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3 FB = [Wp + Wc] -WDW (Eq. 1) F = [W + W ] -W (Eq. 1) 144 4 DR 2.12 WHERE B p c DW Where: FB = buoyant force, lbs/foot of pipe Where: FB = buoyant force, lbs/foot of pipe Chapter 10 385 Marine Installations 2. Wp , the weight of pipe is: 2.0 Wp, the weight of pipe is: 2.0 Wp, the weight of pipe is: W =V P ppp W V W V pp pppp Wp = weight of pipe, lbs/foot of pipe Wp = weight of pipe, lbs/foot of pipe Wc = weight of pipe contents, lbs/foot of pipe Wc = weight of pipe contents, lbs/foot of pipe WDW = weight of the water displaced by the pipe, lbs/foot of pipe WDW = weight of the water displaced by the pipe, lbs/foot of pipe Where V = volume occupied by pipe material per foot of pipe Vp = volume occupiped by pipe material per foot of pipe Where Vp = volume occupied by pipe material per foot of pipe = density of pipe material, lbs/ cu. ft P p = d e n s i t y o f p i p e pm a t e r i a l , l b s / c u . f t p = density of pipe material, lbs/ cu. ft D = mean pipe diameter of the pipe, in m Since Since V D t Since V D t p 144ma p 144ma WHERE Where: D m = Wm e a h n e p r i e p e : d i a m e t e r o f D t h e p = i p e m , i n e a n p i p e d i a m e t e r o f t h e p i p e , i n m ta = average wall thickness, in ta = average wall thickness, in t = average wall thickness, in Where: Do = ouatside pipe diameter, in And since DR Do WHERE m t = minimum wall thickness, in WAndhseinrcee: Do = outside pipe diameter, in m D And since DR tt o= minimum wall thickness, in Where: mD =outsidepipediameter,in moa Then, by assuming thatt the average wall thickness (t ) is 6% larger than the = minimum wall thickness, in m minimum (t ), it can shown that: m t Then, by assuming that the average wall thickness (t ) is 6% larger than the a minimum (t ), it can shown that: Then, by assuming that the average wall thickness (t ) is 6% larger than the 23 o Do = outside pipe diameter, in tm = minimum wall thickness, in m1.06 D 2 a 23 miniWmum (t ), it can sho wDnRt ha1.t0:6 m (Eq. 2) Thepn, by assuming that the average wall thicknepss (ta) is 6% larger than the minimum (tm), it can be shown 144 DR 2 that: 1.06 Do W DR 1.06 (Eq. 2) (2)p 1414.06 πDRD o 2 p Wc, the weighWt of=the pipe conten(tsDiRs e−q1u.0a6l )tρo the volume occupie(dEbqy. 2th)e liquid p 144 DR p inside the pipe times the density of the liquid: Wc, the weight of the pipe contents is equal to the volume occupied by the liquid 3. W , the weight of the pipe contents is equal to the volume occupied by the liquid inside the pipe times the density of the liquid: .0 Wc, the weight of the pipe contents is equal to the volume occupied by the liquid Wc VL L inside the pipe times the density of the liquid: c inside the pipe times the density of the liquid: W V cLLL Where V = the volume occupied by the liquid Wc = VL ρL L = the density of the liquid inside the pipe, lbs/cu ft Where V = the volume occupied by the liquid L WHERE VL = the volume occupied by the liquid, cu ft/linear ft = the density of the liquid inside the pipe, lbs/cu ft V = the volume occupied by the liquid Where If the frPact=iothne doenfstithyeofitnhesliqdueidvinosilduemthepoipfe,tlhbse/cupfitpe (V ) is expressed as R and LLLI = the density of the liquid inside the pipe, lbs/cu ft L a s t h e f o I f r t mh e u f r l a a c t i f o o n r o f t t h h e e i n i n s i d s e i d v o e l u v m o e l o u f t mh e e p i i p s e ( a V s ) i f s o e l x l p o r e w s s s e : d a s R a n d a s t h e f o r m u l a f o r t h e i n s i d e v o l u m e If the fraction of the inside volume of the pipe (V ) is expressed as R and is as follows: I I as the formula for the inside volume is as follows: If the fraction2of the inside volume of the pipe (V ) is expressed as R and DI 1 I as theVfo rmula for the inside volume is as follows: I V DI 1 42 144 I2 4πD144 1 WHERE I Where D = inside diameter of the pipe, in IV= DI = inIside diameter of the pipe, in 4 144 Where D = inside diameter of the pipe, in And also, since D = D – 2t (where t is 1.06 t I o a a m, I as previously assumed) it can then be shown that: And also, since DI = Do – 2ta (whe2re ta is 1.06 tm, as previously assumed) it can Where D = inside diameter of the pipe, in I then be shown that: And also, since DI = Do – 2ta (where ta is 1.06 tm, as previously assumed) it can 2.12 then be shown that: W D 1 1 R (Eq. 3) L L o 2PDF Image | Marine Installations PE
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