Fluid Mechanics - Linear momentum Analysis

[gmy5011]

Homework Statement

Water is flowing into and discharging from a U-shaped pipe section as shown. At flange
(1), 30 kg/s of water flows into the section with the total absolute pressure of 200 kPa. At flange (2), the absolute pressure is 150 kPa. At location (3), 8 kg/s of water discharges to the atmosphere at 100 kPa. Determine the total x and y forces on the flanges connecting the pipe bend. Do not neglect the viscous losses in the pipe bend. Use a momentum-flux correction factor to be 1.03. In your discussion answer the following question: Is it possible to find the force on each flange individually? Why or why not?

This is the given question. What I don't understand is when we are summing the forces, why is the force at flange 2 acting in the positive direction(to the right)? If you see in my solution attempt, I made it negative but in my teachers solution she has it positive. If someone could explain this too me that would be great.

Homework Equations

eqn.1 $\dot{m}$ = $\rho$VA

eqn.2 $\Sigma$F = $\Sigma$_out$\beta$$\dot{m}$V - $\Sigma$_in$\beta$$\dot{m}$V

The Attempt at a Solution

Used eqn.1 to solve for all 3 velocites

then eqn.2 to solve for Frx: Frx + P1V1 - P2V2 = $\beta$$\dot{m}$(-V2) - $\beta$$\dot{m}$V1

 

[Valkarie]

Frx = \beta\dot{m}(V1 - V2) - (P2V2 - P1V1)Used eqn.1 again to solve for Fry: Fry + P3V3 - P2V2 = \beta\dot{m}(-V2) - \beta\dot{m}V3 Fry = \beta\dot{m}(V3 - V2) - (P2V2 - P3V3)Total force on flange 1: F1x = Frx + Fry F1x = \beta\dot{m}(V1 - V2 + V3 - V2) - (P2V2 - P1V1 + P2V2 - P3V3) F1x = \beta\dot{m}(V1 + V3 - 2V2) - (P2V2 - P1V1 - P3V3)Total force on flange 2: F2x = -Frx - Fry F2x = -\beta\dot{m}(V1 - V2 + V3 - V2) + (P2V2 - P1V1 + P2V2 - P3V3) F2x = -\beta\dot{m}(V1 + V3 - 2V2) + (P2V2 - P1V1 - P3V3)