Finding Speed and Volume Flow Rate in a Constricted Pipe

[xray2golf]

Homework Statement

Water moves through a constricted pipe in steady, ideal flow. At one point, where the pressure is 2.50*10^4 Pa, the diameter is 8.0 cm. At another point .5 Meters higher, the pressure is equal to 1.50*10^4 Pa and the diameter is 4.0cm. Find the speed of flow at the lower and the upper sections and also find the volume flow rate through the pipe

Homework Equations

I don't really care about the volume flow rate at this point, am really interested in how to find the speed of the flow at one of the sections, say at the lower section where the altitude would equal zero.

P1+(Rho)(g)(y) + 1/2(rho)(V)^2 = constant

The Attempt at a Solution

Insert pressure at lower point for P1
Zero height eliminates rho g y
Insert density of water for rho
Leave V as V, since that is what we are solving for
Dont know what to insert for constant.

At this point, I am at a loss...constant is a variable? Does that give two variables and how then solve?

 

[Hunterbender]

Recall back to solving problems with energy conservation, this is similar (actually the same, bernoulli's equation is just a by-product of energy conservation...kool isn't it? *insert nerdiness*)

Well, constant usually means the atmospheric pressure (if there are any) or 0.

I tend to do these problems differently. I look at the initial and set it equal to the end. Since energy is conserved (which is what Bernoulli's Law states), then the initial energy must equal to the final energy. (and everything is analogue to what they look it in energy conservation, except m becomes rho - for a reason too)

 

[minger]

You have not written the equation correctly. The equation does not state that the initial conditions + final conditions = constant. It says that any condition = constant. Because of this, we can say that the initial condition = final condition.

So let's look at what you were given. You have pressures at both inlet and outlet, along with a delta z. The only remaining terms are velocity components. So, we have one equation, and two unknowns. We can't solve this directly, so we'll need another equation.

Luckily, the since our flow is incompressible, we can get a relationship between the two velocity terms. From continuity:
\dot{m}_{in} = \dot{m}_{out}
Or:
(\rho A V)_{in} = (\rho A V)_{out}
Since the density remains constant, we can solve for either the inlet or outlet velocity.
V_{in} = \frac{A_{out}}{A_{in}}V_{out}
Plug that into the equation, and now we have one equation with one unknown. Solve.

Hunter made a good point that should be remembered. Bernoulli's equation is derived DIRECTLY from conservation of energy. It basically says the energy at one point is equal to the energy at another point plus any any work that has come in minus any work that the fluid put out. It says that between two points, the energy in the fluid can be transferred between potential energy (z), kinetic energy (velocity), and pressure.

 

[xray2golf]

You have not written the equation correctly. The equation does not state that the initial conditions + final conditions = constant. It says that any condition = constant. Because of this, we can say that the initial condition = final condition.

So let's look at what you were given. You have pressures at both inlet and outlet, along with a delta z. The only remaining terms are velocity components. So, we have one equation, and two unknowns. We can't solve this directly, so we'll need another equation.

Luckily, the since our flow is incompressible, we can get a relationship between the two velocity terms. From continuity:
\dot{m}_{in} = \dot{m}_{out}
Or:
(\rho A V)_{in} = (\rho A V)_{out}
Since the density remains constant, we can solve for either the inlet or outlet velocity.
V_{in} = \frac{A_{out}}{A_{in}}V_{out}
Plug that into the equation, and now we have one equation with one unknown. Solve.

Hunter made a good point that should be remembered. Bernoulli's equation is derived DIRECTLY from conservation of energy. It basically says the energy at one point is equal to the energy at another point plus any any work that has come in minus any work that the fluid put out. It says that between two points, the energy in the fluid can be transferred between potential energy (z), kinetic energy (velocity), and pressure.

Ok, by solving V_{in} = \frac{A_{out}}{A_{in}}V_{out} with the actual data,

V_{in} = (12.6/50.3)V_{out} or V_{in} = .250V_[out]

I assume I plug that into the original P_{}1 + Pgy + 1/2P V^{}2 = constant or into the longer version? (sorry, I am struggling with the Latex)

I will try and work the math on paper and come back with my answer.

Thanks

 

[xray2golf]

OK,

I plugged Vin = .250Vout into the original Bernoulli's equation and ended up with V2 = 3.30.

(25000) + (.5)*(1000)*(.250V2)^2 + (1000)*(9.8)*(0) = (15000) + (.5)*(1000)*(V2)^2 + (1000)*(9.8)(.5)

V2 = 3.30

If this is correct then I will plug that answer back into the equation and solve for the exit velocity.

 

[minger]

Quickly looking at it looks correct; do make sure that your units work out, and tha your sign notation is correct.

 

[xray2golf]

Quickly looking at it looks correct; do make sure that your units work out, and tha your sign notation is correct.

I got V1 to equal .831

When I plug these numbers into the equation for flow rate through the pipe I get very close numbers at both ends...41.8 m^3/sec and 41.5 m^3/sec.

I assume the rounding could be the reason for the difference...the bottom line is the numbers are the same, as the should be throughout the pipe.

Thanks for the assistance.