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]
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 : 20080919
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 misunderstanding. thanks
11.3 que Guest
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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:
1. A1V1=A2V2
where, A= pi/4*d^2
solve for V2
2. P1 + pgh1 + .5pV1^2 = P2 + pgh2 + .5pV2^2
solve for (P1P2).... 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 (P1P2).... 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 p2p1?? 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 P1P2 = .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 P1P2 = .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 P1P2 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 P1P2 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
P1P2 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.
.5pV1^2 + .5pV2^2
And the value you get will be your answer.
DJ Guest
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