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

Answer 2

The speed of the jet-stream if the water-body height is 0.5 m is 10.21 m/s.

The speed of the jet-stream if the body of water-body height is 0.25 m is 3.608 m/s.

The answer is explained as follows.

Given that, the gate-width, $w=1\,m$

The gate-height is, $h_g = 1.2\,m$

Also, the depth of the water body, $h_w=1\,m$

The density of water $\rho = 1000 kg/m^3$

The diameter of the jet-stream, $d = 5\,cm =0.05 \,m$

The height of the jet-stream is 1 m.

Fluid Mechanics

The Force by the water-body at  $h_w= 0.5\,m$ is;

• $F_w = \frac{\rho gh_g}{2}\times h_gw =\frac{1000\,kg/m^3 \times 9.8\,m/s^2 \times (0.5\,m)^2\,1\,m}{2} =1225\,N$

The Force by the water-body at  $h_w= 0.25\,m$ is;  

• $F_w = \frac{\rho gh_g}{2}\times h_gw =\frac{997.9\,kg/m^3 \times 9.8\,m/s^2 \times (0.25\,m)^2\,1\,m}{2} =306.25\,N$

Now, the Force by the jet-stream is:

• $F_j=\rho v^2A_j$

Where $v_j$ is the speed of the jet-stream and $A_j$ is the area.

• $F_j = 1000\times v_j^2\times 3.14 \times (0.025)^2 = 1.96\,v_j^2$

• At the pivot, both the moment-of-force due to the water-body and the jet-stream must be zero.

• The force-applied by the water body acts on one-third of the height from the pivot.

• $F_jh_j = F_w \frac{h_w}{3}$

So, at $h_w= 0.5\,m$ the speed of the jet-stream is;

• $1.96v_j^2 = 1225\times\frac{0.5}{3} =204.166$
• $\implies v_j = \sqrt{\frac{204.166}{1.96}}=10.21\,m/s$

The speed of the jet-stream at $h_w= 0.25\,m$ is;

• $F_jh_j = F_w\frac{h_w}{3}$

• $1.96\,v_j^2 = 306.25\times \frac{0.25}{3}=25.52$
• $\implies v_j^2=\sqrt{\frac{25.52}{1.96}} =3.608\,m/s$

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