Fluid Properties Examples
|
|
|
- Trevor Walker
- 10 years ago
- Views:
Transcription
1 CIVE400: Flui Mecanic Example: nwer Flui Propertie Example. Explain wy te vicoity of a liqui ecreae wile tat of a ga increae wit a temperature rie. Te following i a table of meaurement for a flui at contant temperature. Determine te ynamic vicoity of te flui. u/y (ra - ) (N m - ) CIVE400: Flui Mecanic Example: nwer. Te enity of an oil i 850 kg/m. Fin it relative enity an Kinematic vicoity if te ynamic vicoity i kg/m. oil = 850 kg/m water = 000 kg/m oil = 850 / 000 = 0.85 Dynamic vicoity = = 50 - kg/m Uing Newton' law of vicoity u y were i te vicoity. So vicoity i te graient of a grap of ear tre againt vellocity graient of te above ata, or u y Plot te ata a a grap: Sear tre u/y Calculate te graient for eac ection of te line u/y ( - ) (N m - ) Graient Kinematic vicoity = = / m. Te velocity itribution of a vicou liqui (ynamic vicoity = 0.9 N/m ) flowing over a fixe plate i given by u = 0.68y - y (u i velocity in m/ an y i te itance from te plate in m). Wat are te ear tree at te plate urface an at y=0.4m? t te plate face y = 0m, Calculate te ear tre at te plate face t y = 0.4m, u 0.68y y u 0.68 y y u 0.68 y u N / m y u y te velocity graient i zero at y=0.4 ten te ear tre mut alo be zero. Tu te mean graient = vicoity = 4.98 N / m Example: nwer CIVE400: Flui Mecanic Example: nwer CIVE400: Flui Mecanic
2 CIVE400: Flui Mecanic Example: nwer m of oil weig N. Fin it ma enity, an relative enity,. Weigt = mg Ma m = / 9.8 = kg Ma enity = Ma / volume = / 5.6 = 85 kg/m 85 Relative enity water 5. From table of flui propertie te vicoity of water i given a poie. Wat i ti value in N/m an Pa unit? = poie poie = 0. Pa = 0. N/m = Pa = N/m 6. In a flui te velocity meaure at a itance of 75mm from te bounary i.5m/. Te flui a abolute vicoity Pa an relative enity 0.9. Wat i te velocity graient an ear tre at te bounary auming a linear velocity itribution. = Pa = 0.9 u y u y Pa CIVE400: Flui Mecanic Example: nwer Preure an Manometer. Wat will be te (a) te gauge preure an (b) te abolute preure of water at ept m below te urface? water = 000 kg/m, an p atmopere = 0kN/m. [7.7 kn/m, 8.7 kn/m ] a) pgauge g N / m,( Pa) 7. 7 kn / m,( kpa) b) pabolute pgauge patmoperic ( 770 0) N / m,( Pa) 8. 7 kn / m,( kpa). t wat ept below te urface of oil, relative enity 0.8, will prouce a preure of 0 kn/m? Wat ept of water i ti equivalent to? [5.m,.m] a) water kg / m p g p 0 0 m g of oil b) 000 kg / m 0 0. mof water Wat woul te preure in kn/m be if te equivalent ea i meaure a 400mm of (a) mercury =.6 (b) water ( c) oil pecific weigt 7.9 kn/m () a liqui of enity 50 kg/m? [5.4 kn/m,.9 kn/m,.6 kn/m,.04 kn/m ] a) water kg / m p g N / m Example: nwer CIVE400: Flui Mecanic Example: nwer CIVE400: Flui Mecanic 4
3 CIVE400: Flui Mecanic Example: nwer b) p g N / m c) g p g N / m ) p g N / m.4 manometer connecte to a pipe inicate a negative gauge preure of 50mm of mercury. Wat i te abolute preure in te pipe in Newton per quare metre i te atmoperic preure i bar? [9. kn/m ] 5 patmopere bar 0 N / m pabolute pgauge patmoperic g patmoperic N / m,( Pa) 9. kn / m,( kpa).5 Wat eigt woul a water barometer nee to be to meaure atmoperic preure? [>0m] 5 patmopere bar 0 N / m 5 0 g 5 0 mof water m of mercury (. ). Example: nwer CIVE400: Flui Mecanic 5 CIVE400: Flui Mecanic Example: nwer.6 n incline manometer i require to meaure an air preure of mm of water to an accuracy of +/- %. Te incline arm i 8mm in iameter an te larger arm a a iameter of 4mm. Te manometric flui a enity 740 kg/m an te cale may be rea to +/- 0.5mm. Wat i te angle require to enure te eire accuracy may be acieve? [ 9 ] p p iameter z iameter D θ Scale Reaer Datum line p p mang man g z z z Volume move from left to rigt = z x in D z z x 4 in 4 4 z z x in D D p p man gxin D water g man gxin D water g 074. water gxin x (in 0. ) Te ea being meaure i % of mm = 0.00x0.0 = m Ti % repreent te mallet meaurement poible on te manometer, 0.5mm = m, giving (in 0. ) in [Ti i not te ame a te anwer given on te quetion eet] Example: nwer CIVE400: Flui Mecanic 6 x z
4 CIVE400: Flui Mecanic Example: nwer.7 Determine te reultant force ue to te water acting on te m by m rectangular area B own in te iagram below. [4 560 N,.7m from O] O P CIVE400: Flui Mecanic Example: nwer.8 Determine te reultant force ue to te water acting on te.5m by.0m triangular area CD own in te figure above. Te apex of te triangle i at C. [4.50 N,.8m from P] b B.m.0 m D.0m C.0 m Te magnitue of te reultant force on a ubmerge plane i: R = preure at centroi area of urface R gz N / m Ti act at rigt angle to te urface troug te centre of preure. IOO n moment of area about a line troug O Sc x t moment of area about a line troug O By te parallel axi teorem (wic will be given in an exam), Ioo IGG x, were I GG i te n moment of area about a line troug te centroi an can be foun in table. b I GG Sc x x 45 G G / For a triangle I 86 N / m GG b 6 Dept to centre of gravity i z 0. co m. R gz Ditance from P i x z / co m Ditance from P to centre of preure i I oo Sc x I I x oo GG I GG 5. Sc x. 748 x m G G For a rectangle I GG b te wall i vertical, Sc D an x z, Sc.. 7. m from O Example: nwer CIVE400: Flui Mecanic 7 Example: nwer CIVE400: Flui Mecanic 8
5 CIVE400: Flui Mecanic Example: nwer Force on ubmerge urface. Obtain an expreion for te ept of te centre of preure of a plane urface wolly ubmerge in a flui an incline at an angle to te free urface of te liqui. orizontal circular pipe,.5m iameter, i cloe by a butterfly ik wic rotate about a orizontal axi troug it centre. Determine te torque wic woul ave to be applie to te ik pinle to keep te ik cloe in a vertical poition wen tere i a m ea of fre water above te axi. [76 Nm] Te quetion ak wat i te moment you ave to apply to te pinle to keep te ic vertical i.e. to keep te valve ut? So you nee to know te reultant force exerte on te ic by te water an te itance x of ti force from te pinle. We know tat te water in te pipe i uner a preure of m ea of water (to te pinle).75 CIVE400: Flui Mecanic Example: nwer I GG ' 4 r 4( r ) r. 06m So te itance from te pinle to te line of action of te force i x ' m n te moment require to keep te gate ut i moment Fx kn m. ock gate i to be reinforce wit tree orizontal beam. If te water act on one ie only, to a ept of 6m, fin te poition of te beam meaure from te water urface o tat eac will carry an equal loa. Give te loa per meter. [ N/m,.m, 4.m, 5.47m] Firt of all raw te preure iagram, a below: F Diagram of te force on te ic valve, bae on an imaginary water urface. m, te ept to te centroi of te ic = ept to te centre of preure (or line of action of te force) Calculate te force: F g kn Calculate te line of action of te force,. n moment of area about water urface ' t moment of area about water urface Ioo By te parallel axi teorem n moment of area about O (in te urface) Ioo IGG were I GG i te n moment of area about a line troug te centroi of te ic an I GG = r 4 /4. x R / Te reultant force per unit lengt of gate i te area of te preure iagram. So te total reultant force i R g = N ( per m lengt) lternatively te reultant force i, R = Preure at centroi rea, (take wit of gate a m to give force per m) R g N ( per m lengt) Ti i te reultant force exerte by te gate on te water. Te tree beam oul carry an equal loa, o eac beam carrie te loa f, were R f N f f f Example: nwer CIVE400: Flui Mecanic 9 Example: nwer CIVE400: Flui Mecanic 0
6 CIVE400: Flui Mecanic Example: nwer If we take moment from te urface, DR f f f D f f Taking te firt beam, we can raw a preure iagram for ti, (ignoring wat i below), H/ H F=58860 CIVE400: Flui Mecanic Example: nwer. Te profile of a maonry am i an arc of a circle, te arc aving a raiu of 0m an ubtening an angle of 60 at te centre of curvature wic lie in te water urface. Determine (a) te loa on te am in N/m lengt, (b) te poition of te line of action to ti preure. [ N/m lengt at ept 9.0m] Draw te am to elp picture te geometry, R a 60 We know tat te reultant force, F F gh, o H g R F R y F H g m n te force act at H/, o ti i te poition of te t beam, poition of t beam H. m Taking te econ beam into conieration, we can raw te following preure iagram, F F v =. H H/ f F=58860 f Te reaction force i equal to te um of te force on eac beam, o a before F ( 58860) H 49. m g Te reaction force act at H/, o H=.7m. Taking moment from te urface, ( 58860) ept to econ beam 4. m For te tir beam, from before we ave, ept to tir beam m Example: nwer CIVE400: Flui Mecanic 0in m a 0co m Calculate F v = total weigt of flui above te curve urface (per m lengt) F v g( area of ector - area of triangle) = kn / m Calculate F = force on projection of curve urface onto a vertical plane F g kn / m Te reultant, F F F R v kn / m acting at te angle Fv tan 089. F 9. Example: nwer CIVE400: Flui Mecanic
7 CIVE400: Flui Mecanic Example: nwer ti force act normal to te urface, it mut act troug te centre of raiu of te am wall. So te ept to te point were te force act i, y = 0in 9.=9m.4 Te arc of a brige over a tream i in te form of a emi-circle of raiu m. te brige wit i 4m. Due to a floo te water level i now.5m above te cret of te arc. Calculate (a) te upwar force on te unerie of te arc, (b) te orizontal trut on one alf of te arc. [6.6 kn, 76.6 kn] Te brige an water level can be rawn a:.5m a) Te upwar force on te arc = weigt of (imaginary) water above te arc. Rv gvolume of water volume ( 5. ) m R kn v b) Te orizontal force on alf of te arc, i equal to te force on te projection of te curve urface onto a vertical plane..5.0 m F preure at centroi area g kn CIVE400: Flui Mecanic Example: nwer x = 7.0 m, o = = 9.5. x = 9.5/tan 60 = m. Vertical force = weigt of water above te urface, Fv g x 05. x kn / m Te orizontal force = force on te projection of te urface on to a vertical plane. F g kn / m Te reultant force i FR Fv F kn / m n act at te angle Fv tan F tank wit vertical ie i quare in plan wit m long ie. Te tank contain oil of relative enity 0.9 to a ept of.0m wic i floating on water a ept of.5m. Calculate te force on te wall an te eigt of te centre of preure from te bottom of te tank. [65.54 kn,.5m] Conier one wall of te tank. Draw te preure iagram: 60.5 Te face of a am i vertical to a ept of 7.5m below te water urface ten lope at 0 to te vertical. If te ept of water i 7m wat i te reultant force per metre acting on te wole face? [56.9 kn] Example: nwer CIVE400: Flui Mecanic Example: nwer CIVE400: Flui Mecanic 4
8 CIVE400: Flui Mecanic Example: nwer F f f f enity of oil oil = 0.9 water = 900 kg/m. Force per unit lengt, F = area uner te grap = um of te tree area = f + f + f ( ) f 5974 N f ( ) N ( ) 5. f 09 N F f f f N To fin te poition of te reultant force F, we take moment from any point. We will take moment about te urface. DF f f f D ( ) 09 ( 5. ) D. 47m( from urface) 5. m ( from bae of wall) CIVE400: Flui Mecanic Example: nwer pplication of te Bernoulli Equation. In a vertical pipe carrying water, preure gauge are inerte at point an B were te pipe iameter are 0.5m an 0.075m repectively. Te point B i.5m below an wen te flow rate own te pipe i 0.0 cumec, te preure at B i 475 N/m greater tan tat at. uming te loe in te pipe between an B can be expree a k v were v i te velocity at, g fin te value of k. If te gauge at an B are replace by tube fille wit water an connecte to a U-tube containing mercury of relative enity.6, give a ketc owing ow te level in te two limb of te U-tube iffer an calculate te value of ti ifference in metre. [k = 0.9, m] Part i) = 0.m B = 0.m B Rp 05. m B m Q 0. 0 m / p p 475 N / m B f kv g Taking te atum at B, te Bernoulli equation become: p u p u z z k u B B B g g g g g z 5. z 0 B By continuity: Q = u = u B B u u giving B 0. 0 / m/ 0. 0 / m/ Example: nwer CIVE400: Flui Mecanic 5 Example: nwer CIVE400: Flui Mecanic 6
9 CIVE400: Flui Mecanic Example: nwer Part ii) p p u u z k u B B 000g g g k k 0. 9 pxxl w gzb pb pxxr m grp w gz w grp p pxxl pxxr wgz B pb mgr p wgz wgr p p pb p wgz zbgrpm w R p R m p. Venturimeter wit an entrance iameter of 0.m an a troat iameter of 0.m i ue to meaure te volume of ga flowing troug a pipe. Te icarge coefficient of te meter i uming te pecific weigt of te ga to be contant at 9.6 N/m, calculate te volume flowing wen te preure ifference between te entrance an te troat i meaure a 0.06m on a water U-tube manometer. [0.86 m /] = 0.m CIVE400: Flui Mecanic Example: nwer Wat we know from te quetion: Calculate Q. For te manometer: For te Venturimeter Combining () an () g C g 9. 6 N / m m 0. m u Q/ u Q/ p gz p g z R gr g g p w p p p 9. 6 z z ( ) p u p u z z g g g g g g p p 9. 6 z z 0. 80u ( ) 0. 80u u ieal m/ 0. Qieal m / Q C Q m / iea = 0.m Z Z Rp Example: nwer CIVE400: Flui Mecanic 7 Example: nwer CIVE400: Flui Mecanic 8
10 CIVE400: Flui Mecanic Example: nwer. Venturimeter i ue for meauring flow of water along a pipe. Te iameter of te Venturi troat i two fift te iameter of te pipe. Te inlet an troat are connecte by water fille tube to a mercury U-tube manometer. Te velocity of flow along te pipe i foun to be 5. H m/, were H i te manometer reaing in metre of mercury. Determine te lo of ea between inlet an troat of te Venturi wen H i 0.49m. (Relative enity of mercury i.6). [0.m of water] CIVE400: Flui Mecanic Example: nwer Subtitute in () / Loe L m.4 Water i icarging from a tank troug a convergent-ivergent moutpiece. Te exit from te tank i roune o tat loe tere may be neglecte an te minimum iameter i 0.05m. If te ea in te tank above te centre-line of te moutpiece i.8m. a) Wat i te icarge? b) Wat mut be te iameter at te exit if te abolute preure at te minimum area i to be.44m of water? c) Wat woul te icarge be if te ivergent part of te mout piece were remove. (ume atmoperic preure i 0m of water). [0.075m, 0.066m /, 0.08m /] Z Z H For te manometer: For te Venturimeter p gz p g z H gh w w m p p gz gh gh gz () w w m w p u p u z z Loe g g g g w w wu wu p p w gz wgz Lwg ( ) Combining () an () p u p u z z Loe g g g g w w w Lwg Hgm w u u () but at. From te quetion u 5. H 75. m/ u u 75. u 4 0 u m/ Example: nwer CIVE400: Flui Mecanic 9 From te quetion: pply Bernoulli: If we take te atum troug te orifice: 005. m p minimum preure 44. m g p m p 0 g g p u p u p u z z z g g g g g g z 8. m z z 0 u negligible Between an u g Between an p p u 57. m/ Q u.. m / Example: nwer CIVE400: Flui Mecanic 0
11 CIVE400: Flui Mecanic Example: nwer u 8. g u 599. m/ Q u m If te mout piece a been remove, p p p p u z g g g u gz 599. m/ Q m / 4.5 cloe tank a an orifice 0.05m iameter in one of it vertical ie. Te tank contain oil to a ept of 0.6m above te centre of te orifice an te preure in te air pace above te oil i maintaine at 780 N/m above atmoperic. Determine te icarge from te orifice. (Coefficient of icarge of te orifice i 0.6, relative enity of oil i 0.9). [ m /] CIVE400: Flui Mecanic Example: nwer 780 u 06. o g g u 65. m/ 005 Q m.... /.6 Te icarge coefficient of a Venturimeter wa foun to be contant for rate of flow exceeing a certain value. Sow tat for ti conition te lo of ea ue to friction in te convergent part of te meter can be expree a KQ m were K i a contant an Q i te rate of flow in cumec. Obtain te value of K if te inlet an troat iameter of te Venturimeter are 0.0m an 0.05m repectively an te icarge coefficient i [K=060].7 Venturimeter i to fitte in a orizontal pipe of 0.5m iameter to meaure a flow of water wic may be anyting up to 40m /our. Te preure ea at te inlet for ti flow i 8m above atmoperic an te preure ea at te troat mut not be lower tan 7m below atmoperic. Between te inlet an te troat tere i an etimate frictional lo of 0% of te ifference in preure ea between tee point. Calculate te minimum allowable iameter for te troat. [0.06m] = 0.5m P = 780 kn/m 0.66m oil From te quetion: From te quetion o = 0.05m 05. m Q40m / r m / u Q/ 77. m/ p p 8m 7m g g pply Bernoulli, o o C 06. w Friction lo, from te quetion: pply Bernoulli: f p p 0. g p u p u z z g g g g Take atmoperic preure a 0, Example: nwer CIVE400: Flui Mecanic Example: nwer CIVE400: Flui Mecanic
12 CIVE400: Flui Mecanic Example: nwer p u p u f g g g g p p u u f g g g g u g g u. 46 m/ Q u m.8 Venturimeter of troat iameter 0.076m i fitte in a 0.5m iameter vertical pipe in wic liqui of relative enity 0.8 flow ownwar. Preure gauge are fitte to te inlet an to te troat ection. Te troat being 0.94m below te inlet. Taking te coefficient of te meter a 0.97 fin te icarge a) wen te preure gauge rea te ame b)wen te inlet gauge rea 570 N/m iger tan te troat gauge. [0.09m /, 0.04m /] = 0.076m = 0.5m CIVE400: Flui Mecanic Example: nwer a) p p By continuity: b) p u p u z z g g g g u u z z g g Qu u u u u u 6u g g u 094. m/ 5 Q C u Q m / 4 p p 570 p p u u g g 570 Q g g Q Q 0. 05m / From te quetion: pply Bernoulli: 05. m m m m 800 kg / m C 097. Example: nwer CIVE400: Flui Mecanic Example: nwer CIVE400: Flui Mecanic 4
13 CIVE400: Flui Mecanic Example: nwer Tank emptying 4. reervoir i circular in plan an te ie lope at an angle of tan - (/5) to te orizontal. Wen te reervoir i full te iameter of te water urface i 50m. Dicarge from te reervoir take place troug a pipe of iameter 0.65m, te outlet being 4m below top water level. Determine te time for te water level to fall m auming te icarge to be 075. a gh cumec were a i te cro ectional area of te pipe in m an H i te ea of water above te outlet in m. [5 econ] 50m From te quetion: H = 4m a = (0.65/) = 0.m Q 075. a g 096. In time t te level in te reervoir fall, o Qt t Q Integrating give te total time for level to fall from to. T Q te urface area cange wit eigt, we mut expre in term of. = r But r varie wit. It varie linearly from te urface at H = 4m, r = 5m, at a graient of tan - = /5. r = x = x + 5(4) x = 5 o = ( ) = ( ) Subtituting in te integral equation give x 5 Example: nwer CIVE400: Flui Mecanic 5 r H CIVE400: Flui Mecanic Example: nwer T / / / 764. / 5 / 4 / From te quetion, = 4m = m, o / T / 4 4 / / 5 5 / 4 / ec 4. rectangular wimming pool i m eep at one en an increae uniformly in ept to.6m at te oter en. Te pool i 8m wie an m long an i emptie troug an orifice of area 0.4m, at te lowet point in te ie of te eep en. Taking C for te orifice a 0.6, fin, from firt principle, a) te time for te ept to fall by m b) te time to empty te pool completely. [99 econ, 66 econ].0m.0m Te quetion tell u a o = 0.4m, C = 0.6 pply Bernoulli from te tank urface to te vena contracta at te orifice: p u p u z z g g g g p = p an u = 0. u g We nee Q in term of te eigt meaure above te orifice..6m Q Caou Cao g Example: nwer CIVE400: Flui Mecanic 6 L
14 CIVE400: Flui Mecanic Example: nwer n we can write an equation for te icarge in term of te urface eigt cange: Qt t Q Integrating give te total time for level to fall from to. T Q 68. ( ) a) For te firt m ept, = 8 x = 56, watever te. So, for te firt perio of time: T ec b) now we nee to fin out ow long it will take to empty te ret. We nee te area, in term of. 8L L So T ec Total time for emptying i, T = = 66 ec / / / / CIVE400: Flui Mecanic Example: nwer 4. vertical cylinrical tank m iameter a, at te bottom, a 0.05m iameter arp ege orifice for wic te icarge coefficient i 0.6. a) If water enter te tank at a contant rate of cumec fin te ept of water above te orifice wen te level in te tank become table. b) Fin te time for te level to fall from m to m above te orifice wen te inflow i turne off. c) If water now run into te tank at 0.0 cumec, te orifice remaining open, fin te rate of rie in water level wen te level a reace a ept of.7m above te orifice. [a).4m, b) 88 econ, c) 0.5m/min] Q = m / o = 0.005m From te quetion: Q in = m /, o =0.05m, C =0.6 pply Bernoulli from te water urface () to te orifice (), p u p u z z g g g g p = p an u = 0. u g. Wit te atum te bottom of te cyliner, z =, z = 0 We nee Q in term of te eigt meaure above te orifice. Qout C aou C ao g ( ) For te level in te tank to remain contant: inflow = out flow Q in = Q out m (b) Write te equation for te icarge in term of te urface eigt cange: Example: nwer CIVE400: Flui Mecanic 7 Example: nwer CIVE400: Flui Mecanic 8
15 CIVE400: Flui Mecanic Example: nwer Qt t Q Integrating between an, to give te time to cange urface level T Q / / / / 06 = an = o T = 88 ec c) Q in cange to Q in = 0.0 m / From () we ave Qout Te quetion ak for te rate of urface rie wen =.7m. i.e. Qout m / Te rate of increae in volume i: QQin Qout m / Q = rea x Velocity, te rate of rie in urface i Q u Q 0. 0 u m/ 0. 5m/min orizontal boiler ell (i.e. a orizontal cyliner) m iameter an 0m long i alf full of water. Fin te time of emptying te ell troug a ort vertical pipe, iameter 0.08m, attace to te bottom of te ell. Take te coefficient of icarge to be 0.8. [70 econ] = m m From te quetion W = 0m, D = 0m o = 0.08m C = 0.8 o = 0.08 m Example: nwer CIVE400: Flui Mecanic 9 CIVE400: Flui Mecanic Example: nwer pply Bernoulli from te water urface () to te orifice (), p u p u z z g g g g p = p an u = 0. u g. Wit te atum te bottom of te cyliner, z =, z = 0 We nee Q in term of te eigt meaure above te orifice. Qout C aou C ao g Write te equation for te icarge in term of te urface eigt cange: Qt t Q Integrating between an, to give te time to cange urface level T But we nee in term of Q Surface area = 0L, o nee L in term of L a a ( ) L ( ) L 0 Subtitute ti into te integral term, L.0m a Example: nwer CIVE400: Flui Mecanic 0.0m.
16 CIVE400: Flui Mecanic Example: nwer 0 T / ec 4.5 Two cyliner taning uprigt contain liqui an are connecte by a ubmerge orifice. Te iameter of te cyliner are.75m an.0m an of te orifice, 0.08m. Te ifference in level of te liqui i initially.5m. Fin ow long it will take for ti ifference to be reuce to 0.66m if te coefficient of icarge for te orifice i (Work from firt principle.) [0.7 econ] =.75m =.0m =.5m CIVE400: Flui Mecanic Example: nwer ( ) Qt ( ) From te Bernoulli equation we can erive ti expreion for icarge troug te ubmerge orifice: Q C a g So Integrating o Ca o gt t C a g o T C a g o C a g o ec 4.6 rectangular reervoir wit vertical wall a a plan area of 60000m. Dicarge from te reervoir take place over a rectangular weir. Te flow caracteritic of te weir i Q = H / cumec were H i te ept of water above te weir cret. Te ill of te weir i.4m above te bottom of te reervoir. Starting wit a ept of water of 4m in te reervoir an no inflow, wat will be te ept of water after one our? [.98m] o = 0.08m by continuity, efining, = m 0785m o 008m ao 00050m C ,.. Q t () From te quetion = m, Q = / Write te equation for te icarge in term of te urface eigt cange: Subtituting ti in () to eliminate Example: nwer CIVE400: Flui Mecanic Example: nwer CIVE400: Flui Mecanic
17 CIVE400: Flui Mecanic Example: nwer Qt t Q Integrating between an, to give te time to cange urface level T Q / / From te quetion T = 600 ec an = 0.6m m Total ept = =.98m / / CIVE400: Flui Mecanic Example: nwer Notce an weir 5. Deuce an expreion for te icarge of water over a rigt-angle arp ege V-notc, given tat te coefficient of icarge i 0.6. rectangular tank 6m by 6m a te ame notc in one of it ort vertical ie. Determine te time taken for te ea, meaure from te bottom of te notc, to fall from 5cm to 7.5cm. [99 econ] From your note you can erive: Q 8 C gh 5 5 / tan For ti weir te equation implifie to Q 44. H 5 / Write te equation for te icarge in term of te urface eigt cange: Qt t Q Integrating between an, to give te time to cange urface level T = 0.5m, = 0.075m Q 6 6 / 44. / T ec / / Example: nwer CIVE400: Flui Mecanic Example: nwer CIVE400: Flui Mecanic 4
18 CIVE400: Flui Mecanic Example: nwer 5. Derive an expreion for te icarge over a arp crete rectangular weir. arp ege weir i to be contructe acro a tream in wic te normal flow i 00 litre/ec. If te maximum flow likely to occur in te tream i 5 time te normal flow ten etermine te lengt of weir neceary to limit te rie in water level to 8.4cm above tat for normal flow. C =0.6. [.4m] From your note you can erive: Q Cb / g From te quetion: Q = 0. m /, = x Q =.0 m /, = x were x i te eigt above te weir at normal flow. So we ave two ituation: 0. Cb / / gx 80. bx ( ) / / 0. Cb g x bx ( ) From () we get an expreion for b in term of x b 0. x / Subtituting ti in () give, x x / x x x m So te weir breat i b m / / Example: nwer CIVE400: Flui Mecanic 5 CIVE400: Flui Mecanic Example: nwer 5. Sow tat te rate of flow acro a triangular notc i given by Q=C KH 5/ cumec, were C i an experimental coefficient, K epen on te angle of te notc, an H i te eigt of te uniturbe water level above te bottom of te notc in metre. State te reaon for te introuction of te coefficient. Water from a tank aving a urface area of 0m flow over a 90 notc. It i foun tat te time taken to lower te level from 8cm to 7cm above te bottom of te notc i 4.5econ. Determine te coefficient C auming tat it remain contant uring i perio. [0.65] 8 Te proof for Q C gh CKH 5 5 / 5 / tan i in te note. From te quetion: = 0m = 90 = 0.08m = 0.07m T = 4.5ec So Q =.6 C 5/ Write te equation for te icarge in term of te urface eigt cange: Qt t Q Integrating between an, to give te time to cange urface level T Q 0 C 5/ C / C C / / 5.4 reervoir wit vertical ie a a plan area of 56000m. Dicarge from te reervoir take place over a rectangular weir, te flow caracteritic of wic i Q=.77BH / m /. t time of maximum rainfall, water flow into te reervoir at te rate of 9m /. Fin a) te lengt of weir require to icarge ti quantity if ea mut not excee 0.6m; b) te time neceary for te ea to rop from 60cm to 0cm if te inflow uenly top. [0.94m, 09econ] From te quetion: = m Q =.77 B H / Q max = 9 m / a) Fin B for H = =.77 B 0.6 / B = 0.94m Example: nwer CIVE400: Flui Mecanic 6
19 CIVE400: Flui Mecanic Example: nwer b) Write te equation for te icarge in term of te urface eigt cange: Qt t Q Integrating between an, to give te time to cange urface level T Q B / / B / / T 09ec 5.5 Develop a formula for te icarge over a 90 V-notc weir in term of ea above te bottom of te V. cannel convey 00 litre/ec of water. t te outlet en tere i a 90 V-notc weir for wic te coefficient of icarge i t wat itance above te bottom of te cannel oul te weir be place in orer to make te ept in te cannel.0m? Wit te weir in ti poition wat i te ept of water in te cannel wen te flow i 00 litre/ec? [0.755m,.8m] Derive ti formula from te note: Q 8 C gh 5 5 / tan From te quetion: = 90 C 0.58 Q = 0. m /, ept of water, Z = 0.m giving te weir equation: Q 7. H 5 / a) H i te eigt above te bottom of te V, te ept of water = Z = D + H, were D i te eigt of te bottom of te V from te bae of te cannel. So 5/ Q7. ZD 5/ D D m b) Fin Z wen Q = 0. m / Z Z 8. m 5 / CIVE400: Flui Mecanic Example: nwer 5.6 Sow tat te quantity of water flowing acro a triangular V-notc of angle i 8 5 / Q C tan gh. Fin te flow if te meaure ea above te bottom of te V i 8cm, wen 5 =45 an C =0.6. If te flow i wante witin an accuracy of %, wat are te limiting value of te ea. [0.6m /, 0.77m, 0.8m] Proof of te v-notc weir equation i in te note. From te quetion: H = 0.8m = 45 C = 0.6 Te weir equation become: 5 / Q 47. H 5 / m / Q+% = 0.9 m / 5 / H H 0. 8m Q-% = 0.4 m / 5 / H H 077. m Example: nwer CIVE400: Flui Mecanic 7 Example: nwer CIVE400: Flui Mecanic 8
20 CIVE400: Flui Mecanic Example: nwer pplication of te Momentum Equation 6. Te figure below ow a moot curve vane attace to a rigi founation. Te jet of water, rectangular in ection, 75mm wie an 5mm tick, trike te vane wit a velocity of 5m/. Calculate te vertical an orizontal component of te force exerte on te vane an inicate in wic irection tee component act. [Horizontal.4 N acting from rigt to left. Vertical 4.6 N acting ownwar] CIVE400: Flui Mecanic Example: nwer mm iameter pipeline carrie water uner a ea of 0m wit a velocity of m/. Ti water main i fitte wit a orizontal ben wic turn te axi of te pipeline troug 75 (i.e. te internal angle at te ben i 05). Calculate te reultant force on te ben an it angle to te orizontal. [ kn, 5 9 ] y u 45 5 x From te quetion: a u m/ Q m / a a, o u u m Calculate te total force uing te momentum equation: F T Q u u x co5 co co5 5co N F Q u in 5 u in 45 T y in55in N Boy force an preure force are 0. So force on vane: R F. 44N x t x R F 4. 6N y t y Example: nwer CIVE400: Flui Mecanic 9 u θ From te quetion: a m m m u u m/ Q m / Calculate total force. F Q u u F F F Tx x x Rx Px Bx F Tx co kn F Q u u F F F Ty y y Ry Py By F Ty in kN Calculate te preure force p = p = p = g = = 94. kn/m F p a Tx co pa co co kn F p a in p a in Ty in kn Tere i no boy force in te x or y irection. F F F F Rx Tx Px Bx kn Example: nwer CIVE400: Flui Mecanic 40
21 CIVE400: Flui Mecanic Example: nwer FRy FTy FPy FBy kN Tee force act on te flui Te reultant force on te flui i FR FRx FRy kn FRy tan 5 9' FRx 6. orizontal jet of water 0 mm cro-ection an flowing at a velocity of 5 m/ it a flat plate at 60 to te axi (of te jet) an to te orizontal. Te jet i uc tat tere i no ie prea. If te plate i tationary, calculate a) te force exerte on te plate in te irection of te jet an b) te ratio between te quantity of flui tat i eflecte upwar an tat ownwar. (ume tat tere i no friction an terefore no ear force.) [8N, :] y u x u θ CIVE400: Flui Mecanic Example: nwer Q = a u = 0-5 = 0.0 Q = (a + a ) u Q = a u Q = (a - a )u Calculate total force. FTx Q ux u x FRx FPx FBx F Tx in N Component in irection of jet = 90 in 60 = 8 N tere i no force parallel to te plate F ty = 0 FTy ua ua ua co 0 a a aco 0 a a a a aco aa 4 4a a a a a Tu /4 of te jet goe up, /4 own mm iameter jet of water aving a velocity of 5m/ trike a flat plate, te normal of wic i incline at 0 to te jet. Fin te force normal to te urface of te plate. [.9kN] y u x u From te quetion a = a =x0 - m u = 5 m/ pply Bernoulli, p u p u p u z z z g g g g g g u Cange in eigt i negligible o z = z = z u = u = u =5 m/ By continuity Q = Q + Q an preure i alway atmoperic p = p = p =0. So u θ u a = u a + u a o a = a + a Put te axe normal to te plate, a we know tat te reultant force i normal to te plate. Example: nwer CIVE400: Flui Mecanic 4 From te quetion, jet = 0.075m u =5m/ Q = 5(0.075/) = 0. m / Force normal to plate i Example: nwer CIVE400: Flui Mecanic 4
22 CIVE400: Flui Mecanic Example: nwer F Tx = Q( 0 - u x ) F Tx = ( 0-5 co 0 ) =.9 kn 6.5 Te outlet pipe from a pump i a ben of 45 riing in te vertical plane (i.e. an internal angle of 5). Te ben i 50mm iameter at it inlet an 00mm iameter at it outlet. Te pipe axi at te inlet i orizontal an at te outlet it i m iger. By neglecting friction, calculate te force an it irection if te inlet preure i 00kN/m an te flow of water troug te pipe i 0.m /. Te volume of te pipe i 0.075m. [.94kN at to te orizontal] y p u x p u 45 m CIVE400: Flui Mecanic Example: nwer u u m/ 05. / m/ F T co x N an in te y-irection F Q u u T y y y Qu in in N 4 Calculate te preure force. F P preure force at - preure force at F p co0 p co p p co P x & Draw te control volume an te axi ytem p = 00 kn/m, Q = 0. m / = 45 = 0.5 m = 0. m = 0.77 m = m F p in0 p in p in P y We know preure at te inlet but not at te outlet. we can ue Bernoulli to calculate ti unknown preure. Calculate te total force in te x irection FT Q u u x x x Qu co u by continuity u u Q, o p u p u z z g g g g were f i te friction lo In te quetion it ay ti can be ignore, f =0 Te eigt of te pipe at te outlet i m above te inlet. Taking te inlet level a te atum: z = 0 z = m So te Bernoulli equation become: p p 564. N / m f Example: nwer CIVE400: Flui Mecanic 4 Example: nwer CIVE400: Flui Mecanic 44
23 CIVE400: Flui Mecanic Example: nwer F F P x P y co kn 564. in Calculate te boy force Te only boy force i te force ue to gravity. Tat i te weigt acting in te y irection. F g volume B y N Tere are no boy force in te x irection, F B x 0 6 Calculate te reultant force F F F F T x R x P x B x F F F F T y R y P y B y F F F F R x T x P x B x N CIVE400: Flui Mecanic Example: nwer F F F R R x R y kn n te irection of application i F R y tan tan F R x Te force on te ben i te ame magnitue but in te oppoite irection R F R 6.6 Te force exerte by a 5mm iameter jet againt a flat plate normal to te axi of te jet i 650N. Wat i te flow in m /? [0.08 m /] y u u x F F F F R y T y P y B y N u n te reultant force on te flui i given by F Ry F Reultant φ F Rx From te quetion, jet = 0.05m F Tx = 650 N Force normal to plate i F Tx = Q( 0 - u x ) 650 = 000Q ( 0 - u ) Q = au = ( /4)u 650 = -000au = -000Q /a 650 = -000Q /(0.05 /4) Q = 0.08m / Example: nwer CIVE400: Flui Mecanic 45 Example: nwer CIVE400: Flui Mecanic 46
24 CIVE400: Flui Mecanic Example: nwer 6.7 curve plate eflect a 75mm iameter jet troug an angle of 45. For a velocity in te jet of 40m/ to te rigt, compute te component of te force evelope againt te curve plate. (ume no friction). [R x =070N, R y =5000N own] y x u CIVE400: Flui Mecanic Example: nwer reucing ben, 0.6m iameter uptream, 0.m iameter owntream, a water flowing troug it at te rate of 0.45m / uner a preure of.45 bar. Neglecting any lo i ea for friction, calculate te force exerte by te water on te ben, an it irection of application. [R=4400N to te rigt an own, = 4] y x ρ u u θ ρ u θ From te quetion: a u / m/ m Q m / a a, o u u Calculate te total force uing te momentum equation: F T Q u u x co co45 40 F T y N Q u in in N Boy force an preure force are 0. So force on vane: R F 070N x t x R F 4998N y t y & Draw te control volume an te axi ytem p = N/m, Q = 0.45 m / = 45 = 0.6 m = 0. m = 0.8 m = m Calculate te total force in te x irection FT Q u u x x x Qu co u by continuity u u Q, o u u m/ 06. / m/ Example: nwer CIVE400: Flui Mecanic 47 Example: nwer CIVE400: Flui Mecanic 48
25 CIVE400: Flui Mecanic Example: nwer F T co x 0 N an in te y-irection F Q u u T y y y Qu in in N 4 Calculate te preure force. F P F p co0 p co p p co P x F p in0 p in p in P y preure force at - preure force at CIVE400: Flui Mecanic Example: nwer Te only boy force i te force ue to gravity. Tere are no boy force in te x or y irection, F B x F B y 0 6 Calculate te reultant force F F F F T x R x P x B x F F F F T y R y P y B y F F F F R x T x P x B x N F F F F R y T y P y B y N We know preure at te inlet but not at te outlet. we can ue Bernoulli to calculate ti unknown preure. p u p u z z g g g g were f i te friction lo In te quetion it ay ti can be ignore, f =0 ume te pipe to be orizontal z = z So te Bernoulli equation become: p p 6007 N / m F F P x P y N 600 N co in f n te reultant force on te flui i given by F Ry F Reultant φ F Rx F F F R R x R y kn n te irection of application i F R y 800 tan tan 6. F 45 R x Te force on te ben i te ame magnitue but in te oppoite irection R F R 5 Calculate te boy force Example: nwer CIVE400: Flui Mecanic 49 Example: nwer CIVE400: Flui Mecanic 50
26 CIVE400: Flui Mecanic Example: nwer Laminar pipe flow. 7. Te itribution of velocity, u, in metre/ec wit raiu r in metre in a moot bore tube of 0.05 m bore follow te law, u =.5 - kr. Were k i a contant. Te flow i laminar an te velocity at te pipe urface i zero. Te flui a a coefficient of vicoity of kg/m. Determine (a) te rate of flow in m / (b) te earing force between te flui an te pipe wall per metre lengt of pipe. [6.4x0-4 m /, 8.49x0 - N] Te velocity at itance r from te centre i given in te quetion: u =.5 - kr lo we know: = kg/m r = 0.05m CIVE400: Flui Mecanic Example: nwer 7. liqui woe coefficient of vicoity i m flow below te critical velocity for laminar flow in a circular pipe of iameter an wit mean velocity u. Sow tat te preure lo in a lengt of pipe i um/. Oil of vicoity 0.05 kg/m flow troug a pipe of iameter 0.m wit a velocity of 0.6m/. Calculate te lo of preure in a lengt of 0m. [ 50 N/m ] See te proof in te lecture note for Conier a cyliner of flui, lengt L, raiu r, flowing teaily in te centre of a pipe δr We can fin k from te bounary conition: wen r = 0.05, u = 0.0 (bounary of te pipe) 0.0 =.5 - k0.05 k = 6000 r r R u = r a) Following along imilar line to te erivation een in te lecture note, we can calculate te flow Q troug a mall annulu r: Q ur annulu annulu ( rr) r rr Q r rr b) Te ear force i given by F = (r) 005. Q. 5r6000r r 0 5. r m / r Te flui i in equilibrium, earing force equal te preure force. rl p pr p r L Newton law of vicoity u y, We are meauring from te pipe centre, o u r Giving: p r u L r u p r r L From Newton law of vicoity In an integral form ti give an expreion for velocity, u r u 6000r 000r r F ( 0. 05) N p u L rr Te value of velocity at a point itance r from te centre u r p r C L 4 t r = 0, (te centre of te pipe), u = u max, at r = R (te pipe wall) u = 0; p R C L 4 t a point r from te pipe centre wen te flow i laminar: Example: nwer CIVE400: Flui Mecanic 5 Example: nwer CIVE400: Flui Mecanic 5
27 CIVE400: Flui Mecanic Example: nwer u r p R L 4 r Te flow in an annulu of tickne r Q ur annulu annulu ( rr) r rr p Q R r r r L 4 R p Q Rrr r L p R p L 8 L8 So te icarge can be written p Q 4 L 8 To get preure lo in term of te velocity of te flow, ue te mean velocity: u Q/ p u L Lu p u p per unit lengt b) From te quetion = 0.05 kg/m = 0.m u = 0.6 m/ L = 0.0m p 50 N / m 0. CIVE400: Flui Mecanic Example: nwer 7. plunger of 0.08m iameter an lengt 0.m a four mall ole of iameter 5/600 m rille troug in te irection of it lengt. Te plunger i a cloe fit inie a cyliner, containing oil, uc tat no oil i aume to pa between te plunger an te cyliner. If te plunger i ubjecte to a vertical ownwar force of 45N (incluing it own weigt) an it i aume tat te upwar flow troug te four mall ole i laminar, etermine te pee of te fall of te plunger. Te coefficient of velocity of te oil i 0. kg/m. [ m/] F = 45N = 5/600 m Q plunger cyliner 0.8m Flow troug eac tube given by Hagen-Poieuille equation p Q 4 L 8 Tere are 4 of tee o total flow i 4 4 p 4 ( 5/ 600) Q 4 p p60. 0 L Force = preure area / 600 F 45 p 4 p N / m 0 0. m Q m / Flow up troug piton = flow iplace by moving piton Q = v piton Example: nwer CIVE400: Flui Mecanic = 0.04 v piton Example: nwer CIVE400: Flui Mecanic 54
28 CIVE400: Flui Mecanic Example: nwer v piton = m/ 7.4 vertical cyliner of metre iameter i mounte concentrically in a rum of 0.076metre internal iameter. Oil fill te pace between tem to a ept of 0.m. Te rotque require to rotate te cyliner in te rum i 4Nm wen te pee of rotation i 7.5 rev/ec. uming tat te en effect are negligible, calculate te coefficient of vicoity of te oil. [0.68 kg/m] From te quetion r - = 0.076/ r = 0.075/ Torque = 4Nm, L = 0.m Te velocity of te ege of te cyliner i: u cyl = 7.5 r = =.767 m/ u rum = 0.0 Torque neee to rotate cyliner T urface area 4r L 654. N / m Ditance between cyliner an rum = r - r = = 0.005m Uing Newton law of vicoity: u r u r kg / m ( N / m ) Example: nwer CIVE400: Flui Mecanic 55 CIVE400: Flui Mecanic Example: nwer Dimenional analyi 8. tationary pere in water moving at a velocity of.6m/ experience a rag of 4N. noter pere of twice te iameter i place in a win tunnel. Fin te velocity of te air an te rag wic will give ynamically imilar conition. Te ratio of kinematic vicoitie of air an water i, an te enity of air.8 kg/m. [0.4m/ 0.865N] Draw up te table of value you ave for eac variable: variable water air u.6m/ u air Drag 4N D air 000 kg/m.8 kg/m Kinematic vicoity i ynamic vicoity over enity = u u Te Reynol number = Re Cooe te tree recurring (governing) variable; u,, From Buckingam teorem we ave m-n = 5 - = non-imenional group. u,,, D, 0, 0 a b c u D a b c u eac group i imenionle ten coniering te imenion, for te firt group, : (note D i a force wit imenion MLT - ) a b c M L T LT L ML MLT M] 0 = c + c = - L] 0 = a + b - c + -4 = a + b T] 0 = -a - a = - b = - u D D u Example: nwer CIVE400: Flui Mecanic 56
29 CIVE400: Flui Mecanic Example: nwer n te econ group : a b c M L T LT L ML L T M] 0 = c L] 0 = a + b - c + - = a + b T] 0 = -a - a = - b = - 0 u u So te pyical ituation i ecribe by ti function of nonimenional number,,, 0 D u u For ynamic imilarity tee non-imenional number are te ame for te bot te pere in water an in te win tunnel i.e. air water For air water D D u u air Dair ( ) D N For u air u water uair 6. u 0. 4m/ air air water Example: nwer CIVE400: Flui Mecanic 57 CIVE400: Flui Mecanic Example: nwer 8. Explain briefly te ue of te Reynol number in te interpretation of tet on te flow of liqui in pipe. Water flow troug a cm iameter pipe at.6m/. Calculate te Reynol number an fin alo te velocity require to give te ame Reynol number wen te pipe i tranporting air. Obtain te ratio of preure rop in te ame lengt of pipe for bot cae. For te water te kinematic vicoity wa.0-6 m / an te enity wa 000 kg/m. For air toe quantitie were m / an.9kg/m. [447, 8.4m/, 0.57] Draw up te table of value you ave for eac variable: variable water air u.6m/ u air p p water p air 000 kg/m.9kg/m m m 000 kg/m.8 kg/m 0.0m 0.0m Kinematic vicoity i ynamic vicoity over enity = u u Te Reynol number = Re Reynol number wen carrying water: u Re water To calculate Re air we know, Re water Reair uair u 8. 44m/ air To obtain te ratio of preure rop we mut obtain an expreion for te preure rop in term of governing variable. Cooe te tree recurring (governing) variable; u,, From Buckingam teorem we ave m-n = 5 - = non-imenional group. u,,,, p 0, 0 a b c u a b c u p eac group i imenionle ten coniering te imenion, for te firt group, : a b c M L T LT L ML L T Example: nwer CIVE400: Flui Mecanic 58
30 CIVE400: Flui Mecanic Example: nwer M] 0 = c L] 0 = a + b - c + - = a + b T] 0 = -a - a = - b = - 0 u u n te econ group : (note p i a preure (force/area) wit imenion ML - T - ) a b c M L T LT L ML MT L M] 0 = c + c = - L] 0 = a + b - c - - = a + b T] 0 = -a - a = - b = 0 u p p u So te pyical ituation i ecribe by ti function of nonimenional number,,, p u u 0 For ynamic imilarity tee non-imenional number are te ame for te bot water an air in te pipe. air water air water We are interete in te relationip involving te preure i.e. CIVE400: Flui Mecanic Example: nwer p p u u air water pwater wateruwater pair airuair Sow tat Reynol number, u/, i non-imenional. If te icarge Q troug an orifice i a function of te iameter, te preure ifference p, te enity, an te vicoity, ow tat Q = Cp / / / were C i ome function of te non-imenional group ( / / /). Draw up te table of value you ave for eac variable: Te imenion of tee following variable are ML - u LT - L ML - T - Re = ML - LT - L(ML - T - ) - = ML - LT - L M - LT = i.e. Re i imenionle. We are tol from te quetion tat tere are 5 variable involve in te problem:, p,, an Q. Cooe te tree recurring (governing) variable; Q,, From Buckingam teorem we ave m-n = 5 - = non-imenional group. Q,,,, p 0, 0 a b c Q a b c Q p eac group i imenionle ten coniering te imenion, for te firt group, : a b c M L T L T L ML ML T M] 0 = c + c = - L] 0 = a + b - c - - = a + b T] 0 = -a - a = - b = Example: nwer CIVE400: Flui Mecanic 59 Example: nwer CIVE400: Flui Mecanic 60
31 CIVE400: Flui Mecanic Example: nwer Q Q n te econ group : (note p i a preure (force/area) wit imenion ML - T - ) a b c M L T L T L ML MT L M] 0 = c + c = - L] 0 = a + b - c - - = a + b T] 0 = -a - a = - b = 4 4 Q p 4 p Q So te pyical ituation i ecribe by ti function of non-imenional number, 4,, 0 p Q Q or 4 p Q Q p p Te quetion want u to ow : Q f / / / / Q Take te reciprocal of quare root of : / p Convert by multiplying by ti number / Q a Q p p a ten we can ay / / / a, CIVE400: Flui Mecanic Example: nwer / / / p p / a, a, / 0 or / / / p p Q / 8.4 cyliner 0.6m in iameter i to be mounte in a tream of water in orer to etimate te force on a tall cimney of m iameter wic i ubject to win of m/. Calculate () te pee of te tream neceary to give ynamic imilarity between te moel an cimney, (b) te ratio of force. Cimney: =.kg/m = 60-6 kg/m Moel: = 000kg/m = 80-4 kg/m [.55m/, 0.057] Draw up te table of value you ave for eac variable: variable water air u u water m/ F F water F air 000 kg/m.kg/m kgm kg/m 0.6m m Kinematic vicoity i ynamic vicoity over enity = u u Te Reynol number = Re For ynamic imilarity: Re water Reair 000uwater u 55. m/ water To obtain te ratio of force we mut obtain an expreion for te force in term of governing variable. Cooe te tree recurring (governing) variable; u,, F, From Buckingam teorem we ave m-n = 5 - = non-imenional group. u,,,, F 0, 0 a b c u a b c u F eac group i imenionle ten coniering te imenion, for te firt group, : Example: nwer CIVE400: Flui Mecanic 6 Example: nwer CIVE400: Flui Mecanic 6
32 CIVE400: Flui Mecanic Example: nwer a b c M L T LT L ML ML T M] 0 = c + c = - L] 0 = a + b - c - - = a + b T] 0 = -a - a = - b = - u u i.e. te (invere of) Reynol number n te econ group : a b c M L T LT L ML ML T M] 0 = c + c = - L] 0 = a + b - c - - = a + b T] 0 = -a - a = - b = - u F F u So te pyical ituation i ecribe by ti function of nonimenional number,,, F u u 0 For ynamic imilarity tee non-imenional number are te ame for te bot water an air in te pipe. air water air water To fin te ratio of force for te ifferent flui ue Example: nwer CIVE400: Flui Mecanic 6 CIVE400: Flui Mecanic Example: nwer air water F F u u air water F F u u air water Fair F water 8.5 If te reitance to motion, R, of a pere troug a flui i a function of te enity an vicoity of te flui, an te raiu r an velocity u of te pere, ow tat R i given by ur R f Hence ow tat if at very low velocitie te reitance R i proportional to te velocity u, ten R = kru were k i a imenionle contant. fine granular material of pecific gravity.5 i in uniform upenion in till water of ept.m. Regaring te particle a pere of iameter 0.00cm fin ow long it will take for te water to clear. Take k=6 an =0.00 kg/m. [8min 9.ec] Cooe te tree recurring (governing) variable; u, r, R, From Buckingam teorem we ave m-n = 5 - = non-imenional group. ur,,,, R 0, 0 a b c u r a b c u r R eac group i imenionle ten coniering te imenion, for te firt group, : a b c M L T LT L ML ML T M] 0 = c + c = - L] 0 = a + b - c - - = a + b T] 0 = -a - a = - b = - u r ur i.e. te (invere of) Reynol number Example: nwer CIVE400: Flui Mecanic 64
33 CIVE400: Flui Mecanic Example: nwer n te econ group : a b c M L T LT L ML ML T M] 0 = c + c = - L] 0 = a + b - c - - = a + b T] 0 = -a - a = - b = - u r R R ur So te pyical ituation i ecribe by ti function of nonimenional number,,, R ur ru 0 or R ru ur CIVE400: Flui Mecanic Example: nwer r = m ept =.m 4 mg kru u u5. 0 m/. t min 9. ec ur e quetion ak u to ow R f or R ur f Multiply te LHS by te quare of te RHS: (i.e. (/ ) ) R ru ur R So R ur f Te quetion tell u tat R i proportional to u o te function f mut be a contant, k R k ur R kru Te water will clear wen te particle moving from te water urface reace te bottom. t terminal velocity tere i no acceleration - te force R = mg - uptrut. From te quetion: =.5 o = 500kg/m = 0.00 kg/m k = 6 Example: nwer CIVE400: Flui Mecanic 65 Example: nwer CIVE400: Flui Mecanic 66
p gh 7. 9 10 0. 4 3160
CIVE400: lui Mecanic Preure an Manometer. Wat will be te (a) te gauge preure an (b) te abolute preure of water at ept m below te urface? water = 000 kg/m, an p atmopere = 0kN/m. [7.7 kn/m, 8.7 kn/m ] a)
p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh
IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: [email protected] r J Noakes:[email protected] January 008 Module web site: www.efm.leeds.ac.uk/ive/fluidslevel1
SLOPE STABILITY SLOPE STABILITY. Types of Slope Failure. Types of Slope Failure. Assistant Professor Berrak Teymur
SLPE STABILITY Aitant Profeor Berrak Teymur Slope failure SLPE STABILITY An expoe groun urface that tan at an angle with the horizontal i calle an unretraine lope. The lope can be natural or man-mae. Slope
FLUID MECHANICS. TUTORIAL No.4 FLOW THROUGH POROUS PASSAGES
FLUID MECHANICS TUTORIAL No.4 FLOW THROUGH POROUS PASSAGES In thi tutorial you will continue the work on laminar flow and develop Poieuille' equation to the form known a the Carman - Kozeny equation. Thi
Sections 3.1/3.2: Introducing the Derivative/Rules of Differentiation
Sections 3.1/3.2: Introucing te Derivative/Rules of Differentiation 1 Tangent Line Before looking at te erivative, refer back to Section 2.1, looking at average velocity an instantaneous velocity. Here
Notes: Most of the material in this chapter is taken from Young and Freedman, Chap. 12.
Capter 6. Fluid Mecanics Notes: Most of te material in tis capter is taken from Young and Freedman, Cap. 12. 6.1 Fluid Statics Fluids, i.e., substances tat can flow, are te subjects of tis capter. But
FLUID MECHANICS. TUTORIAL No.7 FLUID FORCES. When you have completed this tutorial you should be able to. Solve forces due to pressure difference.
FLUID MECHANICS TUTORIAL No.7 FLUID FORCES When you have completed this tutorial you should be able to Solve forces due to pressure difference. Solve problems due to momentum changes. Solve problems involving
MECH 2110 - Statics & Dynamics
Chapter D Problem 3 Solution 1/7/8 1:8 PM MECH 11 - Static & Dynamic Chapter D Problem 3 Solution Page 7, Engineering Mechanic - Dynamic, 4th Edition, Meriam and Kraige Given: Particle moving along a traight
Pressure. Pressure. Atmospheric pressure. Conceptual example 1: Blood pressure. Pressure is force per unit area:
Pressure Pressure is force per unit area: F P = A Pressure Te direction of te force exerted on an object by a fluid is toward te object and perpendicular to its surface. At a microscopic level, te force
Physics 110 Spring 2006 Fluid Mechanics Their Solutions
ic Sprin 00 lui Mecanic Teir Solution reure in a flui arie linearl wit ept Wat i te abolute preure at a ept of 00 in te ocean if te enit of ea i 4 k/? t ti ept wat force ut te frae aroun a circular ubarine
Calculating Viscous Flow: Velocity Profiles in Rivers and Pipes
previous inex next Calculating Viscous Flow: Velocity Profiles in Rivers an Pipes Michael Fowler, UVa 9/8/1 Introuction In this lecture, we ll erive the velocity istribution for two examples of laminar
Math 113 HW #5 Solutions
Mat 3 HW #5 Solutions. Exercise.5.6. Suppose f is continuous on [, 5] and te only solutions of te equation f(x) = 6 are x = and x =. If f() = 8, explain wy f(3) > 6. Answer: Suppose we ad tat f(3) 6. Ten
OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS
Unit 41: Fluid Mechanics Unit code: T/601/1445 QCF Level: 4 Credit value: 15 OUTCOME 1 STATIC FLUID SYSTEMS TUTORIAL 1 - HYDROSTATICS 1. Be able to determine the behavioural characteristics and parameters
ACT Math Facts & Formulas
Numbers, Sequences, Factors Integers:..., -3, -2, -1, 0, 1, 2, 3,... Rationals: fractions, tat is, anyting expressable as a ratio of integers Reals: integers plus rationals plus special numbers suc as
SAT Subject Math Level 1 Facts & Formulas
Numbers, Sequences, Factors Integers:..., -3, -2, -1, 0, 1, 2, 3,... Reals: integers plus fractions, decimals, and irrationals ( 2, 3, π, etc.) Order Of Operations: Aritmetic Sequences: PEMDAS (Parenteses
Section 3.3. Differentiation of Polynomials and Rational Functions. Difference Equations to Differential Equations
Difference Equations to Differential Equations Section 3.3 Differentiation of Polynomials an Rational Functions In tis section we begin te task of iscovering rules for ifferentiating various classes of
Hole Detection for Increasing Coverage in Wireless Sensor Network Using Triangular Structure
IJCSI International Journal of Computer Science Iue, Vol. 9, Iue, No, January 0 ISSN (Online): 694-084 www.ijcsi.org 3 Hole Detection for Increaing Coverage in Wirele Senor Network Uing Triangular Structure
L4. Pipes Materials and Loads
The Islamic University of Gaza- Civil Engineering Department SanitaryEngineering- ECIV 4325 L4. Pipes Materials an Loas Base on Dr. Fahi Rabah lecture notes Pipes Materials an loas Materials use for pipes:
v = x t = x 2 x 1 t 2 t 1 The average speed of the particle is absolute value of the average velocity and is given Distance travelled t
Chapter 2 Motion in One Dimenion 2.1 The Important Stuff 2.1.1 Poition, Time and Diplacement We begin our tudy of motion by conidering object which are very mall in comparion to the ize of their movement
New Vocabulary volume
-. Plan Objectives To find te volume of a prism To find te volume of a cylinder Examples Finding Volume of a Rectangular Prism Finding Volume of a Triangular Prism 3 Finding Volume of a Cylinder Finding
Answers to the Practice Problems for Test 2
Answers to the Practice Problems for Test 2 Davi Murphy. Fin f (x) if it is known that x [f(2x)] = x2. By the chain rule, x [f(2x)] = f (2x) 2, so 2f (2x) = x 2. Hence f (2x) = x 2 /2, but the lefthan
Physics 111. Exam #1. January 24, 2014
Phyic 111 Exam #1 January 24, 2014 Name Pleae read and follow thee intruction carefully: Read all problem carefully before attempting to olve them. Your work mut be legible, and the organization clear.
Engineering Bernoulli Equation
Engineering Bernoulli Equation R. Shankar Subramanian Department of Chemical and Biomolecular Engineering Clarkon Univerity The Engineering Bernoulli equation can be derived from the principle of conervation
11 CHAPTER 11: FOOTINGS
CHAPTER ELEVEN FOOTINGS 1 11 CHAPTER 11: FOOTINGS 11.1 Introuction Footings are structural elements that transmit column or wall loas to the unerlying soil below the structure. Footings are esigne to transmit
Linear Momentum and Collisions
Chapter 7 Linear Momentum and Colliion 7.1 The Important Stuff 7.1.1 Linear Momentum The linear momentum of a particle with ma m moving with velocity v i defined a p = mv (7.1) Linear momentum i a vector.
iglidur PRT Slewing Ring Bearing
igliur igliur Slewing Ring Bearing Stanar range Completely maintenance-free Easy installation, intercangeable sliing pas ig wear resistance For ig loa capacity, ig stiffness Available as stainless steel
CE 3500 Fluid Mechanics / Fall 2014 / City College of New York
1 Drag Coefficient The force ( F ) of the wind blowing against a building is given by F=C D ρu 2 A/2, where U is the wind speed, ρ is density of the air, A the cross-sectional area of the building, and
SIMULATION OF DIRECT TORQUE CONTROLLED PERMANENT MAGNET SYNCHRONOUS MOTOR DRIVE
SIMULATION OF DIRECT TORQUE CONTROLLED PERMANENT MAGNET SYNCHRONOUS MOTOR DRIVE Selin Özçıra Nur Bekiroğlu Engin Ayçiçek e-mail: [email protected] e-mail: [email protected] e-mail: [email protected]
Midterm Solutions. mvr = ω f (I wheel + I bullet ) = ω f 2 MR2 + mr 2 ) ω f = v R. 1 + M 2m
Midterm Solutions I) A bullet of mass m moving at horizontal velocity v strikes and sticks to the rim of a wheel a solid disc) of mass M, radius R, anchored at its center but free to rotate i) Which of
The EOQ Inventory Formula
Te EOQ Inventory Formula James M. Cargal Matematics Department Troy University Montgomery Campus A basic problem for businesses and manufacturers is, wen ordering supplies, to determine wat quantity of
Example Optimization Problems selected from Section 4.7
Example Optimization Problems selecte from Section 4.7 19) We are aske to fin the points ( X, Y ) on the ellipse 4x 2 + y 2 = 4 that are farthest away from the point ( 1, 0 ) ; as it happens, this point
f(x + h) f(x) h as representing the slope of a secant line. As h goes to 0, the slope of the secant line approaches the slope of the tangent line.
Derivative of f(z) Dr. E. Jacobs Te erivative of a function is efine as a limit: f (x) 0 f(x + ) f(x) We can visualize te expression f(x+) f(x) as representing te slope of a secant line. As goes to 0,
6. Friction, Experiment and Theory
6. Friction, Experiment and Theory The lab thi wee invetigate the rictional orce and the phyical interpretation o the coeicient o riction. We will mae ue o the concept o the orce o gravity, the normal
EXPERIMENT 11 CONSOLIDATION TEST
119 EXPERIMENT 11 CONSOLIDATION TEST Purpoe: Thi tet i performed to determine the magnitude and rate of volume decreae that a laterally confined oil pecimen undergoe when ubjected to different vertical
oil liquid water water liquid Answer, Key Homework 2 David McIntyre 1
Answer, Key Homework 2 David McIntyre 1 This print-out should have 14 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making
Pipe Flow Calculations
Pipe Flow Calculation R. Shankar Subramanian epartment o Chemical and Biomolecular Engineering Clarkon Univerity We begin with ome reult that we hall ue when making riction lo calculation or teady, ully
Lecture 10: What is a Function, definition, piecewise defined functions, difference quotient, domain of a function
Lecture 10: Wat is a Function, definition, piecewise defined functions, difference quotient, domain of a function A function arises wen one quantity depends on anoter. Many everyday relationsips between
σ m using Equation 8.1 given that σ
8. Etimate the theoretical fracture trength of a brittle material if it i known that fracture occur by the propagation of an elliptically haped urface crack of length 0.8 mm and having a tip radiu of curvature
Heat transfer to or from a fluid flowing through a tube
Heat tranfer to or from a fluid flowing through a tube R. Shankar Subramanian A common ituation encountered by the chemical engineer i heat tranfer to fluid flowing through a tube. Thi can occur in heat
Shell and Tube Heat Exchanger
Sell and Tube Heat Excanger MECH595 Introduction to Heat Transfer Professor M. Zenouzi Prepared by: Andrew Demedeiros, Ryan Ferguson, Bradford Powers November 19, 2009 1 Abstract 2 Contents Discussion
Bob York. Simple FET DC Bias Circuits
Bob York Simple FET DC Bia Circuit Loa-Line an Q-point Conier the effect of a rain reitor in the comnon-ource configuration: Smaller + g D out KL: Thi i the equation of a line that can be uperimpoe on
Section 2.3 Solving Right Triangle Trigonometry
Section.3 Solving Rigt Triangle Trigonometry Eample In te rigt triangle ABC, A = 40 and c = 1 cm. Find a, b, and B. sin 40 a a c 1 a 1sin 40 7.7cm cos 40 b c b 1 b 1cos40 9.cm A 40 1 b C B a B = 90 - A
An Introduction to Fluid Mechanics
0. Contents of the Course Notes For the First Year Lecture Course: An Introduction to Fluid Mechanics School of Civil Engineering, University of Leeds. CIVE1400 FLUID MECHANICS Dr Andrew Sleigh January
du u U 0 U dy y b 0 b
BASIC CONCEPTS/DEFINITIONS OF FLUID MECHANICS (by Marios M. Fyrillas) 1. Density (πυκνότητα) Symbol: 3 Units of measure: kg / m Equation: m ( m mass, V volume) V. Pressure (πίεση) Alternative definition:
A) When two objects slide against one another, the magnitude of the frictional force is always equal to μ
Phyic 100 Homewor 5 Chapter 6 Contact Force Introduced ) When two object lide againt one another, the magnitude of the frictional force i alway equal to μ B) When two object are in contact with no relative
2 Limits and Derivatives
2 Limits and Derivatives 2.7 Tangent Lines, Velocity, and Derivatives A tangent line to a circle is a line tat intersects te circle at exactly one point. We would like to take tis idea of tangent line
Tangent Lines and Rates of Change
Tangent Lines and Rates of Cange 9-2-2005 Given a function y = f(x), ow do you find te slope of te tangent line to te grap at te point P(a, f(a))? (I m tinking of te tangent line as a line tat just skims
Determine the perimeter of a triangle using algebra Find the area of a triangle using the formula
Student Name: Date: Contact Person Name: Pone Number: Lesson 0 Perimeter, Area, and Similarity of Triangles Objectives Determine te perimeter of a triangle using algebra Find te area of a triangle using
Lecture L25-3D Rigid Body Kinematics
J. Peraire, S. Winall 16.07 Dynamics Fall 2008 Version 2.0 Lecture L25-3D Rigi Boy Kinematics In this lecture, we consier the motion of a 3D rigi boy. We shall see that in the general three-imensional
i( t) L i( t) 56mH 1.1A t = τ ln 1 = ln 1 ln 1 6.67ms
Exam III PHY 49 Summer C July 16, 8 1. In the circuit shown, L = 56 mh, R = 4.6 Ω an V = 1. V. The switch S has been open for a long time then is suenly close at t =. At what value of t (in msec) will
Chapter 4. Forces and Newton s Laws of Motion. continued
Chapter 4 Forces and Newton s Laws of Motion continued 4.9 Static and Kinetic Frictional Forces When an object is in contact with a surface forces can act on the objects. The component of this force acting
Homework 8. problems: 10.40, 10.73, 11.55, 12.43
Hoework 8 probles: 0.0, 0.7,.55,. Proble 0.0 A block of ass kg an a block of ass 6 kg are connecte by a assless strint over a pulley in the shape of a soli isk having raius R0.5 an ass M0 kg. These blocks
Derivatives Math 120 Calculus I D Joyce, Fall 2013
Derivatives Mat 20 Calculus I D Joyce, Fall 203 Since we ave a good understanding of its, we can develop derivatives very quickly. Recall tat we defined te derivative f x of a function f at x to be te
1.6. Analyse Optimum Volume and Surface Area. Maximum Volume for a Given Surface Area. Example 1. Solution
1.6 Analyse Optimum Volume and Surface Area Estimation and oter informal metods of optimizing measures suc as surface area and volume often lead to reasonable solutions suc as te design of te tent in tis
Tripping Current and Short Circuit Current of High Voltage Tester
Tripping Current an Short Circuit Current of igh oltage Teter Xu hezhun Wang Keqin Abtract: Tripping current an hort circuit current are iportant technical pecification of high voltage teter, an frequently
Fluid Pressure and Fluid Force
0_0707.q //0 : PM Page 07 SECTION 7.7 Section 7.7 Flui Pressure an Flui Force 07 Flui Pressure an Flui Force Fin flui pressure an flui force. Flui Pressure an Flui Force Swimmers know that the eeper an
1D STEADY STATE HEAT
D SEADY SAE HEA CONDUCION () Pabal alukda Aociate Pofeo Depatment of Mecanical Engineeing II Deli E-mail: [email protected] Palukda/Mec-IID emal Contact eitance empeatue ditibution and eat flow line
MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 1 - BASIC DIFFERENTIATION
MATHEMATICS FOR ENGINEERING DIFFERENTIATION TUTORIAL 1 - BASIC DIFFERENTIATION Tis tutorial is essential pre-requisite material for anyone stuing mecanical engineering. Tis tutorial uses te principle of
13 PERIMETER AND AREA OF 2D SHAPES
13 PERIMETER AND AREA OF D SHAPES 13.1 You can find te perimeter of sapes Key Points Te perimeter of a two-dimensional (D) sape is te total distance around te edge of te sape. l To work out te perimeter
PHY231 Section 2, Form A March 22, 2012. 1. Which one of the following statements concerning kinetic energy is true?
1. Which one of the following statements concerning kinetic energy is true? A) Kinetic energy can be measured in watts. B) Kinetic energy is always equal to the potential energy. C) Kinetic energy is always
Perimeter, Area and Volume of Regular Shapes
Perimeter, Area and Volume of Regular Sapes Perimeter of Regular Polygons Perimeter means te total lengt of all sides, or distance around te edge of a polygon. For a polygon wit straigt sides tis is te
FREESTUDY HEAT TRANSFER TUTORIAL 3 ADVANCED STUDIES
FREESTUDY HEAT TRANSFER TUTORIAL ADVANCED STUDIES This is the third tutorial in the series on heat transfer and covers some of the advanced theory of convection. The tutorials are designed to bring the
Unit 11 Using Linear Regression to Describe Relationships
Unit 11 Uing Linear Regreion to Decribe Relationhip Objective: To obtain and interpret the lope and intercept of the leat quare line for predicting a quantitative repone variable from a quantitative explanatory
CHAPTER 8: DIFFERENTIAL CALCULUS
CHAPTER 8: DIFFERENTIAL CALCULUS 1. Rules of Differentiation As we ave seen, calculating erivatives from first principles can be laborious an ifficult even for some relatively simple functions. It is clearly
Instantaneous Rate of Change:
Instantaneous Rate of Cange: Last section we discovered tat te average rate of cange in F(x) can also be interpreted as te slope of a scant line. Te average rate of cange involves te cange in F(x) over
Fluids and Solids: Fundamentals
Fluids and Solids: Fundamentals We normally recognize three states of matter: solid; liquid and gas. However, liquid and gas are both fluids: in contrast to solids they lack the ability to resist deformation.
Experiment (13): Flow channel
Introduction: An open channel is a duct in which the liquid flows with a free surface exposed to atmospheric pressure. Along the length of the duct, the pressure at the surface is therefore constant and
CE 6303 MECHANICS OF FLUIDS L T P C QUESTION BANK PART - A
CE 6303 MECHANICS OF FLUIDS L T P C QUESTION BANK 3 0 0 3 UNIT I FLUID PROPERTIES AND FLUID STATICS PART - A 1. Define fluid and fluid mechanics. 2. Define real and ideal fluids. 3. Define mass density
PHY231 Section 1, Form B March 22, 2012
1. A car enters a horizontal, curved roadbed of radius 50 m. The coefficient of static friction between the tires and the roadbed is 0.20. What is the maximum speed with which the car can safely negotiate
PHY121 #8 Midterm I 3.06.2013
PHY11 #8 Midterm I 3.06.013 AP Physics- Newton s Laws AP Exam Multiple Choice Questions #1 #4 1. When the frictionless system shown above is accelerated by an applied force of magnitude F, the tension
A new definition of the fractional Laplacian
A new efinition of the fractional Laplacian W. Chen imula Reearch Laboratory, P. O. Box. 34, NO-325 Lyaker, Norway (9 eptember 2002) Keywor: fractional Laplacian, fractional erivative, Riez potential,
As customary, choice (a) is the correct answer in all the following problems.
PHY2049 Summer 2012 Instructor: Francisco Rojas Exam 1 As customary, choice (a) is the correct answer in all the following problems. Problem 1 A uniformly charge (thin) non-conucting ro is locate on the
Incline and Friction Examples
Incline and riction Eample Phic 6A Prepared b Vince Zaccone riction i a force that oppoe the motion of urface that are in contact with each other. We will conider 2 tpe of friction in thi cla: KINETIC
MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS
MECHANICS OF SOLIDS - BEAMS TUTORIAL 2 SHEAR FORCE AND BENDING MOMENTS IN BEAMS This is the second tutorial on bending of beams. You should judge your progress by completing the self assessment exercises.
1. Fluids Mechanics and Fluid Properties. 1.1 Objectives of this section. 1.2 Fluids
1. Fluids Mechanics and Fluid Properties What is fluid mechanics? As its name suggests it is the branch of applied mechanics concerned with the statics and dynamics of fluids - both liquids and gases.
Experiment 3 Pipe Friction
EML 316L Experiment 3 Pipe Friction Laboratory Manual Mechanical and Materials Engineering Department College of Engineering FLORIDA INTERNATIONAL UNIVERSITY Nomenclature Symbol Description Unit A cross-sectional
Here the units used are radians and sin x = sin(x radians). Recall that sin x and cos x are defined and continuous everywhere and
Lecture 9 : Derivatives of Trigonometric Functions (Please review Trigonometry uner Algebra/Precalculus Review on the class webpage.) In this section we will look at the erivatives of the trigonometric
SAT Math Must-Know Facts & Formulas
SAT Mat Must-Know Facts & Formuas Numbers, Sequences, Factors Integers:..., -3, -2, -1, 0, 1, 2, 3,... Rationas: fractions, tat is, anyting expressabe as a ratio of integers Reas: integers pus rationas
Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces. Copyright 2009 Pearson Education, Inc.
Chapter 5 Using Newton s Laws: Friction, Circular Motion, Drag Forces Units of Chapter 5 Applications of Newton s Laws Involving Friction Uniform Circular Motion Kinematics Dynamics of Uniform Circular
Chapter 10: Refrigeration Cycles
Capter 10: efrigeration Cycles Te vapor compression refrigeration cycle is a common metod for transferring eat from a low temperature to a ig temperature. Te above figure sows te objectives of refrigerators
www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x
Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity
Section 2.2 Arc Length and Sector Area. Arc Length. Definition. Note:
Section. Arc Length and Sector Area Arc Length Definition If a central angle, in a circle of a radiu r, cut off an arc of length, then the meaure of, in radian i: r r r r ( in radian) Note: When applying
Backwater Rise and Drag Characteristics of Bridge Piers under Subcritical
European Water 36: 7-35, 11. 11 E.W. Publications Backwater Rise and Drag Characteristics of Bridge Piers under Subcritical Flow Conditions C.R. Suribabu *, R.M. Sabarish, R. Narasimhan and A.R. Chandhru
Description: Conceptual questions about projectile motion and some easy calculations. (uses applets)
Week 3: Chapter 3 [ Edit ] Overview Suary View Diagnotic View Print View with Anwer Week 3: Chapter 3 Due: 11:59p on Sunday, February 8, 2015 To undertand how point are awarded, read the Grading Policy
Angular acceleration α
Angular Acceleration Angular acceleration α measures how rapidly the angular velocity is changing: Slide 7-0 Linear and Circular Motion Compared Slide 7- Linear and Circular Kinematics Compared Slide 7-
Mercury is poured into a U-tube as in Figure (14.18a). The left arm of the tube has crosssectional
Chapter 14 Fluid Mechanics. Solutions of Selected Problems 14.1 Problem 14.18 (In the text book) Mercury is poured into a U-tube as in Figure (14.18a). The left arm of the tube has crosssectional area
XI / PHYSICS FLUIDS IN MOTION 11/PA
Viscosity It is the property of a liquid due to which it flows in the form of layers and each layer opposes the motion of its adjacent layer. Cause of viscosity Consider two neighboring liquid layers A
Chapter H - Problems
Chapter H - Problem Blinn College - Phyic 45 - Terry Honan Problem H.1 A wheel rotate from ret to 1 ê in 3. Aume the angular acceleration i contant. (a) What i the magnitude of the wheel' angular acceleration?
01 The Nature of Fluids
01 The Nature of Fluids WRI 1/17 01 The Nature of Fluids (Water Resources I) Dave Morgan Prepared using Lyx, and the Beamer class in L A TEX 2ε, on September 12, 2007 Recommended Text 01 The Nature of
Unit 1 INTRODUCTION 1.1.Introduction 1.2.Objectives
Structure 1.1.Introduction 1.2.Objectives 1.3.Properties of Fluids 1.4.Viscosity 1.5.Types of Fluids. 1.6.Thermodynamic Properties 1.7.Compressibility 1.8.Surface Tension and Capillarity 1.9.Capillarity
Heat Exchangers. Heat Exchanger Types. Heat Exchanger Types. Applied Heat Transfer Part Two. Topics of This chapter
Applied Heat Transfer Part Two Heat Excangers Dr. Amad RAMAZANI S.A. Associate Professor Sarif University of Tecnology انتقال حرارت کاربردی احمد رمضانی سعادت ا بادی Autumn, 1385 (2006) Ramazani, Heat Excangers
FLUID FLOW Introduction General Description
FLUID FLOW Introduction Fluid flow is an important part of many processes, including transporting materials from one point to another, mixing of materials, and chemical reactions. In this experiment, you
Two Dimensional FEM Simulation of Ultrasonic Wave Propagation in Isotropic Solid Media using COMSOL
Excerpt from the Proceeding of the COMSO Conference 0 India Two Dimenional FEM Simulation of Ultraonic Wave Propagation in Iotropic Solid Media uing COMSO Bikah Ghoe *, Krihnan Balaubramaniam *, C V Krihnamurthy
Solution Derivations for Capa #11
Solution Derivations for Capa #11 1) A horizontal circular platform (M = 128.1 kg, r = 3.11 m) rotates about a frictionless vertical axle. A student (m = 68.3 kg) walks slowly from the rim of the platform
EC201 Intermediate Macroeconomics. EC201 Intermediate Macroeconomics Problem set 8 Solution
EC201 Intermediate Macroeconomics EC201 Intermediate Macroeconomics Prolem set 8 Solution 1) Suppose tat te stock of mone in a given econom is given te sum of currenc and demand for current accounts tat
Fluid Mechanics: Static s Kinematics Dynamics Fluid
Fluid Mechanics: Fluid mechanics may be defined as that branch of engineering science that deals with the behavior of fluid under the condition of rest and motion Fluid mechanics may be divided into three
An inquiry into the multiplier process in IS-LM model
An inquiry into te multiplier process in IS-LM model Autor: Li ziran Address: Li ziran, Room 409, Building 38#, Peing University, Beijing 00.87,PRC. Pone: (86) 00-62763074 Internet Address: [email protected]
