NCEES 2011 No. 502

This question got me all confused. The final part says the inlet head is 10ft above the invert of the inlet end .

I find it hard to understand/visualize the basin as described.

Viessman & Hammer defined h as the elevation of the center-line of the outlet. Now, I would understand if the question says "inlet head is 10ft above the invert of the outlet end"

Am I missing something or maybe i need to brush up my knowledge of discharge from tanks/stormwater storage?

Here are my thoughts:

If the inlet and outlet pipes are at the same elevation (bottom of drain), then h = 10 - (D of outlet*0.5)

The diameter of the inlet does not matter.

makes more sense now.

biofem2011 said:

This question got me all confused. The final part says the inlet head is 10ft above the invert of the inlet end .I find it hard to understand/visualize the basin as described.

Viessman & Hammer defined h as the elevation of the center-line of the outlet. Now, I would understand if the question says "inlet head is 10ft above the invert of the outlet end"

Am I missing something or maybe i need to brush up my knowledge of discharge from tanks/stormwater storage?

Here are my thoughts:

If the inlet and outlet pipes are at the same elevation (bottom of drain), then h = 10 - (D of outlet*0.5)

The diameter of the inlet does not matter.

makes more sense now.

Click to expand...

The key is in the phrase 'short length of pipe'. if the pipe is short, then it would be ok to assume elevation of pipe center at inlet = elev of pipe center at outlet. Therefore the head difference driving flow is 10 ft - D/2 = 8.5 ft