Register to reply 
Water flow true a hole. 
Share this thread: 
#1
Oct1310, 03:42 AM

P: 1

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 xaxis and pressure drop on the yaxis. 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. 


#2
Oct1310, 03:03 PM

P: 102

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:
Have fun, kandelabr 


#3
Oct1310, 06:55 PM

Sci Advisor
HW Helper
P: 2,124

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,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. 


Register to reply 
Related Discussions  
Flow problem with a two diemsional water flow divider  Mechanical Engineering  0  
Tap water flow and water diameter at end of it (Bernoulli eqzn...)  Introductory Physics Homework  4  
Rate of flow through a hole  Advanced Physics Homework  4  
Flow over a hole  Mechanical Engineering  1  
HELP! relationship between rate of flow of water and height of water column  Introductory Physics Homework  1 