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.99 V0 = 8.087 ft/s.In this instance the time to steady flow is 66.4 s.EXTERNAL EQUATCOMMON SFSF=7.5CALL SIMPR(EQUAT, 0.0, 8.09, TIME, 1.0E-04, 30)WRITE(*,*) ' Time = ', TIMEENDCFUNCTION EQUAT(V)COMMON SFRE = 1.6433854E5*VIF (RE.LT.100.) THENFR = 0.64ELSEIF (RE.LT.2100) THENFR = 64.0/REELSE1 SF1 = SFSF = 1.14-2.0*ALOG10(6.3625E-4 + 9.35*SF1/RE)IF(ABS(SF-SF1).GT.1.0E-06) GO TO 1FR = 1.0/SF/SFENDIFDEM = 100.0 - (6.5 + 5000.0*FR)*V**2/64.4IF(DEM.LT.01) THENEQUAT = 31055.9ELSEEQUAT = 310.559/DEMENDIFRETURNENDThe plot of the results from part (c), requested as part (d), shows the velocity increasesrather rapidly until it reaches 80 to 90% of the steady-state value.By that time theacceleration has decreased noticeably, and steady state is approached asymptotically.Aspart (e) shows, only a short program is needed to add more accuracy to the computation,but in this particular example the difference in time to steady state is only four percent.1.00.8o0.6V/V0.40.25101520253035404550t, sec© 2000 by CRC Press LLCFrom these computations we see that pipe friction is the dominant factor in the flowestablishment process when the pipe is sufficiently long.Overlooking this factor wouldbe a severe error, part (a), but for very long pipes the effect of local losses is truly a minoreffect, with the the steady-state velocities and flow-establishment times only differing by afew percent in this problem.* * *The simulation of flow shutdown by use of Eq.7.30 for the physical problem depictedin Fig.7.5 is actually a more difficult problem than the startup problem.The principaldifficulty is in representing correctly the loss coefficient KV for the valve, because it isincorrect to model this coefficient as a constant in this problem.Instead we now have acontinually increasing head loss, and loss coefficient, across the valve which with pipefriction (and, to a minor extent, local losses) causes the flow to decelerate and eventuallystop.We assume that all loss coefficients under unsteady-flow conditions are unchangedfrom steady-flow conditions at the same velocity.The head loss in the system will be thepipe friction loss described by the Darcy-Weisbach equation, the local entrance loss, and thevalve head loss, as the governing equation, Eq.7.30, shows.Since KV varies with thevalve setting, which in turn changes in some predetermined manner with time, a closed-form solution of this ODE is not possible.Thus we must solve this nonlinear equationby numerical methods.Here we choose the fourth-order Runge-Kutta method, described in Appendix A.4.2, asthe numerical solution technique for this problem.Equation 7.30 is rewritten in the formL dV = V 2HR − 1 + KE + KV + f L= F(t,V)(7.33)g dtD 2 gNow the Runge-Kutta method can be applied directly, once the details of computing KVand F(t, V) are set.To complete the setup, we must know the valve operating schedule(percent open P vs.time) and the relation between KV and percent open P.If in addition we wish to know the maximum pressure head to occur (probably at thevalve) as time progresses, we can insert the computed velocity in hL = KVV2/2g to findthis head.However, one complicating factor occurs at the instant of closure; this losscoefficient becomes infinite as the velocity approaches zero, creating an indeterminatepressure head.Even under the best of circumstances, any numerical procedure will produceunreliable results at this point.Fortunately, the maximum pressure usually occurssomewhat before complete valve closure, so the numerical analysis will be terminated afraction of a second before complete closure.Example Problem 7.3 presents the solutionprocess for the reservoir-pipe system of Fig.7.5.Example Problem 7.3The reservoir head on the pipeline in Fig.7.5 is 60 ft.The 12 in-diameter line is3000 ft long with an equivalent roughness e = 0.012 in.Since the valve has been fullyopen for a long time, the flow of water is steady.(a) Calculate the steady-state velocity in the line assuming there is no loss at the valve.Then compute the maximum pressure in the line if the valve closes so that the rateof decrease in velocity is linear in time from its steady-state value to zero in 20 sec.(b) Now assume the valve at the downstream end is a GA Industries 12-in globe valvewhose loss characteristics are given in Appendix C.1 as a function of valve opening.Compute again the steady-state velocity with the valve fully open.Assuming thevalve closes in 20 sec at a rate that is linear in time, find the maximum pressure inthe line.(c) Repeat part (b) but employ a cubic spline interpolation to represent the valve data.© 2000 by CRC Press LLCTo begin part (a) and find the steady-state velocity, we can apply Eq.7.30 directly withdV/dt = 0 and KV = 0:2 VHR =1 + KE + KV + f LD 2 g2 V60 = 1 + 0.5 + 0 + f ( 3000 )1.0 64.4If we solve this equation with the Colebrook-White equation, we obtain V0 = 7.91 ft/sand f = 0.0201.The linearly decreasing velocity creates a constant deceleration so thatL dV = 3000 (−7.91 ) = − 36.8 ftg dt32.220Now we can apply Eq.7.29 from section 1 to 2.Since ( H + V2/2g)1 = HR - KEV2/2gand H2 = z2 + p2/γ = 0
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