Reference: Cameron Hydraulic Data, 1998
I am a US Customary Units kind of guy. So bear with me on a bit of confusion from that.
Pump Hydraulic power = Delta(P) x Q
Where:
Delta(P) = differential pressure across the pump, units are Newton/(meters^2)
Q = Volume, units are meters^3/second
Hydraulic power = (Newtons/meter^2) X (meters^3/second) = (Newton-meters)/second = Watts
Of course, to get electric power input to the motor, one would divide by the pump eff, and divide by the motor eff.
And Meters^3/second = Meters^3/(hour x 3600), and divide by 1000 to get KW.
So, I don't know where the "367" comes from.
Unless it was a mis-type:
kW = (H x Q) / (Pump Eff x (3.6 x 10^6))
motor power = p abs / motor eff, where "p abs" = mechanical input to the pump shaft
And, I am thinking it is highly likely you already knew all this.
So, what is the context?
Ahhhh, yes. Gravity snuck in there. I would not have seen that without your tip.
Did your original post list "where H is in meter"? I'm surprised I missed that. "H" would then be head, and the specific gravity is needed to get the pressure, which of course requires the gravity acceleration constant. Sloppy me for missing that.
So, p1 is a specific gravity = Kg/dm3, kilos/cubic-meter. What is the "d"?