Simulating a Constant Temperature Anemometer
If Vout is 9.2 V then the op-amp output is saturating hard against the 10V rail. This is likely due to the fact that using fixed resistors in a simulation will never resolve the loop into equilibrium i.e. the op-amp will always be unbalanced because RS remains at precisely 1500 ohms and doesn't alter as temperature increases. 

How would it know that localized temperature has increased either: -



There is nothing about RS that makes it anything else other than a perfect 1500 ohm resistor - in reality RS will alter its value with temperature changes and at some amount of output current RS will be heated to a value where the whole circuit acquires equilibrium.

Equilibrium is when the bridge is balanced and, if you shorted out the 140 ohm resistor the bridge attains near-perfect balance i.e. the aim of RS is fall to about 1360 ohms under heat. If air flow increases, RS's resistance rises and this forces the op-amp to drive the transistor harder which warms up the 68R more and forces down the resistance of RS to 1360 ohms.