451 Chapter 6 x Viscous Flow in Ducts 6.27 Let us attack Prob. 6.25 in symbolic fashion, using Fig. P6.27. All parameters are constant except the upper tank depth Z(t). Find an expression for the flow rate Q(t) as a function of Z(t). Set up a differential equation, and solve for the time t0 to drain the upper tank completely. Assume quasisteady laminar flow. Solution: The energy equation of Prob. 6.25, using symbols only, is combined with a control-volume mass balance for the tank to give the basic differential equation for Z(t): Fig. P6.27 energy: h f 32PLV Ugd 2 or: S 4 h Z; mass balance: D2 dZ dt S 4 d ªS 2 S º D Z d2 L » « dt ¬ 4 4 ¼ d 2 V, where V Q S 4 d 2 V, U gd 2 (h Z) 32P L Separate the variables and integrate, combining all the constants into a single “C”: Z ³ Zo dZ hZ t C³ dt, or: Z (h Zo )e Ct h, where C 0 Tank drains completely when Z 0, at t 0 U gd 4 32 P LD2 Z · 1 § ln ¨ 1 o ¸ C © h¹ Ans. Ans.