Engine braking in cold temps drops engine temp?

It gets cold here too (Minnesota), but I've never worried about my engine getting too cold. I do remember seeing cars (mostly commercial trucks) with covers over their radiators in the old days, but I can't say I remember seeing anything like that in modern times. The heater in my taco has worked like a champ down to -20F or so.

 

I've descended those passes in the winter many times. Never noticed the temp gauge dropping.

Thermostat should be closed to the radiator and the fan clutch shouldn't be engaged as the coolant temps drop. Maybe the thermostat is closing a little too late and not enough heat is being produced to bring it back up.

The only time I've ever had the temp gauge drop on a downhill grade it turned out the thermostat was stuck open. This wasn't with a Tacoma.

 

I experience -40* in Alberta and unless the engine just got up to operating temperature, I have the heater on full buck or it's a long hill in those temperatures say 3-5 kms with zero throttle input I have not noticed the temp drop and even when I did I still had heat. I would suspect thermostat and verify that by (preferably with a thermometer gun) check the upper and lower rad hose temps as the engine is warming up to see they don't persistently indicate coolant flow. You may have a to drive a bit to move the gauge, get out and check then again at temperature or ideally with a helper and if you're experienced with cars just put your hand on the upper rad hose and have the helper hold a high idle like 2000-2500 rpm till the engine warms up and feel for a sudden rush of hot coolant through the hose. If the hose warms up at the pace of the engine I would replace the thermostat and recheck for your concern.

 

Normal. No fuel and high rpm/oil pressure means your engine is now air and oil cooled while no combustion takes place.

Remember that we’re talking about water temps. Oil temperature does not change nearly as quickly.

 

OP should have the oil cooler. Therefore oil / water temps should be roughly stable. Air cooling alone isnt going to cool down that much aluminum, steel, water, and oil that quickly

For what its worth I watched my dash gauge vs SGII very close last night in my truck. Theres actually not as much deadband as I thought. You just have to look really damn close at the individual tick mark.

 

Uh, yes. Without combustion taking place it will cool very rapidly. Especially when it’s really cold out because that ambient air gets pumped thru the engine and pushed all around it.

 

Engine braking will still produce heat. If the engine is holding the speed down, kinetic energy is being dissipated. And it's happening in the form of heat transfer to the coolant system and out the exhaust. There is no free lunch.

 

Reread what you said. Yeah I misunderstood.

Still disagree though. The heat from engine braking should maintain temps. As far as air flow around the engine goes he had a freaking grill cover installed. My money is on a bad thermostat.

 

I'm a simpleton and can't see how engine braking creates anywhere near the heat combustion does. You still get some heat from friction and pumping loses but nowhere near what the engine would make if it was producing the power.

Maybe someone smarter than me can explain it to me.

 

^Engine braking can create as much heat as combustion, but with just 5-7% grade, OP likely wasn't doing it aggressively enough (with engine near redline, etc.).

The pumping losses during engine braking are significantly higher than normal operation, because the throttle is closed.

 

I'm not going to pretend to be able to figure out how much heat is being produced at x rpm at a certain grade. I think combustion does produce more heat. But in my experience engine braking produces enough heat to maintain engine temperature and run the heater. I assume the thermostat is mostly closed in these scenarios, so not losing much heat through the radiator.

 

I get the pumping loses are higher because of the much higher vacuum. But I can't see that offsetting the inefficiency of the combustion process. While producing power they waste about 70% of the heat created during combustion out the exhaust and the radiator.

Since theyre not producing power during engine braking they should be able to produce the needed resistance (power) much more efficiently I'd think? I'm probably explaining what I think poorly.

I agree something else is going. I'd think just mass alone of hot aluminum and coolant would be enough to keep heat even if the engine is creating none.

Thermostat is a likely culprit but it should also happen in other low loads I'd think.

Not looking to argue curious here.

 

I think a better way of estimating the heat generation of engine braking is to start by estimating how much engine braking heat generation power is required to slow the truck to the speed it is traveling, taking into account other losses like air resistance and rolling resistance. This seems simpler than comparing combustion efficiency to pumping losses and other forms of friction in the engine. The law of conservation of energy is your friend. I leave the math as an exercise for the reader.

Another thing to consider is how much the OP was using the brakes.

 

Ok, so I nerded the shit out of this one. I can't claim anything I say here is true or accurate, I'm wrong more often than right, but I thought I'd do this for fun anyway. I wanted to a few days ago but was busy with a business trip and now I'm sitting at home and can't sleep so...

I just ran back of the envelope calcs on three thermodynamic states of air: 1-ambient, 2-compressed, 3-expanded. I missed the OP's second post giving specific conditions such as elevation, temp, and speed, so I made assumptions. Also, I couldn't quickly find the compression/expansion ratio's of the Tacoma's Atkinson cycle, so I just took the 8:1 and 13:1 from the Prius. Again, it's back of the envelope stuff here.. I came to the conclusion that dropping coolant temperature of somewhere in the range of 0.2 degrees C per second is REASONABLE. It's probably less, but I'd venture to say this guess is in the ballpark. A 2,500' descent with a 7% grade at 40mph takes 10 minutes, so this equates to 120C temperature drop. Again, the real value is almost certainly LESS than this. I did not account for the heat transfer to the engine block or the oil. Why? Because I'm lazy and it's late. Assuming heat is transferred through the oil cooler to maintain coolant/oil temp equilibrium, you can reduce this by about 30%. The real question is how much heat is transferred to the engine block, as this is a HUGE heat sink? There's also other sources of cooling that effect the temp of the motor block, so I'm just ignoring this altogether. If you don't like my answer ... I dont know, divide it by 5 or something.. the you get a temp drop of 24C (43F) over 10 minutes. I can't pin with certainty how much the Atkinson cycle cools the engine during engine braking, if it even does, I'm just trying to show it's plausible anyways.

My assumption was that the isentropic efficiencies of compression and expansion were both 66%; this is on par with typical reciprocating compressor efficiencies as per Google. I then compared the difference in final temperatures of the air (isentropic vs assumed actual cases) and ASSUMED that the loss of efficiency was due ENTIRELY due to irreversible heat transfer between the air and cylinder walls. This is how I came up with 8.32 kW of heat transfer. The 0.2 K/s was based the heat capacity of water and approximately 9kg of coolant. Again, I'm ignoring heat xfer to the oil and block itself, this assumption is an upper bound.

EDIT: I realize there's a bit of an energy imbalance here. Some of you may point out that gravitational potential energy of the truck is being turned into heat in the engine instead of the brakes, so how can the engine be getting colder? Well... for one, I'm not sure that's the case, I'm just saying its plausible... I think. Also, most of engine braking does not happen due to compression, it is a result of pumping losses as the engine desperate tries to suck air past the closed throttle body. A lot of that heat gets expelled through the exhaust.

EDIT 2: This was fun, but I'd place my bets on your thermostat!

 

Oh boy, Friday is back again.

 

the engine temp gauge on cars are not reliable at all

 

I didn't fully study your analysis, but as I said previously, I think a better approach to the problem is to use energy conservation to estimate the heat generation required to maintain a terminal velocity. That way we can ignore the complexities of thermodynamics and other forms of friction in the engine. Here's a napkin estimate using that approach.

Givens:

terminal velocity = 60 mph ~= 27 m/s
vehicle mass = 4500 lb ~= 2000 kg
incline = 5%

The force of gravity on the truck is:

F = mass * acceleration-of-gravity * incline = 2000 kg * 9.8 m/s^2 * 0.05 = 980 N

Countering this force requires a power of:

P = force * velocity = 980 N * 27 m/s = 26,460 W ~= 26 kW

This gives us an upper bound of the amount of heat generation required to maintain a speed of 60 mph down a 5% incline.

To get a better estimate we could take into account aerodynamic and rolling resistance. I did an estimate for aerodynamic drag force and came up with 190 N. I'll summarize it here.

Givens:

coefficient of drag = 0.4
air density = 1.3 kg / m^3
drag area = 1 m^2

F = 0.5 * air-density * coefficient-of-drag * drag-area * velocity^2 = 0.5 * 1.3 kg/m^3 * 0.4 * 1 m^2 * (27 m/s) ^ 2 = 190 N

Rolling resistance is likely too small to worry about in this napkin estimate, so the adjusted power is:

P = force * velocity = (980 N - 190 N) * 27 m/s = 21,330 W ~= 21 kW

This number is much bigger than yours and I'm not sure why.

Is 21 kW enough to maintain constant engine water temp at 60 mph and 30F? I have no idea. This has been a purely academic exercise.

CAVEAT: I blew through pretty fast and there are likely mistakes.

BTW, I don't see how the Atkinson Cycle has any effect on the answer because 1) we are assuming no fuel is being burned and 2) it's not a given that the Atkinson Cycle mode is enabled during engine braking.

 

To get a rough idea of how much 21 kW is, we can compare it to the heat generated by an idling engine.

Given:

fuel use = 0.5 gallons/hour = 0.53 mL/s
gasoline energy density = 34.2 MJ/L
engine efficiency = 30%

The energy produced per second is:

power = fuel-use * gasoline-energy-density = 0.53 mL / s * 34.2 MJ/L = 0.018 MJ/s = 18 kW

Since the engine is 30% efficient, 70% of this power goes to heat, thus the heat generation is 13 kW.

So it seems that engine braking at 60 mph on a 5% slope generates about twice the heat of an idling engine.

This still doesn't tell us much about the water temp behavior, because the amount of cooling is likely quite different traveling at 60mph compared to sitting still.

 

@arthur106, I re-read your post and realized I completely misunderstood you. I thought you are trying to estimate the heating effects of engine braking using thermodynamics, but I now realize you are trying to estimate the COOLING effects. I had the wrong assumption when I started to read your post. Sorry, for the misread.

That said, I think you are ignoring the heating effects of engine braking. The truck is losing gravitational potential energy but kinetic energy remains constant (otherwise it would continue to accelerate down the incline), thus the truck is losing total energy. That energy loss likely comes in 3 major forms, aerodynamic losses, rolling resistance losses, and engine braking losses. All of these produce waste heat. My calculations suggest that engine braking losses are the biggest of the three.