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Fluid Mechanics - Torque at hinge of closed vessel

by Aerospace
Tags: fluid, hinge, mechanics, torque, vessel
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Aerospace
#1
Jun26-05, 09:22 PM
P: 39
The closed vessel contains water with an air pressure of 10 psi at the water surface. One side of the vessel contains a spout closed by a 6-inch diameter circular gate hinged along one side. Horizontal axis of the hinge is located 10 ft below the water surface. Determine the minimum torque that needs to be applied at the hinge to hold the gate shut.

Okay.
Here's what I am thinking, and please let me know if I am doing this correctly.

First, I'm trying to find the resultant Force = F. Here's where I am slightly confused. Do I take this body as a curved surface or not? because the gate is located at a spout which is circular. If not, then what I did was just

A = pi * r^2 = 9*pi in^2 = pi/16 ft^2
hc (vertical distance from fluid surface to centroid of area) = 10 ft
specific weight = 62.4 lb/ft^3

F = sp wt * hc * A = 122.52 lb

Torque (T) = F * radius = F * (3/12)ft = 30.63 lb-ft

Is that right?
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OlderDan
#2
Jun26-05, 10:56 PM
Sci Advisor
HW Helper
P: 3,031
Quote Quote by Aerospace
The closed vessel contains water with an air pressure of 10 psi at the water surface. One side of the vessel contains a spout closed by a 6-inch diameter circular gate hinged along one side. Horizontal axis of the hinge is located 10 ft below the water surface. Determine the minimum torque that needs to be applied at the hinge to hold the gate shut.

Okay.
Here's what I am thinking, and please let me know if I am doing this correctly.

First, I'm trying to find the resultant Force = F. Here's where I am slightly confused. Do I take this body as a curved surface or not? because the gate is located at a spout which is circular. If not, then what I did was just

A = pi * r^2 = 9*pi in^2 = pi/16 ft^2
hc (vertical distance from fluid surface to centroid of area) = 10 ft
specific weight = 62.4 lb/ft^3

F = sp wt * hc * A = 122.52 lb

Torque (T) = F * radius = F * (3/12)ft = 30.63 lb-ft

Is that right?
The center of the gate is not exactly 10 feet below the surface. The hinge is 10 feet below the surface. The curvature of the spout makes no difference, but the pressure across the surface of the gate is not uniform. To find the torque due to the water accurately you would need to integrate the pressure times the moment arm over the surface of the gate. Since they give you the angle of the gate, I assume they want you to take the pressure variation into account. Even if you did not, the variation in the moment arm as you move away from the hinge would have to be considered.
Dr.Brain
#3
Jun27-05, 04:50 AM
P: 541
Calculate the velocity of efflux from the curved pipe opening using Bernoulli's Theorem. This water-flow will form a force at the circular gate trying to push it outwards, therefore equal and opposite torque is needed to counteract the torque due to water flow. The force due to water flow is given by [itex]dAv^2[/itex] where d is the density of water .The torque will be the moment of this force about the hinge of the circular gate.

BJ

Dr.Brain
#4
Jun27-05, 04:52 AM
P: 541
Fluid Mechanics - Torque at hinge of closed vessel

The above is ofcourse not the case if the water is supposed to be stationary.If you let the liquid flow which is the case, the height of the opening matters and not the shape of the spout.


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