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Pressure and flow rates

by laminar
Tags: flow, pressure, rates
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laminar
#1
Oct7-06, 07:02 PM
P: 15
There is a pipe whose diameter narrows smoothly from 10cm to 5.0cm. The pressure at where this pipe's diameter is 5cm is 50kPa. Given water's density of 1000kg/m^3, and a flow rate of 5.0L/s, what is the pressure at the point where the pipe's diameter is 10cm?

First the continuity equation:

Q=Av

0.005L/s=[pi(0.05/2)^2]v

v=2.546m/s

Then, Bernoulli's equation:

Substitute v1 for A2v2/A1

P1=50000+(.5)(1000)(2.564^2-[(pi*.05^2)(2.564)/(pi*.1^2)])

I left the potential energy out of the equation because it is negligible givien the number of significant figures we have. Even though I did the calculation with adding potential energy it is still wrong.

So I got 79993Pa. I know it is very wrong but I cannot check it since I don't have the answers. Any help please?
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Andrew Mason
#2
Oct7-06, 08:32 PM
Sci Advisor
HW Helper
P: 6,679
Quote Quote by laminar
There is a pipe whose diameter narrows smoothly from 10cm to 5.0cm. The pressure at where this pipe's diameter is 5cm is 50kPa. Given water's density of 1000kg/m^3, and a flow rate of 5.0L/s, what is the pressure at the point where the pipe's diameter is 10cm?

First the continuity equation:

Q=Av

0.005L/s=[pi(0.05/2)^2]v

v=2.546m/s

Then, Bernoulli's equation:

Substitute v1 for A2v2/A1

P1=50000+(.5)(1000)(2.564^2-[(pi*.05^2)(2.564)/(pi*.1^2)])

I left the potential energy out of the equation because it is negligible givien the number of significant figures we have. Even though I did the calculation with adding potential energy it is still wrong.

So I got 79993Pa. I know it is very wrong but I cannot check it since I don't have the answers. Any help please?
Your approach is difficult to follow because you are plugging in numbers too soon. You have made a simple arithmetic error. I am not sure what you did. The second 2.564 should be squared but that does not explain the error.

Do the analysis first then plug in the numbers at the end.

You appear to be using Bernoulli's equation (ignoring gravitational potential):

(1) [tex]P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2[/tex]

and:

(2) [tex]v_1A_1 = \frac{dV}{dt} = v_2A_2[/tex]

so: [itex]v_2 = v_1A_1/A_2[/itex]

and from (1) then:

[tex]P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho \frac{v_1^2A_1^2}{A_2^2} = P_1 - \frac{1}{2}\rho v_1^2\left(1-\frac{A_1^2}{A_2^2}\right)[/tex]

So far, this is what you have done. Now plug in your numbers:

[tex]P_2 = 50000 + .5 * 1000 * 2.546^2(1-.25) = 50000 + 2400 = 52,400 \text{kPa}[/tex]

AM


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