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## Homework Statement

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).

## Homework Equations

Values:

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

Rearrange bernoullis;

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?