Question

A gate is 1 m wide and 1.2 m tall and hinged at the bottom. On one side the gate holds back a 1-m-deep body of water. On the other side, a 5-cm diameter water jet hits the gate at a height of 1 m. What jet speed V is required to hold the gate vertical? What will the required speed be if the body of water is lowered to 0.5 m? What will the required speed be if the water level is lowered to 0.25 m?

Answer 1

Answer:

a) V = 10.2 m/s

b) V = 3.307 m/s

Explanation:

Given:

-Height of water jet h_w= 1m

-Width of the gate w = 1m

- Density p = 997 kg/m^3

- Diameter of water jet D = 0.05 m

Find:

Velocity of jet to keep gate vertical @ different depth of water body:

a) h = 0.5 m

b) h = 0.25 m

Solution:

The Force exerted by the water body is:

                  F_a = p*g*h_a*h_a /2 *w

                  F_a = 997*9.81*1*0.5^2 /2

                  F_a = 1222.571 N

                  F_b = p*g*h_b*h_b /2 *w

                  F_a = 997*9.81*1*0.25^2 /2

                  F_b = 305.643 N

The Force exerted by the water jet is:

                  F_w = p*V^2*A

                  F_w = 997*V^2*pi*D^2 / 4

                  F_w = 1.9576*V^2

The sum of moments about the hinge must be zero. Force exerted by water body acts at 1/3 the height from hinge. Hence,

                  F_w*h_w = F_a*h_a/3

                  1.9576*V_a^2 = 1222.571*0.5/3

                  V_a^2 = 104.0875

                  V_a = 10.2 m/s

                  F_w*h_w = F_b*h_b/3

                  1.9576*V_b^2 = 305.643*0.25/3

                  V_b^2 = 13.01095

                  V_a = 3.607 m/s