Geesh! You could have saved us a lot of trouble, a long time ago!
Here's a quote of the relevant part:
Quote:
If i place my pump inline only the "flowheat" is added to the water and if i place it submerged all consumed energy is added to the water.
The question everybody wants to know is:
How much of the total amount of consumed energy is transformed into flow and how much into heatloss in pump.
----------------------------------------
Here comes a calculation=answer of that question:
(I've used data from my Eheim 1250 as input)
E_tot=28W (total consumed power)
F_max=1200l/h=0.33l/s (maximum waterflow)
h_max=2m (maximum height)
E_flow=? power transformed to flow
E_loss=? powerlosses in pump
g=9.80665N/kg (gravityconstant)
rho=0.998kg/l (density of water in 20C)
(note that F_max and h_max don't occur simultaneusly but using them together gives the "best case senario")
E_flow=F_max*rho*g*h_max=
=0.33(l/s)*1(kg/l)*9.81(N/kg)*2(m)=6.5(Nm/s)=
=6.5W
E_loss=E_tot-E_flow=21.5W
Conclusion:
If the pump was able to pump 1200l/h to a height of 2m the powerinput into the water would be 6.5W and the powerloss in the pump would be 20.5W.
So by placing the pump under water you add 20.5W of unnecessery heat to the water.
|
Now let's run the numbers for a typical flow rate, say 100 gph, for the same Eheim 1250.
1250 curve here!
At 100 gph, the head is about 1.7 meters, or 2.42 psi.
E_tot=28W (total consumed power)
F=100 gph, or 0.10 L/sec
h=1.7m
E_flow=? power transformed to flow
E_loss=? powerlosses in pump
g=9.80665N/kg (gravityconstant)
rho=0.998kg/l (density of water in 20C)
E_flow=F*rho*g*h=
=0.10(l/s)*1(kg/l)*9.81(N/kg)*1.7(m)=1.67(Nm/s)=
=1.67W
E_loss=E_tot-E_flow=26.33W
That doesn't feel right...