Why is P1 not equal to Patm in a nozzle?

In summary, the equations for horizontal momentum in this problem are (ρ)(Q1)(V1) for the rate at which it enters the control volume, (ρ)(Q2)(v2) + (ρ)(Q3)(v3) for the rate at which it exits the control volume, and (ρ)(Q)(v2) + (ρ)(Q3)(v3) - (ρ)(Q1)(V1) for the rate of change of momentum of the fluid within the control volume. The section of the pipe shown in the picture is not exposed to atmosphere and the drawing convention used indicates that the pipe is very long.
  • #1
foo9008
678
4

Homework Statement


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/

Homework Equations

The Attempt at a Solution

 

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  • #2
foo9008 said:

Homework Statement


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 ?
Why should it? Is it exposed to the atmosphere?
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/
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
Chestermiller said:
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?

in the picture , it is exposed to atmosphere , right ? there's a hole at inlet
 
  • #4
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 don't 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
foo9008 said:
in the picture , it is exposed to atmosphere , right ? there's a hole at inlet
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
foo9008 said:
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 don't 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 ?
Answer my questions in post #2, and then you'll understand better.
 
  • #7
Chestermiller said:
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.
this is the hint showing the pipe is very long , and the section is not exposed ti atmosphere?
 
  • #8
foo9008 said:
this is the hint showing the pipe is very long , and the section is not exposed ti atmosphere?
Yes. It's a drawing convention that is always used to indicate that only part of an object is being shown.
 
  • #9
Chestermiller said:
Yes. It's a drawing convention that is always used to indicate that only part of an object is being shown.
thanks for pointing out that , i just knew it
 
  • #10
Chestermiller said:
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?
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
foo9008 said:
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 ?
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)
 
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FAQ: Why is P1 not equal to Patm in a nozzle?

1. What is the force acting in a nozzle?

The force acting in a nozzle is the force exerted by a fluid as it flows through the nozzle. This force is caused by the pressure difference between the inlet and outlet of the nozzle, which accelerates the fluid and propels it out of the nozzle.

2. How does the shape of a nozzle affect the force?

The shape of a nozzle, specifically the converging and diverging sections, can greatly affect the force acting in the nozzle. A converging nozzle increases the velocity of the fluid, resulting in a higher force, while a diverging nozzle decreases the velocity and therefore the force.

3. What factors influence the force in a nozzle?

The force in a nozzle is influenced by several factors, including the fluid's density, viscosity, and velocity, as well as the nozzle's shape and size. Additionally, the pressure difference between the inlet and outlet of the nozzle, as well as any external forces acting on the fluid, can also affect the force.

4. How is the force in a nozzle calculated?

The force in a nozzle can be calculated using the Bernoulli's equation, which takes into account the fluid's density, velocity, and pressure at different points along the nozzle. The resulting force can then be calculated using the equation F=ma, where m is the mass of the fluid and a is the acceleration caused by the pressure difference.

5. What are some real-world applications of the force acting in a nozzle?

The force acting in a nozzle has numerous applications in fields such as aerospace, automotive, and fluid dynamics. It is used in jet engines, rocket propulsion, fuel injectors, and hydraulic systems. It is also important in industries such as agriculture, where nozzles are used to spray pesticides and fertilizers, and in fire fighting, where nozzles are used to control the flow of water from hoses.

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