Is the path loss exponent $\alpha$ is always 2 in friis equation model? In the friis equation,
the reference distance is 1m at the log distance path loss.

Then the formula would be:

$$P_r = P_tA_rA_t(\lambda/(4\pi))^2 /d^\alpha$$

where $d$ is distance between terminals, $P_r$ is received power, $P_t$ is transmission power, $A_r, A_t$ are the rx, tx antenna gains, respectively, and
$\lambda$ is wavelength.

It seems that the path loss exponent $\alpha$ is not mentioned. 

But I haven't seen a case where this $\alpha$ isn't 2.

Is this $\alpha$ always 2 in the friis equation?

And can the simulation environment be used even if this $\alpha$ is not 2?
Path loss exponent is not always equal to 2, only in the free space scenario is this true. In general the received power $P_r$ is proportional to $d^{-\alpha}$, where $\alpha$ is somewhere around 2-6 depending on the situation. The choice of the path loss exponent should match the scenario you're trying to simulate.

For example, for communication at a high altitude with little obstructions then $\alpha$ is probably close to 2 since the situation is approximately free space condition. 

For communication in a building, within the same floor, then $\alpha$ may be around 2-3. This is because walls other obstructions attenuate the signal power further than the free space condition. 

Things can get even worse if we again consider communication in a building but now if the transmitter is on the first floor and receiver is on the third floor, then $\alpha$ could be as high as 4-6 depending on construction of the building, etc.