FE Exam Review - FE Civil/Environmental - Water Resources - Pumps - FE Exam Tutor

[@songitadas7834]

It was a very good example to review. Thank you. W=196 kw

[@directhubfeexam]

Songita Das nicely done Songita! I got the same answer.

[@leizelclara]

Thank you! You’re a big help. I’ve been binge-watching all your videos. 🙏

[@tjmalanga4986]

super helpful videos, hard to find good Environemntal FE references. W=195.94 kW

[@directhubfeexam]

That's right! Well done 👍🏼 I'm glad you're finding the videos helpful. Although I don't focus on Environnmental, many students have used these videos to pass their exam! You will too.

[@shannoncon]

these videos have been so helpful in studying for my FE exam! I see there are two videos in the fluid dynamics playlist that are private, are they able to be made public? thanks!

[@directhubfeexam]

Hi Shannon, I am happy to hear that! Unfortunately, they can't be made public - the videos were created a while back and I noticed they infringe on certain copyrighted questions. Sorry about that.

[@shannoncon]

@@directhubfeexam understood! thanks for the reply :)

[@directhubfeexam]

@@shannoncon Anytime!

[@thomasrezzani2720]

isn't there technically atmospheric pressure of 101.3 kPa, although the terms on both sides of the eqn will cancel out???

[@bonegrubber]

Shoot... what if it's thought of like this. The system only can pump at 78%. Needs to go 50 meters. 50 x 78% = 35. Then subtract that from 50 and you get 15. Add that back and it's 65 required head. Pretty close to the answer.

[@bigiron8748]

That is a coincidence that it works out to be close

[@hsharsha453]

can i know where we have used the pump efficiency?

[@directhubfeexam]

Hi, we would use it when determining the pump brake power.

[@Kate-rq5cg]

Thank you! I got 195.9 kW for w

[@directhubfeexam]

Yes! That's right! Well done. If we wanted to convert the 195.9 kW to horse power we would use page. 2 in FE Handbook 10.1, see what you get for that! Good work.

[@Kate-rq5cg]

@@directhubfeexam I got 262.7 hp

[@directhubfeexam]

@@Kate-rq5cg that’s right! Well done 👍

[@tianyili5809]

Thank you for sharing this video! But why did you add the head loss and minor losses on the right side of the equation instead of minus them?

[@directhubfeexam]

Hi Tianyi, Thank you for watching! The Bernoulli equation (energy equation) is a balance between the energy added to the energy removed between two points or sections (1 and 2). On the left side of the equation, we put the terms that add energy between points 1 and 2. On the right side of the equation, we put the terms that remove energy between points 1 and 2. Therefore, the head loss terms are always on the right side of the equation since they remove energy (losses) from point 1 to 2.

[@supermousa9350]

Just wondering where did you get the hp loses? I can’t find it in the reference...

[@directhubfeexam]

Hi Sinan, yeah unfortunately the equation is not provided in the reference handbook. You can always get the total losses by applying the Bernoulli equation as it was done in the video. The Bernoulli equation is provided on page. 181 in FE Handbook 10.0.1.

[@edgarrosales7923]

Hi, I solved for Hp and got 51.78m and then divided by the 0.78 for the efficiency and got 66.39m how did you get the 63.85m?

[@kabako4485]

I was also getting 51.78, but you need to square your velocity. 7.77 squared will give you the correct answer.

[@edgarrosales7923]

@@kabako4485 I see now, thanks for the reply!

[@MrKota317]

Pump Power (W)=196KW

[@saeedehlotfi2576]

w= 195.95 Kw

[@directhubfeexam]

Very good!

[@mzphat08]

W= (9.81*113.85*0.244)/(.78) =349.38 kw H=200-150+63.85=113.85.... Please let me know if I did it correctly. Thanks

[@directhubfeexam]

Hello, close but I got 195.94 kW. The equation used is the pump power equation in the FE Handbook. W = Q (gamma) h / Efficiency. Therefore, W = 0.244 (9810) (63.85m) / 0.78 = 195940.92 W = 195.94 kW . I think the mistake you made was for the head. This is the required head to be lifted by the pump which we calculated as h_p = 63.85 m.

[@mzphat08]

@@directhubfeexam thanks. I understand where I made a mistake.

[@ssabitha]

@@directhubfeexam here Gamma is 1 g/cm3 right?

[@directhubfeexam]

@@ssabitha That would be the density “row”. Gamma is the unit weight so it would be 9810 N/m^3