# Homework Help: Force acting in nozzle

1. Apr 7, 2016

### foo9008

1. The problem statement, all variables and given/known data
why P1 not equal to P atm ? I can understand that P2= P 3 = P atm , because it is open to the atmospheric pressure... but why not P1 ?

for (0.659-Fx = (10^3)(0.0942)(11.59) + (10^3)(0.053)(10.39) - (10^3)(0.1473)(8.33) = 0.147kN , it's actually (0.659-Fx = (ρ)(Q)(v2) + (ρ)(Q3)(v3) - (ρ)(Q1)(V1)

why not the equation =
(0.659-Fx = (ρ)(Q)(v2x - v1) + (ρ)(Q3)(v3x -v1) ) ? just like the equation in the question that i asked in the previous thread? https://www.physicsforums.com/threads/pressure-in-bent-pipe.865389/
2. Relevant equations

3. The attempt at a solution

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2. Apr 7, 2016

### Staff: Mentor

Why should it? Is it exposed to the atmosphere?
What is the rate at which horizontal momentum enters the control volume?
What is the rate at which horizontal momentum exits the control volume?
What is the rate of change of horizontal momentum of the fluid within the control volume?

3. Apr 7, 2016

### foo9008

in the picture , it is exposed to atmosphere , right ? there's a hole at inlet

4. Apr 7, 2016

### foo9008

i found that by using both 0.659-Fx = (ρ)(Q)(v2x - v1) + (ρ)(Q3)(v3x -v1) ) and (0.659-Fx = (10^3)(0.0942)(11.59) + (10^3)(0.053)(10.39) - (10^3)(0.1473)(8.33) )
i gt 0.659-Fx= 0.417kN , i dont understand the method (0.659-Fx = (10^3)(0.0942)(11.59) + (10^3)(0.053)(10.39) - (10^3)(0.1473)(8.33) ) , can someone explain on this ?

5. Apr 7, 2016

### Staff: Mentor

No. That's the section of the hose near the outlet. They couldn't fit the entire hose in the picture, so they used those curvy lines around the perimeter to show that the hose is continuous there.

6. Apr 7, 2016

### Staff: Mentor

Answer my questions in post #2, and then you'll understand better.

7. Apr 7, 2016

### foo9008

this is the hint showing the pipe is very long , and the section is not exposed ti atmosphere?

8. Apr 7, 2016

### Staff: Mentor

Yes. It's a drawing convention that is always used to indicate that only part of an object is being shown.

9. Apr 7, 2016

### foo9008

thanks for pointing out that , i just knew it

10. Apr 7, 2016

### foo9008

rate at which horizontal momentum enters the control volume =rate at which horizontal momentum exits the control volume
rate of change of horizontal momentum of the fluid within the control volume = 0 ?

11. Apr 7, 2016

### Staff: Mentor

No way. The momentum of the fluid within the control volume is increasing.
Rate at which horizontal momentum enters the control volume = (ρ)(Q1)(V1)
Rate at which horizontal momentum exits the control volume = (ρ)(Q2)(v2) + (ρ)(Q3)(v3)
Rate of change of momentum of fluid within control volume = (ρ)(Q)(v2) + (ρ)(Q3)(v3) - (ρ)(Q1)(V1)