4 replies to this topic

[Blank03]

Hi good day

I would just like to verify if my answer is correct on this one. I do apologize if some of you might think that "oh this is just very simple how could she be so unsure with the answer?"  hahaha. I am, you see, very conscious that MAYBE there might be some hidden question between the question.

So the question is this:

 

 The pressure drop in a 6-in. pipe (I.D. = 6.065 in.) of length L is 1 psi with the flow in the highly

turbulent region. If the same liquid flows at the same volumetric flow rate in a 3-in. pipe (I.D. =

3.068 in.) with length L, the pressure drop (psi) is most nearly: 

 

what i did is the fact that P1/p2=(d2^5)/(d1^5) as i equate it from the original pressure drop formula.

so 1/x=(3.068^5)/(6.065^5). that is 30. 

 

uhm am i correct?

 

thank you very much everyone!

[breizh]

https://bae.okstate....-WeisbachEq.htm

 

Hello,

Apply the Darcy Weisbach equation .

 

calculation is a piece of cake then !

Note : the result is correct  , you should simply write :  Delta P2 =Delta P1 * ( D1/D2)^5 

 

Breizh

Edited by breizh, 19 May 2018 - 09:31 PM.

[fallah]

Blank03,

 

With the conditions you described (same flow) and neglecting the change in f (friction factor) appears your result for pressure drop in 3" pipe equal to 30 psi is correct...

[Pan Nata]

Blank 03,

 

Yes, you are.  The pressure drop in the 3" line shall be around 30 psi empirically, neglecting the friction factor and any line elevation.

 

Good luck.

[breizh]

Hi P N ,

Please read what Naser offers !

 

Your statement is incorrect.

 

Breizh