Recent content by amjid1709

To get the velocities in the pipes I did as follows V1 = 300/250 squared which = 1.44 if you square 1.44 you get V2 = 2.0376 and then to calculate the losess this is the method that was followed: Entry pipe = 0.5 * 2.0376 = 1.0368 Upper pipe 0.028 * 75/ 0.025 * 2.0376 = 17.4182...

I am able to do that at the end when all head losses have been calculated the problem I'm having is where there is the sudden expansion in the pipe how do I calculate the losses at this point. The formula given is V1 - V2 divde by 2 times gravity which is 9.81 x 2, I don't know how to do this...

I still don't understand I have tried your workings but still can't figure it out can you elaborate please.

Two reservoirs have a difference of 25 metres in water levels, a pipe line which of 125m long overall joins them. The upper 75m of the pipeline is 250mm diameter and the lower 50m is 300mm diameter. Local losses at entry and exit are of the form kv2/2g and K has the following values: at...

Hi, I have a question, it is related to flows between two resorvoirs where I need to calculate the discharge between two pipes, with different pipe diameters. The problem I have come across is where there is an expansion in the pipe size from 250mm diameter to 300mm, the formula states...

cheers for that, just what the doctor ordered. Thanks

Thanks for that Mathmate, I will try and do it today a family emergency has risen and I will try and do it tonight. Please keep in contact once I have done it I will get back to you. Many thanks, could'nt have done it without your assistance.

This is what I'm getting over and over again. (832.3072 + 243.84) * (3.2 -11.85)2 + (692.49 +202.88) * 108.16 1076.1472 * 74.8225 + 895.37 *108.16 = 8805887.38 this isn't right and I have not got a clue where I'm going wrong.

Am I correct in saying Yb is the distance of centroid of rectangle B from bottom flange which I have calculated it to be 15.85? and also which you stated Ib=second moment of inertia of rectangle (31.7*6.4) B is this 31.7 *6.4 Xy3 /12 which equals to 692.497

Thank you very much for that, I'm starting to learn it now may be you should consider being a lecturer, you have done more explaining than my own lecturers. I will have a go at that tonight, by the way the link you pasted does not work it says its a unkown file. Would it be possible to...

I'm doing a project on flow rates in a pipe system using the hazen williams formula, does anyone know of a spread sheet or computer software that does this. I have been told that it is possible doing it in Microsoft Excel, does any know?

Second Moment of Area using parallel axis theorem I = Bd3 12 Rectangle A Y = 6.4 + 38.1 -11.85 2 Y= 13.6 Bd3 + (13.6)2 12 = 1017.2mm4 Rectangle B Y = 6.4 + 31.7 -11.85 2 Y= 10.85 Bd3 + (10.85)2 12 = 810.2mm4 Total...

Thanks for the info about the books I have checked the uni library website and they are available to borrow so will do that monday. Did you mean second moment of area?

Thanks mathmate, your guidance has enabled me to crack the question I now have the answer to 11.85 and do undersatnd it now. Just another question how do you calculate the strain at the neutral axis, do i need to use another formula for this?

any body?