PDF Publication Title:
Text from PDF Page: 027
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
PDF Search Title:
Marine Installations PEOriginal File Name Searched:
PE-Marine-Installations.pdfDIY PDF Search: Google It | Yahoo | Bing
Development of a solar powered Electric Ship The Electricship website originally started off as a project to develop a comprehensive renewable, affordable, modular electric ship... More Info
Modular Boat Hull Composite The case for a unsinkable, modular composite hybrid boat hull... More Info
MS Burgenstock Hybrid Electric Catamaran Lake Lucerne Unique shuttle servicing Lucerne to the Burgenstock Resort... More Info
Ground Power Unit GPU Powered by Lithium Ion Batteries The goal of the Ground Power Unit is to provide a readily accessible, modular, ready-to-power solution for remote power... More Info
| CONTACT TEL: 608-238-6001 Email: greg@electricship.com | RSS | AMP |