Question 1. A horizontal pipe system has three branches as...

Name: Question 1. A horizontal pipe system has three branches as shown in the figure. The main pipe (1) is divided into two pipes (2 and 3). Determine the angle ? so that the reaction force in the y direction is zero. Assume that the velocity is uniformly distributed across all segments. D1= 1m, D2= 0.5 m, D3= 0.7 m, P1= 150 kPa, P2= 144.9 kPa, P3= 148.1 kPa, Q1=2 m3 /s, V2=4 m/s, ?= 1000 kg/m.
Uploaded: 2023-07-06T22:17:56-08:00
Duration: 6 min 54 s
Channel: Ethan Deweese
Description: Question 1. A horizontal pipe system has three branches as shown in the figure. The main pipe (1) is divided into two pipes (2 and 3). Determine the angle ? so that the reaction force in the y direction is zero. Assume that the velocity is uniformly distributed across all segments. D1= 1m, D2= 0.5 m, D3= 0.7 m, P1= 150 kPa, P2= 144.9 kPa, P3= 148.1 kPa, Q1=2 m3 /s, V2=4 m/s, ?= 1000 kg/m.

Question 1. A horizontal pipe system has three branches as shown in the figure. The main pipe (1) is divided into two pipes (2 and 3). Determine the angle ? so that the reaction force in the y direction is zero. Assume that the velocity is uniformly distributed across all segments. D1= 1m, D2= 0.5 m, D3= 0.7 m, P1= 150 kPa, P2= 144.9 kPa, P3= 148.1 kPa, Q1=2 m3 /s, V2=4 m/s, ?= 1000 kg/m.

Transcript

00:01 So in this problem we have our triple pipe network and we want to set an angle to keep the force consistent.

00:10 So we have our angle here, inlet one, and then the outlets two and three.

00:18 So we can begin with conservation of mass because there can't be any liquid or fluid piling up in this tube, otherwise it would eventually have to burst.

00:29 And so for conservation of mass we directly know that the inlet is equal to the outlet.

00:32 That is going to be the mass flow, or m dot one is equal to m dot two plus m dot.

00:42 And m dot is rho v1 a1, and we also have the volumetric flow rate q given here, where q is equal to the velocity times the area.

00:59 So rho 2 v2 a2 and rho 3 v3 a3.

01:07 Now we have a density of a thousand kilograms per meter, which is water, and so i kind of suggest that this is incompressible, in which case all of our densities are the same and we can just get rid of them.

01:19 Now the reason why we need to start with this is because we want to look at force, and the force of the outlets is related to the mass flow, m dot, and the velocity.

01:32 And so we need the mass flow and the velocities to get everything.

01:35 And we see we aren't given...

01:36 We know the velocity of two is four meters per second, but we don't know velocity three, and so that's what we want to solve for.

01:46 So we can do that pretty simply.

01:49 V3 v1 a1 minus v2 a2 over a3.

01:58 So v1 a1 is the mass, the volume flow at one, which we're given directly as the two meters cubed per second, but we're given v2 and we're given v2 on its own, which is four meters per second, and the area of two is the area of a circle, so it's pi times the radius squared.

02:20 We know the diameter is 0 .5 meters, so the radius 0 .25 meters squared, and the area of three, we need to do the same thing, pi and 0 .7 meters radius, so 0 .7 diameter, so 0 .35 meters radius.

02:43 I can get rid of that.

02:43 There we go.

02:46 Draw correctly.

02:49 And everything's in base units here, which means we know directly our velocity is in meters per second...

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