Introductory Fluid Mechanics L8 p3 - Example Problem - Conservation of Mass

[@jannesschmunkamp679]

Isn't the calculatet value 4.666m/s for V3 the flow in i direction and the over all outflow adds up to 5.39m/s? Because the when we neclect the angle in the calculations, we should recieve only the pure outflow perpendicular to the area. And so when we include the angle of 30 Degrees afterwards, we can obtain the overall outflow and the j direction? I hope you can help me with that?

[@jannesschmunkamp679]

@Kristopher Carmelo no one gives a damn

[@scotthanon4517]

Since your taking the dot product you need to include the cos(pi/3) term, magnitude of velocity would give you 9.33m/s not 4.67 m/s

[@scotthanon4517]

If your taking A3 to be the area of that pipe, your analysis would make sense if the control volume was drawn correctly (A3 need to be drawn normal to flow)

[@emreyuksel7586]

Hi, we are using degrees not radians.

[@pokkipok5650]

@@scotthanon4517 i also feel the area should be (A3 cos30) to get V3 in the direction parallel to the pipe. that just a roundabout way of doing what u did. so is his way correct or should it be changed?

[@edwinsosa1700]

why is the area vector opposite of the velocity for the first 2 parts but not V3

[@nehemiahstewart]

That is my question also

[@tomasprejda4165]

Hi, it is a little bit too late, but the area vector has the direction OUT OF THE CONTROL VOLUME. So if you draw the box, A1 goes left (left wall), A2 goes up (top wall), A3 goes right (right wall). If you draw a dot in the center of the volume and then draw arrows - pointing from the volume center to the selected inlet/outlet centers, you will get your answer. It does not mean the velocity vector is opposite to the area vector.

[@SpaDerola]

I have a question that has been bugging me: if you have a control volume similar to this, but the control volume is essentially a room where you want the pressure inside to increase, would you have a non zero time rate of change term for the mass conservation? If so, do you have any examples of this kind of calculation?

[@juanlasso8277]

student at the University of Pittsburgh in Pittsburgh, PA. Thank you very much

[@emreyuksel7586]

Thank you so much, it was easy to understand and beneficial :)

[@shadyxd7165]

pls explain last part v= 4.041i-2.333j, how did we get this ? im confused little bit.

[@tomasprejda4165]

Trigonometry: The angle alpha = 60 ° is between the vertical and inclined wall, however the inclined wall therefore also holds 30 ° to the A3 vector, which is actually the x-axis. Let's call this angle beta. Since the angles are below the x-axis, they are negative. Therefore: alpha = - 60 ° and beta = - 30 °. Now we can calculate the right-angled triangle. The v3 we got from the integral equation is not v3, it is the magnitude of v3. v3x = cos(-30 °)*v3_magnitude = cos(-pi/6)*v3_magnitude = (+0.86...) * (4.66...) = 4.04 m/s v3y = sin(-30 °)*v3_magnitude = cos(-pi/6)*v3_magnitude = (-0.5)*(4.66...) = - 2.33 m/s v3 = (4.04, -2.33) m/s .... or written with unit axis vectors: .... v3 = (4.04 i m/s) + (- 2.33 j m/s)

[@iceblade2674]

By using trigonometry. Remember that, when calculating the components of a vector we use the angle that the vector forms with the x axis. So, instead of using 60 we use 30 degrees. Also, the angles would be negative because they are under the x axis.

[@putrahabirun6195]

thank you so much for the clearly explanation