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Aurora09

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  1. hey all, just a quick question on thermo properties: does enthalpy, internal energy, specific value and entropy of a material all change with composition?
  2. Hey guys, I hope you can help me out on understanding the concept of pressure drops in pipes. For the pressure drop equation P2 - P1 = pgHf (where Hf is frictional head loss), does this only apply to pipes when elevations are the same, z1 = z2? Also, if the pipe is sloped at an angle, then do I find the vertical height "h" and calculate the pressure drop as: pgh - phHf ? And finally my third question: If a pipe is vertical, does a pressure drop mean that the height in the water has risen?
  3. Thank you so much!! And your drawing looks perfect to me
  4. Hi @Jbone27 PE. Thanks for taking your time to reply! I should have been more specific in my original post as to what part of the solution confused me, but anyways I was stumped on why they subtracted 15 hrs - 12 hrs = 3 hr. I would have taken the Q value at t = 12 hr. I guess I'm not understanding the concept behind what is actually happening with the rainfall.
  5. @Jbone27 PEhey! sorry for the late response but here is the problem: A 6 hr storm rains on a 25 mi2 (65 km2) drainage watershed. Records from a stream gaging station draining the watershed are shown. (a) Construct the unit hydrograph for the 6 hr storm. (b) Find the runoff rate at t = 15 hr from a two-storm system if the first storm drops 2 in (5 cm) starting at t = 0 and the second storm drops 5 in (12 cm) starting at t = 12 hr. So the average precipitation was calculated as P = V/Area which gives you 4.37 cm. The rest of the solution for Part B was: (b) To find the flow at 15 h, add the contributions from each storm. For the 5 cm storm, the contribution is the 15 h runoff multiplied by its scaling factors; for the 12 cm storm, the contribution is the 15 h - 12 h = 3 h runoff multiplied by its scaling factors: let me know if you understand the solution
  6. Does anyone have a better explanation for the hydrograph problem in Example 20.4, Part B from the CERM? I'm not understanding their solution so if someone can please explain that'd be great!
  7. Hello, I am stuck on the following problem: Water flows in a rectangular channel with depth of 0.40 m, width of 1.0 m and velocity of 6 m/s. Calculate the alternate depth. Here is how the solution manual solved it: E = y1 + v2/2g Plugging in the values above will give you 2.235 m for the initial energy E. Then to solve for y2, they set 2.235 = y2 + Q^2/(2gA^2) in which A equals width (1.0m) times y2. Plugging in the values gives you y2 = 2.17 m. Now here is where my confusion is kicking in: I am trying to solve for y2 using the equation 19.93(a) from the CERM book but I end up getting 1.52 m. Am I not supposed to use this equation?
  8. @Acedeala Thanks for pointing me in the right direction! That's exactly what I ended up doing! I used mannings equation to solve for Qfull (substituting with the Afull value) and then used the ratio of the given flowrate (Q) to Qfull to get the value from the graph for d/D. I also just looked at the example in the CERM that you recommended. I noticed in the problem they mentioned "n varies with depth." In my problem, it doesn't mention if n varies with depth or if n is constant. The hydraulic elements graphs has two different lines for these cases for discharge (Q) and velocity (v). I wonder if on the PE exam they would actually state which case it is; otherwise my answers would be off! 😣
  9. I'm stuck on solving the problem below: A 72 inch diameter vitrified sewer pipe (n=0.014) is laid on a slope of 0.00025 and carries wastewater at a flow of 50 ft^3/s. What is the depth of flow? I think I have to somehow follow the mannings n graph line from the Hydraulics elements graph for Circular Sewers but i'm totally lost here. Can someone provide steps to this solution? thank you!
  10. Thank you - I went ahead and resolved the equation! My answer is still different than the link provided. However, I did notice a mistake in their calculation for K2. Even when I tried correcting their equations, the answer was still different than the way I set it up Anyways, do you think my work below is close enough? If anyone can check my math, that would be great. I've been stuck on this question all weekend.Thank you!
  11. @Slay the P.E. Here is a picture of my work below: (Note: Top reservoir is labeled A and the lower one is labeled B). I wasn't sure if Vb should be 0 so I ended up combing it with the same V in the Darcy head loss equation. My assumption might be wrong here. Also, should I also be adding another hf in the Bernoulli equation to account for one of the branches?
  12. I don't want to create too many topics so I hope its okay to post my additional questions here lol. but anyways, can someone assist me on pipes in parallel (please see question #7 in the following link: http://bu.edu.eg/portal/uploads/Engineering, Shoubra/Civil Engineering/3015/crs-14332/Files/FLUIDS - III.pdf) I used bernoullis eqn between the two reservoirs to find V1 in which I then solved for Q1 = 0.43 m^3/sec. Since the two branches are the same size and have the same friction factor, then Q2 = Q3 = 21.5 m^3/sec. I tried a different method using the darcy weisbach equation Hf = f (L/D) V^2 /2g. I substituted all the variables for the main pipe to solve for V and then multiplied by the Area of the main pipe; I still end up getting Q1 = 0.43 m^3/sec. I really feel like I am solving this correctly but this is driving me crazy 😛
  13. Thanks for your responses! @Audi driver, P.E. @Slay the P.E.
  14. When determining the roughness, e, of a pipe - which value do we use from the range provided in the table? For example, the values of e for cast iron range from 0.0006 - 0.003 (im looking at the table on the moody diagram from the PE handbook). Do I take the average of the highest and lowest value? Just wondering how I should go about this on the actual exam lol
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