A rectangular trough, 1.5 m long, 0.50 m wide, and 0.40 m deep, is completely full of water. One end of the trough has a small drain plug right at the bottom edge. Part APart complete When you pull the plug, at what speed does water emerge from the hole

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

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