Ok the imperial unit for the specific fuel consumption for a jet aircraft is shown below. Its the amount of fuel (in lbs) per hour required to produce 1 pound of thrust.
$$\frac{\frac{lbs}{h}}{lbs}$$
The metric unit is.
$$\frac{kg}{Ns}$$
I tried this.
$$\frac{\frac{lbs}{h}}{lbs}=\frac{lbs}{lbs}\cdot\frac{1}{h}=\frac{2.2\cdot kg}{2.2\cdot9.81\cdot N}\cdot\frac{1}{3600 \cdot s}=\frac{1}{9.81\cdot3600} \left[ \frac{kg}{Ns} \right]$$
But apparently it is.
$$\frac{1}{2.2\cdot9.81\cdot3600} \left[ \frac{kg}{Ns} \right]$$
Best Answer
I think you are correct, and I think whatever source you've found quoting the factor of 2.2 is wrong. In fact a quick Google found this book on aircraft design that gives the conversion factor as 0.283 $\times$ 10$^{-4}$, which is just 1/(9.81*3600) without any rogue factors of 2.2.