Pressure Head

[owillis28]

Haven't done a flow net since college.

Total head loss (headwater to tail water) = 12

Number of equipotential drops= 0.80 foot/drop

Can anyone explain to me how the headwater to Point A = 9.5 FT? (12 * 0.80 = 9.6??)

Did they take 12 * 0.80 * 0.80 = 7.6 (7.68)

I am confused with the solution to this problem. My first answer was C but I knew it wasn't that easy.

I tried to solve this problem with the CERM using Equ. 21.36 but didn't even come close.

 

[Guest]

owillis --

The trick here is to recognize the relationships between elevation head and pressure head. Remember, the water is FLOWING underneath because of the pressure differential through the soil column along the base of the dam.

So, first calculation is to find potentiometric head at point A: ha = #drops/total drops * Difference in Elevation head = 9.5/15*12-ft = 7.6-ft

Now consider what the TOTAL head at A = 12-ft - 7.6-ft = 4.4-ft

So that leaves us with PRESSURE head at A = Total head - Elevation head = 4.4-ft - (-11.0-ft) = 15.4-ft (D)

Let me know if that makes sense, I can try to explain a better if necessary.

JR