Heat Transfer Revisited

This post is going to deal with some tentative calculations on the quantity of heat transferred through the walls of a rocket nozzle I am building. This study is part of my ongoing engine project. To be more specific, I’m interested in finding out how much coolant flow will be required to safely absorb the heat generated by the combustion gases.

It is said that when the engineer who designed the cooling circuit for the Saturn V rocket motor was given his brief, he was told to “make sure it doesn’t melt”. The aims of rocket engine cooling design cannot be put more succinctly than that.

The material I’m using for the nozzle and combustion chamber of my engine is 6082-T6 aluminium. I chose this for a number of reasons, primarily that it has high thermal conductivity, low density combined with high strength and is easily machined.

Strength is one of the properties that I want to conserve since the pressures inside the engine are relatively high. This isn’t easy when you consider that the gases flowing through the chamber and nozzle are somewhere in the region of 2000K (1700 degrees Celsius). Just how exacting a task this is becomes even more apparent when it is realised that the tensile strength of 6082 drops to 90% of the room temperature value at just 420K (about 150 degrees Celsius).

A rocket engine cooling system is really a type of heat exchanger, where the heat from the high temperature gases inside the chamber and nozzle are transferred to the coolant via forced convection and conduction through the chamber and nozzle walls.

The snag is that the coolant is often also the fuel. This means that unlike a normal heat exchanger design in which the coolant flow rate can be made whatever is required to do the job, in a rocket engine we have a flow rate that is already fixed by the mixture ratio of the engine.

I mentioned in my earlier post that I’m using nitrous oxide and isopropanol as propellants in my design. The isopropanol will be the coolant, mixed with about 20% water to improve specific heat capacity.

I’m using an oxidiser to fuel ratio of 2.125. This is very fuel rich but it means that the chamber and nozzle temperatures are much more manageable. To help me in my rocket engine design activities I use a piece of software called Rocket Propulsion Analysis, or RPA. It is produced by Alexander Ponomarenko. This software calculates engine performance parameters based on user inputs from a projected design idea.

So the first thing it would be useful to know in designing the engine cooling system is how much fuel, i.e. coolant flow is there to play with? For an engine design of 8kN (about 1800 lbf) the isopropanol flow rate is 1.4 kg/s. The fuel i.e. coolant flow rate is denoted by the symbol mdot. The highest value of heat flux in the engine is found in the nozzle throat, where the hot gas flow area is smallest.

 Usefully, RPA also gives us nominal engine dimensions and the nozzle throat diameter of the design mentioned above comes out at about 70mm or 0.07m.

In order to calculate the total amount of heat transferred to the coolant, Q, we must first find the average heat transfer rate, q. To find this we need the nozzle hot gas side heat transfer coefficient, or hg. For other parts of the engine this calculation is slightly complicated by the fact that the hot gas stagnation temperature and the temperature of the gas very close to the wall are not exactly the same, due to boundary layer effects.

What is known as the adiabatic wall temperature or Taw has to be found by multiplying the chamber gas stagnation temperature by a so called recovery factor. Fortunately enough the recovery factor at the throat is 1, so the chamber stagnation temperature can be used as the adiabatic wall temperature at this point.

It should be borne in mind now that this adiabatic wall temperature is not the actual wall temperature; if it was the wall would melt! The point of the cooling system is to prevent this. Therefore, a maximum temperature is defined that the wall can be at without being seriously weakened. This is the hot gas wall temperature or Twg.

So, Taw is the temperature of the boundary layer gases adjacent to the wall, whereas Twg is the temperature the wall needs to sit at to retain adequate strength to keep the beast in the cage.

Twg is brought about by the action of the cooling system. Having defined the temperatures inside the nozzle it is time to look at the equations needed to calculate hot gas side heat transfer coefficient hg, average heat transfer rate q, total heat transferred Q and thus the coolant flow rate required, mdot.

At this point the explanation is going to get a bit algebraic. Please bear with me. I know this type of thing looks like gobbledegook and believe me, it is as tedious to type as it is to read. That said an important result should come out of it and one that I hope you, the reader, will be able to help me confirm or deny.

Deep breath, here we go. The equation for the heat transfer coefficient, hg, of a hot gas, in W/m^2K, is:-

hg = 0.026k (p*v/u)^0.8 (1/D)^0.2 (cp*u/k)^0.4  (1)

When:-

k = gas thermal conductivity = 0.2 W/mK

p = gas density at point of interest (throat) = 1.6 kg/m^3

v = gas velocity at point of interest (throat) = 950.7 m/s

u = gas viscosity at point of interest (throat) = 6.9 x 10^-5 kg/m-s

D = throat diameter = 0.07m

cp = gas specific heat at point of interest (throat) = 1.9 x 10^3 J/kg-K

The formula above is taken from Humble, Henry and Larson and is a form of the Colburn equation. All of the above figures were gained from RPA for the engine design in question. Substituting the figures into the formula we get:-

hg = 0.0052 x 749293 x 1.07 x 0.84  (2)

hg = 3502 W/m^2K

Using this heat transfer coefficient we can work out the average heat transfer rate, q, spoken of earlier, and from q can be got the total amount of heat transferred into the coolant, or Q.

Earlier it was mentioned that there are two temperatures of interest in the nozzle flow. These are the adiabatic wall temperature Taw, which can be thought of as the actual gas temperature in the nozzle region. The wall temperature desired to prevent structural failure is Twg.

From RPA:-

Taw = 1940K (1600 degrees Celsius)

And from standard reference works on the high temperature properties of aluminium:-

Twg = 420K (150 degrees Celsius)

This figure for Twg is chosen to give 90% of room temperature UTS for 6082-T6 aluminium, on which the engine chamber and nozzle wall thickness is based.

Given these numbers the average heat transfer rate, q, can be found:-

q = hg x (Taw – Twg)  (3)

Substituting the figures gained so far into the above equation gives:-

q = 3502 x (1940 – 420)

q = 5323040 W/m^2

= 5.32MW/m^2

A lot of rocket engine literature gives these types of figures in Imperial units. I wrestled with these in my early design efforts, but for the sake of my sanity I soon moved over to SI. The figure of 5.32MW/m^2 is about 3.26 BTU/in^2degF in old money. That might sound like a lot, and it is. Remember that the throat has the highest heat flux of any part of the engine and is therefore the most stringent in terms of cooling.

Having discovered q it now becomes possible to calculate what the temperature of the outside wall of the nozzle will be. This is the wall in contact with the coolant and we call this temperature Twc, the temperature of the cooled wall. This temperature is found by evaluating the heat conducted through the nozzle wall.

The formula required is:-

Twc = Twg – q(t/k)  (4)

Where:-

t = wall thickness = 0.003m (3mm wall thickness based on UTS of 6082 at 420K)

k = wall conductivity = 160 W/m^2K (Average value for 6082)

This equation gives a result in Celsius that will be converted to Kelvin by adding 273. Substituting known values into the equation above gives:-

Twc = 470 – 5.32 x 10^6(0.003/160) = 99.75 deg C

Twc = 372.75K

Temperature figures given in Kelvin can be hard to envisage. Considering that 373K is about 100 deg C, and the desired Twg was 150 deg C, then Twc seems about right given the very high thermal conductivity of aluminium.

Knowing this temperature difference it now becomes possible to calculate the total quantity of heat transferred from the hot gas side to the cooled side of the nozzle.

The formula required is:-

Q = kA(dT/t)  (5)

Where:-

A = heat transfer area = area of the inside wall of the nozzle throat

dT = Temperature difference = 470K-372.75K = 97.25

The inside area of the nozzle throat is simply the circumference multiplied by the length. Since the diameter is 0.07m the circumference is 0.219m. The design length of the throat section is 10mm or 0.01m. Hence the area is about 0.0022m.

Putting these values into the above equation gives:-

Q = 160 x 0.0022 x (97.25/0.003)

Q = 11410 W

= 11.4 kW

The coolant flowrate required to remove this quantity of heat can be found by assuming a coolant inlet temperature, Tinlet, of 293K (about 20 degrees Celsius). As previously stated the throat is the trickiest cooling problem; taking the “cold” coolant (fuel) direct from the tank and putting it through the throat jacket would seem to make sense. Assigning a coolant outlet temperature of, say, 320K (about 50 degrees Celsius), the coolant inlet to outlet temperature difference is 30 degrees.

Remember that the coolant or rather the fuel is being expelled from the tanks under pressure. The pressure drop between the tanks and the injector/chamber will be 0.7MPa (100psi). The boiling point of isopropanol at this temperature is in the region of 413K (140 degrees Celsius). So there is plenty of headroom in terms of the coolant temperature rise.

The formula used to find coolant flow rate is then:-

mdot = Q/(cpdT)  (6)

When:-

cp = coolant specific heat

dT = 30

As the coolant flows through the throat jacket a boundary layer will be set up close to the outside throat wall. The convective heat transfer will take place in this boundary layer, so it seems prudent to use a figure for specific heat capacity at the temperature of the coolant at this point. This is known as a film temperature, Tfilm.

This can be thought of as the average of the cooled wall temperature and the inlet temperature. So far the calculations show the cooled wall will be sitting in the region of Twc = 373K (100 degrees Celsius). The inlet temperature, Tinlet, is 293K (20 degrees Celsius).

Then:-

Tfilm = (Twc – Tinlet)/2  (7)

Tfilm = 313K (40 degrees Celsius)

The specific heat of isopropanol at this temperature is 2.8 x 10^3 J/kg-K. Using this figure it is now possible to use equation (6) to calculate mdot, the flow rate required to cool the nozzle throat.

mdot = Q/cpdT

mdot = 11.4 x 10^3/2.8 x 10^3 x 30

mdot = 0.136 kg/sec

Got there in the end! Recall from the earlier discussion of the engine design that the fuel and hence coolant flowrate is 1.4 kg/s.

To summarise what I think these figures show, a flow rate of about 0.136 kg/sec should be adequate to remove around 11.4kW from the nozzle throat area. This should give a hot gas side temperature of around 150 degrees Celsius and a cooled side temperature of 100 degrees Celsius. The coolant temperature rise will be in the region of 30 degrees Celsius.

If the preceding paragraph sounds like I’m hedging, it is because I am! These figures are very much order of magnitude as opposed to exact; remember, I’m trying to “make sure it doesn’t melt”. What I hope is shown, if I have got it right, is that putting the full fuel flow of 1.4 kg/sec through the throat jacket will be more than ample to cool it.

It does get a little more complicated than this. The type of flow is very important. High heat transfer requires turbulent flow and that means a Reynolds number in excess of 10,000. Fortunately other calculations that I’ve done (don’t worry, I’m not going to show them, I think you’ve suffered enough…) show that Reynolds numbers much higher than this are possible with a flow velocity just over 1 m/sec, based on a flow area of about 0.0013m^2.

If the flow is arranged to be perpendicular to the throat area instead of parallel things get even better, as extremely turbulent flow then results with vortex shedding. This so called cylinder in cross flow arrangement is highly beneficial for heat transfer.

A rough surface also helps to increase heat transfer and induce a higher Reynolds number for a lower velocity. This is why golf balls are dimpled. So if, say, I was to put a knurl on the outside surface of the nozzle area that would very likely increase heat transfer. Another aspect of this is something called nucleate boiling. As the fluid film temperature approaches that of the cooled wall, small bubbles start to appear. These rapidly detach and condense in the cooler surrounding liquid. More bubbles are then produced and the cycle repeats.

This process massively increases heat transfer. The surface of a knurl would provide excellent nucleation sites for the bubbles to form. The feed pressure of the fuel could thus be tuned to induce some nucleate boiling.

If you have got this far, thank you for staying with it. Perhaps now you would allow me one last indulgence and tell me if you think I have got it somewhere near right!