big must the surge tank be so that it never runs dry(town loses its water supply)or never overflows(area surrounding the reservoir is flooded ) In general this is a complex design problem but let us look at a simple example to at least illustrate the concept Suppose that under normal conditions the level in the tank is to be half the height of the tank. If the flow rate into the tank becomes zero for a period of time(no rain), how long will it take for the tank to run dry if the outlet flow rate is maintained at its usual value? Specifically, suppose that the cross section area of the tank is 10 ft and its height is 10 ft and the normal outlet flow rate is 12 480 lb/hr First, let us take the volume of liquid in the tank as the control volume. The holdup of water in the control volume will be Holdup= Szp, where p is the density of water( 62.4 Ib/ft) Consider an interval of time At. Suppose that over that time interval the inlet and outlet flow rates are constant but not necessarily equal. Then, by the conservation of mass the change in the holdup will be given by (I-1) Holdup-t- Holdup==Fn△t-Fou△t Now, if we divide both sides of the mass balance equation(Eqn. l-1) by At and take the limit as At->0, we get the differential form of the mass balance, to wit. (-2) d szp/dt= Fin-Fout If we assume that is constant (a reasonable assumption if the temperature is also reasonably constant), then our mass balance equation becomes (-3) dz/dt=(Fin- Fout /S For our problem Fin=0 and Fout=12, 480 lb/hr, both constant. We can calculate dz/dt, that is dz/dt=(0-12480(10)(624)=-20f/h Since the nominal level is 5.0 ft(half the tank height of 10ft), it will take 0. 25 hour or 15 minutes for the tank to run dry Further examples of the application of the principle of conservation of mass, particularly for reacting systems, will be found in the subsequent sections of these notes l1--11- big must the surge tank be so that it never runs dry (town loses its water supply) or never overflows (area surrounding the reservoir is flooded). In general this is a complex design problem but let us look at a simple example to at least illustrate the concept. Suppose that under normal conditions the level in the tank is to be half the height of the tank. If the flow rate into the tank becomes zero for a period of time (no rain), how long will it take for the tank to run dry if the outlet flow rate is maintained at its usual value? Specifically, suppose that the cross section area of the tank is 10 ft2 and its height is 10 ft and the normal outlet flow rate is 12,480 lb/hr. First, let us take the volume of liquid in the tank as the control volume. The holdup of water in the control volume will be Holdup = Szr, where r is the density of water (62.4 lb/ft3 ). Consider an interval of time Dt. Suppose that over that time interval the inlet and outlet flow rates are constant but not necessarily equal. Then, by the conservation of mass the change in the holdup will be given by (I-1) Holdup|t=t - Holdup|t=0 = Fin Dt - Fout Dt Now, if we divide both sides of the mass balance equation (Eqn. I-1) by Dt and take the limit as Dt -> 0, we get the differential form of the mass balance, to wit, (I-2) d[Szr]/dt = Fin - Fout If we assume that  is constant (a reasonable assumption if the temperature is also reasonably constant), then our mass balance equation becomes (I-3) dz/dt = (Fin - Fout)/Sr For our problem Fin = 0 and Fout = 12,480 lb/hr, both constant. We can calculate dz/dt, that is dz/dt = (0 - 12480)/(10)(62.4) = -20 ft/hr. Since the nominal level is 5.0 ft (half the tank height of 10ft), it will take 0.25 hour or 15 minutes for the tank to run dry. Further examples of the application of the principle of conservation of mass, particularly for reacting systems, will be found in the subsequent sections of these notes