The diagram above shows a large tank of water (open to the air above) from which
ID: 1957961 • Letter: T
Question
The diagram above shows a large tank of water (open to the air above) from which
water is owing at a steady rate. Point 1 is at an elevation of 15.0 m above the ground,
and points 2 and 3 are both 2.0 m above the ground. The cross-sectional area at point 2
is 480 cm and the cross-sectional are at point 3 is 160 cm . You may assume that the
area of the tank is much larger than either of these.
1. What is the speed of the water at point 3, and what is the ow rate of water out of
the system?
2. If I insert a tube into the system at point 2 (as illustrated in the drawing above),
which is open to the air above, a column of water will ll it from the system. What
will be its height (relative to the point 2)?
Explanation / Answer
H = z + p/ + v2/2g = 15m
Assuming point 3 is open to the atmosphere:
15 = 2 + 0 + v2/19.6
v=15.96m/s
Q=v*A=15.96 m/s* 0.016 m2 = 0.255 m3/s
Point 2:
Q is the same but v will be slower: 0.255 = 0.048m2 * v
v=5.31 m/s
H = 15 = 2 + p/ + v2/2g
The rise in the tube will be p/
p/ = 15-2-5.312/19.6 = 11.56m