Pressure In a Horizontal Pipe -

[needhelpfast]

Pressure In a Horizontal Pipe - Please Help!

Homework Statement

The pressure in a section of horizontal pipe with a diameter of 2.40 cm is 143 kPa. Water flows through the pipe at 2.90 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should the diameter of the constricted section be?

Homework Equations

I have been using
v1 = IV/A = (2.8 × 10 × 4/π × 4 × 10) =
v2 = v1^2 + 2(P1 − P2)/ρW]1/2
d2 = 2.4(v1/v2)1/2 cm = 1.68 cm

The Attempt at a Solution

See equations above. This problem is making me crazy! Please help!

 

[Xerxes1986]

I have to admit, this one kind of stumped me...but luckily the best internet resource for physics helped, hyperphysics!

I would review this page on hyper physics: http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html#bcal"

The equation you need (assuming the 2 pipes are at the same height) is:

$$P_1+\frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2$$

where \rho is the density of the fluid (which you did not give). This is essentially an energy conservation equation. The KEY point in this calculation is that you have to make several simplifying assumptions (read the hyperphysics article) including laminar flow and flow velocity. You find that the kinetic energy term is actually:

$$\frac{1}{2} \rho \frac{v_{max}^{2}}{3}$$

Where v_max is given by:

$$v_{max}=2 v_{effective} = 2 \frac{Flow rate}{Cross-section Area}$$

You should be able to solve this now.

 

[tak08810]

I'm in the same class and I have the same problem and I don't really understand your explanation.

Using the equation you provided, and setting density to 1000 kg * m^3 (the density to water), velocity should be 9.887 L/S. Then plugging in the values to V1*A1 = V2 * A2 I found what the cross section area should be for the constricted section, and then determined what the diameter should be. That did not give me the answer.

Here's my version of the question: The pressure in a section of horizontal pipe with a diameter of 2.00 cm is 147 kPa. Water flows through the pipe at 2.40 L/s. If the pressure at a certain point is to be reduced to 101 kPa by constricting a section of the pipe, what should the diameter of the constricted section be?

Submission # Try Submitted Answer
1 Incorrect. (Try 1) 3.93 cm
2 Incorrect. (Try 2) 0.005091 m
3 Incorrect. (Try 3) 16.9 cm
4 Submission not graded. Use more digits. 0.0012 cm
5 Incorrect. (Try 4) 0.00118 m
6 Incorrect. (Try 5) 0.00236 m
7 Incorrect. (Try 6) .236 cm
8 Incorrect. (Try 7) 0.985 cm
9 Incorrect. (Try 8) 2.00 cm
10 Submission not graded. Use more digits. 1.9 cm
11 Incorrect. (Try 9) 1.90 cm
12 Incorrect. (Try 10) 1.65 cm
13 Incorrect. (Try 11) 1.66 cm

 

[needhelpfast]

Yeah I was pretty sure it was 1.65 cm, but apparently that's not right! Let me know if you make any progress! I am going to keep trying

 

[tak08810]

bump!

also xerxes the link you posted doesn't work

 

[tak08810]

I figured it out. You have to convert L to m/s. First convert L to m^3, which is done simply by dividing by 1000. Then divide that by the cross section area (d/2)^2*pi to get the velocity. The rest is then done as before.

https://www.physicsforums.com/showthread.php?t=195437 - the thread that helped me out