Water flows steadily through the pipe fitting shown below,...

Water flows steadily through the pipe fitting shown below, and discharges to the atmosphere at point 3. The pipe fitting is held in place so that it does not move, and you will need to calculate the forces required to hold it, using the information provided below. $ ho_{water} = 1000 ext{ kg/m}^3$. $ heta = 77^circ$. $P_1 = 41 ext{ kPa gauge}$. $D_1 = 8.4 ext{ cm}$. $V_1 = 7 ext{ m/s}$. $P_2 = 26 ext{ kPa gauge}$. $D_2 = 6.1 ext{ cm}$. $V_2 = 3.4 ext{ m/s}$. $D_3 = 5.8 ext{ cm}$.

Transcript

00:01 Hello students there is a pipe setup here so this is the pipe setup closed pipe with one end here and the pipe has an opening here as well then we have the second opening with a closed end here and then it goes in the curvature of theta and then this is the third opening so this is the first and this is the second this theta there is a theta here angle so we are given that the density of the water is equal to 1000 kilograms and then we have theta is equal to 77 degrees p1 which is the pressure of the inlet here p1 is equal to 41 kilopascal diameter of the first opening is equal to 6 .1 sorry 8 .1 8 .4 centimeter and then we have velocity at the first one is equal to 7 meter per second now we go to p2 right p2 will be equal to 26 that is the second point velocity this is the p2 right and then we have d2 which is equal to 6 .1 centimeter and then v2 is equal to 3 .4 centimeter meter per second and d3 which is the third diameter this is d3 d3 so d3 will be equal to 5 .8 centimeter so we need to calculate the force required to hold this whole thing right so this pipe has to be steady so that means there are there is a force in this direction has to balance this force and this force at this force so there are forces acting in this direction so we need to hold this steady using these forces right so what we can do this force and this force and this force and this force sum equally so that it must be balancing each other so there are no other position to hold so this is the position to hold see this will be cancelling each other because the pressure is the same at these two points so we are interested in the p1 in this direction and this direction right so we can write down the continuity equation so we can say a1 v1 must be equal to a3 v3 right so v3 velocity 3 must be equal to a1 so that is equal to pi times d1 by 2 the whole square divided by times v1 by a3 which is equal to pi times d3 by 2 the whole square so let's substitute…

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