Energy Conservation (Bernoulli’s Equation)Bernoulli’s Equation (Continued)PowerPoint PresentationBernoulli’s Eqn/Energy Conservation (cont.)Slide 5Energy conservation(cont.)Energy conservation (cont.)Frictional losses in piping systemSlide 9Slide 10Losses in Pipe FlowsMinor Loss through flow entranceEnergy Conservation (cont.)Slide 14Slide 15Conservation of Energy – ApplicationSlide 17Energy Conservation (Bernoulli’s Equation)If one integrates Euler’s eqn. along a streamline, between two points ,  &Which gives us the Bernoulli’s EquationConstant 2222221211 gzVpgzVpFlow work + kinetic energy + potential energy = constant0212121gdzVdVdp0 gdzVdVdpRecall Euler’s equation:Also recall that viscous forces were neglected, i.e. flow is invisicdWe get :Bernoulli’s Equation (Continued)pAxtWAVppAVtxAptxpAtW1,Flow Work (p/) : It is the work required to move fluid across the control volume boundaries.Consider a fluid element of cross-sectional area A with pressure p acting on the control surface as shown.Due to the fluid pressure, the fluid element moves a distance x within time t. Hence, the work done per unit time W/t (flow power) is:Flow work per unit massFlow work or Power1/mass flow ratepvpFlow work is often also referred to as flow energyt)unit weighper (energy g where,2222221211 zgVpzgVpVery Important: Bernoulli’s equation is only valid for :incompressible fluids, steady flow along a streamline, no energy loss due to friction, no heat transfer.Application of Bernoulli’s equation - Example 1:Determine the velocity and mass flow rate of efflux from the circular hole (0.1 m dia.) at the bottom of the water tank (at this instant). The tank is open to the atmosphere and H=4 mH12p1 = p2, V1=0)/(5.69)85.8()1.0(4*1000)/(85.84*8.9*22)(22212skgAVmsmgHzzgVBernoulli’s Equation (Cont)Bernoulli’s Eqn/Energy Conservation (cont.)Example 2: If the tank has a cross-sectional area of 1 m2, estimate the time required to drain the tank to level 2. h(t)12First, choose the control volume as enclosedby the dotted line. Specify h=h(t) as the waterlevel as a function of time.0 20 40 60 80 10001234time (sec.)water height (m)42.5e-007h( )t1000 tsec 90.3 t 0.0443t- 204hEnergy exchange (conservation) in a thermal system12112zgVp22222zgVpEnergy added, hA(ex. pump, compressor)Energy extracted, hE(ex. turbine, windmill)Energy lost, hL(ex. friction, valve, expansion)pumpturbineheat exchangercondenserhEhAhL, friction lossthrough pipeshLloss throughelbowshLloss throughvalvesEnergy conservation(cont.)2222121122 zgVphhhzgVpLEAExample: Determine the efficiency of the pump if the power input of the motoris measured to be 1.5 hp. It is known that the pump delivers 300 gal/min of water.Mercury (m=844.9 lb/ft3)water (w=62.4 lb/ft3)1 hp=550 lb-ft/sNo turbine work and frictional losses, hence: hE=hL=0. Also z1=z26-in dia. pipe4-in dia.pipeGiven: Q=300 gal/min=0.667 ft3/s=AVV1= Q/A1=3.33 ft/s V2=Q/A2=7.54 ft/skinetic energy head gain V Vgft2212 2 227 54 3 332 32 20 71( . ) ( . )* .. ,p z z p z zp p zlb ftw o m w o wm w1 22 129 62 1 25 97813            ( )(844. .4)* . . /pumpZ=15 in12zoIf energy is added, removed or lost via pumps turbines, friction, etc.then we useExtended Bernoulli’s EquationLooking at the pressure term:Energy conservation (cont.)Example (cont.)Pressure head gain:pump work p pfthp p V VgftwAw2 12 1 2212978136215 67216 38 ..4. ( ). ( )Flow power delivered by pumpP =Efficiency =PPwinputQhft lb shp ft lb sP hpA    ( .4)( . )( . ). ( / )/.... .62 0 667 16 38681 71 5501 241 241 50 827 82 7%Frictional losses in piping systemloss head frictional22 equation, sBernoulli' Extended2122221211LLEAhpppzgVphhhzgVpP1P2Consider a laminar, fully developed circular pipe flowpP+dpwDarcy’s Equation:R: radius, D: diameterL: pipe lengthw: wall shear stress[ ( )]( ) ( ) ,,p p dp R R dxdpRdxp p phgLDfLDVgwwLw    FHIKFHIKFHGIKJ   21 222242Pressure force balances frictional force integrate from 1 to 2where f is defined as frictional factor characterizingpressure loss due to pipe wall shear stresswf VFHIKFHGIKJ4 22242VfwgVDLfDLghwL242 When the pipe flow is laminar, it can be shown (not here) that by recognizing that as Reynolds numberTherefore, frictional factor is a function of the Reynolds numberSimilarly, for a turbulent flow, f = function of Reynolds number also. Another parameter that influences the friction is the surfaceroughness as relativeto the pipe diameter DSuch that D Pipe frictional factor is a function of pipe Reynoldsnumber and the relative roughness of pipe.This relation is sketched in the Moody diagram as shown in the following page.The diagram shows f as a function of the Reynolds number (Re), with a series ofparametric curves related to the relative roughness DfVDVDff Ff F FHIKFHIK6464, Re ,Re,(Re).Re, :.DFfRe,DLosses in Pipe FlowsMajor Losses: due to friction, significant head loss is associated with the straight portions of pipe flows. This loss can be calculated using the Moody chart or Colebrook equation.Minor Losses: Additional components (valves, bends, tees, contractions, etc) inpipe flows also contribute to the total head loss of the system. Their contributionsare generally termed minor losses.The head losses and pressure drops can be characterized by using the loss coefficient,KL, which is defined asOne of the example of minor losses is the entrance flow loss. A typical flow pattern for flow entering a sharp-edged entrance is shown in the following page. A vena contracta region is formed at the inlet because the fluid can not turn a sharp corner.Flow separation and associated viscous effects will tend to decrease the flow energy;the phenomenon is fairly