Example of Rate Equation Derivation - Storage Tank
Description: Flow into and out of a tank
A common situation in industry involves filling and draining a (right circular cylinder) storage tank simultaneously as depicted in the figure at the right. At time equal zero, fluid is pumped into the tank at a constant rate, Qin (m3 s -1). Simultaneously, the outlet valve is opened, permitting fluid to exit the tank at a rate Qout (m3 s -1). To a first approximation, flow through a valve is proportional to the pressure drop across the valve. Since both the tank interior and the outlet side of the valve are at atmospheric pressure, the pressure drop across the valve is equal to the liquid head (gauge pressure) inside the tank. Thus,
where Cv is the valve constant ( m3 Pa -1 s -1), r is the fluid density ( kg m –3 ), g is gravitational acceleration (m s –2 ), and h is the fluid level in the tank at any time (m).
Conservation of mass
In situations such as this, we are typically interested in knowing the fluid level at any given time, and in particular, whether the tank will overflow. The situation obviously involves conservation of mass because only mass is flowing into and out of the tank. Thus, we start with the conservation of mass equation
Since we are dealing with a constant density fluid, the mass of fluid in the tank is found by multiplying the volume of fluid in the tank by the density of the fluid. The rate of mass gain is given by the density times the inflow rate. Likewise, the rate of mass loss is given by the density times the outflow rate. Thus, the conservation equation becomes
where A is the tank cross-sectional area (m2), t is the time (s), and Dt is the time increment (s).
Substituting the valve discharge expression for Qout , dividing both sides by Dt and taking the limit as (Dt ® 0), produces the rate equation
Since both density and cross-sectional area are constant, we can take them outside the derivative and then divide both sides by rA to get
Note that the proper time to divide through by “constants” is after setting up the conservation balance and taking the limit to pass to the differential. This protects your mathematical development in such situations where things you think might be constant really are not.
The final rate equation also has an initial condition (i.c.) specified. Rate equations are not completely specified unless an initial condition is given. As implied with the initial condition shown, the initial condition need not be at time equal to zero; it could be at any time. The initial condition reports the state of the system, in this case, the initial fluid level in the tank, just prior to the (sudden) external changes which cause the system to respond.
The rate equation for fluid flow into and out of the tank also illustrates a common occurrence. Even though we used conservation of mass to derive the rate equation, the final equation is not an equation about mass! Rather, it turns out to be an equation for the quantity which we were really interested in: the fluid level in the tank.
Approach to steady-state
The rate equation describing flow into and out of a tank can be rearranged to
from which we find
Note that this steady-state solution is the same as that found previously. This is expected because the steady-state solution for a problem depends only on the physical situation characterized by the problem and not the mathematical method used determine the steady-state.
Consider flow into and out of a tank where
A = 10 m2; Cv = 7.5×10-4 m3/(s×Pa); h0 = 1 m; H = 10 m;
Qin = 60 m3/s; and r = 1000 kg/m3
Will the tank overflow? Approximately how long will the tank take to reach a new steady-state? What will the fluid level in the tank be after two time constants? Note that if the tank overflows, the fluid level will reach a new steady-state which is equivalent to the height of the tank! To answer the first question, we need to evaluate the steady-state solution,
The tank does not overflow since the steady-state height is less than the tank height. To answer the second question, we need to evaluate the time constant,
Since it takes approximately four time constants to approach steady-state, only about 5.4 s will be required to approach the steady-state height in the tank. Finally, the fluid height in the tank after two time constants is found by noting from the time response table that the level after two time constants is approximately
Consider the same problem where Qin is 100 m3/s. Now h ss = 13.6 m and the tank will overflow. We would like to estimate how long the tank will take to overflow. To answer this question, we can sketch a graph of the approach to steady-state similar to the generic sketch. The time constant is still 1.36 s since it does not depend on flow rate. The table below tabulates the change in height as a function of time (values calculated according to the generic approach to steady-state table).
Note that d = yss - y0 is used to calculate the expected height of liquid even though we know that the tank will overflow. This is because the rate equation applies until the time the tank overflows and the liquid level will change as if the tank does not overflow. The data in the table are plotted in the figure below to obtain a curve of liquid height in the tank as a function of time. The time to overflow is estimated by finding the tank height on the ordinate, 10 m, moving horizontally across to the response curve and then vertically down to the abscissa. Thus, the estimated time until the tank overflows is 1.7 s after the valves are opened.
Copyright © 2001, John F Patzer II
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