiator.

I am enjoying the discussion.

Turtlevette: Are you an M.E. if so you should be ablle to come to the samej conclusion as me...

 

[A_Harman]

Lots of discussion about heat capacity, and none about convective heat transfer: Q = h * A * dT, where Q is heat transfer, h is the convective heat transfer coefficient, A is the area over which heat is being transferred, and dT is the temperature difference from the surface to the fluid. It's a simple looking equation, but determining h to solve it is where the true complexity of heat transfer comes in. The convective heat transfer coefficient depends on other fluid properties than just specific heat. Density, viscosity, and thermal conductivity also matter because they affect the nature of the interaction between the fluid and the heat transfer surface. It can't be assumed that the heat transfer out of an engine running at a given speed and load will be the same with 50/50 antifreeze and pure water because of a different combination of fluid properties.

 

[Al]

Originally Posted By: A_Harman
It can't be assumed that the heat transfer out of an engine running at a given speed and load will be the same with 50/50 antifreeze and pure water because of a different combination of fluid properties.

Yea I really explained it without listing the actual equation. That was the basis of my argument instead of using Heat = flow x specific Enthalpy x Delta T

But you are exactly right. That is the only relavant equation which really doesn't change significantly whether you are using water or antifreeze.

ok.I will try to stay out of this thread..carry on guys.

 

[SteveSRT8]

Originally Posted By: A_Harman
Lots of discussion about heat capacity, and none about convective heat transfer: Q = h * A * dT, where Q is heat transfer, h is the convective heat transfer coefficient, A is the area over which heat is being transferred, and dT is the temperature difference from the surface to the fluid. It's a simple looking equation, but determining h to solve it is where the true complexity of heat transfer comes in. The convective heat transfer coefficient depends on other fluid properties than just specific heat. Density, viscosity, and thermal conductivity also matter because they affect the nature of the interaction between the fluid and the heat transfer surface. It can't be assumed that the heat transfer out of an engine running at a given speed and load will be the same with 50/50 antifreeze and pure water because of a different combination of fluid properties.


Don't forget dwell time, that can vary significantly depending on duty cycle...

 

[Gokhan]

Originally Posted By: turtlevette
I think both of you have some things right and some things wrong.

The specific heat is less important than the heat conductivity which I mentioned in a thread below.

It's all about how fast you can move the heat out of the block. delta T

for an example check out the heat conductivity of Cu vs Al. Then check out the specific heat of Cu vs Al.

Although Al can "hold" more heat, Cu is a better heat conductor.

In this case, the water antifreeze mixture has both better heat conductivity and higher specific heat, so Gokhan's math still works.

It's a good point about the heat conductivity. If the heat conductivity of the coolant is less, it will add up to the heat resistance of the entire heat circuit, both at the engine and at the radiator. This will not only increase the engine temperature but also decrease the radiator efficiency, resulting in a higher average coolant temperature.

So, after taking the thermal conductivity into account, not only T_outlet - T_inlet for the coolant will be higher but also the average temperature of the coolant (T_outlet + T_inlet)/2 will be higher with antifreeze (ethylene glycol). The former is thanks to higher specific heat and the latter is thanks to higher heat conductivity of the water.

So, what everyone says is not wrong. Water is a better coolant than antifreeze as far as the thermal properties (heat conduction and convection) are concerned. Of course, you still need 50% antifreeze to protect the engine and cooling system.

Incidentally, I also know this from experience. I once drove on the freeway with distilled water only when I was flushing the cooling system. The engine ran ice-cold and took a much longer time to warm up!