# Ch 11.3 Help

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## Ch 11.3 Help

**Question 1**The flow rate of a fluid through a pipe with a cross sectional area of 18.19 cm

^{2}is 18.32 liters per second. What is the speed of the fluid (see sheet 25). Indicate with a positive (negative) sign whether the expression you use is valid for a fluid that cannot be (can easily be) compressed.

__How To Solve__

Answer is

**Question 2**Water flows through a pipe vertically upward. The pipe has the same cross sectional area everywhere. You mark two points on the pipe, one 1.129 m higher than the other. What is the pressure difference between bottom and top points inside the pipe (see sheet 29; hint: what does the equation on sheet 25 tell you about the velocities?)? Indicate with a positive (negative) sign whether your result depends (does not depend) on the velocity of the water flow.

__How To Solve__

Answer is

**Question 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).

__How To Solve__

Answer is

**Question 4**A fluid with a given viscosity flows through a cylindrical vessel. You reduce the vessel radius by a factor 1.456 (divide by this factor). If the pressure difference is the same as before what is the ratio of the new flow rate over the old flow rate (see sheet 36,38) ? Indicate with a negative (positive) sign whether the pressure difference has to be increased (decreased) in order to maintain the same flow rate as before the radius reduction.

__How To Solve__

Answer is

*[Note: Currently being worked on]*

**Guest01**- Posts : 133

Join date : 2008-09-19

## Question 1

Question 1

The flow rate of a fluid through a pipe with a cross sectional area of 18.19 cm2 is 18.32 liters per second. What is the speed of the fluid (see sheet 25). Indicate with a positive (negative) sign whether the expression you use is valid for a fluid that cannot be (can easily be) compressed.

How To Solve

Answer is

First:

Convert your values into SI units so 18.19cm^2=.001819m^2

1 liter per second=.001 cubic meters per second

so therefore 18.32 liters per second=.01832 cubic meters per second.

Then we divide our flow rate (.01832) by our cross sectional area to get velocity.

Answer is positive.

The flow rate of a fluid through a pipe with a cross sectional area of 18.19 cm2 is 18.32 liters per second. What is the speed of the fluid (see sheet 25). Indicate with a positive (negative) sign whether the expression you use is valid for a fluid that cannot be (can easily be) compressed.

How To Solve

Answer is

First:

Convert your values into SI units so 18.19cm^2=.001819m^2

1 liter per second=.001 cubic meters per second

so therefore 18.32 liters per second=.01832 cubic meters per second.

Then we divide our flow rate (.01832) by our cross sectional area to get velocity.

Answer is positive.

**Guest24**- Guest

## Question 4

Question 4

A fluid with a given viscosity flows through a cylindrical vessel. You reduce the vessel radius by a factor 1.456 (divide by this factor). If the pressure difference is the same as before what is the ratio of the new flow rate over the old flow rate (see sheet 36,3Cool ? Indicate with a negative (positive) sign whether the pressure difference has to be increased (decreased) in order to maintain the same flow rate as before the radius reduction.

How To Solve

Answer is

So we can set up a proportion here.

Delta p1/Delta p2=r2^4/r1^4

So numerically we divide our given number above (1.456) into 1. -->1/1.456=.6868131868

We take this number and raise this to the fourth power so .6868131868^4=.2225

This is our answer

Answer is negative

A fluid with a given viscosity flows through a cylindrical vessel. You reduce the vessel radius by a factor 1.456 (divide by this factor). If the pressure difference is the same as before what is the ratio of the new flow rate over the old flow rate (see sheet 36,3Cool ? Indicate with a negative (positive) sign whether the pressure difference has to be increased (decreased) in order to maintain the same flow rate as before the radius reduction.

How To Solve

Answer is

So we can set up a proportion here.

Delta p1/Delta p2=r2^4/r1^4

So numerically we divide our given number above (1.456) into 1. -->1/1.456=.6868131868

We take this number and raise this to the fourth power so .6868131868^4=.2225

This is our answer

Answer is negative

**Guest24**- Guest

## question 2 and 3

I have no idea where to even start for number 2 and 3. can some give some hints? thanks!

**student**- Guest

## Re: Ch 11.3 Help

can anyone pls post what they did for question 2 and three...the equation for 2 above is a little bit mis-understanding. thanks

**11.3 que**- Guest

## fsdfa

Question 2

Water flows through a pipe vertically upward. The pipe has the same cross sectional area everywhere. You mark two points on the pipe, one 1.129 m higher than the other. What is the pressure difference between bottom and top points inside the pipe (see sheet 29; hint: what does the equation on sheet 25 tell you about the velocities?)? Indicate with a positive (negative) sign whether your result depends (does not depend) on the velocity of the water flow.

the question is asking for the different between bottom and top points which is delta P

delta P=density of water * g * factor factor=1.129

density of water =1000

g=9.81

delta P=1000*9.81*1.129, that is your answer negative

Water flows through a pipe vertically upward. The pipe has the same cross sectional area everywhere. You mark two points on the pipe, one 1.129 m higher than the other. What is the pressure difference between bottom and top points inside the pipe (see sheet 29; hint: what does the equation on sheet 25 tell you about the velocities?)? Indicate with a positive (negative) sign whether your result depends (does not depend) on the velocity of the water flow.

the question is asking for the different between bottom and top points which is delta P

delta P=density of water * g * factor factor=1.129

density of water =1000

g=9.81

delta P=1000*9.81*1.129, that is your answer negative

**chen**- Guest

## Question 3 Possible Solution

I haven't entered it into my maple ta yet, but i think these are the steps:

where, A= pi/4*d^2

solve for V2

solve for (P1-P2).... the pgh's obv cancel out because there is no height difference.

oh, and make sure your units are all SI

Like I said, I haven't checked my answer on maple ta, but i think my logic is correct.

**1. A1V1=A2V2**where, A= pi/4*d^2

solve for V2

**2. P1 + pgh1 + .5pV1^2 = P2 + pgh2 + .5pV2^2**solve for (P1-P2).... the pgh's obv cancel out because there is no height difference.

oh, and make sure your units are all SI

Like I said, I haven't checked my answer on maple ta, but i think my logic is correct.

**S**- Guest

## Re: Ch 11.3 Help

The possible solution for question 3 above comes out correct. Answer is negative.

**krolik**- Guest

## Re: Ch 11.3 Help

i'm still having trouble with question 3 I get V2 then with the second part i'm having trouble getting the actual answer, I get this huge number which obviously isn't correct. are we just subtracting p2-p1?? thanks for anyone's help!

**????**- Guest

## question 3 with numbers

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

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

**me**- Guest

## re: student

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

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

**k**- Guest

## Re: Ch 11.3 Help

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

And the value you get will be your answer.

**.5pV1^2 + .5pV2^2**And the value you get will be your answer.

**DJ**- Guest

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