Ch 11.3 Help

please help me with number 3

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).

A=pi/4*d^2
So, A1=pi/4*(.07937)^2
A2= pi/4*(.0159)^2
V1=.0746
v2= X (what youre solving for)

A1*V1 = A2*V2
Solve for V2

Then do

P1 + pgh1 + .5pV1^2 = P2 + pgh2 + .5pV2^2
The phh's cancel so then its just P1 + .5pV1^2 = P2 + .5pV2^2
Bring everything to the other side and you have P1-P2 = .5pV1^2 + .5pV2^2
Where V2 was determined in the first step, and p = 1000 because its the density of water.
and remember its negative

get guys

what is P1 and P2 equal to?

P1-P2 ??

how would u convert velocity 9.548cm/s into SI unit?

i am keep getting #3 wrong...please help me

multiply velocity in cm/s * .01 to put it in meters per second
And once you have P1-P2 on one side and a value on the other, youre done. You don't have to solve for the individual P values or anything, youre just looking for the difference

P1-P2 IS the difference (same as DeltaP). Hence, you just solve the right side of the equation:

.5pV1^2 + .5pV2^2

And the value you get will be your answer.

It should be P1 -P2 = 0.5 * rho * (V2^2 - V1^2) not +

I still am not getting #3...HELP please

Have you converted units properly?

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.

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