Pumps TDH for slurry

[equimo]

I am designing a pump system for slurry which is a mixture of water, clay, and sand. The slurry has Specific gravity (Sg) 1.5. The manufacturer says that we don't have to calculate the Total Dinamic Head (TDH) affected by 1.5 due that the pump itself assume that difference: The head in meters of water of his characteristic curve is divided by a factor called ER= 0.94 the result is head of slurry according with the mechanical eng. But when I see the formula of the pump potency given by the manufacter, it is P=Q x TDH of water x Sg / (3960 xERx efficency). Then, here he is affecting the TDH by Sg , so at the end of the day the pump is pumping a Head times Sg
My question is: How can a pump tested a determinate RPM and tested with water multiply 1.5 times the head shows in its curve? Why I cannot say i.e. TDH required by pump= 50 x 1.5= 75 m. and this 75 m shall be fall within the curve? Any feedback is very appreciate

 

[Artisi]

equimo
"Why I cannot say i.e. TDH required by pump= 50 x 1.5= 75 m. .............."
You cannot say it because it is not true!

The head seen as a measurement in metres at the pump discharge is the static head includng any friction losses etc divided by the ER of 0.94 (I cannot confirm 0.94 as I don't have the necessary data to hand but sounds to be in the right area).
ie, if the static head is 75m then the effective head the pump must generate to pump 75m of slurry is 75/0.94 = 79.8m
you therefore need to run the pump at the speed required to generate 80m as the curves are drawn for water performance
The power calculation is correct - it must includes the SG as this impacts on the power required. The formula for power calculation isn't saying that the head is increased by 1.5.
The standard calculation for water is
P = flow x head / (3960 x efficiency)
but as you are pumpng slurry you need to include the ER (deratng factor) and the SG as the product in heavier than water.
If you are working in metric terms the formula becomes
l/s x m x SG / (102 x ER x efficiency) = Kw

 

[equimo]

Thanks for your reply, Artisi. I continue a bit confused: The standard formula for water include the Sg of water which is 1, and the slurry pump also has Sg, in this case 1.5. The pressure generated by water is head times 1, and the pressure generated by slurry pump is head times Sg=1.5. So the discharge pipe shall be set up over this resultant pressure. Though that the slurry pump needs to have more power to generate this pressure. Is it right?

 

[TenPenny]

The pump will generate "x" feet of head pumping water or slurry; the measured pressure in, say, psi, will vary depending on the SG of the liquid; similarly, the power required will vary with the SG of the liquid.

 

[djack77494]

equimo - "Is it right?"
Yes, it is right. With SG=1.5, the pressure (in psi, bar, kg/cm^2, or kPA, etc.) will be 1.5 * the pressure IF the fluid had a SG=1. Likewise, the pump's (motor's) power requirement will also be 1.5 * the power IF the fluid had a SG=1.
Doug