
#1
Oct706, 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? 



#2
Oct706, 08:32 PM

Sci Advisor
HW Helper
P: 6,561

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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