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!
