- #1
LT72884
- 335
- 49
- Homework Statement
- A hydraulic jump (see Video V10.11) is in place downstream
from a spillway as indicated in Fig. P5.10. Upstream of
the jump, the depth of the stream is 0.6 ft and the average stream
velocity is 18 ft/s. Just downstream of the jump, the average stream
velocity is 3.4 ft/s. Calculate the depth of the stream, h, just downstream
of the jump.
- Relevant Equations
- Q1=Q2 or V1A1=V2A2 where A is the cross sectional area of the water.
I have 90% of this done. When i looked in my manual i noticed an odd jump they made and i don't get why. that is where i need the help
V1(b1h1) = V1(b2h2)
Im solving for h2.
What i can't figure out is what to use for the base measurement of the cross section of the water. i have no idea how wide the reservoir channel is. It never gave me that number and we can't just "assume" that the base is the same in the smaller stream than it is in the larger stream. But the solution manual made that assumption so that the bases cancel out. I disagree with that. Since the stream IS SMALLER, therefore it has a smaller base than the larger stream at the bottom... Canceling out the bases can't work.
thanks for the help:)
V1(b1h1) = V1(b2h2)
Im solving for h2.
What i can't figure out is what to use for the base measurement of the cross section of the water. i have no idea how wide the reservoir channel is. It never gave me that number and we can't just "assume" that the base is the same in the smaller stream than it is in the larger stream. But the solution manual made that assumption so that the bases cancel out. I disagree with that. Since the stream IS SMALLER, therefore it has a smaller base than the larger stream at the bottom... Canceling out the bases can't work.
thanks for the help:)
