Velocity of Water from Overhead Tank: Formula & Calculation

[sudeep_s26]

What would the velocity of water be at the end of pipe of inner diameter 1 inch coming from the base of a overhead tank 10 meters high. Ignore the frictional losses. Also please post a formula for the same.

 

[Mindscrape]

Could you post a diagram? Also, do you have any thoughts of your own?

 

[sudeep_s26]

Image attached.

I Googled for it, but could not find anything usefull.
let me know your comments.

 

[Mindscrape]

So, if this is a homework problem I can't just give you the answer, but I'll point you in the right direction. The equation you want to use is an equation well known to fluid dynamics, and it is called Bernoulli's equation. For an understanding and derivation of the equation, remember the work energy theorem

\Delta T + \Delta U = W_{ext}

So, think of a tube, it could be your tube or some other tube. Work will be done by the pressure surrounding the fluid. At one end of the tube the work done by the pressure will be

W = \int_c F_1 \cdot dr

assume that the forcing pressure is constant

W = F_1 \cdot r_1

then at any point in the tube, the orientation is such that the force and displacement will be parallel (the water can't shoot out of the tube, and the air pressures around the tube cancel out)

W = F_1 \Delta r_1 = (p_1 A_1) \Delta r_1 = p_1 V

The work for the other end can be described similarly, except the force vector points opposite to the displacement vector (antiparallel) so the dot product introduces a negative and

W_{ext} = W_1 + W_2 = p_1 V - p_2 V

So, now let's piece together the other parts of the equation. The gravitational potential energy gives the familiar

\Delta U = mgy_2 - mgy_1

But we care about the volume of fluid, and we know that the mass density is mass per volume, so if we multiply the top and bottom of each term by volume we will get

\Delta U = \rho V g y_2 - \rho V g y_1

Rho represents the fluid's density, as described in the last paragraph. The kinetic energy can similarly be derived

\Delta T = \frac{1}{2}mv_2^2 - \frac{1}{2} mv_1^2 = \frac{1}{2} \rho V v_2^2 - \frac{1}{2} \rho V v_1^2

So the energy equation in terms of fluids is now Bernoulli's equation

\frac{1}{2} \rho V v_2^2 - \frac{1}{2} \rho V v_1^2 + \rho V g y_2 - \rho V g y_1 = p_1 V - p_2 V

which is often written as

\frac{1}{2} \rho V v_2^2 + \rho V g y_2 + p_2 V = \frac{1}{2} \rho V v_1^2 + \rho V g y_1 + p_1 V

take out all the volumes too

\frac{1}{2} \rho v_2^2 + \rho g y_2 + p_2 = \frac{1}{2} \rho v_1^2 + \rho g y_1 + p_1

Here is the equation you can use, and how/why it works. A quick hint, define your zero potential energy to be zero at the bottom of the mountain.