Recent content by mstephens88

Ok yes think I've got the correct answer, i do apologise it was the derivative of pressure. thank you all for your help. MS

Does the equation in the attachment look correct?

What does tex n frac mean?

sorry Dickfore but that doesn't make n e sense to me?

The equation is for flow rate down a circular tube, in my question i am asked to rearrange the equation to find the pressue drop(dp). I have attached a picture of the equation. Thanks

No the dp = drop in pressure. Does that make a lot of difference. Thanks MS

I am trying to rearrange to work out dp. To make dp the subject.

[a]1. How do I make dp the subject? Q=[(dp/dx)/2k]^(1/n).[pi/(1/n)+3].R^(1/n)+3 ^= power of Please Help I also have attached the equation. Thanks M.Stephens