Recent content by Struggling

yes it does help. I sort of figured out how to do it, made a spread sheet but then realized i had estimated my demand wrong. We have estimated each household uses 42.5 L/hr during the peak hour(8-9am). now the new problem i have is working out the flow rate in m^3/s because 42.5 L/hr is...

Its from the AS 3500 Plumbing & Drainage Standards, the reason I am sort of worried about that formula is it says its for rapid sizing of pipes for residential areas. i also just realized i can't use that formula because H = h*100/L*1.5 and h is head loss which i can't work out without...

Hi, Im currently doing an assignment where i have to design a water reticulating system for a small suburb. I have planned out all my distribution mains and my reticulation mains and services I also have the water demand(Q) for the houses using the Australian Standards for plumbing and...

Argghhhhh i don't know how to do it. my heads aching lol. its very late maybe i should sleep on it. any other info or help on how to convert it would be much appreciated. thanks! thanks Doc Al, youve been a lot of help :smile:

the only data i have is 19.23mm(0.1923 m) diameter and the flow rate which is lit/min. to convert this into m^3/sec is totally baffaling me. We got a rotameter reading in mm, looked at a graph which gave us the flow rate in lit/min. why does this have to be changed? unless it doesn't i had got...

ah ok so its... A = pi/4*19.23^2 = 290.435 ? so for a rate of flow of 0.3 lit/min the answer would be 0.3 = 290.435*v 0.3/290.435 = v v = 0.00103m/s

or is the area supposed to be A = pi x diameter x diameter. ?

we did some tests using rotameters. The rotameters would return results such as 50mm on rotameter 10 (i forget what the sizes mean) we then had a graph for the size rotameter we used and we would look at the graph and find 50mm to have a flow rate of 0.3 (lit/min). so eg for a size 18 rotameter...

do you mind if i ask, if my flow rate is 0.3 lit/min, and my cross sectional area is A = pi/4(19.23) = 15.103 mm^2 V = av v = a/V v = 15.103/0.3 = 50.34? or was it ment to be v = V/a? my basic math is shocking :blushing: *** never mind I am 90% sure its v = V/a so the answer would be 0.02m/s...

i know those formulas but i struggle to see where my data fits into the equation. i think i must be lost with the meaning of the rotameter and flow rates given. *** edit thought about it longer sorry hang on is it saying Flow rate = cross sectional area x velocity?

Hi, i have a rotameter reading, flow rate (lit/min), Head Loss(mm) and diameter of a pipe. from this i need to find out the Velocity of the water flowing through the pipe. iam confused as to what formula to use to find this out, i have searched through 2 textbooks and cannot find anything...

hi can anyone explain to me how to get the H value for runge kutta second method? I've searched everywhere online but i just don't understand it. if found h = tn - to/n?? i know what value of "to" is but no clue what values to put in for n and tn? thanks

hi all just doing some questions and come along these two which have me stumped. There multiple choice questions but id rather know the working out to them as i have exams in a couple of weeks. thanks. 1. The mass of a hot-air balloon and its fully loaded basket is 500kg excluding the air in...

Hi for school i need to design a balsa wood beam. Winner is judged be strength to weight ratio. i have searched around and found all the stress and strain values for balsa wood at different densitys etc... now my dilemma is that i have to use formulas to design this beam before it is...

oh alright well wat i ended up doing this morning was: Izz = (1/12)6.4x31.7^3 = 16989.34 Iaa = (1/12)6.4x38.1^3 = 29496.72 Iaa = Izz +Ay^2 therfore (Square Root)Iaa-Izz/A = Y (square root) 29496.72 - 16989.34/ 446.72 = Y Y = 5.29mm and i just tried the formula you gave me and i...