Please refer to the figure below as a part of the pipe flow...

Name: Please refer to the figure below as a part of the pipe flow in 2D. Derive the algebraic equation for mass conservation for cell 5. Use a similar procedure for cell 1 that we did in class. Since density is a constant and will cancel out of all terms, you can consider only the volume flow rate in your derivation.
Uploaded: 2022-03-04T11:10:33-08:00
Duration: 5 min 29 s
Channel: Panna S
Description: Please refer to the figure below as a part of the pipe flow in 2D. Derive the algebraic equation for mass conservation for cell 5. Use a similar procedure for cell 1 that we did in class. Since density is a constant and will cancel out of all terms, you can consider only the volume flow rate in your derivation.

Please refer to the figure below as a part of the pipe flow in 2D. Derive the algebraic equation for mass conservation for cell 5. Use a similar procedure for cell 1 that we did in class. Since density is a constant and will cancel out of all terms, you can consider only the volume flow rate in your derivation.

Transcript

00:01 Hello everyone.

00:02 In this question, the figure is given.

00:06 So, you can see in this figure, this wider portion is having the area of cross -section 40 centimeter square and this is more narrower is having area of cross -section as 10 centimeter square.

00:20 So i can write it as a1 is equal to 40 centimeter square and this is equals to 0 .004 meter square.

00:29 Similarly, i can write a2 that is 10 centimeter square and this is equals to 0 .001 meter square.

00:38 Also the rate of flow of given is dv by d t is given 6 into 10 raised to the power minus 3 meter cube per second.

00:49 Now we will use the equation of the rate of flow that is dv by d t this is equals to a 1 v1 and this is is equal to a 2 v2.

00:59 Now we will just substitute the value and will calculate the values of v1 and v2.

01:04 So let us first calculate the value of v1.

01:07 That means a1 v1, this is equal to dv by d t.

01:12 That means we can say that 0 .004 v1, this is equal to 6 into 10 dash to the power minus 3.

01:22 That means our v1 is 3 by 2 meter per second or 1 .5 meter per second.

01:33 Similarly, a2 v2 is also equals to dv by dt.

01:39 We can say that v2 is equal to 6 into 10 raised to the power minus 3 upon a2 is 0 .001.

01:50 That means it is equals to 6 meter per second.

01:55 So the velocities are 1 .5 meter per second and 6 meter per second.

02:00 This v1 is from the wider portion and this v2 is from the narrow portion.

02:09 Now in the b part we have to calculate the pressure difference.

02:16 So for that we will write the equation p1 plus half row v1...

Question

Please give Ace some feedback