Answer:
V = 2.801 m/s
Explanation:
Given.
- The dimension of the tank = 1.5 x 0.5 x 0.4 m
- A plug at the bottom of the tank
Find:
When you pull the plug, at what speed does water emerge from the hole
Solution:
- We can compute the solution either by an energy balance or Bernoulli's equation as follows:
P_1 + 0.5*p*V_1^2 + p*g*z_1 = P_2 + 0.5*p*V_2^2 + p*g*z_2
Where,
P: Pressure of the fluid (gauge)
V: Velocity of the fluid
p: Density of the fluid
z: The elevation of fluid from datum(free surface)
g: The gravitational acceleration constant
- We will consider the two states by setting a datum at the bottom of the surface.
State 1 (Free surface of the fluid at level):
P_1 = 0 (gauge - atm)
V_1 = 0 (Free surface)
z_1 = 0.40 m
State 2 (Plug level @ bottom):
P_2 = 0 (gauge - atm)
V_2 = unknown (need to calculate)
z_1 = 0 ... (datum)
- Lets plug the values at each state in the equation above:
0 + 0 + p*g*z_1 = 0 + 0.5*p*V_2^2 + 0
V^2 = 2*g*z_1
V = sqrt ( 2*g*z_1)
V = sqrt ( 2*9.81*0.4)
V = 2.801 m/s