# How can I determine the fuel temperature in the tanks of an Airbus A320 during climb?

I know that, for example, at 10000ft the outer temperature is around $$-10\, ^\circ\! C$$.
How can I calculate the fuel temperature in the tanks?
In technical notes I find on the internet for Airbus there are mainly min max limitations that trigger the sensors.

• without having the initial conditions and more info about the current situation, I don't see how you would calculate anything. Commented Jan 25, 2023 at 15:09
• The temp would be different if the fuel started at 30°C than if it started at 15°C, so you're going to have to come up with more parameters than you've provided. Commented Jan 25, 2023 at 19:35
• @MichaelHall While frictional heating and cooling to ambient conditions can be calculated as a physics problem, you cannot use physics alone to determine the number one source of fuel heating: the engine fuel/oil heat exchangers. Commented Jan 25, 2023 at 23:24
• Airbus has software to calculate this, called FTP, available to operators. If there's a legitimate need, you'll have to work with them, or find somebody at a carrier to run it for you. Commented Jan 25, 2023 at 23:31
• To all commentators mentioning how complicated and variable: it is an engineering issue, with more than one input indeed. The answer can show the methodology, then set a start temperature as an example for a specific aeroplane. The question is about the methodology. Commented Jan 26, 2023 at 1:11

When a body at a certain temperature $$T_i$$ is blown upon by air at temperature $$T_{air}$$ there's an exchange of heat per second $$\dot{Q}$$ (in Watt) which is:

$$\dot{Q}=hA(T_i -T_{air})$$

where:

• $$A$$ is the surface of the body involved in the heat exchange; and
• $$h$$ is the convective heat transfer coefficient which tells us how good the flow of air is exchanging heat with the body; for example, moving air exchanges heat better than still air (think about a ventilator): in the former case $$h$$ is bigger; air blowing on the entire body's surface exchanges heat better than if it's blowing only on a small part of it: in the former case $$h$$ is bigger.

Now, when the body exchange some heat $$Q$$ with the surroundings, its temperature changes of:

$$T_f-T_i=\frac{Q}{mc_p}$$

where:

• $$m$$ is the mass of the body; and
• $$c_p$$ is the heat capacity which tells us how much conductive the body is; for example, air is less conductive than water: if you are invested from air at 200°C when you open your oven, it's not a big deal but just a couple of drops of water at some 80° will make you scream: air il less conductive than water and its $$c_p$$ is lower (1:4 more or less).

Now, if we equal these two equations and solve in respect of $$T$$ we get:

$$T_f = T_{air} + (T_i-T_{air})e^{-bt}$$

where:

• $$b=\frac{hA}{mc_p}$$;
• $$t$$ is time; and
• $$T_f$$ is the final temperature of the body after $$t$$ seconds.

That's all!

Let's do the math for our kerosene flying at -50°C:

• $$h \rightarrow$$ for a flat surface blown upon air at high speed, its value is some 100W/kgK;
• $$A \rightarrow$$ let's suppose that the tanks occupies some 70% of the entire wing surface; for an A320 this makes $$A=0.7 \times 124=85m²$$;
• $$m \rightarrow 27'000l\times 0.8kg/l=21'000kg$$;
• $$c_p \rightarrow$$ for kerosene it is 2'200J/kgK;
• $$T_{air} \rightarrow -50°C=220K$$;
• $$T_i \rightarrow$$ this is the initial temperature of the kerosene, we suppose to takeoff on a sunny Sardinian day 30°C=300K;
• $$t \rightarrow$$ let's see what happens after 1 hour of flight $$t= 1 \times 3'600s$$

Substituting:

• $$b=\frac{100 \times 85}{21'000 \times 2'200}=0,000184s^{-1}$$
• $$T_f=220 + (300-220)e^{-0.000184 \times 3'600}= 220 + 41= 261K=-13°C$$

So, after having been blown upon with air at -50°C for one hour, our kerosene has gone from +30 to -13°C.

We introduced some simplification:

• we have supposed that the whole mass of kerosene interacts with the -50°C of the air; obviously only the layers of kerosene on the surface exchange heat, the innermost part is not in direct contact with the -50°C and it doesn't change its temperature so fast as the outer layers;
• kerosene is used as coolant in the various heat exchangers which can be found on a jetliner; this heat exchange raises its temperature and has to be taken into account.

Anyway as a first conservative estimation that equation should be enough.

• +1 for explaining the underlying physics! However, a temperature change of 43°C in one hour seems rather excessive to me. If you look at the plots in Airbus FAST 36 (PDF, page 8), you'll see temperature changes of 20 to 30°C (depending on which tank) during the first 1000 NM of the flight (about 2 hours). Commented Jan 28, 2023 at 9:59
• @Bianfable: thanks 🖖 As said at the end of the answer, this is a conservative estimation, the real value is going to be less (drop of temperature) as you actually shown. Anyway note that the plots from the pdf use 5° as initial temperature and -60°C for outside air temperature. And from the picture of the wing my 70% seems to be a bit too high. I try to redo the math with the values of that plot, thanks for sharing it 👍 Commented Jan 28, 2023 at 10:27
• I think this is the opposite of a conservative estimate. This gives a sort of upper bound, and a conservative estimate usually gives a lower bound. Commented Jan 28, 2023 at 15:51
• @Federico: well "conservative" means "closer to (or even past) the worst case". Here the worst case is frozen fuel, which is the exactly the direction given by the calculation (-43°C temperature drop instead of some -30°C as per plot provided by Bianfable). So, the calculation has simplifications but they are conservative. Commented Jan 28, 2023 at 16:02
• Math is interesting and fun to follow. Since both top and bottom of wing exchange heat, A value might be 2x. Also, a crude volume to surface area ratio might bring these "ball park" numbers even closer. +1. Commented Jan 28, 2023 at 22:11

The fuel temperatures in the wing fuel tanks are displayed to the pilots on the ECAM (electronic centralized aircraft monitor) FUEL page below the fuel quantity indications. Pilots can select this page manually and the ECAM can also call up this page automatically if fuel temperature gets too low.

You can see the fuel temperature indications in this video:

The actual fuel temperature inside the fuel tanks depends on the static and total air temperatures (TAT), the fill level of the tank and other major factors. For example: The fuel is also heated by the engine oil heat exchanger. Flying faster can increase the TAT and heat up the wing surfaces and the fuel inside the wings by heat exchange processes (as others already have shown in their replies). But also increasing the airspeed increases the drag of the aircraft so you need a higher engine thrust to keep the high airspeed. Higher thrust requires a higher engine RPM, increasing engine oil temperature over time. And because of the engine oil heat the with fuel that is going back to the fuel tanks is heated whist the engine oil is cooled.

So on an almost empty tank, during the descent with the engine at idle and at high altitude after a hours of flying in cold air the fuel is very likely going to reach the coldest temperature in the fight. Sometimes it can remain cold enough so that during the descent or even after landing ice can form on the upper and lower surface of the wing, known as cold soak effect.

Here is an example of the cold soak effect: https://www.airliners.net/photo/Air-Canada/Airbus-A320-211/991791/L