Відео

For this problem, the focus is on calculating the waste sludge flow rate (Q_w) associated with the secondary clarifier and aeration tank processes. The sludge from the primary clarifier is typically removed separately as "primary sludge" and is not mixed with the activated sludge in the secondary treatment system. If the problem required finding the total sludge flow rate from both the primary and secondary clarifiers, it would explicitly state something like: "Determine the total sludge flow rate (MGD)..."

Thank you! Keep working through lots of practice problems! Definitely focus on mastering the easier topics first, as it’s impossible to be an expert in every subject. In my opinion, you can completely bomb one topic and still pass! You’ve got this!

I'm happy to know this helped! Thank you for practicing with me!

Question. What type of retaining wall? Is the retaining wall allowed to move away from the soil? If so, the active case is ok but this assumes the wall moves a tiny bit away from the soil.

@@jaysonbolanos8063 I appreciate it man!!! This made my day. For real, it means so much to me to know I’m helping you master these analysis skills for your FE exam. Thank you so much for the best wishes and for this comment 🙏🏼

Hi Fernando! Good question! First of all, for this FE type question (basic fundamental question) since the does not provide information about changes in the velocity of the gas or the height of the system. Without these values, we cannot say the KE or PE changes are substantial enough. Also in a tight controlled unit like a piston-cylinder assembly, the motion of the gas and the piston is usually very small and very controlled. The gas particles inside the cylinder do not experience big changes in velocity or height therefore we can often assume changes in kinetic and potential energy negligible.

My pleasure!

The Hazen Williams equation applies in this case since the table for the Hazen Williams coefficients is given. I hope this makes sense.

@@silindilenkosi1355 you’re welcome !

is that because the direction of the headless? in the reference book "Multipath Pipeline Problems" section, the head loss in the parallel pipe with same flow direction pipes are equal. but in this question, they are reverse

Close! It's still true that the headloss of parallel pipes is the same. This concept should be remember when dealing with parallel pipes that start at the same junctoin (node) and end at the same junction (node). But why is the headloss equal to zero in a looped system? It all goes back to CONSERVATION OF ENERGY. In each closed water distribution loop, there can be no net gain or loss of energy (head). Imagine being a water particle going around the loop from the start to the end. You will end up where you started therefore the energy level must be the same. If we have a head loss in one pipe in the loop it must be balanced by energy head gains or losses in other pipes. This makes the total head loss around the loop equal to zero. For the water distriubtion system to be in equilibrium, each pipe path around the loop must have the same head loss, so there is no “extra” energy at any point. What we do in real life is adjust the flow distriubtion to ensure the head loss is equal to zero in a looped system. By applying the condition that the sum of head losses around each loop is zero, the iterative Hardy Cross method is applied to adjust the flow distribution in the network until all loops satisfy this balance. KEY WORD: ADJUST FLOW DISTRIUBTION. The flowrate into each junction will be different. But in the end, the headloss in one pipe is balance by the headloss in another.

Good one!

That's right! Nice work!!!

I don't know much about PPI especially for the Mechanical FE exam. They've been in this FE prep space for a long time now. Just make sure they have updated resources that cover the style of the latest 2024 FE exam which includes a balance of alternative type questions and the classic multiple choice questions.

Older videos I have are good to cover for the basics. But newer videos are more similar to real FE problems you may see.

Thank you so much!

Hey! It's normal to fail this exam. All my students do before finally hitting the passing mark. Reflect on and learn from each failed attempt. Come up with a new game plan that will switch up your previous study techniques. Feel free to email me if you need help!

check it in gd&t section which is right below the limit and fit

Haha, I loved reading this!!! I’m so glad this helped. Thank you 😊

We would need to convert L to cubic meters then atm to pascals separately then we would get joules. We are still using page.3 for convert these. 1 L = 0.001 cubic meters (m^3) 1 atm = 101328 pascals = 101326 N/m^2 Multiply the two: 1L - atm = 0.001 m^3 - 101326 N/m^2 = 101.328 N - m = 101.328 Joules

@@farooquenadeem5300 Not this long. But definitely know the concept here.

More to come!

if im not mistaken, the hf is the enthalpy of the fluid when it is exactly at a saturated liquid state (meaning the fluid is 100% liquid, 0% gas). At state 4, it is not the same enthalpy because the fluid at state 4 is not at a saturated liquid state. It is a mixture of vapor and gas (state 4 is inside the liquid-vapor dome). So you need to find liquid-vapor ratio of state 4 by finding the quality (x=0.873 meaning the fluid at state 4 is 87.3% gas). Then you can estimate the enthalpy between the hf (h at saturated liquid state) and hg (h at superheated vapor state) using the enthalpy-quality formula.

Keep cooking! Remember to always take breaks. You got this!

Here it is: www.directhub.net/wp-content/uploads/2022/08/ideal-gas-mixture-2-solution-1661186754.1069.png

@@ole2445 Didn’t say this in video, but “gamma” = unit weight of water = 9810 already accounts for the gravity value.

Show me your work . Or explain

The “free bodies” in this case are the beams. Therefore we would strictly focus on drawing the free body diagrams for the beams while neglecting the reactions at A since they do not impact either beam.

Easy.....lol

Thank you!