Determine efficiency, voltage, current from motor characteristic curves
As voltage varies torque vs speed, does the efficiency curve vary ? 


Yes


Motor Specs    
Max Eff.= 52.2% @ 5.916 A, 2999 RPM , 0.236 N.m    
Max Power 45.07%@ 9.427 A ,2226 RPM , 0.436 N.m  Pe=226.2W, Pm= 101.6W


Your operating specs:  


Torque: 0.35 N.m 
RPM   : 1200
What are V,I,Eff.? :   
Using Max Eff.,    


\$\dfrac {0.236 Nm}{0.436}(=54\%) \cdot 24V = 13V, ~~2999 ~RPM \cdot 54\% =1619~RPM\$


\$\dfrac{T_{spec}}{T_{max}}= \dfrac{0.35Nm}{0.436}= 80.3\%\$
which is much greater than torque ratio 54%  at peak eff.     


0.35 N.m = 80.3% of max T thus current
Iss=80.3%*9.427 = 7.57 A  and power Pe = 98.4W = 7.57A⋅ 13V   
yet RPM has dropped to 54% of max power at 80.3% T so mechanical power has dropped to 80% of 54% = 43.2% of 101.6W= Pm=43.9W mechanical power (P2) 

thus efficiency is 44.7%  about the same as the max T eff. (45.07%)


You will need a tach to regulate the RPM if you use this to vary the BLDV voltage or some other servo method.