Water flow true a hole.

Hello every one.
This is my first post here so be gentle if i placed my topic in the wrong section.

My problem is as following.

I am trying to calculate the speed of the water wich exits true a metal plate full of cirkulare holes.

The metal plate is square and and has 14 holes each with a diameter of 14mm.
On top of the plate is a watercolum with a total hight of 50cm.
The water runs true the plate in to free air so there is no pressure working against the flow.

I want to use the calculated speed of the water to make a graph wich shows water mass on the x-axis and pressure drop on the y-axis.

I have come across following formulas:

Volume = surfuce area of holes * speed of water

press loss = zeta * (1/2)*zeta*water speed^2

I am not sure that the last calculation is correct and i have bin unable to find the zeta value in any books of mine.
im guessing its a value wich relates to the size of the hole and the dynamics of water.

I have that some of you can give me a push in the right direktion.
Sorry for spelling errors.

Best regards.
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 This is how I would do it: In the following pictures, there are the following quantities: p - pressure [N/m^2] rho - density of water [kg/m^3] v - speed of flow [m/s] z - height of a water particle before and after the hole (the difference is approx. 0) The weird greek letter for which I can't remember the name at the moment is a coefficient of losses for this very case ( = 1.8). I got that from my engineering handbook. Don't forget the following:At the moment the water level in the reservoir starts to fall, the velocity will begin to decrease, which is obvious from the equation. If holes are pointing downwards, the velocity that you get from this equation is only correct near the wall. As the water continues to fall, it accelerates and the jet gets narrower. I would re-check the calculations just in case I messed something up, which wouldn't be the first time it happened to me. Have fun, kandelabr
 Recognitions: Homework Help Science Advisor ZalienMC: Assuming you have atmospheric pressure above the water surface in the tank, the drainage hole stream exit velocity, and the total flow rate, as a function of tank water depth, are as follows. v = [2*g*z/(1 + ko)]^0.5 = Cd*(2*g*z)^0.5 = 2.6572*z^0.5, Q = (14 holes)*v*A*1000 = 2.1551*v = 5.7265*z^0.5,where v = drainage hole stream exit velocity (m/s), z = tank water depth (m), Q = total flow rate (L/s), ko = 1.778, Cd = 0.60, g = 9.8067 m/s^2, and d = drainage hole diameter = 0.014 m. E.g., if z = 0.500 m, you have v = 2.6572*0.500^0.5 = 1.879 m/s, Q = 5.7265*0.500^0.5 = 4.049 L/s.