In addition, I've seen quite a few internet posts claiming that a mechanically-driven fan is 100% efficient in its power conversion. There are two problems with this logic:
A typical fan clutch states that it slips 20% when unlocked and 80% when locked. I've never seen anyone state at which engine speed this occurs at, so I'll take the conservative approach and assume this occurs at high idle (1000 RPM). For simplicity for this example, I'm also going to assume that water pump RPM equals engine RPM. If you change this RPM value based on a different pulley ratio, you will note that it does not affect the rest of the calculations. In fact, you could substitute in any variable of your choice and it will ultimately not change the final answer of the following calculations.
So how much power does an unlocked fan + clutch pull at 1000 RPM? We'll have to make a few assumptions here. Imagine if you were to take a typical household box fan and duct tape it behind the radiator. Due to the restrictions of the radiator, condenser, and grille, it would have to be roughly at 50W power consumption. 50 watts isn't much in the realm of things. Way lower than our electric fan at full power. But in order to spin this fan at the unlocked speed, the clutch slips to 20% of shaft speed. Torque on both ends of the clutch is the same. But power is proportional to torque times RPM. So where does the extra power off the water pump go? 80% of the energy leaving the water pump is lost as heat in the fan clutch! This means we are using 250 watts, 200 of which is being wasted as heat! That's 1/3 of a horsepower we are using to drive this fan and clutch.
But what if the fan locks up? Surely it becomes more efficient! Assuming the 20%/80% rules apply here, the fan spins at 4X the speed when the clutch locks up. According to the fan affinity laws, power scales with the cube of shaft RPM. Head increases with the square of shaft RPM. Airflow scales linearly with the shaft RPM. The head assumption is perfectly valid in this case, as the force required to pull air through the radiator increases with the square of the airspeed as per the drag equations. But in order to spin our fan at 4X the speed, we need to provide 64X more power! Hence our fan is suddenly demanding 3,200 watts, or roughly 4.3 horsepower! In addition, the fan clutch is still burning up 20% of the energy as heat, while passing 80%. So our fan + clutch combination is now burning 4,000 watts (5.4 HP) of engine power! And we have 800 watts being converted straight to heat! (Ever wonder why fan clutches have so many heat fins?)
So what happens when we start revving our engine higher? If it were a mechanically locked fan, power would increase by the cube of the input RPM, and we would be burning an insane amount of horsepower at redline! Fortunately the mechanical clutch helps us out here, but perhaps by not as much as we might like.
The torque transmitted by a fluid coupling is proportional to the square of the slip RPM. Hence, the faster it slips, the more torque it can transmit. Or in other words, as torque increases, slip RPM increases by the square root of the applied torque. Given the fan affinity laws, the torque on the impeller shaft is proportional to the square of the RPM driving it. So (to my actual surprise), it turns out that the slip ratio between input and output speed is actually roughly constant!
So what happens when we double engine RPM to 2000? Our fan/clutch combinations are now taking 8X as much power! Our unlocked fan is now taking 2000 watts (2.7 HP) of power, and our locked fan is taking 32,000 watts (43.2 HP)!
That said, few fan clutches will remain locked past a few thousand engine RPM. In addition, as the fan clutch oil heats up, the fan begins to lose viscosity, and the fan slip ratio actually increases. Realistically, you will never see a clutch fan take more than about 40 HP off the engine when locked at high RPM.
But our unlocked fan still takes a lot of power. Even if we assume the slip ratio doubles to 90% (10% shaft speed), at redline (5300 RPM), that fan is pulling 4,650 watts, or 6.2 horsepower! 90% of that loss immediately becomes heat! And if somehow it is still locked, you could see a loss as high as 40 horsepower! (These numbers line up well with a lot of dyno results you can find on various racing forums.)
So there are really two things that prevent the engine clutch fan from being efficient mechanically. (1), a large amount of power is converted to heat in the clutch. You've probably noticed that a fan clutch has many heat fins. All that heat is wasted energy. (2), with a lack of control over the input shaft RPM, the power demands increase rapidly and unnecessarily. The fan has to be sized to cool at idle, and as such at high RPM it is way overpowered.
Meanwhile, the brushless electric fan always has a loss coefficient of about 50% (roughly 60% alternator efficiency multiplied by the 85% efficiency ratio of the electric motor), and only uses the power it needs to when it needs to.
TL;DR Fan clutches are great and save us a lot of fuel over directly driven fans, but they are still very inefficient and a well set-up electric fan will absolutely dominate the clutch fan in terms of power (and as a result, fuel) conservation.
- For a fan driven with a viscous clutch, this is simply not true. A significant portion of the power taken off the water pump shaft is lost as heat. More details later.
- For a fan solidly connected to the water pump shaft, this is technically true. However, as per the fan affinity laws, the fan requires a stupid amount of power to run at high RPMs. More on this later.
A typical fan clutch states that it slips 20% when unlocked and 80% when locked. I've never seen anyone state at which engine speed this occurs at, so I'll take the conservative approach and assume this occurs at high idle (1000 RPM). For simplicity for this example, I'm also going to assume that water pump RPM equals engine RPM. If you change this RPM value based on a different pulley ratio, you will note that it does not affect the rest of the calculations. In fact, you could substitute in any variable of your choice and it will ultimately not change the final answer of the following calculations.
So how much power does an unlocked fan + clutch pull at 1000 RPM? We'll have to make a few assumptions here. Imagine if you were to take a typical household box fan and duct tape it behind the radiator. Due to the restrictions of the radiator, condenser, and grille, it would have to be roughly at 50W power consumption. 50 watts isn't much in the realm of things. Way lower than our electric fan at full power. But in order to spin this fan at the unlocked speed, the clutch slips to 20% of shaft speed. Torque on both ends of the clutch is the same. But power is proportional to torque times RPM. So where does the extra power off the water pump go? 80% of the energy leaving the water pump is lost as heat in the fan clutch! This means we are using 250 watts, 200 of which is being wasted as heat! That's 1/3 of a horsepower we are using to drive this fan and clutch.
But what if the fan locks up? Surely it becomes more efficient! Assuming the 20%/80% rules apply here, the fan spins at 4X the speed when the clutch locks up. According to the fan affinity laws, power scales with the cube of shaft RPM. Head increases with the square of shaft RPM. Airflow scales linearly with the shaft RPM. The head assumption is perfectly valid in this case, as the force required to pull air through the radiator increases with the square of the airspeed as per the drag equations. But in order to spin our fan at 4X the speed, we need to provide 64X more power! Hence our fan is suddenly demanding 3,200 watts, or roughly 4.3 horsepower! In addition, the fan clutch is still burning up 20% of the energy as heat, while passing 80%. So our fan + clutch combination is now burning 4,000 watts (5.4 HP) of engine power! And we have 800 watts being converted straight to heat! (Ever wonder why fan clutches have so many heat fins?)
So what happens when we start revving our engine higher? If it were a mechanically locked fan, power would increase by the cube of the input RPM, and we would be burning an insane amount of horsepower at redline! Fortunately the mechanical clutch helps us out here, but perhaps by not as much as we might like.
The torque transmitted by a fluid coupling is proportional to the square of the slip RPM. Hence, the faster it slips, the more torque it can transmit. Or in other words, as torque increases, slip RPM increases by the square root of the applied torque. Given the fan affinity laws, the torque on the impeller shaft is proportional to the square of the RPM driving it. So (to my actual surprise), it turns out that the slip ratio between input and output speed is actually roughly constant!
So what happens when we double engine RPM to 2000? Our fan/clutch combinations are now taking 8X as much power! Our unlocked fan is now taking 2000 watts (2.7 HP) of power, and our locked fan is taking 32,000 watts (43.2 HP)!
That said, few fan clutches will remain locked past a few thousand engine RPM. In addition, as the fan clutch oil heats up, the fan begins to lose viscosity, and the fan slip ratio actually increases. Realistically, you will never see a clutch fan take more than about 40 HP off the engine when locked at high RPM.
But our unlocked fan still takes a lot of power. Even if we assume the slip ratio doubles to 90% (10% shaft speed), at redline (5300 RPM), that fan is pulling 4,650 watts, or 6.2 horsepower! 90% of that loss immediately becomes heat! And if somehow it is still locked, you could see a loss as high as 40 horsepower! (These numbers line up well with a lot of dyno results you can find on various racing forums.)
So there are really two things that prevent the engine clutch fan from being efficient mechanically. (1), a large amount of power is converted to heat in the clutch. You've probably noticed that a fan clutch has many heat fins. All that heat is wasted energy. (2), with a lack of control over the input shaft RPM, the power demands increase rapidly and unnecessarily. The fan has to be sized to cool at idle, and as such at high RPM it is way overpowered.
Meanwhile, the brushless electric fan always has a loss coefficient of about 50% (roughly 60% alternator efficiency multiplied by the 85% efficiency ratio of the electric motor), and only uses the power it needs to when it needs to.
TL;DR Fan clutches are great and save us a lot of fuel over directly driven fans, but they are still very inefficient and a well set-up electric fan will absolutely dominate the clutch fan in terms of power (and as a result, fuel) conservation.
