FLUID FLOW Mechanical Energy Balance V gz 2 2 dp potential expansion energy change work losses Wo F kinetic work added/ sum of energy change subtracted by friction pumps or compressors Note that the balance is per unit mass. In differential form Rewrite as follows dp g dz V dV F Wo divide by dL (L is the length of pipe) dp dz dV F W g V o dL Tot dL dL L L or : dp dp dp dp dL Tot dL elev dL accel dL frict ( Wo is usually ignored, as the equation applies to a section of pipe ) L The above equation is an alternative way of writing the mechanical energy balance. It is not a different equation. The differential form of the potential energy change is dL dZ g dZ g sin dL What about the friction losses? 1) Fanning or Darcy-Weisbach equation (Often called Darcy equation) 2V 2 f F dL D This equation applies for single phase fluids !!! The friction factor is obtained from the Moody Diagram (see P&T page 482). Friction factor equations. (Much needed in the era of computers and excel) 16 Re 0.046 f Re a f Laminar Flow smooth pipes: a=0.2 Iron or steel pipes a=0.16 1 2.51 2 log10 f 3.7 D Re f Colebrook equation for turbulent flow. Equivalent length of valves and fittings. Pressure drop for valves and fittings is accounted for as equivalent length of pipe. Please refer to P&T for a table containing these values (page 484). SCENARIO I Piping is known. Need pressure drop. (Pump or compressor is not present.) Incompressible Flow a) Isothermal (is constant) dp dV F dZ g V dL dL Tot dL L for a fixed V constant dV = 0 L F 2 V 2 f D L p g Z 2V 2 f D F b) Nonisothermal It will not have a big error if you use (Taverage), v(Taverage) Compressible Flow (Gasses) a) Relatively small change in T (known) For small pressure drop (something you can check after you are done) can use Bernoulli and fanning equation as flows V2 g dz v dp d F 2 g 1 V F dz dp 2 dV 2 2 v v v v V v but G A V = Velocity v = Specific volume (m3/Kg) G = Molar flow (Kg/hr) A = Cross sectional area g 1 F G dV G dL dz dp 2 2 f 2 A v A D v v v 2 Now put in integral form dp G g 2 v A v dz Assume 2 2 dV G 1 2 f dL A D V Tin Tout 2 fin fout 2 (Tin , Pin ) (Tout , Pout ) 2 f (Tin , Pin ) f (Tout , Pout ) 2 Tav fav fav The integral form will be g z 2 av Now use Then out in dp G Vout L G ln 2 f av A v A Vin D 2 pv Z RT M v Z av RTav pM 2 M; Molecular weight dp 1 v Zav RTav p dp 2 Z 1 2 pout pin2 av RTav Therefore ; 2av g z 2 2 V M L G G 2 2 pout pin ln out 2 f av A A 2 Zav RTav D Vin but, Vout Zout Tout pin Vin Zin Tin pout Z RT 2 pout pin 2 av av M 1 Zout Tout pin L G G 2 2 2 f ln g z av av A D A ZinTin pout 2 2 This is an equation of the form p out F( p out ) Algorithm a) Assume p (out1) b) Use formula to get a new value p (out2 ) F p (out1) c) Continue using p (outi 1) F p (outi) until i 1) i) p (out p (out i) p (out OR BETTER: USE Solver in EXCEL, or even better use PRO II, or any other fluid flow simulator. CAN THIS BE APPLIED TO LONG PIPES. What is the error ? ===> If pout pin 0.2 0.3 you will be OK. What to do if not. Use shorter pin sections of pipe. What if temperature change is not known Use total energy balance as your second equation g dz d ( pv) VdV du q w o Potential Energy rate of work done on fluid element by pressure forces change of internal energy Energy Kinetic Energy work done on fluid element by external forces Heat Note : friction is not present ! external temp q UTo T DdL } dA 1 G Flow rate Flux Then, (ignore wo ,will not use when pumps or compressors are not present) UTo TDdL V2 gdz dh d G 2 Integrate and solve for hout (use Tav in the heat transfer equation) hout hin 2 U To Tav DL Vout Vin2 g z 2 z1 G 2 But Vout v out RTav G G Z av A pout M A Tout Tout p out , h out Procedure : a) Assume Tout, pout b) Use mechanical energy balance to obtain p (out1) c) Use total energy balance to obtain h (out1) (1) d) get temperature Tout e) continue until convergence Heat Balance Subtract mechanical energy balance from total energy balance to get dh vdp q F Integrate to get the result (use averages as before) h out h in UTo Tav D G 2 1 G 1 L f av p out p in 2 av A av D How is it done in simulators? Pipe is divided in several "short" segments and either averaging is done, or the inlet temperature is used. SCENARIO II Have turbine or Compressor/pump need Wo Easy : use total energy with q = 0 and z = 0 dh dw o dV 2 2 Joule/sec wo V2 W h G 2 Joule/g g/sec V2 W G h 2 Positive for compressor Negative for turbine for turbines, this is about 0. h is known for turbines but not for compressors. Therefore we need to go back to the Mechanical Energy equation for pumps/compressors. Indeed, the Bernoulli equation gives neglect (?) V 2 W G vdp 2 Pumps ( is constant) 1 vdp p d ps W G p For compressors pvn = constant (The evolution is nearly isentropic) n = Cp/Cv (Ideal gas) n Cp/Cv (Real gas) 1 Substitute v p s n 1 vs 1 n integrate to get n 1 n pd n W G 1 p s v s n 1 p s