Fluid mechanics -- flow through a pipe (Bernoulli equation)

[jinro]

Homework Statement

water flows steadily with negligible viscious effects through this pipe.

the 4-inch diameter section of the thin walled tubing will collapse if the pressure within is at 6 psi below atmospheric pressure. determine maximum h so that the tube won't collapse.

final answer for h to be in feet.

known data:
p1 = 0, V1 = 0, Z2 = 0, p2 = -6 psi, specific weight of water = 62.4 lb/ft^3, gravity = 32.2ft/s^2

Homework Equations

bernoulli equation
p1/gamma + z1 + V1^2/2*g = p2/gamma + z2 + V2^2/2*g
gamma is specific weight of water, z is height of water, V is velocity of water, g is gravity, p is pressure

to convert from lb/in^2(psi) to lb/ft^2, multiple by 144in^2/ft^2

The Attempt at a Solution

using bernoulli equation and known data, p2 = -6*144 = -864lb/ft^2

with z1 = 4ft
4ft = (-864)/(62.4lb/ft^3)+V2^2/(2*32.2ft/s^2)
-->V2 = 33.90~~ ft/s

from point 1 to 3

p1/gamma + z1 + V1^2/2*g = p3/gamma + z3 + V3^2/2*g
p3 = 0, z3 = -h, V3 = A2/A3*V2 = (D2/D3)^2*V2 = 15.0672 ft/s

4ft = -h + (15.0672ft/s)^2/(2*32.2ft/s^2)
4ft = -h + 3.525
h = -0.4748ft?

a negative value for h would not make sense!

thanks in advance

 

[gmax137]

You set P2=-6 psi. This means you are using gauge pressure (ie, pressure relative to atmospheric pressure). When using Bernoulli equation, you must use absolute pressures, not gauge pressures.

 

[haruspex]

You set P2=-6 psi. This means you are using gauge pressure (ie, pressure relative to atmospheric pressure). When using Bernoulli equation, you must use absolute pressures, not gauge pressures.

Even if the density is constant? How will adding 1atm/gamma to both sides of every occurrence of the equation change the outcome?

 

[haruspex]

a negative value for h would not make sense!

It makes physical sense, but implies the diagram is misleading, and suggests a calculation error by either you or the question setter.

 

[gmax137]

Even if the density is constant? How will adding 1atm/gamma to both sides of every occurrence of the equation change the outcome?

Hmm I will have to think about that one.

 

[Chestermiller]

You set P2=-6 psi. This means you are using gauge pressure (ie, pressure relative to atmospheric pressure). When using Bernoulli equation, you must use absolute pressures, not gauge pressures.

This is not correct. For incompressible fluid, use of gauge pressure is perfectly acceptable.

 

[gmax137]

Sorry! @haruspex and @Chestermiller are correct. I will have to do the algebra myself to see if/where the OP went wrong.

 

[gmax137]

The OP's setup has zero elevation at the restriction.

The OP solved by equating the head at the top of the tank to the head at the restriction, to find the velocity at the restriction. And then used this to find the head at the exit, allowing the height h to be calculated.

I can't quite explain why this is incorrect, except to note that the flow rate (and hence the velocity at the restriction) depends in part on the elevation difference between the top of the tank and the exit. The greater the height h, the greater the flow through the piping. Somehow the OP approach fails to take this into account; s/he determined the restriction velocity without finding h. I don't think that is possible.

I solved this by setting the exit at zero elevation, the restriction at elev h and the top of the tank at (4+h). Then equate the head at the top of the tank to the head at the exit, yielding a relationship between h and the exit velocity. Using that relationship along with the continuity (Vrestriction vs. Vexit), I was able to solve for h. I got 3.93 feet.

 

[haruspex]

s/he determined the restriction velocity without finding h

It was determined on the basis of the pressure in the central section being the critical value. You seem to be arguing that it has to calculated in a causal way, i.e. based on the exit height. I see no need for that. The equations should turn out the same.

If you would be so kind as to post your working, I will endeavour to pinpoint the reason for the different answer.

 

[gmax137]

It was determined on the basis of the pressure in the central section being the critical value. You seem to be arguing that it has to calculated in a causal way, i.e. based on the exit height. I see no need for that. The equations should turn out the same.

If you would be so kind as to post your working, I will endeavour to pinpoint the reason for the different answer.

OK, this helps me, with the question I have on the OP method. I will check over my work to see if I can spot any blunders before I pass it along.

 

[gmax137]

Thanks @haruspex ! In the course of transcribing my scribbled calcs into LaTex, I corrected an error and obtained the same result as the OP's simpler approach. For posterity () here is my updated calculation:

Bernoulli:
$$H = z + \frac {144~P} {\rho}~+~\frac {v^2} {2g}$$
where
H = head (feet)
z = elevation (feet)
P = gauge pressure, psig
$rho$ = fluid weight density, lbf/ft^3
v = fluid velocity, ft/sec^2
g = grav acceleration, 32.2 ft/sec^2

I am setting z=0 at the 6-inch diameter exit. Then the elevation of the 4-inch section is h, and the top of the fluid in the tank is at elev (4+h).

I call the top of the fluid location "1", the 4-inch section location "2", and the exit location "3". I will determine the total head at these locations as H1, H2, H3. The density $\rho$ is constant at 62.4 lb/ft^3.

At location 1, the elevation is (4+h), the pressure is atmospheric (P=0 psig), and the velocity is assumed negligible (i.e., the tank area is >> the pipe area).

$$H_1~=~z_1+ \frac {144~P_1} {\rho}~+~\frac {v_1^2} {2g}$$
$$=~(4+h)~+~\frac {144~*~0} {\rho}~+~\frac {0^2} {2g}$$
$$=~(4+h)$$

At location 2, the elevation is h, and the pressure is -6 psig
$$H_2~=~z_2+ \frac {144~P_2} {\rho}~+~\frac {v_2^2} {2g}$$
$$=~h~+~\frac {144~*~(-6)} {\rho}~+~\frac {v_2^2} {2g}$$
$$=~h~+~\frac {-864} {\rho}~+~\frac {v_2^2} {2g}$$

At location 3, the elevation is zero, and the pressure is atmospheric (P=0 psig).
$$H_3~=~z_3+ \frac {144~P_3} {\rho}~+~\frac {v_3^2} {2g}$$
$$=~0~+~\frac {144~*~0} {\rho}~+~\frac {v_3^2} {2g}$$
$$=~\frac {v_3^2} {2g}$$

Equate H1 and H3, and solve for v3
$$=~(4+h)~=~\frac {v_3^2} {2g}$$
$$v_3^2~=~(4+h)2g$$
$$v_3~=~\sqrt {(4+h)2g}$$

Equate H2 and H3
$$h~+~\frac {-864} {\rho}~+~\frac {v_2^2} {2g}=~\frac {v_3^2} {2g}$$

by continuity,
$$A_2~v_2~=~A_3~v_3$$
$$v_2~=~ \frac {A_3} {A_2} ~v_3$$
And
$$v_2^2~=~( \frac {A_3} {A_2} )^2~v_3^2$$

Then
$$h~+~\frac {-864} {\rho}~+~\frac {( \frac {A_3} {A_2} )^2~v_3^2} {2g}=~\frac {v_3^2} {2g}$$
Solve for h
$$h~=~\frac {v_3^2} {2g} ~-~\frac {v_3^2} {2g} (\frac {A_3} {A_2})^2~+~\frac {864} {\rho}$$
$$h~=~\frac {v_3^2} {2g} ~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}$$

Insert v3^2 from above:
$$h~=~(4~+h) ~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}$$
$$h~=~4~(1~-~(\frac {A_3} {A_2})^2)~+~h~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}$$
$$h~=~\frac {4~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}} {1~-~(1~-~(\frac {A_3} {A_2})^2)}$$

$$(1~-~(\frac {A_3} {A_2})^2)~=~(1~-~(\frac {6^2} {4^2})^2)~=~-4.0625$$
so
$$h~=~\frac {4~(-4.0625)~+~\frac {864} {62.4}} {1~-~(-4.0625)}$$
$$h~=~-0.4748$$

Showing that "my" approach yields the same result as the OP.

 

[haruspex]

Thanks @haruspex ! In the course of transcribing my scribbled calcs into LaTex, I corrected an error and obtained the same result as the OP's simpler approach. For posterity () here is my updated calculation:

Bernoulli:
$$H = z + \frac {144~P} {\rho}~+~\frac {v^2} {2g}$$
where
H = head (feet)
z = elevation (feet)
P = gauge pressure, psig
$rho$ = fluid weight density, lbf/ft^3
v = fluid velocity, ft/sec^2
g = grav acceleration, 32.2 ft/sec^2

I am setting z=0 at the 6-inch diameter exit. Then the elevation of the 4-inch section is h, and the top of the fluid in the tank is at elev (4+h).

I call the top of the fluid location "1", the 4-inch section location "2", and the exit location "3". I will determine the total head at these locations as H1, H2, H3. The density $\rho$ is constant at 62.4 lb/ft^3.

At location 1, the elevation is (4+h), the pressure is atmospheric (P=0 psig), and the velocity is assumed negligible (i.e., the tank area is >> the pipe area).

$$H_1~=~z_1+ \frac {144~P_1} {\rho}~+~\frac {v_1^2} {2g}$$
$$=~(4+h)~+~\frac {144~*~0} {\rho}~+~\frac {0^2} {2g}$$
$$=~(4+h)$$

At location 2, the elevation is h, and the pressure is -6 psig
$$H_2~=~z_2+ \frac {144~P_2} {\rho}~+~\frac {v_2^2} {2g}$$
$$=~h~+~\frac {144~*~(-6)} {\rho}~+~\frac {v_2^2} {2g}$$
$$=~h~+~\frac {-864} {\rho}~+~\frac {v_2^2} {2g}$$

At location 3, the elevation is zero, and the pressure is atmospheric (P=0 psig).
$$H_3~=~z_3+ \frac {144~P_3} {\rho}~+~\frac {v_3^2} {2g}$$
$$=~0~+~\frac {144~*~0} {\rho}~+~\frac {v_3^2} {2g}$$
$$=~\frac {v_3^2} {2g}$$

Equate H1 and H3, and solve for v3
$$=~(4+h)~=~\frac {v_3^2} {2g}$$
$$v_3^2~=~(4+h)2g$$
$$v_3~=~\sqrt {(4+h)2g}$$

Equate H2 and H3
$$h~+~\frac {-864} {\rho}~+~\frac {v_2^2} {2g}=~\frac {v_3^2} {2g}$$

by continuity,
$$A_2~v_2~=~A_3~v_3$$
$$v_2~=~ \frac {A_3} {A_2} ~v_3$$
And
$$v_2^2~=~( \frac {A_3} {A_2} )^2~v_3^2$$

Then
$$h~+~\frac {-864} {\rho}~+~\frac {( \frac {A_3} {A_2} )^2~v_3^2} {2g}=~\frac {v_3^2} {2g}$$
Solve for h
$$h~=~\frac {v_3^2} {2g} ~-~\frac {v_3^2} {2g} (\frac {A_3} {A_2})^2~+~\frac {864} {\rho}$$
$$h~=~\frac {v_3^2} {2g} ~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}$$

Insert v3^2 from above:
$$h~=~(4~+h) ~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}$$
$$h~=~4~(1~-~(\frac {A_3} {A_2})^2)~+~h~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}$$
$$h~=~\frac {4~(1~-~(\frac {A_3} {A_2})^2)~+~\frac {864} {\rho}} {1~-~(1~-~(\frac {A_3} {A_2})^2)}$$

$$(1~-~(\frac {A_3} {A_2})^2)~=~(1~-~(\frac {6^2} {4^2})^2)~=~-4.0625$$
so
$$h~=~\frac {4~(-4.0625)~+~\frac {864} {62.4}} {1~-~(-4.0625)}$$
$$h~=~-0.4748$$

Showing that "my" approach yields the same result as the OP.

Thanks for sorting that out.

 

[gmax137]

Thanks for sorting that out.

So, unless I've blundered again, the negative value of h means the restriction is below the exit, with the top of the tank less than 4 feet above the exit. The diagram is misleading at best.

For me, the take-away from the exercise:

1. the pressure terms can be taken from an arbitrary reference value
2. the elevation terms can be taken from an arbitrary reference value

 

[haruspex]

the negative value of h means the restriction is below the exit, with the top of the tank less than 4 feet above the exit. The diagram is misleading at best.

Quite so.

the pressure terms can be taken from an arbitrary reference value

Yes, as long as the density is constant.