Derivation of Exit Velocity from Bernoulli without the Torricelli Assumptions

P = Pressure	r = density	Subscripts: 1 = top, 2 = exit
y1 = height of water from bottom	y2 = height of opening from bottom	v = velocity
g = acceleration of gravity	R = radius	h = y1 - y2

 

 

The following derivation is for a cylinderical tank with variables and subscripts as defined above.

 

Assumptions:

(1)  Both the top and the hole are open to atmospheric pressure.

(2)  The fluid is incompressible

 

In the following derivation I use Bernoulli’s equation and the equation of continuity.

 

Bernoulli’s equation P1 + ½ r1v12 +  r1gy1 = P2 + ½ r2v22 +  r2gy2

Equation of Continuity r1A1v1 = r2A2v2

 

From Bernoulli’s equation

P₁ – P₂ = ½ rv₂² +  rgy₂  -  ½ rv₁² -  rgy₁ = 0                               since P₁ = P₂ from assumption 1 above

 

½ rv₂² +  rgy₂  = ½ rv₁² + rgy₁   ]   rv₂² +  2rgy₂  = rv₁² + 2rgy₁  

 

Dividing through by r leaves                                           Assumption 2 above, r₁ = r₂

 

v₂² +  2gy₂  = v₁² + 2gy₁  

 

2gy₁  - 2gy₂ = v₂² – v₁² ]  2g(y₁ – y₂) = 2gh = v₂² – v₁²

 

where h is the distance from the top of the tank (the liquid) to the center of the hole

 

2gh = v₂² – v₁²    ]  v₂² = 2gh + v₁² 

 

From the Equation of Continuity (using incompressibility of the fluid)         

A₁v₁ = A₂v₂  ] v₁ = A₂v₂/ A₁  = (A₂/A₁)v₂ = (pR₂²/pR₁²)v₂ = (R₂/R₁)²v₂  

v₁ = (R₂/R₁)²v₂

v₂² = 2gh + v₁²  = 2gh + (R₂/R₁)⁴v₂²

 

v₂² - R₂/R₁)⁴v₂² = 2gh =

 

v₂² (1 - R₂/R₁)⁴ = 2gh

 

v₂ = ((2gh/(1 - R₂/R₁)⁴)^1/2

 

 

This shows that the exit velocity is equal to the square root of 2gh divided by 1-R₂/R₁)⁴

 

What are the implications for an experiment?