7.1 Pressure Distribution

Transcription

1 7.1 Preure Ditribution The atmopheric preure i impreed on the free urface boundar. Hence, the reference preure on the free urface i taken a zero preure. The preure ditribution in free urface flow i governed b the acceleration including gravit. Thu Euler' equation in and n direction can be written a ( p + γ Z) = ρa ( p + γ Z) = ρa n m direction binormal normal plane rectifing plane tangent tream line direction n direction (Principal normal) oculating plane. The direction of the normal to direction i toward the plane Centre of curvature i conidered a poitive. v Thu the acceleration a n i given b an = r in which v i the velocit of flow along the treamline, r i the radiu of curvature of the treamline. (i) If a n i zero then (a) v =, no flow and (b) r, the treamline are traight line. ( p+ γ z) = (a) v =, then p + z = contant. γ

2 h =h Hdrotatic preure ditribution in parallel flow t free urface p γ =, hence contant = z 1 Therefore, at an point x below the free urface, the preure p x px = the ditance from the free urface a 'h' γ p = γ h x Ho h X Straight Gravit Dam P= γηο Thu, the preure varie linearl with depth from free urface and i known a hdro tatic preure ditribution. h Hdrotatic an γ g

3 (ii) In general, when the flow i in the channel with mall lope bed θ, then the treamline are nearl parallel to the bed. The vertical depth and the depth normal to boundar are nearl ame. Hence, one can aume the hdrotatic preure ditribution to be valid. (iii) In cae of large channel lope, expreion for preure can be written a Preure at a point x can upport the weight of the fluid. P x = γ xcoθ x or P x = γ coθ ' ' h = co θ B B' c Preure ditribution on 'C θ Preure ditribution in parallel flow in channel of large lope If h i the total depth normal to the boundar, then the vertical depth d can be related to h = dcoθ p h coθ d co θ γ = = Thu the hdraulic grade line doe not match with the water urface. (iv) Preure ditribution over curved boundarie. In field ituation when the flow ha to pa over a pillwa, mooth curve are provided near the cret. Similarl for energ diipation the bucket are provided. The treamline have a large curvature. Hence, preure ditribution require to be converted. The curve could be either convex or concave. Theoreticall thi flow i known a curvilinear flow. The curvature introduce appreciable acceleration component or centrifugal force normal to the direction of flow. Thu the connection

4 for the hdrotatic preure ditribution i to be introduced and thu it can be written a h = h + c h = h c for convex. β r ο concave and convex profile on pillwa

5 h c h γc B B' h = h - c h c h h = h + c B γc B' Convex urface: Centrifugal force oppoing Gravit force Example: Spillwa Cret Non Hdrotatic Preure ditribution n Concave urface: Centrifugal force in the ame direction of Gravit force Example: Flip Bucket Non Hdrotatic Preure ditribution p an γ g p an + z = r + c, for Concave = 1 for Convex ection γ g For a Concave vertical ec tion p an = 1+ γ γ an thu h= h ± c in which c = g In a curvilinear flow a v = r v c = gr If the variation of v w.r.t to r i known, then acceleration could be evaluated. The following three ituation arie in the field (i) v = contant and equal to mean velocit. (ii) v = c (free vortex) r (iii) v = rc (forced vortex) v v (iv) =, R.5 i the radiu of curvature at the mid depth. (r + d ) R.5

6 Problem: Show that for a circular pillwa bucket having a radiu of curvature R the effective preure ditribution i (a) If the velocit i contant over the depth it can be hown that the preure at an point r and θ i p v r ( ) γ (b) Effective piezometric head. = r Rc + coβ + ln g Rc h = Z + coβ + cp v 1 + ln R c 1 R c h grc co β R c V β 1 R c Flow in a bucket

7 Example: Compute the overturning moment due to preure on a retaining wall oln: (i) ume θ to be mall 3 P γ Force acting on the retaining wall, P = rea of preure triangle. = 1 γ γ = Overturning moment = P * ditance from the bae at which P i acting = γ * = γ 6 3 (iii) If θ i not negligible, = co θ ( co ) γ θ γ P= = co 4 θ 3 4 γ 4 γ over turning moment = co θ * = co θ 6 3 6

8 7. Preure correction coefficient Show that β z γ = d 1 β = z hd ( ) but h = h ± C 1 β = ( h ± C) d z 1 1 = h d + C d z z 1 β = 1+ C d z 1 1 α = hvd 1 cvd = + in which α i the preure ditribution coefficient. dv c=, d i the depth of flow in the ection. g r Solution: h=h +C Head recorded in a curve = tatic preure preure = Ma of water * depth = ρg v d h lo preure i α Qρg Thu α Qρg = ρg v d 1 1 α = v d h v d h C = α = h v d + 1 α = 1+ v d C v d C for uniform flow h v d = Q. ( ) ± correction factor.