Ratio of wetted perimeter to partial volume/area
Ratio of wetted perimeter to partial volume/area
(OP)
I am looking for an equation which can relate the ratio of the wetted perimeter to the partial area (WP:PA) of a horizontal pipe part filled with liquid.
I am familiar with the usual partial volume equations and use them regularly however the starting point with these is that you typically know 'h' (height of liquid) and 'D' (internal diameter). Whilst I can simply determine the ratio WP:PA if I use trial and error for 'h' for a given 'D' it is not an elegant solution which can go into a spreadsheet cell. I have tried and failed miserably to rearrange the formulas in GPSA to give theta as a function of the liquid area and even MathCAD won't turn the equations around for me.
I have created a lookup table to do this in excel (attached) but again it is not elegant.
Is there a simple equation (or even a complicated equation which I could at least put in a spreadsheet cell)?
I am familiar with the usual partial volume equations and use them regularly however the starting point with these is that you typically know 'h' (height of liquid) and 'D' (internal diameter). Whilst I can simply determine the ratio WP:PA if I use trial and error for 'h' for a given 'D' it is not an elegant solution which can go into a spreadsheet cell. I have tried and failed miserably to rearrange the formulas in GPSA to give theta as a function of the liquid area and even MathCAD won't turn the equations around for me.
I have created a lookup table to do this in excel (attached) but again it is not elegant.
Is there a simple equation (or even a complicated equation which I could at least put in a spreadsheet cell)?





RE: Ratio of wetted perimeter to partial volume/area
Both are functions of the radius R and the central angle θ expressed in radians:
Area, A = R2(θ - sinθ)
Perimeter, P = 2πR(θ/2π) = Rθ
A/P = R[1- (sinθ/θ)]
Besides, if needed:
the chord C = 2R sin(θ/2)
the sagitta S = R[1-cos(θ/2)]
I took these equations from old notes. I may be wrong, please check and confirm.
RE: Ratio of wetted perimeter to partial volume/area
The problem I am trying to solve is this:
knowing the volumetric flows of gas and liquid in a horizontal pipe of known diameter, assuming no slip (i.e. the flow area of the liquid is equivalent to the liquid volumetric flow fraction) determine the height of liquid in the pipe and thus the wetted surface area. This is then used in a fire scenario for blowdown.
I have found more info here http:/
Using the example excel sheet in that link I was able to construct a single (monster) formula in excel that combined 3 steps of the numerical solution and thus give my height of liquid to sufficient accuracy.
Thanks again!
RE: Ratio of wetted perimeter to partial volume/area
A B/D condition seems to me not to be sufficiently (if at all) laminar to warrant such an assumption. Even with an horizontal flow in thelaminar regime the wetted perimeter may gradually drop due to ΔP or head differences, thus it appears the result obtained from a volume flow rate would rather be an average. I'd appreciate to have your enlightening comment.
RE: Ratio of wetted perimeter to partial volume/area
I appreciate it is somewhat of a simplification - once the blowdown valves open the liquid levels will be all over the place with the turbulent gas flow. In addition any vertical legs would drain back into the horizontal legs. But I need a conservative approach and it was the most reasonable I could come up with?
But that aside, I was puzzled at how to turn the equations around to make them explicit in Theta. I now understand a numerical solution is required and it wasn't just my lack of math skills.