Q. A large water tank, open at the top, is situated at the top of a hill. A pipe leads downhill from
the bottom of the tank. The pipe has a diameter 5.0 cm where it is attached to the tank at point
A, but gradually narrows to a diameter of 3.0 cm at a point B which is 12.0 m below the
surface level of the water in the tank.
Velocity at A is 4.0 ms.
(ii) Calculate the absolute pressure at the lower level (point B).
p1 = pO = 1.01 * 10^5
P(density) = 1.00 * 10^3
v1 = 4.0 (back to this in a second)
y1 = 0 (assume, since the question indicates nothing else)
p2 = ?
v2 = (v1A1)/A2 = 11 ms
y2 = 12 m
p2 = p1 + 1/2P(v1^2 - v2^2) + Pg(y1 - y2)
Sub in, calculate to get answer.
The answer given in the exam booklet gives v1 as 0, while y1 = 12 and y2 = 0.
I'm pretty sure the y values were swapped because it essentially achieves the same purpose, ie. the distance between the two points.
However, I have no idea as why v1 would be 0. Any help?
The booklet also states that we assume the tank to be large, so v1 ~ 0. Why can't we use the v1 = 4 value, however?