Siphon pressure and velocity problem

In summary, Bernoulli equation states that the velocity of a fluid is proportional to the square of the pressure differential across the boundary. The point on the water's surface and point D have the same pressure, so the velocity at point D is the same as the velocity at point O. Using the continuity equation, the pressures at points A, B, and C must all be negative since the Bernoulli equation states that the pressure at a point is proportional to the velocity at that point. However, all of the pressures are incorrect, so the student is not sure what they are doing wrong.
  • #1
angelala
6
0

Homework Statement


ut.jpg


Homework Equations


Bernoulli equation.
Continuity equation

The Attempt at a Solution



Area of pipe is constant, so v_a = v_b = v_c = v_d.[/B]

Using point on water surface and point D.
Point on water surface: z = 8, v = 0, P = 0 (P_atm)
Point D: z = -2, P = 0 (P_atm) v = ?

Using Bernoulli, v_d = \sqrt {2*10*g} = 25.4 ft/s

P_a = -4018.56 psi
P_b = -20092.8 psi
P_c = -24111.36 psi

All are gauge pressures. Since they are all negative, I am not sure if I am doing something wrong?
 

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  • #2
angelala said:

Homework Statement


View attachment 237644

Homework Equations


Bernoulli equation.
Continuity equation

The Attempt at a Solution



Area of pipe is constant, so v_a = v_b = v_c = v_d.[/B]

Using point on water surface and point D.
Point on water surface: z = 8, v = 0, P = 0 (P_atm)
Point D: z = -2, P = 0 (P_atm) v = ?

Using Bernoulli, v_d = \sqrt {2*10*g} = 25.4 ft/s

P_a = -4018.56 psi
P_b = -20092.8 psi
P_c = -24111.36 psi

All are gauge pressures. Since they are all negative, I am not sure if I am doing something wrong?
These pressures are all incorrect. Please show your work.
 
  • #3
Chestermiller said:
These pressures are all incorrect. Please show your work.

Bernoulli
P/rho*g + v^2/2g + z = P/rho*g + v^2/2g + z

For point of water surface O and D,

P_atm/rho*g + 0/2g + 8 = P_atm/rho*g + v^2/2g + -2
8 = v^2/2g -2
v = sqrt ( (8+2) *2g ) = 25.4 ft/sFor point O and A
P/rho*g + v^2/2g + z = P/rho*g + v^2/2g + z
P_atm/rho*g + 0 + 8 = P/rho*g + 25.4^2/2g + 0
0 + 0 + 8 = P/rho*g + 25.4^2/2g
P/rho*g = -2
P = -4018.56

For point O and B
P/rho*g + v^2/2g + z = P/rho*g + v^2/2g + z
P_atm/rho*g + 0 + 8 = P/rho*g + 25.4^2/2g + 8
0 = P/rho*g + 25.4^2/2g
-10 = P/rho*g
P = -20092.8
 
  • #4
angelala said:
P/rho*g = -2
P = -4018.56
The problem is not that the numbers are negative but that they are so huge.
I do not understand the details of that calculation. This should be (minus) the pressure resulting from a 2ft head of water, i.e. more like -1psi.
 
  • #5
haruspex said:
The problem is not that the numbers are negative but that they are so huge.
I do not understand the details of that calculation. This should be (minus) the pressure resulting from a 2ft head of water, i.e. more like -1psi.

P/rho*g = -2

rho is 62.4 lbm/ft^3
g = 32.2 ft / s^2
g_c = 32.2

P = -2 * (62.4/32.2) *32.2 = -4 psf.
 
  • #6
angelala said:
P/rho*g = -2

rho is 62.4 lbm/ft^3
g = 32.2 ft / s^2
g_c = 32.2

P = -2 * (62.4/32.2) *32.2 = -4 psf.
I do not know where you got the /32.2 from.

psi is pounds weight per square inch, psf is pounds weight per square ft.
If ρ is 62.4 lbm/ft^3 then ρg is 62.4 pounds weight per cubic foot.
P = (-2 ft)(62.4 lbw/ft3) = -124.8 psf = -124.8/144 psi.
 
  • #7
So rho*g is gamma = 62.4 lbf / ft^3 right?
I was just converting from lbm to lbf using the gravitational constant g_c
 
  • #8
angelala said:
So rho*g is gamma = 62.4 lbf / ft^3 right?
Yes, if that's what gamma means in this context.
angelala said:
I was just converting from lbm to lbf using the gravitational constant g_c
Ok, I see.. I should have written, I don't know how you got 4 from that.
 
Last edited:
  • #9
angelala said:
P/rho*g = -2

rho is 62.4 lbm/ft^3
g = 32.2 ft / s^2
g_c = 32.2

P = -2 * (62.4/32.2) *32.2 = -4 psf.
This should be -124.8 psf = -0.87 psi
 
  • #10
In all you calculations using Bernoulli to get the pressures at A, B, and C, you should have used point D as the other point. Then the kinetic energy terms would have canceled out.
 
  • #11
Chestermiller said:
In all you calculations using Bernoulli to get the pressures at A, B, and C, you should have used point D as the other point. Then the kinetic energy terms would have canceled out.

okay, but does point O work as well?
 
  • #12
angelala said:
okay, but does point O work as well?
Sure. It’s just easier using point D.
 
  • #13

Related to Siphon pressure and velocity problem

What is siphon pressure and velocity?

Siphon pressure and velocity refers to the flow of liquid through a tube or pipe due to the difference in pressure between two points. This pressure difference creates a vacuum that pulls the liquid from a higher point to a lower point.

How does siphon pressure and velocity work?

Siphon pressure and velocity work by utilizing the principles of gravity and atmospheric pressure. When the liquid in the higher container is released, it creates a downward force that pushes the liquid out of the tube, while the atmospheric pressure at the lower point pulls the liquid down the tube.

What factors affect siphon pressure and velocity?

The main factors that affect siphon pressure and velocity are the height difference between the two points, the diameter and length of the tube, and the density and viscosity of the liquid being siphoned. The greater the height difference and the longer and narrower the tube, the faster the liquid will flow.

What are the applications of siphon pressure and velocity?

Siphon pressure and velocity have many practical applications, such as in plumbing systems, aquariums, and irrigation systems. It is also used in laboratory experiments and industrial processes for transferring liquids from one container to another.

What are the safety precautions when working with siphon pressure and velocity?

Some safety precautions to keep in mind when working with siphon pressure and velocity include ensuring that the liquid being siphoned is not harmful or toxic, using appropriate tubing and containers, and avoiding ingesting the liquid accidentally. It is also essential to monitor the siphon process and release the pressure if necessary to prevent spills or accidents.

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