Re: Thermostats
Posted: Thu Apr 16, 2020 10:13 pm
Rosco,
"Fools rush in . . . " Here goes. Applied Thermodynamics 101.
The coolant has a certain specific heat (and specific heat is expressed in units of energy divided by volume divided by temperature. For water it's 4.2 kilojoules per litre per degree C.)
When the coolant flows through a radiator, there's a temperature difference between the inlet and the outlet. Outlet is cooler than the inlet, and heat energy is dumped into the air. If you have a flow rate of say n litres per second, and a temperature difference of T degrees, the amount of heat energy removed by the radiator is 4.2 kilojoules/litre/degree C times n litres per second times T degrees C. The "litres" units cancel, the "degrees C" units cancel, and what you are left with is 4.2 times n times T kilojoules per second, a.k.a. kilowatts. If the radiator has to dissipate a constant thermal power, the flow rate and coolant temperature drop must be inversely related to each other.
Let's plug some numbers in. For convenience, let's say that the radiator has to dissipate 42 kW. If your coolant flows at a litre per second, the coolant's temperature drop will be 10 degrees C. (4.2 x 1 x 10 = 42.) If your coolant flows at ten litres per second, the coolant's temperature drop will be 1 degree C. (4.2 x 10 x 1 = 42.) The faster the coolant flows through the radiator, the smaller the temperature difference between outlet and inlet. Conversely, the more slowly the coolant flows through the radiator, the greater the temperature difference between outlet and inlet.
Note that I am talking about the coolant temperature difference between the radiator's outlet and inlet here.
There is a second quantity to be considered: the radiator's thermal resistance. This is expressed in units of degrees C per kilowatt. It tells you how hot the coolant will get if the radiator has to dissipate a certain amount of thermal power.
More numbers. Let's say the radiator has a thermal resistance of 1 degree per kilowatt, and the engine it's cooling is dumping 42 kW into the cooling system. 1 degree per kilowatt times 42 kilowatts gives a temperature rise of 42 degrees above ambient. Using the example above, if the coolant flows at a litre per second, the water will come out of the radiator ten degrees cooler than it went in. If the coolant flows at ten litres per second, the coolant will come out of the radiator only a degree cooler than it went in, but still about 42 degrees above ambient temperature.
Finally, let's consider the effect that an 80 degree C thermostat will have on our cooling system on a stinking hot 40-degree day. You still have 42 kW to get rid of, the water pump can move ten litres per second, and the poor old radiator can only dissipate a kilowatt per degree C. The temperature rise above ambient (40 degrees) will be 42 degrees, giving a coolant temperature of 82 degrees. The thermostat will be fully open and the cooling system is operating at maximum flow, 82 degrees in and 81 degrees out. Now let's make it a gentle autumn day: 20 degrees. The coolant temperature would be 62 degrees if there were no thermostat (20 + 42 = 62). But the thermostat regulates the coolant temperature leaving the engine to 80 degrees by restricting the flow. The coolant temperature leaving the radiator will be still be 62 degrees (a bit of a fudge to preserve the 42 degrees above ambient), so the coolant temperature drop is 18 degrees, and because of the inverse relationship between coolant temperature drop and flow rate, the flow rate will be 0.55 litres per second.
It is all just a question of what the cooling system has to do to get rid of the engine's heat. Generally the limiting factor is neither the coolant's specific heat, nor the water pump's maximum flow rate, but the radiator's thermal resistance.
I am now running my FC without a thermostat. At highway speeds, the coolant is about 50 degrees above ambient, and the temperature drop across the radiator is about 12 degrees. Fifty degrees above ambient does not leave a great deal of headroom if I (say) go for a drive up to Brisbane in the summer.
Rob
"Fools rush in . . . " Here goes. Applied Thermodynamics 101.
The coolant has a certain specific heat (and specific heat is expressed in units of energy divided by volume divided by temperature. For water it's 4.2 kilojoules per litre per degree C.)
When the coolant flows through a radiator, there's a temperature difference between the inlet and the outlet. Outlet is cooler than the inlet, and heat energy is dumped into the air. If you have a flow rate of say n litres per second, and a temperature difference of T degrees, the amount of heat energy removed by the radiator is 4.2 kilojoules/litre/degree C times n litres per second times T degrees C. The "litres" units cancel, the "degrees C" units cancel, and what you are left with is 4.2 times n times T kilojoules per second, a.k.a. kilowatts. If the radiator has to dissipate a constant thermal power, the flow rate and coolant temperature drop must be inversely related to each other.
Let's plug some numbers in. For convenience, let's say that the radiator has to dissipate 42 kW. If your coolant flows at a litre per second, the coolant's temperature drop will be 10 degrees C. (4.2 x 1 x 10 = 42.) If your coolant flows at ten litres per second, the coolant's temperature drop will be 1 degree C. (4.2 x 10 x 1 = 42.) The faster the coolant flows through the radiator, the smaller the temperature difference between outlet and inlet. Conversely, the more slowly the coolant flows through the radiator, the greater the temperature difference between outlet and inlet.
Note that I am talking about the coolant temperature difference between the radiator's outlet and inlet here.
There is a second quantity to be considered: the radiator's thermal resistance. This is expressed in units of degrees C per kilowatt. It tells you how hot the coolant will get if the radiator has to dissipate a certain amount of thermal power.
More numbers. Let's say the radiator has a thermal resistance of 1 degree per kilowatt, and the engine it's cooling is dumping 42 kW into the cooling system. 1 degree per kilowatt times 42 kilowatts gives a temperature rise of 42 degrees above ambient. Using the example above, if the coolant flows at a litre per second, the water will come out of the radiator ten degrees cooler than it went in. If the coolant flows at ten litres per second, the coolant will come out of the radiator only a degree cooler than it went in, but still about 42 degrees above ambient temperature.
Finally, let's consider the effect that an 80 degree C thermostat will have on our cooling system on a stinking hot 40-degree day. You still have 42 kW to get rid of, the water pump can move ten litres per second, and the poor old radiator can only dissipate a kilowatt per degree C. The temperature rise above ambient (40 degrees) will be 42 degrees, giving a coolant temperature of 82 degrees. The thermostat will be fully open and the cooling system is operating at maximum flow, 82 degrees in and 81 degrees out. Now let's make it a gentle autumn day: 20 degrees. The coolant temperature would be 62 degrees if there were no thermostat (20 + 42 = 62). But the thermostat regulates the coolant temperature leaving the engine to 80 degrees by restricting the flow. The coolant temperature leaving the radiator will be still be 62 degrees (a bit of a fudge to preserve the 42 degrees above ambient), so the coolant temperature drop is 18 degrees, and because of the inverse relationship between coolant temperature drop and flow rate, the flow rate will be 0.55 litres per second.
It is all just a question of what the cooling system has to do to get rid of the engine's heat. Generally the limiting factor is neither the coolant's specific heat, nor the water pump's maximum flow rate, but the radiator's thermal resistance.
I am now running my FC without a thermostat. At highway speeds, the coolant is about 50 degrees above ambient, and the temperature drop across the radiator is about 12 degrees. Fifty degrees above ambient does not leave a great deal of headroom if I (say) go for a drive up to Brisbane in the summer.
Rob