Pure thought experiment I wanted to run through. What happens if we try to open-loop cool an electric motor used as a propulsor? Can the coolant flow be used as a steam rocket to give useful additional thrust?
Temperature differential is going to be our friend for minimizing the amount of coolant used, so let’s assume our motor uses SmCo magnets, maybe grade 26. These have a Curie temperature of 441 °C, but are usually specified for operation up to 350 °C. We will specify a coolant pressure of 100 atm (!!!), corresponding to a boiling point of 302 °C, which should give us comfortable temperature margin on the motor. (We will ignore, for now, the corrosion implications of running water through a heat exchanger at these temperatures and pressures.)
For a traditional BLDC motor, we might target an efficiency of 96%. However, with the use of SmCo instead of Nd, and with the copper windings running at ~80% higher resistance due to temperature, targeting an efficiency of 92% is reasonable. We will analyze, for concreteness, a motor of 100 kW (134.1 hp) output power, so 8.7 kW of excess input power is dissipated as heat, and that heat is extracted by the cooling system.
Although the heat capacity of water is 4.18 kJ/(kg*K) at low temperature, up at 300 *C the isobaric specific heat increases to 5.76 kJ/(kg*K). We will indeed run this as an isobaric system — we will have a fixed pressure relief valve at 100 atm, and pump in pressurized cool water at a variable rate to maintain the coolant exit temperature at a target point. With a maximum water inlet temperature of 40 °C, our mid-temperature heat capacity is around 4.4 kJ/(kg*K), and we will use this value rather than integrating over temperature for now.
Our 262 K temperature rise thus corresponds to about 27 kg/hour of water flow, or 7.2 gph. For comparison, a piston engine of this power with a SFC of 0.45 lbs/hp*hr would burn 27.4 kg/hour of gasoline, or 10 gph (due to gasoline’s lower density). So compared to a piston engine, our consumable flow is the same. On the one hand, water is cheaper than gas; on the other hand, we still need to carry batteries!
Now, there are two remaining questions. The first is to figure out what it takes to push 7.2 gph of water across a 100 atm pressure gradient. The second is to determine the results of the same 7.2 gph of water, now superheated, coming out of the pressure relief valve and flashing to steam.
Taking the second question first, it is obvious that this superheated and pressurized steam should be run through a nozzle and converted to thrust. Our pressure ratio — 100 : 1 — is much, much greater than the choked flow condition (~1.9 : 1), so we will have choked flow. The speed of sound in 350 °C, 5 MPa steam is around 571 m/s, so we’ll use this as a decent approximation of our throat velocity. The density of steam at our chamber condition is around 0.055 g/cm^3; our throat area is a tiny 0.24 mm^2. The 27 kg/hr of flow at throat velocity of 571 m/s only contributes 1 lbf of thrust; but we do have thermal expansion through a nozzle still to accelerate the flow further.
Water as a propellant has gamma (ratio of specific heats) of about 1.3, increasing with decreasing temperature; this is a bit higher than the typical 1.2 of combustion products. Unfortunately, this means we can only accelerate the exhaust to Mach 1.9 or so before its temperature falls below 100 °C, putting us at risk of condensation in the nozzle. This corresponds to an exit pressure at the end of the nozzle of 15.3 atm — we’re massively underexpanding — under which conditions the water will have already condensed out. It’s not clear how much we can actually accelerate the water with a nozzle here, because the phase change from condensation screws up the standard rocket equations.
To get an optimistic estimate, let’s return to the 8.7 kW of energy that we added to our 27 kg of water each hour. By energy conservation, this can accelerate the water to at most 1077 m/s, corresponding to 1.8 lbf of thrust. This is a good match to our Mach 1.9 exit… but also, useless. To close this out… at 150 kts, this corresponds to about 1 hp equivalent, assuming a typical propeller efficiency.
I choose to leave as an exercise for the reader the analysis of the pump power needed to achieve the required coolant pressure.
In summary: This is a bad idea. Don’t do this. Use a closed-loop cooling system and coolant-to-air radiator. Run the numbers for SmCo vs Nd, and don’t be surprised if for a closed loop system the lower temperature differential for Nd (and thus, the order of magnitude higher coolant flow needed) might outweigh the efficiency hit and engineering challenges of going to SmCo.
Temperature differential is going to be our friend for minimizing the amount of coolant used, so let’s assume our motor uses SmCo magnets, maybe grade 26. These have a Curie temperature of 441 °C, but are usually specified for operation up to 350 °C. We will specify a coolant pressure of 100 atm (!!!), corresponding to a boiling point of 302 °C, which should give us comfortable temperature margin on the motor. (We will ignore, for now, the corrosion implications of running water through a heat exchanger at these temperatures and pressures.)
For a traditional BLDC motor, we might target an efficiency of 96%. However, with the use of SmCo instead of Nd, and with the copper windings running at ~80% higher resistance due to temperature, targeting an efficiency of 92% is reasonable. We will analyze, for concreteness, a motor of 100 kW (134.1 hp) output power, so 8.7 kW of excess input power is dissipated as heat, and that heat is extracted by the cooling system.
Although the heat capacity of water is 4.18 kJ/(kg*K) at low temperature, up at 300 *C the isobaric specific heat increases to 5.76 kJ/(kg*K). We will indeed run this as an isobaric system — we will have a fixed pressure relief valve at 100 atm, and pump in pressurized cool water at a variable rate to maintain the coolant exit temperature at a target point. With a maximum water inlet temperature of 40 °C, our mid-temperature heat capacity is around 4.4 kJ/(kg*K), and we will use this value rather than integrating over temperature for now.
Our 262 K temperature rise thus corresponds to about 27 kg/hour of water flow, or 7.2 gph. For comparison, a piston engine of this power with a SFC of 0.45 lbs/hp*hr would burn 27.4 kg/hour of gasoline, or 10 gph (due to gasoline’s lower density). So compared to a piston engine, our consumable flow is the same. On the one hand, water is cheaper than gas; on the other hand, we still need to carry batteries!
Now, there are two remaining questions. The first is to figure out what it takes to push 7.2 gph of water across a 100 atm pressure gradient. The second is to determine the results of the same 7.2 gph of water, now superheated, coming out of the pressure relief valve and flashing to steam.
Taking the second question first, it is obvious that this superheated and pressurized steam should be run through a nozzle and converted to thrust. Our pressure ratio — 100 : 1 — is much, much greater than the choked flow condition (~1.9 : 1), so we will have choked flow. The speed of sound in 350 °C, 5 MPa steam is around 571 m/s, so we’ll use this as a decent approximation of our throat velocity. The density of steam at our chamber condition is around 0.055 g/cm^3; our throat area is a tiny 0.24 mm^2. The 27 kg/hr of flow at throat velocity of 571 m/s only contributes 1 lbf of thrust; but we do have thermal expansion through a nozzle still to accelerate the flow further.
Water as a propellant has gamma (ratio of specific heats) of about 1.3, increasing with decreasing temperature; this is a bit higher than the typical 1.2 of combustion products. Unfortunately, this means we can only accelerate the exhaust to Mach 1.9 or so before its temperature falls below 100 °C, putting us at risk of condensation in the nozzle. This corresponds to an exit pressure at the end of the nozzle of 15.3 atm — we’re massively underexpanding — under which conditions the water will have already condensed out. It’s not clear how much we can actually accelerate the water with a nozzle here, because the phase change from condensation screws up the standard rocket equations.
To get an optimistic estimate, let’s return to the 8.7 kW of energy that we added to our 27 kg of water each hour. By energy conservation, this can accelerate the water to at most 1077 m/s, corresponding to 1.8 lbf of thrust. This is a good match to our Mach 1.9 exit… but also, useless. To close this out… at 150 kts, this corresponds to about 1 hp equivalent, assuming a typical propeller efficiency.
I choose to leave as an exercise for the reader the analysis of the pump power needed to achieve the required coolant pressure.
In summary: This is a bad idea. Don’t do this. Use a closed-loop cooling system and coolant-to-air radiator. Run the numbers for SmCo vs Nd, and don’t be surprised if for a closed loop system the lower temperature differential for Nd (and thus, the order of magnitude higher coolant flow needed) might outweigh the efficiency hit and engineering challenges of going to SmCo.
