Hemodialysis Home Modeling
Hemodialysis Therapy
Derivation of
Rate Equations Properties
of Rate Equations
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
and
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.
Hemodialysis Home Modeling
Hemodialysis Therapy
Derivation of
Rate Equations Properties
of Rate Equations
Copyright ©
2001, John F Patzer II
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