Fluid Mechanics- inviscid fluid pressure in a pipe

In summary: I'm not sure what you mean, so your saying let Pout equal to Patm as there is no gauge pressure on exit?My understanding is that if there is a gauge pressure on exit you would need to have been told that in order to calculate PinIn the absolute pressure scale, a perfect vacuum has P = 0 psia and atmospheric pressure is P = 14.7 psia.P = 0 psig (psi gauge) is equivalent to P = 14.7 psia (psi absolute)If you are dealing with pressure differentials, it makes no difference if both pressures are absolute or gauge, but Bernoulli's equation is written assuming the pressures are measured on the absolute scale.Your answers for
  • #1
Lee Harrington
6
0

Homework Statement



Water flows through a circular pipe with a 180° horizontal elbow and exits to the atmosphere through a nozzle as shown in Fig. Q3. The diameter of the pipe is 300 mm and the diameter of the nozzle exit is 160 mm. The density of water is 999 kg/m3 . The mass flow rate of water is 140 kg/s. Incompressible, inviscid flow may be assumed. Determine:
(i) the velocity of the water at sections 1 and 2 respectively,
(ii) the gauge pressure at section 1, and
(iii) the magnitude and direction of the force Rx exerted by the water on the elbow

Homework Equations


Bernoulli's Equation?

The Attempt at a Solution


I have determined that Vin=1.98m/s and Vout=6.97m/s and i know how to determine Rx, but i have no idea how to get the gauge pressure

Do you use bernoulli's equation and let P2 equal atmospheric pressure? But does that mean you let the exit velocity equal zero?
 

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  • #2
Lee Harrington said:

Homework Statement



Water flows through a circular pipe with a 180° horizontal elbow and exits to the atmosphere through a nozzle as shown in Fig. Q3. The diameter of the pipe is 300 mm and the diameter of the nozzle exit is 160 mm. The density of water is 999 kg/m3 . The mass flow rate of water is 140 kg/s. Incompressible, inviscid flow may be assumed. Determine:
(i) the velocity of the water at sections 1 and 2 respectively,
(ii) the gauge pressure at section 1, and
(iii) the magnitude and direction of the force Rx exerted by the water on the elbow

Homework Equations


Bernoulli's Equation?

The Attempt at a Solution


I have determined that Vin=1.98m/s and Vout=6.97m/s and i know how to determine Rx, but i have no idea how to get the gauge pressure

Do you use bernoulli's equation and let P2 equal atmospheric pressure? But does that mean you let the exit velocity equal zero?
I don't see any other way to analyze this pipe than to use the Bernoulli equation. I would not assume that if the gauge pressure is zero, the exit velocity is zero, which you have found is clearly not the case. If the velocity were zero, where would all the water go which enters the pipe at location 1?

Remember, gauge pressure = absolute pressure - atmospheric pressure, if that helps.
 
  • #3
ok, so using bernoulli's equation:

pin = ρ((Vout2 - Vin2)/2) ?

And that's the gauge pressure?
 
  • #4
Also for the next part, do you continue to assume that pout=0 ?
 
  • #5
Lee Harrington said:
ok, so using bernoulli's equation:

pin = ρ((Vout2 - Vin2)/2) ?

And that's the gauge pressure?

It would be better if you showed your complete calculations.
 
  • #6
SteamKing said:
It would be better if you showed your complete calculations.
converting mass flow rate to volume flow rate and using Q=VA:
Vin=1.98m/s ; Vout=6.97m/s

pin=999((6.972 - 1.982)/2)=22308 Pa
 
  • #7
Lee Harrington said:
converting mass flow rate to volume flow rate and using Q=VA:
Vin=1.98m/s ; Vout=6.97m/s

pin=999((6.972 - 1.982)/2)=22308 Pa
Yeah, but where's your Bernoulli equation for this problem?
 
  • #8
SteamKing said:
Yeah, but where's your Bernoulli equation for this problem?
sorry, being lazy

pin/ρ + Vin2/2 + gz = pout/ρ + Vout2/2 + gz

pout=0
pin/ρ + Vin2/2 = + Vout2/2

isolating pin
pin = ρ((Vout2 - Vin2)/2)
 
  • #9
Lee Harrington said:
sorry, being lazy

pin/ρ + Vin2/2 + gz = pout/ρ + Vout2/2 + gz

pout=0
pin/ρ + Vin2/2 = + Vout2/2

isolating pin
pin = ρ((Vout2 - Vin2)/2)
Since you are not exhausting to a vacuum, Pout is clearly a gauge pressure.

In the Bernoulli equation, you should use absolute pressures in your calculations. Any results can be converted to gauge readings afterwards.
 
  • #10
SteamKing said:
Since you are not exhausting to a vacuum, Pout is clearly a gauge pressure.

In the Bernoulli equation, you should use absolute pressures in your calculations. Any results can be converted to gauge readings afterwards.

I'm not sure what you mean, so your saying let Pout equal to Patm as there is no gauge pressure on exit?

My understanding is that if there is a gauge pressure on exit you would need to have been told that in order to calculate Pin
 
  • #11
Lee Harrington said:
I'm not sure what you mean, so your saying let Pout equal to Patm as there is no gauge pressure on exit?

My understanding is that if there is a gauge pressure on exit you would need to have been told that in order to calculate Pin
In the absolute pressure scale, a perfect vacuum has P = 0 psia and atmospheric pressure is P = 14.7 psia.

P = 0 psig (psi gauge) is equivalent to P = 14.7 psia (psi absolute)

If you are dealing with pressure differentials, it makes no difference if both pressures are absolute or gauge, but Bernoulli's equation is written assuming the pressures are measured on the absolute scale.
 
  • #12
Your answers for the gauge pressures in and out of the section of pipe (22308 and 0) are correct. Now, what are your ideas on how to do part iv?

Chet
 

FAQ: Fluid Mechanics- inviscid fluid pressure in a pipe

1. What is an inviscid fluid?

An inviscid fluid is a fluid that has no viscosity, meaning it has no internal resistance to flow and will not experience any frictional forces when in motion.

2. How does pressure change in a pipe with an inviscid fluid?

In an inviscid fluid, pressure remains constant along a streamline, meaning it does not change as the fluid flows through a pipe. This is due to the lack of internal resistance to flow.

3. What is the Bernoulli's equation for inviscid fluids in a pipe?

The Bernoulli's equation for inviscid fluids in a pipe states that the total pressure at any point along a streamline is equal to the sum of the static pressure and dynamic pressure. This equation is written as P + 1/2ρv^2 = constant, where P is pressure, ρ is density, and v is the fluid's velocity.

4. How does the velocity of an inviscid fluid change in a pipe?

In an inviscid fluid, the velocity remains constant along a streamline, meaning it does not change as the fluid flows through a pipe. This is due to the lack of internal resistance to flow.

5. Can an inviscid fluid experience turbulence in a pipe?

No, an inviscid fluid cannot experience turbulence in a pipe because it has no viscosity, which is necessary for the formation of turbulent flow. Instead, the fluid will flow in smooth, streamlined paths.

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