Respuesta :
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
The speed at which water emerge from the hole is 2.8 m/s.
Based on the given information,
• The length of the rectangular trough is 1.5 m, the width of the trough is 0.50 m, and the height of the trough is 0.40 m.
Now the speed of the efflux can be determined by using the formula,
[tex]V = \sqrt{2gh} \\V = \sqrt{2*9.8*0.40} \\V = 2.8 m/s[/tex]
Thus, the speed of the water emerging from the hole is 2.8 m/s.
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