# How to calculate the fuel flow using the thrust specific fuel consumption?

Using the following parameters:

• Static Thrust : $T_0 = 12 KN$
• Thrust specific fuel consumption: $C_T=0.7 N/NH$

I know that the fuel flow is linked to thrust with the following formula:

$F=C_TT$.

And I need to calculate the fuel flow in Kg/s.

I started by converting $C_T$ in Kg/Ns:

$0.7 N/NH = 0.7 \frac{1}{9.80665} \frac{1}{3600} = 1.9828* 10^{-5} \frac{Kg}{Ns}$

Then I multiplied this value with the thrust:

$F = 12000* 1.9828* 10^{-5} = 0.2379 \frac{Kg}{s}$

However this does not seem to be the correct answer. What am I doing wrong?

• Your units look wrong. TSFC should be g or kg per N and h to begin with. Your conversion should not be needed. If you use the same unit for weight and thrust (like lbs in Imperial Units), you get the specific impulse which is popular in rocketry, but unusual in aviation. The number of your TSFC looks like the TSFC in kg/Nh for a low-bypass engine. Maybe you should check the source of that figure and use it without the conversion. – Peter Kämpf Sep 4 '16 at 5:44

## 1 Answer

As Peter Kämpf said, the units should be in kg per N per second/hour.
In your case assuming it's a typo, and assuming the formula is $\text{C}_\text{T}$ = 0.7 kg/h per Newton.

Then in static condition,

Fuel flow $= 12 \cdot 10^3 \cdot 0.7 \cdot \frac{1}{60} \frac{kg}{s}$

gives 140 kg of fuel per second. Seems correct.

As a side note TSFC is not a fixed value but a variable of velocity as well. So have to make sure this CT is for static condition.