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I did a Civil Engineering course some years ago and from my textbook I had a question on a pipe with a bend. Now I think I have the part of the question on force correct, but I am not sure how the torque will be calculated in this situation. Can someone please help and give me an idea as to how to work this out?

I have the question as an image with my calculations below.

enter image description here

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It looks like you know how to calculate the forces due to a fluid flow? At the outlet, you must simply consider the momentum equation (Newton's Second Law), to find the axial force at said outlet - the torque then is simply this force multiplied by 1.7 (the perpendicular distance from the pivot, 'A').

Torque radius diagram

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  • $\begingroup$ Jonathon R Swift I have tried to reply to you, but system does not let me do this. I want to clarify about torque, as you have only covered part (a). From question is flow going down the vertical main and into pipe D and pipe C? Why do we use r = 2.4 * sin45 = 1.7 to multiply by force to get torque for part (a)? I can see that if I take my 1453.17 N and multiply by 2.4 * sin 45, I get torque of 2466 N. If I do it with the Force Answer of 1454 N, I get torque of 2468 N which agrees with answer. You have not shown me how to get torque for part (b) when pipe tapers from 150mm to 100mm, $\endgroup$ – Rob Wilkinson Sep 26 '19 at 6:24
  • $\begingroup$ In both cases you need to calculate the force as a result of the fluid leaving the exit of pipe C - because it is tapered in case b), the fluid will exit with less momentum compared to case a), meaning that there is a lower torque. $\endgroup$ – Jonathan R Swift Sep 26 '19 at 12:44
  • $\begingroup$ Jonathon R Swift Thanks for that, I thought it is flowing down the vertical and leaving exit at pipe C. You haven't answered my part about about why force is multiplied by r = 2.4 * sin45 in part (a)? Also Why is the torque less than the force in part (b) ? Is the distance calculated differently for part (b) ? Is torque also calculated in a slightly different way because of the tapering in part (b) ? $\endgroup$ – Rob Wilkinson Sep 26 '19 at 12:54
  • $\begingroup$ The distance is identical - it’s the momentum of the fluid exiting through the smaller orifice that has changed! $\endgroup$ – Jonathan R Swift Sep 27 '19 at 5:50
  • $\begingroup$ Hi Jonathon R Swift Can you please show me in step by step calculation for part (b) how this works to find torque ? I am confused because torque is less than force for part (b) $\endgroup$ – Rob Wilkinson Sep 29 '19 at 2:45
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Question 28

I am just working on the torque for part (b) and will have the rest of this answer up , as soon as I can.

The force to calculate torque is the force in the 45 degree direction from B to C.

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