Recent content by Xengine

Ah, i forgot that the velocity wasnt the same on both sides apparentely, so i find 150991.10 Pa using your equation. I also think i know what I'm doing wrong for the power... I think I have to add both losses to the the total head, is that correct? so P=...

Allright, when changing the signs of both the 3.66 and 12.2, i find an outlet pressure of 31229.22 Pa. Do you have any idea what I am doing wrong with calculating the necessary power? Do i have to work with only the highest head of 12.2m and not with the sum of both maybe? Thanks! P.S.: sorry...

Thank you for your reply! from 0+0+0-1.83 = p2/(1000*9.81)+3.05+3.0579^2/(2*9.81) (i calculated the speed by dividing the flow by the cross sectional area: 0.057/(1.864*10^-2) i find p2= -52548.18 Pa I wonder, if this is correct, if I have to substitute this value in absolute value or not...

But there isn't any data given about the exact height of the outlet point for example, i think the arrow just indicates the inlet and outlet pipe rather than the specific point that the pressure needs to be calculated on, or am I missing something?

Homework Statement http://imgur.com/ktfgCAL flow rate: 0.057 m^3/s cross sectional area of all pipes: 1.864*10^-2 m² frictional losses inlet pipe: 1.83m frictional losses outlet pipe: 3.66m Asked: pressure at inlet pressure at outlet necessary pumping powerHomework Equations Bernouilli...

Homework Statement http://imgur.com/ktfgCAL flow rate: 0.057 m^3/s cross sectional area of all pipes: 1.864*10^-2 m² frictional losses inlet pipe: 1.83m frictional losses outlet pipe: 3.66m density: 1000 kg/m^3 g= 9.81 N/kg Asked: pressure at inlet pressure at outlet necessary pumping...