Siphon pressure and velocity problem

[angelala]

Homework Statement

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?

 

[Chestermiller]

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.

 

[angelala]

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

 

[haruspex]

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.

 

[angelala]

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.

 

[haruspex]

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.

 

[angelala]

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

 

[haruspex]

So rho*g is gamma = 62.4 lbf / ft^3 right?

Yes, if that's what gamma means in this context.

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.

 

[Chestermiller]

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

 

[Chestermiller]

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.

 

[angelala]

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?

 

[Chestermiller]

okay, but does point O work as well?

Sure. It’s just easier using point D.

 

[angelala]

Sure. It’s just easier using point D.

Thanks @Chestermiller and @haruspex !