Ch 11.3 Help

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Question 3

Post  periwinkle on Sun Nov 16, 2008 10:38 pm

You mark two points on a water pipe which is horizontally positioned with water flowing through it from left to right. The diameter of the pipe at the left mark is 7.937 cmand at the right mark is 1.59 cm. What is the pressure difference between left and right (see sheet 29,25)if the velocity of the fluid on the left is 7.46 cm/s ? Now tilt the pipe such that the right mark is higher than the left mark and indicate with a negative (positive) sign whether the pressure difference increases (decreases).

d1 = 0.07937 m V1 = 0.0746 m/s
d2 = 0.0159 m V2 = ?

A1V1 = A2V2

A1 = (pi/4)*(d1^2) = 0.004947
A2 = (pi/4)*(d2^2) = 1.9855e-4

A1V1 = 0.004947*0.0746 m/s = 3.6904e-4
A2V2 = 1.9855e-4*V2

DIVIDE!!! --> A1V1 / A2 = V2 = 1.8587

Then...

.5*rho*(V2^2 - V1^2)
.5*1000*(1.8587^2 - 0.0746^2) = 1724.600265

I think this should be correct.

periwinkle

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Re: Ch 11.3 Help

Post  meg on Mon Nov 17, 2008 3:33 am

OO ok i got it now..thank you!!

meg
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