There was a hiccup in the stream at about 57:00 where the session dropped the connection for about a minute. I apologize for the drop, but hopefully it doesn't take too much away from the entire session or that particular problem. If you have any questions, please drop a comment below. I hope the session helps in your FE review!
I'm taking the FE exam (Civil) at the end of the month and haven't taken any Water Resources or Environmental Engineering courses. I'm so happy to find these videos you put up. Thank you!
I know that this is a lazy way to arrive to the answer for question 6, but I think the question was designed such that math is not necessary. If you want to supply pressure with a pump to a place that is Verticaly above you, your going to have to supply greater pressure than the amount needed at that higher elevation. only one answer fits that so youd be able to immediatly pick that.
Hii Mark, I just wanted to thank you for the videos and let you know that i recently passed the FE after being out of school for 2 years thanks to you and these reviews. I appreciate all the work you do. Thank you!
Hi, can anyone help me to understand this confusion. in question 6 how you can say the P1 is at atm and mark it as zero. the pipe goes into the reservoir right, so it has some depth to it and thus gives some static pressure value. I always have doubt relating to how to consider static pressure term as zero, in which cases we should say zero or neglect this term in energy equation?
The head developed by a pump as hp = -w/g, but work term w is negative for a pump, therefore the head developed by a bump hp, is always positive. I didn't see it in the manual either, just have to account for it in the energy equation.
quick question, in number 6 math you have started with P1 being 0 whereas if we had taken the P1 as "x" which is the required pressure at pump, we would get x/gamma=159' and solved "x" for psi then we could get the same answer. Hp was a very confusing bit for me.
Hey, Mark! Thanks for sharing the material and putting time and effort on this :) It's very helpful. Quick question: for problem 4, you chose the option C which is 150 m3/sec, but the formula gave you a 105 m3/s flow. Was it a mistake or there is an explanation that I am missing for that? Thank you!!
In Q9 L is the length according to the Handbook, while the problem gives you width and it doesn't say it is rectangular. Also Q is 125 cfs as given in the question, why did you use Q as 90 cfs?
Hey Mark, for question 7, I believe that residual pressure and pitot pressure should be two different things. Usually, in the field, we take a separate measurement for residual pressure. It seems like those two terms are used as the same thing in the video which is a little strange, considering the fact that they are two different concepts.
@Morgan Teng thanks so much for the clarification! I updated the problem sheet so that the given residual and pitot pressures are both listed as 45 psi. You are correct, they are indeed two separate measurements and two concepts. Thanks!
If it's only 80% purity, you need to divide to get the total amount you need. When you divide by 80%, the larger amount will be what you need. If you take 80% of the larger amount, you have the right amount.
I cannot find euler's constant in the manual. I suppose I'll just have to memorize that, but it isn't within the calculator or anywhere in the manual. I even searched by the number itself
You're correct that "P = fraction of wastewater sample volume to total combined volume." I should have written 294 mL in blue. Because it's a 300 mL bottle, the 6 mL is diluted in a total of 300 mL. You can only add 294 more mL to a 300 mL bottle. I didn't write that very clearly in the hand written portion. Thanks for the clarification!
Mark, I don't understand, why you divide 92.4mg/l by 80 present? I know that we use only 80 presents. But what type of relation you use to reach on this decision? thank you?
I looked it up and dividing by the percentage (as a decimal) is the correct procedure. I just do not understand the logic. If only 80% of the solute was available, I would think you have to multiply the result by 80%.
hello Mark. Thanks for the explanation. For the question 5, shouldn't we add another head loss? There is vertical pipe between Res. 1 and Res. 2 after pump. So it means same Circular Pipe Head Loss Equation with L = 235-210?. Thank you.
for Q6, how did you get headloss=13.82 ft?? I've tried re-entering it into my calculator several different ways and continue to get 26,224.25... Please help.
I got the same answer at first then I forgot to divide the gallon by 60 for getting it in seconds. so, its 100gallon per minutes to 100/(7.481 x 60) to get 0.222 cubic feet per second.
Do the exam solutions usually use all the given data in one way or another? i.e on question 16 there is a hint that we need a second equation since the given k value had not been used in the first equation we solved. I know some exams will throw out "red herring" info that isn't needed to solve. It seems like FE questions normally use all the given data? general thoughts feelings on this?
I have no idea why I wrote down 90 cfs. I'm sorry for the confusion! The flow in the problem is 125 cfs and that's what I used for the solution. Trying to do the problems, respond to comments and produce the videos can make things messy at times (there's a lot to think about). Thanks for asking!
Mark, On my previous fe they gave me a problem with two weirs with 2 different heights. If this so happens to be on my fe again, would you do 2 weir formulas and then the solutions to both formulas together to get the total? Thanks
A hydraulic jump by definition occurs where the flow transitions from supercritical to subcritical. If both are the same, both super or both sub, there is not jump. Does that make sense?
On problem #1, why is the intensity dependent on time of concentration? The clouds don’t care about the time of concentration of some place it’s passing by, what am I missing here? Would appreciate very much your input
As long as the storm lasts long enough to get to the time of concentration, that's the peak flow (intensity). The whole watershed will be contributing to the outlet at that time. Here's a reference: "A storm with duration equal to the time of concentration will have the full watershed contributing to the runoff at the outlet, (and thus reach peak flow rate) just before the storm ends. A storm of duration greater than the time of concentration, and the same recurrence interval, will be less intense." www.cedengineering.com/userfiles/Rational%20Method%20with%20Excel-R1.pdf
yeah, like energy equation. oh my dear lord. pressure at point 1 zero? then you just throw h_p in equation although its not in manual, but that is well known.. imgonnakillmyself
@@MarkMattsonPE hi sir good morning, is there any online FE Civil reviewer problems online ? when i add to cart reviewer is cheap $29 compare to the shipment it is coat $99.99? What you can recommend? thanks .
I'm not sure what program you're talking about, but I'd talk with classmates or colleagues to get their first hand recommendations on additional practice problems.
First of all, I really appreciate your videos, but I have one issue for watching them. Whenever you sigh, I felt you are tired and then I just wanted to turn off them. So, my wish is that I don't want to hear your sigh in your new videos. Thanks a lot!
There was a hiccup in the stream at about 57:00 where the session dropped the connection for about a minute. I apologize for the drop, but hopefully it doesn't take too much away from the entire session or that particular problem. If you have any questions, please drop a comment below. I hope the session helps in your FE review!
9
A year later coming back to say thank you for ALL the study material to help me pass my FE last week! You're greatly appreciated Mark!
For number 9 you can use the calculators "num solve" function to calculate H quickly. A very useful tool in my opinion
Thanks Kyle Long! A great tip that will help someone for sure. Getting familiar with the calculator can save so much time.
Thanks Kyle, that's a super helpful tip!
Try it. Shift solve with raise to 3/2 equation? There are alot of answer if you do that.
@@Whacker409 The calculator tells you how many answers there are with the L-R=number when you put in this equation it only has one answer
I'm taking the FE exam (Civil) at the end of the month and haven't taken any Water Resources or Environmental Engineering courses. I'm so happy to find these videos you put up. Thank you!
@1:16:15 another good option is to use the "num-solv" on the TI-36x pro calculator .
You bring wide concepts and touch essential topics. Thank you!
Hey Mark, Thank you for your amazing lectures! Super beneficial and I can't wait to take My FE on the 11th of June and pass it .
hi, did you pass?
Did you pass??
For question 8, if you have the TI-36X pro (NCEES approved), you can use the num-solv function to solve for H I believe.
i am getting totally different numbers ((
I know that this is a lazy way to arrive to the answer for question 6, but I think the question was designed such that math is not necessary. If you want to supply pressure with a pump to a place that is Verticaly above you, your going to have to supply greater pressure than the amount needed at that higher elevation. only one answer fits that so youd be able to immediatly pick that.
true
you mean the Answer to this question is wrong? I am confused why he didn't add one more head loss which is difference in elevation 235-210= 25ft
@@jamshidpishgam3630 That is picked up in the z terms. Z1=210ft & Z2=235ft. Z1 gets subtracted when solving for Hp.
Hii Mark, I just wanted to thank you for the videos and let you know that i recently passed the FE after being out of school for 2 years thanks to you and these reviews. I appreciate all the work you do. Thank you!
You're incredible as always Mark. Thanks so much
this guy could break RUclips if he did a PE Civil Review. EET would probably send Agents to his house to take it down lol.
Appreciate you Mark! What you do goes underappreciated.
Great series very happy to find this series. Taking the FE in July.
The Greatest Review ever !!!
1:52:24 should it be 12.1 Mg in the denominator ?
yeah
Hi, can anyone help me to understand this confusion. in question 6 how you can say the P1 is at atm and mark it as zero. the pipe goes into the reservoir right, so it has some depth to it and thus gives some static pressure value. I always have doubt relating to how to consider static pressure term as zero, in which cases we should say zero or neglect this term in energy equation?
48:00 where is he getting the hp from? its not in the manual.
The head developed by a pump as hp = -w/g, but work term w is negative for a pump, therefore the head developed by a bump hp, is always positive. I didn't see it in the manual either, just have to account for it in the energy equation.
You add Hp whenever there is pump involved. It should be a separate equation but it's not in FE.
quick question, in number 6 math you have started with P1 being 0 whereas if we had taken the P1 as "x" which is the required pressure at pump, we would get x/gamma=159' and solved "x" for psi then we could get the same answer. Hp was a very confusing bit for me.
Hey, Mark! Thanks for sharing the material and putting time and effort on this :) It's very helpful. Quick question: for problem 4, you chose the option C which is 150 m3/sec, but the formula gave you a 105 m3/s flow. Was it a mistake or there is an explanation that I am missing for that? Thank you!!
In Q9 L is the length according to the Handbook, while the problem gives you width and it doesn't say it is rectangular. Also Q is 125 cfs as given in the question, why did you use Q as 90 cfs?
Hey Mark, for question 7, I believe that residual pressure and pitot pressure should be two different things. Usually, in the field, we take a separate measurement for residual pressure. It seems like those two terms are used as the same thing in the video which is a little strange, considering the fact that they are two different concepts.
@Morgan Teng thanks so much for the clarification! I updated the problem sheet so that the given residual and pitot pressures are both listed as 45 psi. You are correct, they are indeed two separate measurements and two concepts. Thanks!
Hello Mark! Thank you or this video. please what textbook do you recommend for an in depth study of water resources for the FE?
Hi! I had a question on Problem 15.. Why do you divide it by 80% purity instead of multiplying it?
If it's only 80% purity, you need to divide to get the total amount you need. When you divide by 80%, the larger amount will be what you need. If you take 80% of the larger amount, you have the right amount.
Hi Mark, why can you use the energy equation in question 6? In the manual it states no pumps. Thank you!
On the first question, i use I as 5.3 and i got the Q volume is 21.2, which answer should i choose at this point if i am doing this at FE?
Me too
I think ate accurate point for 11 minutes it's 5.3
And the correct answer changes to D-22 cfs
I cannot find euler's constant in the manual. I suppose I'll just have to memorize that, but it isn't within the calculator or anywhere in the manual. I even searched by the number itself
1:45:49 Hi Mark I thought it said that P is the fraction of ww sample volume to total combined volume. Does it mean that it's 306 ml instead?
You're correct that "P = fraction of wastewater sample volume to total combined volume." I should have written 294 mL in blue. Because it's a 300 mL bottle, the 6 mL is diluted in a total of 300 mL. You can only add 294 more mL to a 300 mL bottle. I didn't write that very clearly in the hand written portion. Thanks for the clarification!
@@MarkMattsonPE oh I see. Thanks for confirming !
For questions 12 it would have been faster if you were to use D/4 for the hydraulic radius of a circular pipe
Hi Mark! I think you forget the fitting loss. Thank you.
Mark, I don't understand, why you divide 92.4mg/l by 80 present? I know that we use only 80 presents. But what type of relation you use to reach on this decision? thank you?
Same question. I do not understand why we divide by 80% instead of multiply.
I looked it up and dividing by the percentage (as a decimal) is the correct procedure. I just do not understand the logic. If only 80% of the solute was available, I would think you have to multiply the result by 80%.
hello Mark. Thanks for the explanation. For the question 5, shouldn't we add another head loss? There is vertical pipe between Res. 1 and Res. 2 after pump. So it means same Circular Pipe Head Loss Equation with L = 235-210?. Thank you.
for Q6, how did you get headloss=13.82 ft?? I've tried re-entering it into my calculator several different ways and continue to get 26,224.25... Please help.
Q=0.223,D=0.20,C=135,L=200 Try again with these values....
I had the same issue but I mistakenly read the pipe diameter as 1/2" instead of 2-1/2"
I got the same answer at first then I forgot to divide the gallon by 60 for getting it in seconds. so, its 100gallon per minutes to 100/(7.481 x 60) to get 0.222 cubic feet per second.
Thanks for the video!
Do the exam solutions usually use all the given data in one way or another? i.e on question 16 there is a hint that we need a second equation since the given k value had not been used in the first equation we solved. I know some exams will throw out "red herring" info that isn't needed to solve. It seems like FE questions normally use all the given data? general thoughts feelings on this?
Problem 1 i = 5.5
For question 9, where did Q=90 cfs come from?
I have no idea why I wrote down 90 cfs. I'm sorry for the confusion! The flow in the problem is 125 cfs and that's what I used for the solution.
Trying to do the problems, respond to comments and produce the videos can make things messy at times (there's a lot to think about). Thanks for asking!
Mark,
On my previous fe they gave me a problem with two weirs with 2 different heights. If this so happens to be on my fe again, would you do 2 weir formulas and then the solutions to both formulas together to get the total?
Thanks
I thank God for you
Hi Mark, for Q15. Why did you divide by 80% instead of multiplying by 80%?
Hi Mark, Thanks for your best video. in the 1st question each area has different rainfall intensity. why do you use same rainfall intensity for both?
I'm here for the Environmental FE review.
Mark,
My answer to question #6 is 280 psi! I believe you made a mistake somewhere! please review it and advise!
For question 5, do we get a hydraulic jump if either the upstream wasn't supercritical or downstream wasn't subcritical?
A hydraulic jump by definition occurs where the flow transitions from supercritical to subcritical. If both are the same, both super or both sub, there is not jump. Does that make sense?
On problem #1, why is the intensity dependent on time of concentration? The clouds don’t care about the time of concentration of some place it’s passing by, what am I missing here? Would appreciate very much your input
As long as the storm lasts long enough to get to the time of concentration, that's the peak flow (intensity). The whole watershed will be contributing to the outlet at that time.
Here's a reference: "A storm with duration equal to the time of concentration will have the full watershed contributing to the runoff at the outlet, (and thus reach peak flow rate) just before the storm ends. A storm of duration greater than the time of concentration, and the same recurrence interval, will be less intense."
www.cedengineering.com/userfiles/Rational%20Method%20with%20Excel-R1.pdf
why not use Hazel Williams instead of Manning's for question #4? Thanks!
Manning's is typically used for open channel flow. Hazen Williams is typically used for pressurized flow.
Thank you so much.
maaaaan, question 6 is too frustrating. just throwing so many concepts to get final solution, no cohesion, no proper connection, hated it
yeah, like energy equation. oh my dear lord. pressure at point 1 zero? then you just throw h_p in equation although its not in manual, but that is well known.. imgonnakillmyself
until when they will finished the FE review?
The plan for this semester's review is to hold new review sessions each Tuesday through the end of April.
@@MarkMattsonPE hi sir good morning, is there any online FE Civil reviewer problems online ? when i add to cart reviewer is cheap $29 compare to the shipment it is coat $99.99? What you can recommend? thanks .
I'm not sure what program you're talking about, but I'd talk with classmates or colleagues to get their first hand recommendations on additional practice problems.
My exam is Wednesday 27/2023. after two days.
how did you do brody?
Thanks!
Thank you, this exam seems quite easy tbh.
Love the jokes!
Thanks mark! I believe Q 6 could be solved in seconds without calc.
How?
send solution
hey, I believe he solved Q6 more complicated than it should. Do you have any other way you can share?
@@shatoojalal2319 you can try to ignore the headloss due to pipe friction (which took most of the time) and see that the answer is ~63feet
First of all, I really appreciate your videos, but I have one issue for watching them. Whenever you sigh, I felt you are tired and then I just wanted to turn off them. So, my wish is that I don't want to hear your sigh in your new videos. Thanks a lot!
I appreciate the feedback. I'll do my best and probably try to drink more coffee :) Thanks!
Take it like it is, sacker. The entitlement is great of the woke
Thank you very much ❤